Analysis Spm Additional Mathematics

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1 (a) Onetoonerelation. (b) {(1,b),(2,a),(3, c)}

2 (a)

–2

2

4

6

4

6

8

SetA SetB

(b) (i) Manytoonerelation. (ii) Objects=–2,2,4,6 Images=4,6,8

3 (a) (i) Domain={a,b,c,d} Codomain={e,f,g} (ii) Range={e,f,g} (b) Manytomanyrelation.

4 (a) Domain={–1,1,3,5} Codomain={1,9,25} (b) 9 (c) 5 (d) Range={1,9,25}

5 (a)

SetQ

SetP

7

5

3

2

8 9 12 25 49

(b) (i) 3 (ii) 25

6 (a) Manytoonerelation. (b) (i) Images=p,q Range={p,q} (ii) Noobject

7 (a) Codomain={a,b,c} Range={b} (b) 3,6,9

8 (a)

Multiple of

2

3

4

6

8

9

12

SetP SetQ

(b) {(2,6),(2,8),(2,12),(3,6), (3,9),(3,12),(4,8),(4,12)}

(c)

Set Q

Set P

12

9

8

6

2 3 4

9 (a) Notafunction. (b) Notafunction. (c) Afunction. (d) Afunction.

10 (a) Onetoonerelation. (b) Manytoonerelation. (c) Onetomanyrelation. (d) Manytomanyrelation.

11 (a) Afunction. (b) Notafunction. (c) Notafunction. (d) Afunction.

12 (a) Onetoonefunction. (b) Notonetoonefunction. (c) Notonetoonefunction. (d) Notonetoonefunction.

13 (a) Yes (b) Yes (c) No (d) Yes

14 (a) f:xa x (b) f:xax2

15 (a) (i) f(–1)= 5–2(–1) = 7 (ii) f(2)+f(–2) = [5–2(2)]+[5–2(–2)] = 1+9 = 10

(iii) g( 27 ) = 7(2

7 )+4

= 6

(iv) g( 17 )–g(– 1

7 ) = [7(1

7 )+4]–[7(– 17 )+4]

= 5–3 = 2

(b) (i) f(x) = g(x) 5–2x = 7x+4 9x = 1

x =19

(ii) f(x) = 9 5–2x = 9 2x = –4 x = –2

16 (a) f(3)+g(4)

= ( 33

+2)+(34

(4)–1) = 3+2 = 5 (b) 2f(6)–3g(8)

= 2(63

+2)–3[34

(8)–1] = 8–15 = –7 (c) f(–12)–g(–12)

= (–123

+2)–[ 34

(–12)–1] = –2+10 = 8 (d) f(–3)–g(–4)

= (– 33

+2)–[ 34

(–4)–1] = 1+4 = 5

17 (a) f(0) = –7 m(0)+n = –7 n = –7 f(3) = 2 3m+n = 2 3m–7 = 2 3m = 9 m = 3 \m+n=3+(–7)=–4 (b) f(x) = 3x–7 f(3) = 3(3)–7 = 2 (c) f(x) = 2 3x–7 = 2 3x = 9 x = 3

18 f(3) = 10 a(3)2+b = 10 9a+b = 10… 1 f(–2) = –10 a(–2)2+b = –10 4a+b = –10… 2 1 – 2 : 5a = 20 a= 4 Substitutea=4into 1 : 9(4)+b = 10 b = 10–36 = –26 \ a

b = – 4

26

= – 213

19 f(x)=3x–2 (a) f(–2) = 3(–2)–2 = 8 f(1) = 3(1)–2 = 1

© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

�© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

(b) f(x) = 8 3x–2 = 8 3x–2 = 8 or 3x–2= –8 3x = 10 3x = –6 x = 10

3 x = –2

(c) f(x) = x 3x–2 = x 3x–2 = x or 3x–2 = –x 2x = 2 4x = 2 x = 1 x = 1

220 (a) k=1 (b) 0<f(x)<3

21 (a) f(x) = x2–2x+3 f(–1) = (–1)2–2(–1)+3 = 6 \m=6 f(4) = 42–2(4)+3 = 11 \n=11 (b) 2<f(x)<11

22 (a) h(3)–h(2) = 8–4 = 4 (b) f(2)+g(12) = 9+8 = 4 (c) h(2)3h(1) = 4(4) = 16

(d)h(3)h(1)

=84

= 2

23 (a) s (b) s (c) u (d) w (e) u (f) w

24 (a) 4 (b) –3 (c) 3 (d) –1

25 (a) f=g 3x+4= x2+6 x2–3x+2= 0 (x–1)(x–2)= 0 x=1orx=2 (b) fg = gf f(x2+6) = g(3x+4) 3(x2+6)+4 = (3x+4)2+6 3x2+22 = 9x2+24x+22 6x2+24x = 0 6x(x+4) = 0 x=0orx=–4

26 (a) f 2=g 2(2x+3)+3 = 3x+6 4x+9 = 3x+6 x = –3 (b) 3(3x+6)+6 = 2x+3 9x+24 = 2x+3 7x = –21 x = –3

27 fg(a) = 8 f(a–3) = 8 (a–3)2+4 = 8 a2–6a+5 = 0 (a–1)(a–5) = 0 a=1ora=5

28 fg(–4)+2 = gf(2) f(2)+2 = g(4–2m) 4–2m+2 = 3(4–2m)+14 6–2m = 26–6m 4m = 20 m = 5

29 (a) gf(x) = x+5 4f(x)–1 = x+5 4f(x) = x+6

f(x)=x+6

4 (b) gf(x) = x

f(x)

1–f(x) = x

f(x) = x–xf(x) f(x)+xf(x) = x f(x)[1+x] = x

f(x) =x

1+x ,x ≠ –1

(c) fg(x) =2x2–13x+22 f(x–3) =2x2–13x+22 f(k) =2(k+3)2–13(k+3)+22 =2(k2+6k+9)–13k –39+22 =2k2+12k+18–13k –17 =2k2–k+1 \ f(x)=2x2–x+1 (d) fg(x)=4x2+12x f(2x+1) = 4x2+12x

f(k) = 4(k–12 )2

+12(k–12 )

= 4( k2–2k+14 )+

6(k–1) = k2–2k+1+6k–6 = k2–4k–5 f(x)=x2–4x–5

30 (a) fg(x) = 4x2+2x 2g(x)+4 = 4x2+2x 2g(x) = 4x2+2x–4 g(x) = 2x2+x–2 \g(2) = 2(2)2+2–2 = 8

(b) fg(x) = x–2

f (2x+1x ) = x–2

f(k) =1

k–2–2

f(x) =1

x–2–2

\f(4) =1

4–2–2

= –112

31 f 2(x) = 16x–15 f (ax+b) = 16x–15 a(ax+b)+b = 16x–15 a2x+ab+b = 16x–15 a2 =16 and ab+b = –15 a =±4 4b+b = –15 =4(0) 5b = –15 b = –3

\a+b = 4+(–3) = 1

32 g(2) = 5 4a+b = 5 … 1 gf(1) = –1 g(–1) = –1 a+b = –1 … 2 1 – 2 : 3a = 6 a = 2 Substitutea=2into 1 : 8+b = 5 b = –3 \a=2,b=–3

33 (a) –2 (b) f(a) = 8 a+5 = 8 a = 3

34 (a) a (b) q (c) a (d) b

35 (a) y = 3x–2 y+2 = 3x

x =y+2

3

\f –1:xa x+2

3

(b) y =5x+23x–2

3xy–2y = 5x+2 3xy–5x = 2y+2 x(3y–5) = 2y+2

x =2y+23y–5

\f –1:xa 2x+23x–5 ,x ≠

53

(c) y =3x+2

4x 4xy = 3x+2 4xy–3x= 2 x(4y–3) = 2

x =2

4y–3

\f –1:xa 2

4x–3 ,x ≠ 34

(d) y =x–14–x

4y–xy = x–1 x+xy = 4y+1 x(1+y) = 4y+1

x =4y+1y+1

\f –1:xa 4x+1x+1 ,x ≠ –1

(e) y =3x

5x–2 5xy–2y = 3x x(5y–3) = 2y

x =2y

5y–3

\f –1:xa 2x

5x–3 ,x ≠ 35


(f) y =2x–5

2 2y = 2x–5 2x = 2y+5

x =2y+5

2

\ f –1:xa 2x+5

2

36 (a) 7–5y = x 5y = 7–x

y =7–x

5

\ f :xa 7–x

5

(b)–5y–13y–2 = x

–5y–1 = 3xy–2x 3xy+5y = 2x–1 y(3x+5) = 2x–1

y =2x–13x+5

\ f :xa 2x–13x+5 ,x ≠ –

53

(c)y+4

3 = x

y+4 = 3x y = 3x–4 \ f :xa 3x–4

(d)3y

y–3 = x

3y = xy–3x xy–3y = 3x y(x–3) = 3x

y = 3x

x–3

\ f :xa 3x

x–3 ,x ≠ 3

(e)23 y = x

y =32 x

\ f:x a 32

x

(f)5y = x

y =5x

\ f : x a 5x

,x ≠ 0

37 (a) f(6) = 5

12+k

3 = 5

12+k = 15 k = 3

(b) y =2x+3x–3

xy–3y = 2x+3 xy–2x = 3y+3 x(y–2) = 3y+3

x =3y+3y–2

f –1(x)=3x+3x–2 ,x ≠ 2

f –1(–7) =3(–7)+3

–7–2

=–18–9

= 2

38 y =x+2x–3

xy–3y = x+2 xy –x = 3y+2 x(y–1) = 3y+2

x =3y+2y–1

f –1(x) = 3x+2x–1

,x ≠ 1

f –1(4) = 3(4)+24–1

=143

39 y = 1x +3

xy+3y = 1 xy = 1–3y

x = 1–3yy

f –1(x)=1–3xx

,x ≠ 0

f –1(m) = 3

1–3mm

= 3

1–3m = 3m 6m = 1

m =16

40 fg(x) = 1+4x

2x–3

f(x) = 1+4x2x–3

,x ≠ 32

y = 1+4x2x–3

2xy–3y = 1+4x x(2y–4) = 3y+1

x = 3y+12y–4

\f –1:xa 3x+12x–4

,x ≠ 2

41 y–15

= x

y = 5x+1 f(x) = 5x+1

3–y2

= x

3–y = 2x y = 3–2x g(x) = 3–2x fg(x) = f(3–2x) = 5(3–2x)+1 = 16–10x y = 16–10x 10x = 16–y

x = 16–y10

( fg)–1(x) = 16–x10

( fg)–1(6) = 16–610

= 1

42 g–1(x) = 1–xx

g–1(y) = x

1–yy

= x

1–y = xy xy+y = 1 y(x+1) = 1

y = 1x+1

g(x) = 1x+1

,x ≠ – 1

gf(x) = g( 1x–1 )

=

1

( 1x–1 )+1

=1

( xx–1 )

= x–1x

,x ≠ 0

h(x) = gf(x)

= x–1x

,x ≠ 0

y = x–1x

xy = x–1 1 = x–xy 1 = x(1–y)

x = 11–y

\h–1:xa 11–x

,x ≠ 1

1 (a) f:xax+4 (b) a=–1

2 (a) f(x–2) = (x–2)2–5 = x2–4x–1 (b) 5f(–2) = 5[(–2)2–5] = 5(–1) = –5

3 f(m2–7m+6)= g(m+2) m2–7m+6+2= (m+2)2–7(m+2) +6 m2–7m+8= m2+4m+4–7m –14+6 m2–7m+8= m2–3m–4 4m = 12 m = 3

4 (a) (i) Objects=3,2,–3 Range={4,9} (ii) 2 (b) Notafunctionbecausef –1isnot

onetoonefunction.

5 (a) f(2) = 1

k

2–k= 1

k = 2–k 2k = 2 k = 1


(b) y =1

x–1

xy–y = 1 xy = y+1

x =y+1

y

f –1(x) =x+1

x

f –1(–3) =–3+1

–3

=23

6 (a) f(–1) = –12

24

–p+q= –12

–2 = –p+q p–q = 2… 1 f(1) = 6

24

p+q = 6

p+q = 4… 2 1 + 2 : 2p = 6 p = 3 Substitutep=3into 1 : 3–q = 2 q = 1 \p=3,q=1

(b) f(x) = x

24

3x+1 = x

24 = 3x2+x 3x2+x–24 = 0 (3x–8)(x+3) = 0

x=83

orx=–3

7 (a) Atx-axis,y=0, 0= 2x–1

x =12

\k=12

(b) |2x–1|=3 2x–1= 3 or 2x–1= –3 2x = 4 2x = –2 x = 2 x = –1 \–1<x<2

8 y =3–x2x

2xy+x = 3 x(2y+1) = 3

x =3

2y+1

f–1(x) =3

2x+1

3

2x+1 =

3–x2x

6x = 3+5x–2x2

2x2+x–3 = 0 (2x+3)(x–1) = 0

x=–32

orx=1

9 f(mx+n) = m(mx+n)+n = m2x+mn +n

Comparewithf 2(x)=4x–9 m2= 4 m = ±2 Whenm=2, 3n = –9 n = –3 Whenm=–2, –n = –9 n = 9

10 (a) f(3x+2) = 3(3x+2)+2 f 2(x) = 9x+8

(b) f 2(2) = 9(2)+8 = 26 f 2(–1) = 9(–1)+8 = –1 Range:–1<f 2(x)<26

11 (a) gf = g(2x+1) = (2x+1)2–1 = 4x2+4x (b) 4x2+4x = x2–1 3x2+4x+1 = 0 (3x+1)(x+1) = 0

x=–13

orx=–1

12 (a) f(x)=x2andg(x)=x+1 gf = g(x2) = x2+1 \gf:xax2+1

(b) gf(3) = 32+1 = 10

13 (a) 25 (b) 2

14 (a) (i) 3 (ii) 7

(b) g(x2) = x2–2 g(k) = ( k )2–2 = k–2 \g:xax–2

15 fg(x) = x2+4x+5 [g(x)]2+1 = x2+4x+5 [g(x)]2 = x2+4x+4 = (x+2)2

g(x) = x+2 \g:xax+2

16 (a) f (1)= 3(1)+51

= 8

(b) 3x+5x

= 8

3x2–8x+5= 0 (3x–5)(x–1)= 0

x=53

orx=1

17 h(x) = gf(x) = g(2–3x) = 2–3x+4 = 6–3x y = 6–3x

x =6–y

3

h–1(x) =6–x

3

\h–1(3) =6–3

3

= 1

18 (a) f –1(y) = x

2y–13–y

= x

2y–1 = 3x–xy 2y+xy = 3x+1 y(2+x) = 3x+1

y =3x+12+x

f(x) =3x+12+x

f(x) = 2

3x+12+x

= 2

3x+1 = 4+2x x = 3

(b) gf(2) = g[ 3(2)+12+2 ]

= g(74)

= 4(74)

= 7

19 y= 2x–m

x=y+m

2

f –1(x)=12

x+m2

Comparewithf –1(x)=nx+

52

\ m=5,n=12

20 (a) f –1(y) = x

y–k

2 = x

y–k = 2x y = 2x+k f(x) = 2x+k f(–1) = 1 –2+k = 1 k = 3

(b) f (12) = 2(1

2)+3

= 4

21 (a) gf(5) = 10 g(2) = 10 4m–2 = 10 4m = 12 m = 3

(b) f ( 4x–3 ) =

4

( 4x–3 )–3

=4

( 13–3xx–3 )

=

4x–1213–3x

Comparewithax–bc–dx

.

\a=4,b=12,c=13,d=3


22 (a) A(32

,–4) (b) –4<f(x)<3

(c) |2x–3|–4 = –2 |2x–3| = 2 2x–3 = 2 or 2x–3 = –2 2x = 5 2x = 1

x =52

x =12

23 (a) y = 3x–2

x =y+2

3

\f –1:x a13

x +23

(b) 3x–2 =13

x+23

9x–6 = x+2 8x = 8 x = 1 Whenx=1, y= 3(1)–2 = 1 \P(1,1)

(c) –2<f(x)<7,0<f –1(x)<3

24 (a) f 2 = f ( 1x+1 )

=1

( 1x+1 )+1

=x+1x+2

,x ≠ –2

gf = g( 1x+1 )

= 3( 1x+1 )+1

=x+4x+1

,x ≠ –1

(b)x+1x+2

=x+4x+1

x2+2x+1 = x2+6x+8 4x = –7

x = – 74

25 (a) f(x) =18x

+8x

f (x) = 0

–18x2 +8 = 0

18x2 = 8

x2 =94

x = ±32

=32

(0)

Range:32

<f(x)<30

(b) y = 8–3x

x =8–y

3

f –1(x) =8–x

3

g(8–3x) = 8–3(8–3x) = 9x–16

8–x

3 = 9x–16

8–x = 27x–48 28x = 56 x = 2

(c) h(x)=ax–b At(3,2), 2= 3a–b b = 3a–2… 1 y = ax–b

x =y+b

a

h–1(x) =x+b

a Forthepointofintersection

wherex=4,

ax–b =x+b

a a2x–ab = x+b 4a2–ab = 4+b… 2 4a2–a(3a–2) = 4+3a–2 a2–a–2 = 0 (a–2)(a+1) = 0 a = 2(0) When a=2,b= 3(2)–2 = 4

26 (a) (i) y = 5–6x

6x

= 5–y

x =6

5–y

\f –1:xa 6

5–x,x ≠ 5

(ii)6

5–x = 5–

6x

6

5–x+

6x

= 5

30

x(5–x) = 5

30 = 25x–5x2

5x2–25x+30 = 0 x2–5x+6 = 0 (x–2)(x–3) = 0 x=2orx=3

(b) (i) y =x–3

2 x = 2y+3 \g–1:xa2x+3

(ii) 2x+3 =x–3

2 4x+6 = x–3 3x = –9 x = –3

27 (a) gf (b) g–1f (c) f –1g

28 (a) f(x) = –x2+6x+1 = –(x2–6x–1) = –(x2–6x+(–3)2–(–3)2–1] = –[(x–3)2–10] = 10–(x –3)2

y

y=f(x)

xO 3

10

1

(b) Becauseforx>3,itsonetoonefunction.

(c) y = 10–(x–3)2

(x–3)2 = 10–y

x–3 = 10–y

x = 10–y +3

\f –1:xa 10–x +3

29 (a) f(–3) =–4 g(–4) = 6 –3a+5 =–4 b

–4+6 = 6 3a =9b = 12 a =3

\a=3,b=12

(b) gf(–3)=6,f –1g–1(6)=–3

(c) gf(x) = g(3x+5)

=12

3x+5+6

=12

3x+11,x ≠ –

113

30 (a) f(2) = –1 p–2q = –1… 1 g(2) = 2

6

2q–1 = 2

6 = 4q–2 4q = 8 q = 2 Substituteq=2into 1 : p–2(2)= –1 p = –1+4 = 3 \ p=3,q=2

(b) (i) f(y) = 3–2y x = 3–2y 2y = 3–x

y =3–x

2

f –1(x) =3–x

2

\f –1:y a3–y

2

(ii) gf –1(y) = g( 3–y2 )

=6

2( 3–y2 )–1

=6

2–y ,y ≠ 2

gf –1:ya6

2–y ,y ≠ 2

31 (a) (i) f 2 = f(6–3x) = 6–3(6–3x)

�

= 9x–12

(ii) y = 9x–12

x =y+12

9

\( f 2)–1(x)=x+12

9

(b) y = 6–3x

x =6–y

3

f –1(x) =6–x

3

(f –1)2 = f –1( 6–x3 )

=

6–( 6–x3 )

3 =

x+129

\( f –1)2=( f 2)–1

(c)

y

x24

3O

6

12

Range:0<y<12

32 (a) y

xO 2

1y=2x–4

y=x +1

4

(b) Numberofsolutions=2

33 (a) (i) f(2) = 5 4a–b = 5… 1 f(–3) = 15 9a–b = 15… 2 2 – 1 : 5a = 10 a = 2 Substitutea=2into 1 : 4(2)–b = 5 b = 3

(ii) f(x) = 2x2–3 f(1) = 2(1)2–3 = –1

(b) 2x2–3 = x 2x2–x–3 = 0 (2x–3)(x+1) = 0

x=32

orx=–1

34 (a) f 2(x)= f ( xm–nx )

=

( xm–nx )

m–n( xm–nx )

=( xm–nx )

( m2–mnx–nxm–nx )

=

xm2–mnx–nx

=

xm2–(mn+n)x

Comparewithf 2(x)=

x4–3x

m2= 4 or mn+n = 3 m = ±2 3n = 3 = 2(0) n = 1 \m=2,n=1

(b) y =x

2–x 2y–xy = x 2y = x(1+y)

x =2y

1+y

f –1(x)=2x

1+x ,x ≠ –1

\f –1(12) =

2(12)

1+12

=

23

35 (a) f(2) = 9 4a+1 = 9 4a = 8 a = 2 gf(2) = 25 6(2)2+b = 25 24+b = 25 b = 1 \a=2,b=1

(b) g(2x2+1) = 6x2+1

g(k) = 6( k–12 )2

+1

= 6( k–12 )+1

= 3k–2 \g:xa3x–2



Quadratic Equations

1 (a) No (b) No (c) Yes (d) Yes

2 (a) 3x2 – 12x = 0, a = 3, b = –12, c = 0 (b) x2 – 25 = 0, a = 1, b = 0, c = –25 (c) x2 – 8x + 8 = 0, a = 1, b = –8,

c = 8 (d) x2 + x – 5 = 0, a = 1, b = 1,

c = –5

3 (a) –3, 8 (b) –1, 5

(c) –2

12

, 4 (d) 0, 5

4 (a) –4, 0 (b) 4, 9

(c) 7 (d) –4,

52

5 (a) –2, 12

(b) –5, –4

(c) –4, 2 (d) –3, 2

6 x2 + 9x + 2p = 0 When x = –5, (–5)2 + 9(–5) + 2p = 0 2p = 20 p = 10

7 mx2 – 5x – 3 = 0 When x = 3, m(3)2 – 5(3) – 3 = 0 9m = 18 m = 2

8 (a) ax2 + 5x = 12 When x = 2, a(2)2 + 5(2) = 12 4a = 2

a = 12

(b)

12

x2 + 5x = 12

x2 + 10x – 24 = 0 (x + 12)(x – 2) = 0 x = –12 or x = 2 \ x = –12

9 x2 + (2p – 1)x + (p2 – 3p – 4) = 0 When x = 0, p2 – 3p – 4 = 0 (p + 1)(p – 4) = 0 p = –1 or p = 4

10 x2 – 2x = 5 When x = a, a2 – 2a = 5 a(a – 2) = 5

a – 2 = 5a

1a

= a – 2

5 (shown)

11 (a) (x + 6)(x + 2) = 32 x2 + 8x – 20 = 0 (x + 10)(x – 2) = 0 x = –10 or x = 2 (b) (x + 4)2 – 3(x + 4) = 10 x2 + 8x + 16 – 3x – 12 = 10 x2 + 5x – 6 = 0 (x + 6)(x – 1) = 0 x = – 6 or x = 1 (c) 8x2 – 16 = 7x 2

x2 – 16 = 0 (x + 4)(x – 4) = 0 x = – 4 or x = 4

(d)

x + 6

x(x + 3) =

12

2x + 12 = x2 + 3x x2 + x – 12 = 0 (x + 4)(x – 3) = 0 x = – 4 or x = 3 (e) 4y2 – 12y + 9 = 6y + 1 4y2 – 18y + 8 = 0 2y2 – 9y + 4 = 0 (2y – 1)(y – 4) = 0

y = 12

or y = 4

(f) 2x2 + 3x – 2 = x2 + 16 x2 + 3x – 18 = 0 (x + 6)(x – 3) = 0 x = – 6 or x = 3

12 (a) x2 + 8x = 20 x2 + 8x + (4)2 = 20 + (4)2

(x + 4)2 = 36 x = 36 – 4 or x = – 36 – 4 = 2 = –10 \ x = 2 or x = –10 (b) 5y2 – 30y = 18

y2 – 6y = 185

y2 – 6y + (–3)2 =

185

+ (–3)2

(y – 3)2 =

635

y =

635

+ 3 or y = –

635

+ 3

= 6.55 = – 0.55 \ y = 6.55 or y = – 0.55 (c) 2y2 – 12y = 15

y2 – 6y =

152

y2 – 6y + (–3)2 =

152

+ (–3)2

(y – 3)2 =

332

y =

332

+ 3 or y = –

332

+ 3

= 7.062 = – 1.062 \ y = 7.062 or y = –1.062

(d) x2 – 4x = 21 x2 – 4x + (–2)2 = 21 + (–2)2

(x – 2)2 = 25 x = 5 + 2 or x = –5 + 2 = 7 = –3 \ x = 7 or x = –3 (e) 3y2 + 6y = 2

y2 + 2y =

23

y2 + 2y + (1)2 =

23

+ (1)2

(y + 1)2 =

53

y =

53

– 1 or y = –

53

– 1

= 0.291 = – 2.291 \ y = 0.291 or y = –2.291 (f) 5y2 – 30y = 18

y2 – 6y =

185

y2 – 6y + (–3)2 =

185

+ (–3)2

(y – 3)2 =

635

y =

635

+ 3 or y = –

635

+ 3

= 6.5496 = – 0.5496 \ y = 6.5496 or y = –0.5496

13 (a) 2x2 + 5x – 3 = 0

x =

–5 52 – 4(2)(–3)

2(2)

=

–5 494

=

–5 + 7

4 or

–5 – 7

4

=

12

or –3

\ x = 12

or x = –3

(b) x2 + 3x – 28 = 0

x =

–3 32 – 4(1)(–28)

2(1)

=

–3 1212

=

–3 + 11

2 or

–3 – 11

2 = 4 or –7 \ x = 4 or x = –7 (c) x2 – 2x – 8 = 0

x =

2 (–2)2 – 4(1)(–8)

2(1)

=

2 362

=

2 + 6

2 or

2 – 6

2


2© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

= 4 or –2 \ x = 4 or x = –2 (d) 12x2 – 7x – 12 = 0

x =

7 (–7)2 – 4(12)(–12)

2(12)

=

7 62524

=

7 + 25

24 or

7 – 25

24

= 1

13

or – 34

\ x = 1 13

or x = – 34

(e) 5x2 – 6x –8 = 0

x =

6 (–6)2 – 4(5)(–8)

2(5)

=

6 19610

=

6 + 14

10 or

6 – 14

10

= 2

or –

45

\ x = 2 or x = – 45

(f) 4x2 – 8x –5 = 0

x =

8 (–8)2 – 4(4)(–5)

2(4)

=

8 1448

=

8 + 12

8 or

8 – 12

8

= 2

12

or – 12

\ x = 2 12

or x = – 12

14 (a) x2 – 3x – 5 = 0 a + b = 3 ab = –5 (b) 2x2 + 6x = 5x + 9 2x2 + x – 9 = 0

x2 + x2

– 92

= 0

a + b = – 12

ab = –

92

(c) 2x2 + 2 = 3x 2x2 – 3x + 2 = 0

x2 – 32

x + 1 = 0

a + b = 32

ab = 1

(d)

1

x(x + 1) =

14

x2 + x – 4 = 0 a + b = –1 ab = –4

15 (a) x2 – (6 + 10)x + (6)(10) = 0 x2 – 16x + 60 = 0

(b) x2 – (–10 + 7)x + (–10)(7) = 0 x2 + 3x – 70 = 0

(c) x2 –

( 23

+ 12 )x +

( 23 )( 1

2 ) = 0

x2 –

76

x + 26

= 0

6x2 – 7x + 2 = 0 (d) x2 – (–7 – 3)x + (–7)(–3) = 0 x2 + 10x + 21 = 0

(e) x2 – ( 1

3 +

13 )x +

( 13 )( 1

3 ) = 0

x2 –

23

x + 19

= 0

9x2 – 6x + 1 = 0 (f) x2 – (0 + 4)x + (0)(4) = 0 x2 – 4x = 0

16 3x2 – 5x + 2 = 0

a + b =

53

ab =

23

(a)

1a

+ 1b

=

a + bab

=

( 53 )

( 23 )

=

52

(b)

ab

+ ba

=

a 2 + b 2

ab =

(a + b)2 – 2abab

=

( 53 )2

– 2( 2

3 )23

=

(139 )

( 23 )

=

136

(c) 2a + 2b = 2(a + b)

= 2( 5

3 )

=

103

(d)

3a

+ 3b

=

3(a + b)ab

=

3( 53 )23

=

152

17 2x2 + 4x – 7 = 0 a + b = –2

ab = –

72

(a) (a – 1) + (b – 1) = (a + b) – 2 = –2 – 2 = –4

(a – 1)(b –1) = ab – a – b + 1 = ab – (a + b) + 1

= –

72

– (–2) + 1

= –

12

\ x2 –(– 4)x +

(– 12 )

= 0

x2 + 4x –

12

= 0

2x2 + 8x – 1 = 0

(b) 2a +

2b =

2(a + b)ab

=

2(–2)

(– 72 )

=

87

( 2a)( 2

b ) =

4

ab

=

4

(– 72 )

= –

87

\ x2 –

87

x – 87

= 0

7x2 – 8x – 8 = 0 (c) 4a + 4b = 4(a + b) = 4(–2) = –8 (4a)(4b) = 16ab

= 16(– 72 )

= –56 \ x2 + 8x – 56 = 0 (d) a2 + b 2 = (a + b)2 – 2ab

= (–2)2 – 2(–

72 )

= 4 + 7 = 11 (a2)(b 2) = (ab)2

=

(– 72 )2

=

494

\ x2 – 11x +

494

= 0

4x2 – 44x + 49 = 0

(e)

a5

+ b5

=

a + b5

= –

25

(a5 )( b

5 ) =

ab25

=

(– 72 )

25

= –

750

\ x2 +

25

x –

750

= 0

50x2 + 20x – 7 = 0


(f) a2b + ab2 = ab(a + b)

= –

72

(–2)

= 7 (a2b)(ab2) = (ab)3

=

(– 72 )3

= –

3438

\ x2 – 7x –

3438

= 0

8x2 – 56x – 343 = 0

18 x2 – 2(m + 1)x + m(m + 1) = 0 a + 2a = 2(m + 1) 3a = 2m + 2

a =

2m + 2

3 … 1

a(2a) = m(m + 1) 2a2 = m(m + 1) … 2 Substitute 1 into 2 :

2( 2m + 23 )2

= m2 + m

2(4m2 + 8m + 4) = 9m2 + 9m 8m2 + 16m + 8 = 9m2 + 9m m2 – 7m – 8 = 0 (m + 1)(m – 8) = 0 m = –1 or m = 8

19 px2 + (p – 1)x + 2p + 3 = 0

x2 + (p – 1)

p x + 2p + 3

p = 0

a + b = – (p – 1)

p ab =

2p + 3p

(a) a + (–a) = – (p – 1)

p 0 = 1 – p p = 1

(b) a( 1a) =

2p + 3p

p = 2p + 3 p = –3

20 a + (a + 2) = –(2 – h) 2a + 2 = h – 2 2a = h – 4

a = h – 4

2 … 1

a(a + 2) = h a2 + 2a = h … 2

Substitute 1 into 2 :

(h – 42 )2

+ 2(h – 4

2 ) = h

h2 – 8h + 164

+ h – 4 = h

h2 – 8h + 16 + 4h – 16 = 4h h2 – 8h = 0 h(h – 8) = 0 h = 0 or h = 8

Substitute h = 8 into 1 :

a =

8 – 42

= 2 a + 2 = 2 + 2 = 4 (a) The roots are 2 and 4. (b) h = 8

21 2(x2 + 3x + 2) = n 2x2 + 6x + 4 – n = 0 a + (a + 3) = –3 2a = –6 a = –3

a(a + 3) =

4 – n

2 … 1

Substitute a = –3 into 1 :

–3(–3 + 3) =

4 – n2

0 = 4 – n n = 4

22 x2 + 3k2 – 5kx = 1 – 2k x2 – 5kx + 3k2 + 2k – 1 = 0 a + 4a = 5k 5a = 5k a = k a(4a) = 3k2 + 2k – 1 4a2 = 3k2 + 2k – 1… 1 Substitute a = k into 1 : 4k2 = 3k2 + 2k – 1 k2 – 2k + 1 = 0 (k – 1)(k – 1) = 0 k = 1

23 x2 – 3x – 2 = 0 a + b = 3 ab = –2 x2 – 6x + p = 0

ha

+ hb

= 6

h(a + b) = 6ab 3h = –12 h = –4

( ha)( h

b ) = p

h2

ab = p

h2 = p(–2) (– 4)2 = –2p 2p = –16 p = –8 \ h = –4, p = –8

24 3x2 + 5x + 1 = 0

a + b = –

53

ab =

13

mx 2 – 4x + n = 0

a + 3 + b + 3 = 4m

a + b + 6 = 4m

– 53

+ 6 =

4m

133

=

4m

m =

1213

(a + 3)(b + 3) =

nm

ab + 3(a + b) + 9 =

nm

13

+ 3(– 53 )

+ 9 =

1312

n

133

=

1312

n

n = 4

\ m = 1213

, n = 4

25 2x2 + kx + 24 = 0

a + b = –

k2

… 1

ab = 12 … 2 Substitute a = 4 + b into 2 : b(4 + b) = 12 b 2 + 4b – 12 = 0 (b + 6)(b – 2) = 0 b = –6 or b = 2 Substitute b = –6 into 1 :

4 + b + b = –

k2

–8 = –

k2

k = 16 Substitute b = 2 into 1 :

4 + b + b = –

k2

8 = –

k2

k = –16 \ k = –16 ( 0)

26 x2 – 9x + 2k = 0 a + b = 9 … 1 ab = 2k … 2 Substitute a = 2b into 1 : 3b = 9 b = 3 Substitute b = 3 into 2 : (2b)(b) = 2k 2k = 18 k = 9

27 mx2 + 7x – n = 0

x2 +

7m

x – nm

= 0

a + b = –

7m

ab = – nm

– 4 +

12

= – 7m

– 72

= – 7m


m = 2

– 4( 12 )

= – nm

–2 = –

n2

n = 4 \ m = 2, n = 4

28 x2 – hx – k = 0 a + b = h ab = –k –3 + 5 = h h = 2 –3(5) = –k k = 15 \ h = 2, k = 15

29 p + 1 + q + 1 = 1 p + q = –1 p = –1 – q … 1 (p + 1)(q + 1) = –12 pq + p + q + 1 = –12 … 2

Substitute 1 into 2 : q(–1 – q) – 1 – q + q + 1 = –12 q2 + q – 12 = 0 (q + 4)(q – 3) = 0 q = – 4 or q = 3 When q = – 4, p = –1 – (– 4) = 3 When q = 3, p = –1 – 3 = – 4 \ p = 3, q = – 4; p = – 4, q = 3

30 a + 3a = –p 4a = –p

a = – p4 … 1

a(3a) = q 3a2 = q … 2

Substitute 1 into 2 : 3(–

p4 )2

= q

3p2

16 = q

3p2 = 16q (shown)

31 (a) 2x2 + 5x – 3 = 0 b2 – 4ac = 52 – 4(2)(–3) = 25 + 24 = 49 ( 0) \ Two distinct roots. (b) 2 – 4x – x2 = 0 b2 – 4ac = (–4)2 – 4(–1)(2) = 16 + 8 = 24 ( 0) \ Two distinct roots. (c) x2 – 6x + 11 = 0 b2 – 4ac = (–6)2 – 4(1)(11) = 36 – 44 = –8 ( 0) \ No roots. (d) x2 + x – 4 = 0 b2 – 4ac = 12 – 4(1)(–4) = 1 + 16 = 17 ( 0) \ Two distinct roots.

(e) x2 – 6x + 9 = 0 b2 – 4ac = (–6)2 – 4(1)(9) = 36 – 36 = 0 \ Two equal roots. (f) x2 – 4x + 4 = 0 b2 – 4ac = (–4)2 – 4(1)(4) = 0 \ Two equal roots.

32 (a) 4x2 – kx + 25 = 0 (–k)2 – 4(4)(25) = 0 k2 – 400 = 0 (k + 20)(k – 20) = 0 k = –20 or k = 20 (b) (k + 1)x2 + 3k – 2(k + 3) = 0 (k + 1)x2 + k – 6 = 0 02 – 4(k + 1)(k – 6) = 0 – 4(k2 – 5k – 6) = 0 k2 – 5k – 6 = 0 (k + 1)(k – 6) = 0 k = –1 or k = 6 (c) k2x2 – (k + 2)x + 1 = 0 (–k – 2)2 – 4(k2)(1) = 0 k2 + 4k + 4 – 4k2 = 0 3k2 – 4k – 4 = 0 (3k + 2)(k – 2) = 0 k = –

23

or k = 2

(d) (4k – 14)2 – 4(2k + 3)(16k + 1) = 0 16k2 – 112k + 196 – 4(32k2 + 50k + 3) = 0 112k2 + 312k – 184 = 0 14k2 + 39k – 23 = 0 (7k + 23)(2k – 1) = 0

k = – 237

or k = 12

(e) (k – 3)x2 + 4k + 1 = 4kx – 6x (k – 3)x2 + 6x – 4kx + 4k + 1 = 0 (6 – 4k)2 – 4(k – 3)(4k + 1) = 0 36 – 48k + 16k2 – 16k2 + 44k + 12 = 0 –4k + 48 = 0 4k = 48 k = 12

33 (a) 2x2 + 4c + 3 = 0 02 – 4(2)(4c + 3) 0 –32c – 24 0 32c + 24 0

c – 34

(b) cx2 + 5x + 1 = 0 52 – 4c(1) 0 –4c –25

c

254

(c) x2 + 2cx + c2 – 5c + 7 = 0 (2c)2 – 4(1)(c2 – 5c + 7) 0 4c2 – 4(c2 – 5c + 7) 0 20c – 28 0

c 75

(d) x2 – 2x + 2 – c = 0 (–2)2 – 4(1)(2 – c) 0 4c – 4 0 c 1

34 (a) (m + 2)x2 – 2mx + m – 5 = 0 (–2m)2 – 4(m + 2)(m – 5) 0 4m2 – 4m2 + 12m + 40 0 12m + 40 0 m –

103

(b) 3x2 – 6x + m = 0 (–6)2 – 4(3)m 0 36 – 12m 0 –12m –36 m 3 (c) 5x2 + 4x + 6 – 3m = 0 16 – 4(5)(6 – 3m) 0 16 – 120 + 60m 0 60m 104

m 2615

(d) x2 – 2mx + m2 + 5m – 6 = 0 (–2m)2 – 4(1)(m2 + 5m – 6) 0 4m2 – 4(m2 + 5m – 6) 0 –20m + 24 0 –20m –24

m 65

(e) x2 + 2mx + (m – 1)(m – 3) = 0 (2m)2 – 4(1)(m – 1)(m – 3) 0 4m2 – 4(m2 – 4m + 3) 0 16m – 12 0 16m 12

m 34

35 (a) x + p = 7x – px2

px2 – 6x + p = 0 (–6)2 – 4(p)(p) = 0 36 – 4p2 = 0 4p2 = 36 p2 = 9 p = 3 \ p = –3 or p = 3 (b) x + py = 10

y = 10 – x

p

Substitute y =

10 – x

p into x2 + y2 = 10:

x2 +

(10 – xp )2

= 10

x2 + 100 – 20x + x2

p2 = 10

p2x2 + x2 – 20x + 100 = 10p2

(1 + p2)x2 – 20x + 100 – 10p2 = 0 (–20)2 – 4(1 + p2)(100 – 10p2) = 0 400 – 4(1 + p2)(100 – 10p2) = 0 400 – 4(100 + 90p2 – 10p4) = 0 –360p2 + 40p4 = 0 p2 – 9 = 0 (p + 3)(p – 3) = 0 \ p = –3 or p = 3 (c) 2y + x = p

y = p – x

2

Substitute y =

p – x

2 into y2 + 4x = 20:

( p – x2 )2

+ 4x = 20


p2 – 2px + x2

4 + 4x = 20

x2 – 2px + p2 + 16x = 80 x2 + (16 – 2p)x + p2 – 80 = 0 (16 – 2p)2 – 4(1)(p2 – 80) = 0 256 – 64p + 4p2 – 4p2+ 320 = 0 576 – 64p = 0 64p = 576 p = 9 (d) px – 1 = x2 – 2x + 3 x2 – 2x – px + 4 = 0 x2 + (–p – 2)x + 4 = 0 (–p – 2)2 – 4(1)(4) = 0 p2 + 4p + 4 – 16 = 0 p2 + 4p – 12 = 0 (p + 6)(p – 2) = 0 p = –6 or p = 2 (e) (p – 2)y = 3x

y =

3x

p – 2

Substitute y =

3x

p – 2 into xy =

1 – x:

x( 3xp – 2)

= 1 – x

3x2 = (1 – x)(p – 2) 3x2 = p – 2 – px + 2x 3x2 + (p – 2)x + 2 – p = 0 (p – 2)2 – 4(3)(2 – p) = 0 p2 – 4p + 4 – 24 + 12p = 0 p2 + 8p – 20 = 0 (p + 10)(p – 2) = 0 p = –10 or p = 2

36 (a) x + 2y = 3

y =

3 – x

2

Substitute y =

3 – x

2 into x2 + y2 = h:

x2 +

( 3 – x2 )2

= h

x2 +

9 – 6x + x2

4 = h

4x2 + 9 – 6x + x2 = 4h 5x2 – 6x + 9 – 4h = 0 (–6)2 – 4(5)(9 – 4h) 0 36 – 180 + 80h 0 80h 144

h

95

(b) x – h = hx2 + 2hx hx2 + (2h – 1)x + h = 0 (2h – 1)2 – 4(h)(h) 0 4h2 – 4h + 1 – 4h2 0 –4h –1

h 14

(c) 2x + h = x2 – x + 1 x2 – 3x + 1 – h = 0 (–3)2 – 4(1)(1 – h) 0 9 – 4(1 – h) 0 9 – 4 + 4h 0 4h –5 h –

54

(d) 2h + 1 = x +

h2

x 2hx + x = x2 + h2

x2 – 2hx – x + h2 = 0 x2 + (–2h – 1)x + h2 = 0 (–2h – 1)2 – 4(1)(h2) 0 4h2 + 4h + 1 – 4h2 0 4h –1

h – 14

(e) 3x + h = 5 – 3x – 2x2

2x2 + 6x + h – 5 = 0 62 – 4(2)(h – 5) 0 36 – 8(h – 5) 0 36 – 8h + 40 0 –8h –76

h 192

37 (a) 2a – x = (x – 1)2 + a 2a – x = x2 – 2x + 1 + a x2 – x + 1 – a = 0 (–1)2 – 4(1)(1 – a) 0 –3 + 4a 0 4a 3

a

34

(b) ax – 3 = x + 1x

ax2 – 3x = x2 + 1 (a – 1)x2 – 3x – 1 = 0 (–3)2 – 4(a – 1)(–1) 0 9 + 4(a – 1) 0 9 + 4a – 4 0 4a –5

a – 54

(c) x + 3y = a

y =

a – x

3

Substitute y =

a – x

3 into y2 =

2x + 3:

( a – x3 )2

= 2x + 3

a2 – 2ax + x2 = 18x + 27 x2 + (–2a – 18)x + a2 – 27 = 0 (–2a – 18)2 – 4(1)(a2 – 27) 0 4a2 + 72a + 324 – 4a2 + 108 0 72a + 432 0 a –6 (d) Substitute y = ax + 6 into 2x2 – xy

= 3: 2x2 – x(ax + 6) = 3 2x2 – ax2 – 6x – 3 = 0 (2 – a)x2 – 6x – 3 = 0 (–6)2 – 4(2 – a)(–3) 0 36 + 12(2 – a) 0 36 + 24 – 12a 0 –12a –60 a 5 (e) 5x – a = x 2 + 3x + 3 x2 – 2x + 3 + a = 0 (–2)2 – 4(1)(3 + a) 0 4 – 4(3 + a) 0 –8 – 4a 0

– 4a 8 4a –8 a –2

38 kx2 + 3hx + 2h = 0 (3h)2 – 4(k)(2h) = 0 9h2 – 4k(2h) = 0 9h2 = 8hk h =

89

k

39 3cx2 – 7dx + 3c = 0 (–7d)2 – 4(3c)(3c) = 0 49d2 – 4(3c)(3c) = 0 49d2 – 36c2 = 0 36c2 = 49d2

( cd )

2

=

4936

cd

= 76

\ c : d = 7 : 6

40 (a) x2 + kx + 2k – 4 = 0 b2 – 4ac = k2 – 4(1)(2k – 4) = k2 – 8k + 16 = (k – 4)2

b2 – 4ac 0 for all values of k. Hence, the roots are real. (b) x2 + (1 – k)x – k = 0 b2 – 4ac = (1 – k)2 – 4(1)(–k) = 1 – 2k + k2 + 4k = k2 + 2k + 1 = (k + 1)2

b2 – 4ac 0 for all values of k. Hence, the roots are real.

1 4x – 53

=

x2 + 8x

4x2 – 5x = 3x2 + 24 x2 – 5x – 24 = 0 (x – 8)(x + 3) = 0 x = 8 or x = –3

2 3x2 – x – 10 = x2 – 2x 2x2 + x – 10 = 0 (2x + 5)(x – 2) = 0

x = – 52

or x = 2

3 x2 + 10x + 25 – 3x – 12 = 15 x2 + 7x – 2 = 0

x =

–7 72 – 4(1)(–2)

2(1)

=

–7 49 – 4(–2)

2

=

–7 57

2 = 0.275 or –7.275 \ x = 0.275 or x = –7.275

4 x2 + 2x – 8 = 6 x2 + 2x = 14


x2 + 2x + (1)2 = 14 + (1)2

(x + 1)2 = 15 x = 15 – 1 or – 15 – 1 = 2.873 or – 4.873 \ x = 2.873 or x = – 4.873

5 2x2 – 4x – 3 = 0 a + b = 2

ab = –

32

a + 3 + b + 3 = a + b + 6 = 2 + 6 = 8 (a + 3)(b + 3) = ab + 3(a + b) + 9

= –

32

+ 6 + 9

=

272

\ x2 – 8x +

272

= 0

2x2 – 16x + 27 = 0

6 mx2 + (m – 5)x – 20 = 0

a + (–a) = –(m – 5)

m 0 = 5 – m m = 5

7 2x2 + px + p – 1 = 0

a + b = –

p2

ab =

p – 12

a2 + b2 = 4 (a + b)2 – 2ab = 4 (–

p2 )2

– 2( p – 1

2 ) = 4

p2

4 – p + 1 = 4

p2

4 – p = 3

p2 – 4p – 12 = 0 (p – 6)(p + 2) = 0 p = 6 or p = –2

8 3x2 – x + 2 = 0

a + b =

13

ab =

23

a2 + b 2 = (a + b)2 – 2ab

=

( 13 )2

– 2( 2

3 )

=

19

– 43

= – 119

9 2x2 – 9k = 2 – 3k

2( 12 )2

– 9k = 2 – 3k

12

– 2 = 6k

– 32

= 6k

k = – 14

10 b2 – 4ac = 121 (–9)2 – 4(2)(h) = 121 81 – 4(2)(h) = 121 –8h = 40 h = –5

11 a( 1a)

=

m – 53

3 = m –5 m = 8

12 a + 3a = – 4 4a = – 4 a = –1 a(3a) = c – 4 When a = –1, 3 = c – 4 c = 7

13 (a) Substitute x = 5 into x2 + (m + 1)x + 3m + 2: 25 + (m + 1)(5) + 3m + 2 = 0 8m + 32 = 0 8m = –32 m = – 4 (b) x2 – 3x – 10 = 0 (x – 5)(x + 2) = 0 x = 5 or x = –2 \ x = –2

14 a + 2a =

3m

3a =

3m

a =

1m

… 1

2a2 =

1

2m … 2


2( 1m)2

=

1

2m

2

m2 =

1

2m m2 – 4m = 0 m(m – 4) = 0 m = 0 or m = 4 \ m = 4

15 4x2 – px + 25 = 0 (–p)2 – 4(4)(25) = 0 p2 – 400 = 0 (p + 20)(p – 20) = 0 p = –20 or p = 20

16 x2 – 5x + 4p – 2 = 0 a + b = 5 … 1 ab = 4p – 2 … 2 a – b = 9 … 3 1 + 3 : 2a = 14 a = 7 Substitute a = 7 into 1 : 7 + b = 5 b = –2 Substitute a = 7 and b = –2 into 2 : 7(–2) = 4p – 2

4p = –12 p = –3

17 p + q = –p 2p = –q

p = –

q2

… 1

p – q = 6 … 2 Substitute 1 into 2 :

– q2

– q = 6

–3q = 12 q = –4

When q = –4, p =

–(–4)

2 = 2 \ pq = 2(–4) = –8

18 x2 + (2p – 1)x + p2 + 3p – 4 = 0 (2p – 1)2 – 4(1)(p2 + 3p – 4) 0 4p2 – 4p + 1 – 4p2 – 12p + 16 0 17 – 16p 0 16p 17

p 1716

19 k + 3x = 5 – 3x – 2x2

2x2 + 6x + k – 5 = 0 (6)2 – 4(2)(k – 5) = 0 36 – 8(k – 5) = 0 76 – 8k = 0

k = 9 12

20 a + 3a = – m4

4a = – m4

a = – m

16 … 1

3a2 = n4

… 2


3(– m16)2

=

n4

3m2

256 =

n4

n = 3m2

64

21 (5)2 – 4(1)(3 – c) 0 25 – 4(3 – c) 0 13 + 4c 0 4c –13

c

– 13

4

22 q2 – 4p(–9) = 0 q2 + 36p = 0

p = – 136

q2

23 x2 + (2x + k)2 = 5 x2 + 4x2 + 4kx + k2 = 5 5x2 + 4kx + k2 – 5 = 0


(4k)2 – 4(5)(k2 – 5) = 0 16k2 – 20(k2 – 5) = 0 – 4k2 + 100 = 0 k2 = 25 k = 5

24 a + b = 2

ab =

h2

a2 + b 2 = –k (a + b)2 – 2ab = –k

(2)2 – 2( h

2 ) = –k

4 – 2( h2 )

= –k

h – k = 4 … 1 (ab)2 = 16

h2

4 = 16

h2 = 64 h = 8 ( 0) Substitute h = 8 into 1 : 8 – k = 4 k = 4 \ h = 8, k = 4

25 2c – x – (x – 1)2 = c c – x – (x2 – 2x + 1) = 0 x2 – x + 1 – c = 0 (–1)2 – 4(1)(1 – c) 0 1 – 4(1 – c) 0 –3 + 4c 0 4c 3

c

34

26 a + (a + 3k) = 8 2a = 8 – 3k

a =

8 – 3k

2 … 1

a(a + 3k) = h … 2 Substitute 1 into 2 :

(8 – 3k2 )(8 – 3k

2 + 3k)

= h

(8 – 3k2 )(8 + 3k

2 ) = h

64 – 9k2 = 4h

h =

64 – 9k2

4

27 x2 – 2mx + m2 + 5m– 6 = 0 (a) (–2m)2 – 4(1)(m2 + 5m – 6) = 0 4m2 – 4(m2 + 5m – 6) = 0 24 – 20m = 0

m = 65

(b) 24 – 20m 0 20m 24

m 65

(c) 24 – 20m 0 20m 24

m 65

28 2x2 – 5x – 4 = 0

a + b =

52

ab = –2 (a) a2 + b2 = (a + b)2 – 2ab

=

( 52 )2

– 2(–2)

=

254

+ 4

= 10 14

(b)

2ab

+ 2ba

=

2(a2 + b 2)ab

=

2(414 )

–2

= –

414

(2ab )( 2b

a ) =

4abab

= 4

\ x2 +

414

x + 4 = 0

4x2 + 41x + 16 = 0

29 3x2 + 2mx – m = 0 (a) (2m)2 – 4(3)(–m) = 0 4m2 + 12m = 0 4m(m + 3) = 0 m = 0 or m = –3 \ m = –3

(b) a( 1

a ) = – m

3 3 = –m m = –3

(c) a + 3a =

– 2m

3

4a =

– 2m

3

a = –

16

m

… 1

3a2 =

– m

3

a2 = – m

9 … 2


m2

36 =

– m

9 m = – 4

30 (a) 2(x2 – x – 2) = p 2x2 – 2x – 4 – p = 0 a + b = 1 … 1 a – b = 5 … 2 1 + 2 : 2a = 6 a = 3 Substitute a = 3 into 1 : 3 + b = 1 b = –2

ab = – 4 – p

2

– 6 = – 4 – p

2 –12 = – 4 – p p = 8

(b) mx2 + nx – k = 0

a + b = –

nm

ab = –

km

kx2 – mx – 1 – n = 0

a + b =

mk

ab =

–1 – n

k

mk

= – nm

so, m2 = – kn

k = – m2

n … 1

and –1 – nk

= – km

– m – mn = – k2

k2 = m(1 + n) … 2


(– m2

n )2

= m(1 + n)

m4

n2 = m(1 + n)

m3 = n2(1 + n)

31 x2 + hx + k = 0 (a) –2 + 6 = –h h = – 4 –2(6) = k k = –12 \ h = – 4, k = –12 (b) x2 – 4x – 12 – p = 0 (– 4)2 – 4(1)(–12 – p) 0 16 – 4(–12 – p) 0 64 + 4p 0 p –16

32 (a) 2x2 + kx – 12 = 0 When x = – 4, 2(– 4)2 + k(– 4) – 12 = 0 32 – 4k – 12 = 0 4k = 20 k = 5 (b) 2x2 + 5x – 12 = 0 (2x – 3)(x + 4) = 0

x = 32

or x = – 4

33 (a) 2x2 – 12x + k = 0

x2 – 6x +

k2

= 0

a + 5a = 6 6a = 6 a = 1

5a2 =

k2

k = 10 (b) 2x2 – 12x + 10 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x = 1 or x = 5 \ The roots are 1 and 5.


37 (a)

12

[(x + 3) + (2x – 2)](x – 1) = 112

12

(3x + 1)(x – 1) = 112

3x2 – 2x – 1 = 224 3x2 – 2x – 225 = 0 (b) (3x + 25)(x – 9) = 0

x = – 253

(Reject) or x = 9

The length of QR = 2x – 2 = 2(9) – 2 = 16 cm

38 (a) (3x + 2)2 + (2x + 2)2 = (5x)2

9x2 + 12x + 4 + 4x2 + 8x + 4 = 25x2

12x2 – 20x – 8 = 0 3x2 – 5x – 2 = 0 (shown) (b) (3x + 1)(x – 2) = 0

x = –

13

(Reject) or x = 2

\ x = 2

(c) Area =

12

[3(2) + 2][2(2 + 1)]

=

12

(8)(6)

= 24 cm2

39 (a) 2x2 – 4x + 5 = 0 a + b = 2

ab =

52

(b) 2a + 1 + 2b + 1 = 2(a + b) + 2 = 2(2) + 2 = 6 (2a + 1)(2b + 1) = 4ab + 2(a + b) + 1

= 4( 5

2 ) + 2(2)

+ 1 = 15 \ x2 – 6x + 15 = 0

40 (a) 2x2 – mx + 3 = 0

a + b =

m2

ab =

32

9x2 – 52x + 4 = 0

1a2

+ 1b 2

=

529

a2 + b2

(ab)2 =

529

(a + b)2 – 2ab

(ab)2 =

529

(m2 )2

– 2( 3

2 )( 3

2 )2

=

529

m2 – 12

9 =

529

m2 = 64 m = 8 \ m = 8 (b) 2x2 – 6x – 3 = 0 a + b = 3

ab = –

32

(i) a2 + b 2 = (a + b)2 – 2ab

= (3)2 – 2(–

32 )

= 9 – 2(–

32 )

= 12 (ii) (a – b)2 = a2 – 2ab + b2

= a2 + b2 – 2ab

= 12 – 2(– 32 )

= 15 a – b = 15 \ a – b = 15 (a > b)

41 (a) ( x2 + (17 – x)2)( x2 + (17 – x)2) = 169 x2 + 289 – 34x + x2 = 169 2x2 – 34x + 120 = 0 x2 – 17x + 60 = 0(shown) (b) (x – 5)(x – 12) = 0 x = 5 or x = 12

34 (a) x2 – 4x – 1 = 2mx – 10m x2 + (– 4 – 2m)x + 10m – 1 = 0 (– 4 – 2m)2 – 4(1)(10m – 1) = 0 16 + 16m + 4m2 – 40m + 4 = 0 4m2 – 24m + 20 = 0 m2 – 6m + 5 = 0 (m – 1)(m – 5) = 0 m = 1 or m = 5 (b) 2x2 + 5x + 3 – k = 0 (5)2 – 4(2)(3 – k) 0 25 – 8(3 – k) 0 1 + 8k 0

k – 18

35 1 – c = 3x2 – 2x 3x2 – 2x + c – 1 = 0 (a) (–2)2 – 4(3)(–1) 0 4 – 12(c – 1) 0 16 – 12c 0 12c 16

c 43

(b) 16 – 12c = 0 12c = 16

c = 43

(c) 16 – 12c 0 12c 16

c 43

36 3mx2 – 7nx + 3m = 0 (a) (–7n)2 – 4(3m)(3m) = 0 49n2 – 36m2 = 0 49n2 = 36m2

m2

n 2 =

4936

(mn )2

=

4936

mn

= 76

\ m : n = 7 : 6 (b) 3(7)x2 – 7(6)x + 3(7) = 0 21x2 – 42x + 21 = 0 x2 – 2x + 1 = 0 (x – 1)(x – 1) = 0 x = 1

�

1 (a) Quadraticfunction,a=1,b=–4,c=5 (b) Quadraticfunction,a=–3,b=2,c=7 (c) f(x)=x2–7x Quadraticfunction,a=1,b=–7,c=0 (d) f(x)=6+4x–x3

Notaquadraticfunction.

(e) f(x)=x–4x

+2

Notaquadraticfunction. (f) Notaquadraticfunction.

2 (a)

(b)

f(x)

xO–2 –1 1 2 3 4 5

–5

–10

–15

–20

–25

–35

3 (a) x –2 –1 0 1 2 3 4 5

f(x) –15 –3 5 9 9 5 –3 –15

f(x)

10

5

O–1–2 1 2 3 4 5x

–5

–10

–15Axisofsymmetryx=1.5

(1.5,9.5) Maximumpoint

(b) x –2 –1 0 1 2 3 4 5 6

f(x) 6 2.5 0 –1.5 –2 –1.5 0 2.5 6

f(x)

6

5

4

3

2

1

O–1–2 1 2 3 4 5 6x

–1

–2(2,-2) Minimumpoint

Axisofsymmetryx=2

4 (a) Twodistinctroots. (b) Noroots. (c) Twoequalroots. (d) Twodistinctroots.

5 (a) b2–4ac = (–7)2–4(2)(9) = 49–4(2)(9) = –23(0) ∴Noroots. (b) b2–4ac = (–8)2–4(1)(16) = 64–4(16) = 0 ∴Twoequalroots.

5

f(x)

25

20

15

10

5

O–3 –2 –1 1 2 3 4x

–5



(c) b2–4ac = (–4)2–4(–1)(3) = 16–4(–1)(3) = 16+12 = 28(0) ∴Twodistinctroots. (d) f(x) = x2–4x+1 b2–4ac = (–4)2–4(1)(1) = 16–4 = 12(0) ∴Twodistinctroots.

6 (a) f(x)=x2+mx+2m–3 m2–4(1)(2m–3) = 0 m2–8m+12 = 0 (m–2)(m–6) = 0 m=2orm=6 (b) (–4m–2)2–4(m+2)(4m) = 0 16m2+16m+4–16m2–32m = 0 4–16m = 0

m =14

(c) f(x) = x2–4x–2mx+10m–1 = x2+(–4–2m)x+10m–1 (–4–2m)2–4(1)(10m–1)= 0 16+16m+4m2–40m+4= 0 4m2–24m+20= 0 m2–6m+5= 0 (m–1)(m–5)= 0 m=1orm=5 (d) (4m–8)2–4(m+1)(2m) = 0 16m2–64m+64–8m2–8m = 0 8m2–72m+64 = 0 m2–9m+8 = 0 (m–1)(m–8) = 0 m=1orm=8

7 (a) (–2k)2–4(1)(3k) 0 4k2–12k 0 4k(k–3) 0 k0ork3 (b) (4k)2–4(5)(k2–20) 0 16k2–20k2+400 0 400–4k2 0 k2–100 0 (k+10)(k–10) 0 –10k10 (c) (2k2)–4(1)(k2–5k+7) 0 4k2–4(k2–5k+7) 0 20k–28 0 20k 28

k 75

(d) f(x)=kx2+2kx+k–4 (2k)2–4(k)(k–4)0 4k2–4k(k–4)0 16k0 k0

8 (a) f(x) = 2x2–px2–6x–3 = (2–p)x2–6x–3 (–6)2–4(2–p)(–3)0 36+12(2–p)0 60–12p0 12p–600 12p60 p5 (b) f(x) = x2+4x2+4px+p2–5 = 5x2+4px+p2–5

(4p)2–4(5)(p2–5)0 16p2–20(p2–5)0 –4p2+1000 4p2–1000 p2–250 p–5orp5 (c) (–4p)2–4(p)(4p–5) 0 16p2–4p(4p–5) 0 20p 0 p 0 (d) f(x)=(p+2)x2–2px+p–5 (–2p)2–4(p+2)(p–5)0 4p2–4(p2–3p–10)0 12p+400

p–103

9 (5)2–4(2)(3–k) 0 25–8(3–k) 0 25–24+8k 0 1+8k 0 8k –1

k –18

(shown)

10 (2m)2–4(1)(m–1)(m–3) 0 4m2–4(m2–4m+3) 0 16m–12 0

m 34

(shown)

11 (a) Minimumvalue=5,x=3

(b) Maximumvalue=4,x=12

(c) Maximumvalue=–3,x=6

(d) Minimumvalue=–7,x=–32

(e) Minimumvalue=1,x=12

(f) Maximumvalue=23

,x=–1

12 (a) y = 2(x2–2x+52 )

= 2[x2–2x+(–1)2+52

– (–1)2] = 2[(x–1)2+

32 ]

= 2(x–1)2+3 ∴Minimumvalue=3,x=1 (b) y = –x2–2x+3 = –(x2+2x–3) = –[x2+2x+(1)2–3–(1)2] = –[(x+1)2–4] = 4–(x+1)2

∴Maximumvalue=4,x=–1 (c) y = x2+3x+4

= x2+3x+( 32 )2

+4–(32 )2

= (x+32 )2

+74

∴Minimumvalue=74

,x=–32

(d) y = 3(x2+2x–43 )

= 3[x2+2x+(1)2–43

–(1)2] = 3[(x+1)2–

73 ]

= 3(x+1)2–7 ∴Minimumvalue=–7,x=–1 (e)

y = –2(x2–52

x–12 )

= –2[x2–52

x+(– 54 )2

–12

–(– 54 )2]

= –2[(x–54 )2

–3316 ]

=338

–2(x–54 )2

∴Maximumvalue=338

,x=54

(f) y = –(x2–4x–5) = –[x2–4x+(–2)2–5–(–2)2] = –[(x–2)2–9)] = 9–(x–2)2

∴Maximumvalue=9,x=2

13 y = 4(x2+2x–94 )

= 4[x2+2x+(1)2–94

–(1)2] = 4[(x+1)2–

134 ]

= 4(x+1)2–13 ∴Minimumvalue=–13,x=–1

14 f(x) = 2(x2–3x+52 )

= 2[x2–3x+(– 32 )2

+52

–(– 32 )2]

= 2[(x–32 )2

+14 ]

= 2(x–32 )2

+12

(a) a=2,p=32

,q=12

(b) ( 32

,12 )

15 f(x) = x2+5x–6

= x2+5x+( 52 )2

–6–( 52 )2

= (x+ 52 )2

–494

∴Theleastvalue=–494

,x=–52

16 f(x) = x2–4x–k = x2–4x+(–2)2–k–(–2)2

= (x–2)2–k–4 ∴–k–4 = –9 k+4 = 9 k = 5

17 f(x) = –2x2+px+10

= –2(x2–p2

x–5) = –2[x2–

p2

x+(– p4 )2

–5–(– p4 )2]

= –2[(x–p4 )2

–5–p2

16] =

80+p2

8–2(x–

p4 )2

∴ 80+p2

8 = 18

80+p2 = 144 p2 = 64 p =±8


18 f(x) = –x2+4x+12 = –(x2–4x–12) = –[x2–4x+(–2)2–12–(–2)2] = –[(x–2)2–16] = 16–(x–2)2

∴Thegreatestvalueoff(x)is16.

19 y =12

(x2+10x+25+x2–14x+49)

=12

(2x2–4x+74)

= x2–2x+37 = x2–2x+(–1)2+37–(–1)2

= (x–1)2+36 ∴Minimumvalue=36,x=1

20 f(x) = x2+6mx+144 = x2+6mx+(3m)2+144–(3m)2

= (x+3m)2+144–9m2

∴h=3mandk=144–9m2

21 (a) y = –(x2+x–6)

= –[x2+x+( 12 )2

–6–( 12 )2]

= –[(x+12 )2

–6–14 )]

=254 –(x+

12 )2

y

x2O–3

6

–12

254

Maximumpoint=(– 12

,254 )

(b) y = 2(x2+72

x–2) = 2[x2+

72 x+( 7

4 )2

–2–( 74 )2]

= 2[(x+74 )2

–8116 ]

= 2(x+74 )2

–818

y

xO

–4

–4

–818

12

–74

Minimumpoint=(– 74

,–818 )

(c) y = –(x2–4x–5) = –[x2–4x+(–2)2–5–(–2)2] = –[(x–2)2–9] = 9–(x–2)2

y

x

9

5

–1 O 2 5

Maximumpoint=(2,9)

(d) y = –3(x2–4x–53 )

= –3[x2–4x+(–2)2–53 –(–2)2]

= –3[(x–2)2–173 ]

= 17–3(x–2)2

y

x

17

O 2

5

Maximumpoint=(2,17)

22 y = –(x2+3x–12)

= –[x2+3x+( 32 )2

–12–( 32 )2]

= –[(x+32 )2

–12–94 ]

=574 –(x+

32 )2

y

x

574

O–

32

12

23 (a) y=a(x–1)2+3 At(0,5), 5 = a+3 a = 2 ∴y=2(x–1)2+3 (b) y=a(x–1)2+8 At(0,6), 6 = a+8 a = –2 y=–2(x–1)2+8 (c) y=a(x–2)2–1 At(1,0), 0 = a–1 a = 1 ∴y=(x–2)2–1 (d) y=a(x+2)2+18 At(0,10), 10 = 4a+18 4a = –8 a = –2 y=–2(x+2)2+18

24 y = 2(x2+4x+52 )

= 2[x2+4x+(2)2+52

–(2)2]

= 2[(x+2)2–32 ]

= 2(x+2)2–3 ∴Theminimumvalueis–3.

25 y = –x2–4x+5 = –(x2+4x–5) = –[x2+4x+(2)2–5–(2)2] = –[(x+2)2–9] = 9–(x+2)2

∴Themaximumvalueis9.

26 b2–4ac = 16–4(–1)(–6) = 16–24 = –8(0) Thecurveisaparabolawitha

maximumpointanddoesnotintersectthex-axis,soyisalwaysnegativeforallrealvaluesofx.

27 b2–4ac= 16–4(3)(2) = –8(0) Thecurveisaparabolawitha

minimumpointanddoesnotintersectthex-axissoyisalwayspositiveforallvalueofx.

28 y = a(x+2)2+5 7 = a+5 a = 2 y = 2(x+2)2+5

–2x

O

13

5

y

29 y = 3(x2–2x+113 )

= 3[x2–2x+(–1)2+113 –(–1)2]

= 3[(x–1)2+83 ]

= 3(x–1)2+8

y

xO 1

8

11

Turningpoint=(1,8) y-intercept=11

30 (a) y = 4(x2–3x–74 )

= 4[x2–3x+(– 32 )2

–74 –(– 3

2 )2] = 4[(x–

32 )2

–74 –

94 ]

= 4(x–32 )2

–16


y

x

x-intercept

72

32

–12–16

x-intercept

y-intercept

O–7

Minimumpoint=( 32

,–16) (b) y = –2(x2–

12

x–212 )

= –2[x2–12

x+(– 14 )2

–212

–(– 14 )2]

= –2[(x–14 )2

–16916 ]

=1698

–2(x–14 )2

y

x14

72

–3 O

1698

21

x-intercept

y-intercept

x-intercept

Maximumpoint=( 14

,1698

)31 (a) 2x2–x–60 (2x+3)(x–2)0

–

32

2x

∴x–32

orx2

(b) 2x2–3x–20 (2x+1)(x–2)0

–

12

2x

∴x–12

orx2

(c) x2–x–60 (x–3)(x+2)0

–2 3

x

∴x–2orx3 (d) 2x2+5x–120 (2x–3)(x+4)0

–4 32

x

∴–4x32

(e) 2x2+7x–220 (2x+11)(x–2)0

–

112

2x

∴–112

x2

(f) 2x2–7x–40 (2x+1)(x–4)0

–

12

x4

∴x–12

orx4

(g) 4(4x2–12x+9)x2

15x2–48x+360 5x2–16x+12 0 (5x–6)(x–2) 0

65

x2

∴65

x2

(h) x2+4x+4 2x+7 x2+2x–3 0 (x–1)(x+3) 0

–3 1 x

∴x–3orx1

32 (a) x2–4x–5 = 0 (x+1)(x–5) = 0 x=–1orx=5 ∴A(–1,0);B(5,0) (b) (i) x–1orx5 (ii) –1x5

33 2x2–7x+30 (2x–1)(x–3)0

12

3x

∴x12

orx3

34 3x2–5x+1–x2

4x2–5x+10 (4x–1)(x–1)0

14

1x

∴x14

orx1

35 –1x2–4x+27 x2–4x+2 –1 x2–4x+3 0 (x–1)(x–3) 0 x1orx3

x2–4x+27 x2–4x–50 (x+1)(x–5)0 –1x5 ∴–1x1or3x5

36 (a) (x+3)(x–5)0 x2–2x–150 x2–2x15 ∴a=2,b=15 (b) (x+2)(x–4) 0 x2–2x–8 0 x2–8 2x 2x2–16 4x ∴a=–16,b=4

37 x2+xy+8 = 0

x2+x(k–x2 )+8 = 0

2x2+kx–x2+16 = 0 x2+kx+16 = 0 k2–4(1)(16) 0 k2–64 0 (k+8)(k–8) 0 k–8ork8

38 (–m–4)2–4(1)(1) 0 m2+8m+12 0 (m+6)(m+2) 0

–6 –2x

∴m–6orm–2

39 (4–k)2–4(2–3k)(2) 0 16–8k+k2–16+24k 0 k2+16k 0 k(k+16) 0 ∴–16k0

40 (c–4)2–4(1)(1) 0 c2–8c+12 0 (c–2)(c–6) 0 ∴2c6

41 x2–x(2x–p)+(2x–p)2 = 1 x2–2x2+px+4x2–4px+p2–1 = 0 3x2–3px+p2–1 = 0 (–3p)2–4(3)(p2–1) 0 9p2–12(p2–1) 0 9p2–12p2+12 0 12–3p2 0 3p2–12 0 3p2 12 p2 4 ∴p–2orp2

42 hx–9= x2+3x x2+(3–h)x+9 = 0 (3–h)2–4(1)(9)0 9–6h+h2–360 h2–6h–270 (h+3)(h–9)0 ∴h–3orh9

43 x2+(mx+2)2 = 2 x2+m2x2+4mx+2 = 0


(1+m2)x2+4mx+2 = 0 16m2–4(1+m2)(2)0 16m2–8–8m20 8m2–80 m21 ∴m–1orm1

44 x2+4x2–20x+25

a2 = 5

a2x2+4x2–20x+25–5a2 = 0 (a2+4)x2–20x+25–5a2 = 0 (–20)2–4(a2+4)(25–5a2) 0 400–4(25a2–5a4+100–20a2) 0 20a4–100a2+80a2 0 20a4–20a2 0 a2–1 0 (a+1)(a–1) 0 ∴–1a1

45 (–p)2–4(1)(p+3)0 p2–4(p+3)0 p2–4p–120 (p+2)(p–6)0 ∴p–2orp6(shown)

1 (a) y=a(x–1)2–18 At(–2,0), 0 = a(–2–1)2–18 0 = a(–3)2–18 9a = 18 a = 2 ∴a=2,p=1,q=18 (b) y=2(x–1)2–18 AtA,x=0, y = 2(–1)2–18 = –16 ∴A(0,–16)

2 b2–4ac0 (m–1)2–4(1)(4)0 m2–2m+1–160 m2–2m–150 (m+3)(m–5)0 ∴m–3orm5

3 (p–1)2–4(1)(–p+4)0 p2–2p+1+4p–160 p2+2p–150 (p+5)(p–3)0 ∴–5p3

4 (a) y=a(x–1)2+4 At(3,0), 0 = a(3–1)2+4 4a = –4 a = –1 ∴a=–1,p=1,q=4 (b) x=1

5 (a) x=2 (b) y=a(x–2)2+1 At(0,9), 9 = 4a+1 4a = 8 a = 2 ∴y=2(x–2)2+1

6 p=–2 At(0,3), 3 = p2+q 3 = (–2)2+q q = –1 ∴p=–2,q=–1

7 (–m–3)2–4(–1)(3–4m) 0 m2+6m+9+12–16m 0 m2–10m+21 0 (m–3)(m–7) 0 ∴3m7

8 y = –x2+4x+1 = –(x2–4x–1) = –[x2–4x+(–2)2–1–(–2)2] = –[(x–2)2–5] = 5–(x–2)2

∴Maximumvalueofyis5and x=2.

9 x2+5x+6x+6 x2+4x0 x(x+4)0

–4 0x

∴x–4orx0

10 x2–7x+100 (x–2)(x–5)0 2x5 ∴p=2,q=5

11 (a) A(–1,3) (b) y=a(x+1)2+3 At(0,1), 1= a+3 a = –2 ∴y=–2(x+1)2+3

12 (a) f(x)=x2–8x+12 Atx-axis, f(x) = 0 x2–8x+12 = 0 (x–2)(x–6) = 0 x=2orx=6 Aty-axis,x=0 f(x) = 02–8(0)+12 = 12 ∴a=2,b=6,c=12 (b) f(x)=(x–4)2–4

13 f(x)= –x2+nx+m = –(x2–nx–m)

= –[x2–nx+(– n2 )2

–m–(– n2 )2]

= –[(x–n2 )2

–m–n2

4 ] = 4m+n2

4–(x–n

2 )2 ∴ n

2= 2 and 4m+(4)2

4= 5

n = 4 4m+16 = 20 4m = 4 m = 1 ∴m=1,n=4

14

f(x)=k–(x–h)2

x(3,0)(1,0)

x=2

O

∴h=2 At(1,0), 0= k–(1–2)2

0= k–1 k = 1 ∴h=2,k=1

15 f(x)=ax2+bx+c At(0,10), 10= a(0)2+b(0)+c c = 10

f(x) = a(x2+ba

x+ ca )

= a[x2+ba

x+( b2a)2

+ ca

–( b2a)2]

= a[(x+ b2a)2

+4ac–b2

4a2 ] = a(x+ b

2a)2

+4ac–b2

4a

– b2a

= 2

–b = 4a b = –4a… 1

4ac–b2

4a = 18

40a–b2 = 72a… 2 Substitute 1 into 2 : 40a–(–4a)2 = 72a 40a–16a2 = 72a 16a2+32a = 0 16a(a+2) = 0 a=0(reject)ora=–2 Substitutea=–2into 1 : b = –4(–2) = 8 ∴a=–2,b=8,c=10

16 Minimumvalue=3 Whenx=1

f(x)

4

3

O x1

17 (a) h=1,k=–4 (b) f(x) = (x+1)2–4 = x2+2x–3 Atx-axis, x2+2x–3 = 0 (x+3)(x–1) = 0 x=–3orx=1 ∴a=–3,b=1 Aty-axis, f(x) = 02+2(0)–3 = –3 ∴a=–3,b=1,c=–3

�

18 (a) p=2,q=18 Aty-axis, 10 = 18–a(0–2)2

10 = 18–4a 4a = 8 a = 2 ∴a=2,p=2,q=18 (b)

y=f(x)

f(x)

xO

–10

(2,–18)

19 (a) f(x) = x2–x+7

= x2–x+(– 12 )2

+7–(– 12 )2

= (x– 12 )2

+274

∴p= 12

,q=274

(b) x= 12

20 (k–4)2–4(1)(1) 0 k2–8k+12 0 (k–2)(k–6) 0 ∴k2ork6

21 (a) k=5 (b) y=a(x–3)2+b At(0,–10),–10=9a+b… 1 At(1,0),0=4a+b… 2 1 – 2 : –10= 5a a= –2 Substitutea=–2into 1 : –10= –18+b b = 8 ∴y=–2(x–3)2+8

22 f(x)=x2+px+2p–3 p2–4(1)(2p–3) = 0 p2–8p+12 = 0 (p–2)(p–6) = 0 p=2orp=6

23 mx+4= x2–4x+5 x2+(–4–m)x+1= 0 (–4–m)2–4(1)(1) 0 16+8m+m2–4 0 m2+8m+12 0 (m+6)(m+2) 0 ∴–6m–2

24 Minimumvalue=4whenx= 13

25 (a) p=1,q=5 f(x)=a(x–1)2+5 At(0,7), 7= a+5 a = 2 ∴a=2,p=1,q=5 (b) y=–2(x–1)2–5

26 y=a(x–1)2+4 At(3,0), 0 = a(3–1)2+4 4a = –4 a = –1 ∴y=4–(x–1)2

27 (a) (4m)2–4(m+1)(9) = 0 16m2–36(m+1) = 0 16m2–36m–36 = 0 4m2–9m–9 = 0 (4m+3)(m–3) = 0

m=– 34

orm=3

(b) mx–5= x2–1 x2–mx+4= 0 (–m)2–4(1)(4)0 m2–160 (m+4)(m–4)0 ∴m–4orm4

28 f(x) = x2–4x+3 = x2–4x+(–2)2+3–(–2)2

= (x–2)2–1 ∴p=–2,q=–1 (a) Minimumvalue=–1 (b) x=2

y

x31O

–1

3

2

29 (a) x2+x– 34

0

4x2+4x–3 0 (2x+3)(2x–1) 0

x– 32

orx 12

(b) Minimumvalue=4whenx=3

y

3

4

O x

13

30 (a) y=p–(x–2)2

At(1,0), 0 = p–(1–2)2

0 = p–1 p = 1 ∴p=1,q=2 (b) Maximumvalue=1 (c) y=1–(x–2)2

Aty-axis, y = 1–(–2)2

= –3 ∴A(0,–3) (d) y=(x–2)2–1

31 (a) y= –x2–px+7 = –(x2+px–7)

= –[x2+px+(p2 )2

–7–(p2 )2]

= –[(x+p2 )2

–7–p2

4 ] = 28+p2

4–(x+p

2 )2

∴ 28+p2

4= 16

28+p2= 64 p2= 36 p = 6(0) q = p

2

= 62

= 3 ∴p=6,q=3 (b) f(x)=16–(x+3)2

Maximumvalue=16when x=–3 (c) f(x)0 –x2–6x+70 x2+6x–70 (x+7)(x–1)0 ∴–7x1

32 f(x) = x2+hx+7

= x2+hx+(h2 )2

+7–(h2 )2

= (x+h2 )2

+7–h2

4

∴ h2

= 2

h = 4 Whenh=4,theminimumpoint

= 7–(4)2

4 = 7–4 = 3 (a) f(x)=x2+4x+7=(x+2)2+3 Aty-axis, f(x) = 02+4(0)+7 = 7 ∴A(0,7);B(–2,3) (b) h=4

33 (a) A(–2,0) Aty-axis, y = 12+p(0)–02

= 12 ∴B(0,12) ∴A(–2,0);B(0,12) (b) y = –x2+px+12 = –(x2–px–12)

= –[x2–px+(– p2 )2

–12–(– p2 )2]

= –[(x–p2 )2

–12–p4

2] = 48+p2

4–(x–p

2 )2

∴ p2

= 2

p = 4

q = 48+(4)2

4

= 16

34 (a) f(x)=2(x2– 72

x+ 32 )

= 2[x2– 72

x+(– 74 )2

+32

–(– 74 )2]

= 2[(x– 74 )2

+32

–4916 ]

= 2[(x– 74 )2

–2516 ]


�

= 2(x– 74 )2

–258

∴a=2,b= 74

,c=258

(b) Minimumvalue=– 258

(c)

y

x

y=| f(x)|

O 12

74

3 4

7

2583

35 (a) f(x)=16–4x–2x2

Atx-axis,16–4x–2x2 = 0 2x+4x–16 = 0 x2+2x–8 = 0 (x+4)(x–2) = 0 x=–4orx=2 ∴A(2,0);B(–4,0) (b) f(x)= –2x2–4x+16 = –2(x2+2x–8) = –2[x2+2x+(1)2–8–(1)2] = –2[(x+1)2–9] = 18–2(x+1)2

∴p=1 (c) Forf(x)=18–2(x+1)2, themaximumpointis(–1,18). Forf(x)=9–(x+1)2, themaximumpointis(–1,9).

36 f(x) =2(x2–4x+ 32 )

=2[x2–4x+(–2)2+32

–(–2)2] =2[(x–2)2– 5

2 ] =2(x–2)2–5 (a) (2,–5) (b)

y

x52O

–5–1

3

13

(c) 2x2–8x+3 k 2x2–8x+3–k 0 64–4(2)(3–k) 0 40+8k 0 8k –40 k –5

37 (a) mx–5 = 6x–6–x2

x2+(m–6)x+1 = 0 (m–6)2–4(1)(1) 0 m2–12m+32 0 (m–4)(m–8) 0 ∴m4orm8 (b) (–2)2–4(1)(p+3) = 0 4–4(p+3) = 0 4–4p–12 = 0 4p = –8 p = –2

38 (a) y=a(x+1)2–18 At(2,0), 0 = a(3)2–18 9a = 18 a = 2 ∴a=2,b=–18,c=–1 (b) A(–4,0) (c) f(x)=|2(x+1)2–18| Whenx=3,f(x)= |2(16)–18| = 14 Whenx=1,f(x)= |2(4)–18| = 10 ∴Range:0f(x)14

39 (a) A(4,0) (b) f(x)=8–a(x–2)2 At(4,0), 0 = 8–a(4) 4a = 8 a = 2 ∴a=2,p=–2,q=8 (c) f(x)=8–2(x+2)2

40 (a) f(x) = x2+2hx+(h)2+4h–(h)2

= (x+h)2+4h–h2

∴ 4h–h2 = 3 h2–4h+3 = 0 (h–1)(h–3) = 0 h=1orh=3 (b)

f(x)

x

12

43

O–1–3

(c) (–1,3)and(–3,3)

41 (a) f(x)=x2–3x+5 Aty-axis,f(x) = 02–3(0)+5 = 5 ∴P(0,5)

(b) f(x) = x2–3x+(– 32 )2

+5

–(– 32 )2

= (x– 32 )2

+114

∴Thecoordinatesofminimum

point=(32

,114 )

42 (a) f(x)=a(x–2)2–6 Aty-axis, –4 = a(0–2)2–6 4a = 2

a = 12

∴a= 12

,p=–2,q=–6

(b) x=2

(c) f(x)=12

(x+2)2–6


�

1 (a) y=3–x… 1 x2–3x+y2=5… 2 Substitute 1 into 2 : x2–3x+(3–x)2 = 5 x2–3x+9–6x+x2 = 5 2x2–9x+4 = 0 (2x–1)(x–4) = 0

x=12

orx=4

Substitutex=12

into 1 :

y = 3–12

= 212

Substitutex=4into 1 : y = 3–4 = –1

∴x=12

,y=212

;x=4,y=–1

(b) y=14–2x… 1 2x2–y2+6=2xy… 2 Substitute 1 into 2 : 2x2–(14–2x)2+6 =2x(14–2x) 2x2–(196–56x+4x2)+6 =28x–4x2

2x2+28x–190 =0 x2+14x–95 =0 (x+19)(x–5) =0 x=–19orx=5 Substitutex = –19into 1 : y = 14–2(–19) = 52 Substitutex = 5into 1 : y = 14–2(5) = 4 ∴x=–19,y=52;x=5,y=4

(c) y=2x+3… 1 2x2+y2–4x=39… 2 Substitute 1 into 2 : 2x2+(2x+3)2–4x = 39 2x2+4x2+12x+9–4x = 39 6x2+8x–30 = 0 3x2+4x–15 = 0 (3x–5)(x+3) = 0

x=53

orx=–3

Substitutex=53

into 1 :

y = 2(53 )+3

= 613

Substitutex=–3into 1 : y = 2(–3) + 3 = –3

∴x=53

,y=613

;x=–3,y=–3

(d) x=10–2y… 1 2y2–7y+x=0… 2 Substitute 1 into 2 : 2y2–7y+10–2y = 0 2y2–9y+10 = 0 (2y–5)(y–2) = 0

y=52

ory=2

Substitutey=52

into 1 :

x = 10–2(52 )

= 5 Substitutey=2into 1 : x = 10–2(2) = 6

∴x=5,y=52

;x=6,y=2

(e) x=4y–11… 1 y2–2x=7… 2 Substitute 1 into 2 : y2–2(4y–11) = 7 y2–8y+22 = 7 y2–8y+15 = 0 (y–3)(y–5) = 0 y=3ory=5 Substitutey=3into 1 : x = 4(3)–11 = 1 Substitutey=5into 1 : x = 4(5)–11 = 9 ∴x=1,y=3;x=9,y=5

(f) 2y=3x–1

y=3x–1

2… 1

9x2+y=7… 2 Substitute 1 into 2 :

9x2+3x–1

2 = 7

18x2+3x–1 = 14 18x2+3x–15 = 0 6x2+x–5 = 0 (6x–5)(x+1) = 0

x=56

orx=–1

Substitutex=56

into 1 :

y =

3(56)–1

2

=34

Substitutex=–1into 1 :

y =3(–1)–1

2 = –2

∴x=56

,y=34 ;x=–1,y=–2

2 (a) 3x=2y+1

x=2y+1

3… 1

4x2+9y2=15xy… 2 Substitute 1 into 2 :

4(2y+13 )2

+9y2 = 15y(2y+13 )

4(4y+4y+19 )+9y2 = 5y(2y+1)

16y2+16y+4+81y2 = 90y2+45y 7y2–29y+4 = 0 (7y–1)(y–4) = 0

y=17

ory=4

Substitutey=17

into 1 :

x =

2(17)+1

3

=37

Substitutey=4into 1 :

x =2(4)+1

3 = 3

∴x=37

,y=17

;x=3,y=4

(b) 2x=10–3y

x=10–3y

2 … 1

2y+3x=5xy… 2 Substitute 1 into 2 :

2y+3(10–3y2 ) = 5y(10–3y

2 ) 4y+30–9y = 50y–15y2

15y2–55y+30 = 0 3y2–11y+6 = 0 (3y–2)(y–3) = 0

y=23

ory=3

Substitutey=23

into 1 :

x =10–3(2

3 )2

= 4 Substitutey=3into 1 :

x =10–3(3)

2

=12

∴x=4,y=23

;x=12

,y=3

(c) y=3x–2… 1 2x2+y2=3xy… 2 Substitute 1 into 2 : 2x2+(3x–2)2 = 3x(3x–2) 2x2+9x2–12x+4 = 9x2–6x



2x2–6x+4 = 0 x2–3x+2 = 0 (x–1)(x–2) = 0 x=1orx=2 Substitutex=1into 1 : y = 3(1)–2 = 1 Substitutex=2into 1 : y = 3(2)–2 = 4 ∴x=1,y=1;x=2,y=4

(d) y=7–2x… 1 4y–3x=xy… 2 Substitute 1 into 2 : 4(7–2x)–3x = x(7–2x) 28–8x–3x = 7x–2x2

2x2–18x+28 = 0 x2–9x+14 = 0 (x–2)(x–7) = 0 x=2orx=7 Substitutex=2into 1 : y = 7–2(2) = 3 Substitutex=7into 1 : y = 7–2(7) = –7 ∴x=2,y=3;x=7,y=–7

(e) y=3–3x… 1 3x+2y=6xy… 2 Substitute 1 into 2 : 3x+2(3–3x) = 6x(3–3x) 3x+6–6x = 18x–18x2

18x2–21x+6 = 0 6x2–7x+2 = 0 (3x–2)(2x–1) = 0

x=23

orx=12

Substitutex=23

into 1 :

y = 3–3(23 )

= 1

Substitutex=12

into 1 :

y = 3–3(12 )

=32

∴x=23

,y=1;x=12

,y=32

(f) 3y = 2x+2

y =2x+2

3… 1

8x–9y=xy5

40x–45y=xy… 2 Substitute 1 into 2 :

40x–45( 2x+23 ) = x( 2x+2

3 ) 40x–30x–30 =

13

(2x2+2x)

3(10x–30) = 2x2+2x 2x2–28x+90 = 0 x2–14x+45 = 0 (x–5)(x–9) = 0 x=5orx=9

Substitutex=5into 1 :

y =2(5)+2

3 = 4 Substitutex=9into 1 :

y =2(9)+2

3

=203

∴x=5,y=4;x=9,y=203

3 (a) y=5–2x… 1 x2+y2=5… 2 Substitute 1 into 2 : x2+(5–2x)2 = 5 x2+25–20x+4x2 = 5 5x2–20x+20 = 0 x2–4x+4 = 0 (x–2)(x–2) = 0 x = 2 Substitutex=2into 1 : y = 5–2(2) = 1 ∴x=2,y=1

(b) 2x+3y=7

x=7–3y

2… 1

x2+xy+y2=7… 2 Substitute 1 into 2 :

( 7–3y2 )2

+y( 7–3y2 )+y2 =7

49–42y+9y2

4+

7y–3y2

2+y2 =7

49–42y+9y2+14y–6y2+4y2 =28 7y2–28y+21 =0 y2–4y+3 =0 (y–1)(y–3) =0 y=1ory=3 Substitutey=1into 1 :

x =7–3(1)

2 = 2 Substitutey=3into 1 :

x =7–3(3)

2 = –1 ∴x=2,y=1;x=–1,y=3

(c) 5x+3y=2x+y+1 2y=1–3x

y=1–3x

2… 1

3x2–y2=2x+y+1… 2 Substitute 1 into 2 :

3x2–(1–3x2 )2

= 2x+1–3x

2+1

3x2–( 1–6x+9x2

4 )= 2x+1–3x

2+1

12x2–1+6x–9x2= 8x+2–6x+4 3x2+4x–7= 0 (3x+7)(x–1)= 0

x=–73

orx=1

Substitutex=–73

into 1 :

y =1–3(– 7

3)2

= 4 Substitutex=1into 1 :

y =1–3(1)

2 = –1

∴x=–73

,y=4;x=1,y=–1

(d) 2x+y = 1 y = 1–2x… 1 4x2+12x+y2=1… 2 Substitute 1 into 2 : 4x2+12x+(1–2x)2 = 1 4x2+12x+1–4x+4x2 = 1 8x2+8x = 0 x2+x = 0 x(x+1) = 0 x=0orx=–1 Substitutex=0into 1 : y = 1–2(0) = 1 Substitutex=–1into 1 : y = 1–2(–1) = 3 ∴x=0,y=1;x=–1,y=3

(e) y=7–2x… 1 x2–xy+y2=7… 2 Substitute 1 into 2 : x2–x(7–2x)+(7–2x)2 = 7 x2–7x+2x2+49–28x+4x2 = 7 7x2–35x+42 = 0 x2–5x+6 = 0 (x–2)(x–3) = 0 x=2orx=3 Substitutex=2into 1 : y = 7–2(2) = 3 Substitutex=3into 1 : y= 7–2(3) = 1 ∴x=2,y=3;x=3,y=1

(f) 2(x+y) = 10 x+y = 5 x = 5–y… 1 x2–y+y2 = 10… 2 Substitute 1 into 2 : (5–y)2–y+y2 = 10 25–10y+y2–y+y2 = 10 2y2–11y+15 = 0 (2y–5)(y–3) = 0

y=52

ory=3

Substitutey=52

into 1 :

x = 5–52

=52

Substitutey=3into 1 : x = 5–3 = 2

∴ x=52

,y=52

;x=2,y=3


4 (a) x=3–2y… 1 xy+y2=1… 2 Substitute 1 into 2 : y(3–2y)+y2 = 1 3y–2y2+y2 = 1 y2–3y+1 = 0

y =–(–3)±(–3)2–4(1)(1)

2(1)

=3±5

2

= 2.618or0.382 Substitutey=2.618into 1 : x = 3–2(2.618) = –2.236 Substitutey=0.382into 1 : x = 3–2(0.382) = 2.236 ∴x=–2.236,y=2.618; x=2.236,y=0.382

(b) y=2–x… 1 x2–3x–y=4… 2 Substitute 1 into 2 : x2–3x–(2–x)= 4 x2–2x–6= 0

x =–(–2)±(–2)2–4(1)(–6)

2(1)

=2±28

2

= 3.6458or–1.6458 Substitutex=3.6458into 1 : y = 2–3.6458 = –1.6458 Substitutex=–1.6458into 1 : y = 2–(–1.6458) = 3.6458 ∴x=3.646,y=–1.646; x=–1.646,y=3.646

(c) x=y+4 … 1 6y+x=3xy… 2 Substitute 1 into 2 : 6y+y+4 = 3y(y+4) 7y+4 = 3y2+12y 3y2+5y–4 = 0

y =–5±52–4(3)(–4)

2(3)

=–5±73

6 = 0.5907or–2.2573 Substitutey=0.5907into 1 : x = 0.5907+4 = 4.5907 Substitutey=–2.2573into 1 : x = –2.2573+4 = 1.7427 ∴x=4.591,y=0.591; x=1.743,y=–2.257

(d) x=2y+3… 1 y2+2x2=5xy… 2

Substitute 1 into 2 : y2+2(2y+3)2 =5y(2y+3) y2+2(4y2+12y+9) =10y2+15y y2+8y2+24y+18 =10y2+15y

y2–9y–18 =0

y =–(–9)±(–9)2–4(1)(–18)

2(1)

=9±153

2 = 10.6847or–1.6847 Substitutey=10.6847into 1 : x = 2(10.6847)+3 = 24.3694 Substitutey=–1.6847into 1 : x = 2(–1.6847)+3 = –0.3694 ∴x=24.369,y=10.685; x=–0.369,y=–1.685

(e) y=1–3x… 1 x2+2xy+2y2=10… 2 x2+2x(1–3x)+2(1–3x)2=10 x2+2x–6x2+2(1–6x+9x2)=10 x2+2x–6x2+2–12x+18x2=10 13x2–10x–8=0

x =–(–10)±(–10)2–4(13)(–8)

2(13)

=10±516

26 = 1.2583or–0.4891 Substitutex=1.2583into 1 : y = 1–3(1.2583) = –2.7749 Substitutex=–0.4891into 1 : y = 1–3(–0.4891) = 2.4673 ∴x=1.258,y=–2.775; x=–0.489,y=2.467

(f) x=5–3y… 1 x2+y2=6x–4y… 2 (5–3y)2+y2 = 6(5–3y)–4y 25–30y+9y2+y2 = 30–18y–4y 10y2–8y–5 = 0

y =–(–8)±(–8)2–4(10)(–5)

2(10)

=8±264

20 = 1.2124or–0.4124 Substitutey=1.2124into 1 : x = 5–3(1.2124) = 1.3628 Substitutey=–0.4124into 1 : x = 5–3(–0.4124) = 6.2372 ∴x=1.363,y=1.212; y=6.237,y=–0.412

5 (a) y=–2x–2… 1

y+2x=xy2

2y+4x=xy… 2 Substitute 1 into 2 : 2(–2x–2)+4x = x(–2x–2) –4x–4+4x = –2x2–2x 2x2+2x–4 = 0 x2+x–2 = 0 (x+2)(x–1) = 0 x=–2orx=1

Substitutex=–2into 1 : y = –2(–2)–2 = 2 Substitutex=1into 1 : y = –2(1)–2 = –4 ∴A(–2,2),B(1,–4)

(b) 2x = –3y+2

x =–3y+2

2 … 1

2xy = –1… 2 Substitute 1 into 2 :

2y(–3y+22 ) = –1

–3y2+2y = –1 3y2–2y–1 = 0 (3y+1)(y–1) = 0

y=–13

ory=1

Substitutey=–13

into 1 :

x =–3(– 1

3 )+2

2

=32

Substitutey=1into 1 :

x =–3(1)+2

2

= –12

∴A(– 12

,1),B(32

,–13 )

6 x=3y–3… 1 2x–3y=6xy… 2 Substitute 1 into 2 : 2(3y–3)–3y =6y(3y–3) 6y–6–3y =18y2–18y 18y2–21y+6 =0 6y2–7y+2 =0 (3y–2)(2y–1) =0

y=23

ory=12

Substitutey=23

into 1 :

x = 3(23 )–3

= –1

Substitutey=12

into 1 :

x = 3(12 )–3

= –32

∴(–1,23 )and(– 3

2,

12 )

7 (a) x=2y+2… 1 x2+y2= 2xy+1… 2 Substitute 1 into 2 : (2y+2)2+y2=2y(2y+2)+1 4y2+8y+4+y2=4y2+4y+1 y2+4y+3=0 (y+3)(y+1)=0 y=–3ory=–1


Substitutey=–3into 1 : x = 2(–3)+2 = –4 Substitutey=–1into 1 : x = 2(–1)+2 = 0 ∴(–4,–3)and(0,–1)

(b) Midpoint =[–4+02

,–3+(–1)

2 ] =(–2,–2)

8 2x=y+1 At(m,3), 2m =3+1 m =2 3x2+nxy+y2=9 At(2,3), 3(2)2+n(2)(3)+(3)2 = 9 12+6n = 0 6n = –12 n = –2 ∴m=2,n=–2 y=2x–1… 1 3x2–2xy+y2=9… 2 Substitute 1 into 2 : 3x2–2x(2x–1)+(2x–1)2 = 9 3x2–4x2+2x+4x2–4x+1 = 9 3x2–2x–8 = 0 (3x+4)(x–2) = 0

x=–43

orx=2

Substitutex=–43

into 1 :

y = 2(– 43 )–1

= –113

Theothersolutionis(– 43

,–113 ).

9 4x+4y= 48 x+y=12 x=12–y… 1 x2+y2=80… 2 Substitute 1 into 2 : (12–y)2+y2 = 80 144–24y+y2+y2 = 80 2y2–24y+64 = 0 y2–12y+32 = 0 (y–4)(y–8) = 0 y=4ory=8 Substitutey=4into 1 : x = 12–4 = 8 Substitutey=8into 1 : x = 12–8 = 4 ∴x=4,y=8

10 2x = 1–3y

x =1–3y

2 … 1

3y2 = x2+2… 2 Substitute 1 into 2 :

3y2 = (1–3y2 )2

+2

3y2 =1–6y+9y2

4 +2

12y2 = 1–6y+9y2+8 3y2+6y–9 = 0 y2+2y–3 = 0 (y+3)(y–1) = 0 y=–3ory=1 Substitutey=–3into 1 :

x =1–3(–3)

2 = 5 Substitutey=1into 1 :

x =1–3(1)

2

= –1 Thepointsofintersectionare(5,–3)

and(–1,1). ∴Distance = [1–(–3)]2+(–1–5)2

= 16+36

= 52

=7.21units

1 y=3–2x … 1 x2+y2=2 … 2 Substitute 1 into 2 : x2+(3–2x)2 = 2 x2+9–12x+4x2 = 2 5x2–12x+7 = 0 (5x–7)(x–1) = 0

x=75

orx=1

Substitutex=75

into 1 :

y = 3–2(75 )

=15

Substitutex=1into 1 : y = 3–2(1) = 1

∴x=75

,y=15

;x=1,y=1

2 x=1–3y… 1 x2–3y2=2xy… 2 Substitute 1into 2 : (1–3y)2–3y2 = 2y(1–3y) 1–6y+9y2–3y2 = 2y–6y2

12y2–8y+1 = 0 (6y–1)(2y–1) = 0

y=16

ory=12

Substitutey=16

into 1:

x = 1–3(16 )

=12

Substitutey=12

into 1:

x = 1–3(12 )

= –12

∴x=12

,y=16

;x=–12

,y=12

3 y=1–2x… 1 x2–2y2=4xy… 2 Substitute 1into 2: x2–2(1–2x)2 = 4x(1–2x) x2–2(1–4x+4x2) = 4x–8x2

x2–2+8x–8x2 = 4x–8x2

x2+4x–2 = 0

x =–4±42–4(1)(–2)

2(1)

=–4±24

2

= 0.4495or–4.4495 Substitutex=0.4495into 1 : y = 1–2(0.4495) = 0.101 Substitutex=–4.4495into 1 : y = 1–2(–4.4495) = 9.899 ∴x=0.450,y=0.101; x=–4.450,y=9.899

4 y=5–x… 1 4x2–6y=24… 2 Substitute 1 into 2 : 4x2–6(5–x) = 24 4x2–30+6x = 24 4x2+6x–54 = 0 2x2+3x–27 = 0 (2x+9)(x–3) = 0

x=–92

orx=3

Substitutex=–92

into 1 :

y = 5–(– 92 )

=192

Substitutex=3into 1 : y = 5–3 = 2

∴x=–92

,y=192

;x=3,y=2

5 2x–2y= x+y–1 x= 3y–1… 1 2x2–11y2=x+y–1… 2 Substitute 1 into 2 : 2(3y–1)2–11y2 = 3y–1+y–1 2(9y2–6y+1)–11y2= 4y–2 18y2–12y+2–11y2 = 4y–2 7y2–16y+4 = 0 (7y–2)(y–2) = 0

y=27

ory=2

Substitutey=27

into 1 :

x = 3(27 )–1

= –17

Substitutey=2into 1 : x = 3(2)–1 = 5

x=–17

,y=27

;x=5,y=2


6 [2(6–3y2 )+1]2

+6(y–2)2 = 49

(7–3y)2+6(y–2)2 = 49 49–42y+9y2+6(y2–4y+4) = 49 15y2–66y+24 = 0 5y2–22y+8 = 0 (5y–2)(y–4) = 0

y=25

ory=4

Substitutey=25

intox=6–3y

2:

x =

6–3(25 )

2

=125

Substitutey=4intox=6–3y

2:

x =6–3(4)

2 = –3

∴x=125

,y=25

;x=–3,y=4

7 x–y=4 y=x–4… 1 2x2–y2=17… 2 Substitute 1 into 2 : 2x2–(x–4)2 = 17 2x2–(x2–8x+16) = 17 x2+8x–33 = 0 (x+11)(x–3) = 0 x=–11orx=3 Substitutex=–11into 1 : y = –11–4 = –15 Substitutex=3into 1 : y = 3–4 = –1 ∴x=–11,y=–15;x=3,y=–1 (–11,–15);(3,–1)

8 y=3–5x… 1 x2+y2–3x=2… 2 Substitute 1 into 2 : x2+(3–5x)2–3x = 2 x2+9–30x+25x2–3x = 2 26x2–33x+7 = 0 (26x–7)(x–1) = 0

x=726

orx=1

Substitutex=726

into 1 :

y = 3–5( 726 )

=4326

Substitutex=1into 1 : y = 3–5(1) = –2

∴x=7

26,y=

4326

;x=1,y=–2

( 726

,4326 );(1,–2)

9 (a) x=1–3y… 1 2x2+xy=y2+36… 2 2(1–3y)2+y(1–3y) = y2+36 2(1–6y+9y2)+y–3y2 = y2+36

14y2–11y–34 = 0 (14y+17)(y–2) = 0

y=–1714

ory=2

Substitutey=–1714

into 1 :

x = 1–3(– 1714 )

=6514

Substitutey=2into 1 : x = 1–3(2) = 1–6 = –5

∴M(–5,2)andN (6514

,–1714 )

(b) Midpoint

= [ –5+6514

2,

2+(– 1714 )

2 ] = (– 5

28,

1128 )

10 px+qy=2 At(1,2), p(1)+2q =2 p =2–2q… 1 qx+p2y=10 At(1,2), q(1)+2p2=10 q+2p2=10… 2 Substitute 1 into 2 : q+2(2–2q)2 = 10 q+2(4–8q+4q2) = 10 8q2–15q–2 = 0 (8q+1)(q–2) = 0

q=–18

orq=2

Substituteq=–18

into 1 :

p = 2–2(– 18 )

=94

Substituteq=2into 1 : p = 2–2(2) = –2

∴p=94

,q=–18

;p=–2,q=2

11 x=5–2y… 1 2x+y=2xy… 2 2(5–2y)+y = 2y(5–2y) 10–4y+y = 10y–4y2

4y2–13y+10 = 0 (4y–5)(y–2) = 0

y=54

ory=2

Substitutey=54

into 1 :

x = 5–2(54 )

=52

Substitutey=2into 1 : x = 5–2(2) = 1

∴P(1,2)andQ(52

,54 )

12 (a) y=–2x–5… 1 y2+(2x+3)2=10… 2 Substitute 1 into 2 : (–2x–5)2+(2x+3)2 = 10 4x2+20x+25+4x2+12x+9 = 10 8x2+32x+24 = 0 x2+4x+3 = 0 (x+3)(x+1) = 0 x=–3orx=–1 Substitutex=–3into 1 : y = –2(–3)–5 = 1 Substitutex=–1into 1 : y = –2(–1)–5 = –3 ∴A(–3,1)andB(–1,–3) (b) Midpoint

= [ –3+(–1)2

,1+(–3)

2 ] = (–2,–1)

13 y=3x–7… 1 x2+y2–xy=7… 2 Substitute 1 into 2 : x2+(3x–7)2–x(3x–7) = 7 x2+9x2–42x+49–3x2+7x = 7 7x2–35x+42 = 0 x2–5x+6 = 0 (x–2)(x–3) = 0 x=2orx=3 Substitutex=2into 1 : y = 3(2)–7 = –1 Substitutex=3into 1 : y = 3(3)–7 = 2 ∴x=2,y=–1;x=3,y=2 R(2,–1);S(3,2)

LengthofRS: [2–(–1)]2+(3–2)2

= 32+12

= 10 = 3.162units

14 xy=70… 1

π(x2 )–y = 1

227 (x

2 )–y = 1

11x7

–y = 1

y =11x7

–1… 2


x(117

x–1) = 70

11x2–7x = 490 11x2–7x–490 = 0 (11x+70)(x–7) = 0 x = 7(0) Substitutex=7into 1 : xy = 70 7y = 70 y = 10 ∴x=7,y=10


Indices and Logarithms

1 (a) (–3)4 = –3 –3 –3 –3 = 81 (b) 64

23 = (43)

23

= 42

= 16 (c) (–125)

13 = [(–5)3]

13

= –5 (d) (–27)

– 23 = [(–3)3]

– 23

= (–3)–2

=

1

(–3)2

=

19

(e)

(2581)3

2

=

[( 59 )2]3

2

= ( 5

9 )3

=

125729

(f)

(1258 )–

43

=

1

(1258 )4

3

=

1

[( 52 )3]4

3

=

1

(52 )4

=

16625

2 (a) 913 9

16 = 9

13 +

16

= 912

= 9 = 3 (b) 27

12 3

12 = 3

32 3

12

= 332 +

12

= 32

= 9 (c) 2–3 16

34 = 2–3 (24)

34

= 2–3 23

= 2–3 + 3

= 20

= 1 (d) 36

16 6

23 = (62)

16 6

23

= 613 +

23

= 6 (e) 16

12 64

– 23 = (42)

12 (43)

– 23

= 4 4–2

= 41 + (–2)

= 4–1

=

14

(f) 53 2512 = 53 (52)

12

= 53 + 1

= 54

= 625

3 (a) 3235 16

14 = (25)

35 (24)

14

= 23 21

= 23 – 1

= 22

= 4 (b) 25

12 125

– 23 = (52)

12 (53)

– 23

= 51 5–2

= 51 – (–2)

= 53

= 125 (c) 8

– 23 4

12 = (23)

– 23 (22)

12

= 2–2 21

= 2–2 – 1

= 2–3

=

18

(d) 27–

43 81

– 14 = (33)

– 43 (34)

– 14

= 3–4 3–1

= 3–4 – (–1)

= 3–3

=

127

(e) 8–3 2–5 = (23)–3 2–5

= 2–9 – (–5)

= 2–4

=

116

(f) 4914 7

34 = (72)

14 7

34

= 712

– 34

= 7–

14

=

1

714

4 (a) ( 1

27)– 43

81

14 3–1

= (3–3)–

43 (34)

14 3–1

= 3 4 3 3–1

= 34 + 1 – (–1)

= 36

= 729 (b) 8

23 4–1 512

– 43

= (23)23 (22)–1 (29)

– 43

= 22 2–2 2–12

= 22 – 2 – 12

= 2–12

=

1

212

=

1

4096

(c) 713 49

14 7

– 16 = 7

13 (72)

14 7

– 16

= 713 +

12 – (–

16)

= 71

= 7

(d)

823

4 12847

=

(23)23

22 (27)47

=

22

22 24

=

22

26

=

124

=

116

(e)

313 9

13

2723

=

313 (32)

13

(33)23

=

3

13 +

23

32 =

31

32

=

13

(f)

27

– 43 81

14

9 =

(33)

– 43 (34)

14

32

= 3–4 31

32

=

3–3

32

=

135

=

1

243

5 (a) (a2b–3)5 = a2 5b–3 5

= a10b–15

(b)

(a–2

b4 )2

=

a–2 2

b4 2

=

a–4

b8

=

1

a4b8

(c) 16a

– 52 4a

– 32 =

164

a–

52 – (–

32)

= 4a–1

=

4a

(d) x–8y–4 = (x–8)12(y–4)

12

= x–4y–2

=

1

x4y2

(e) (a5b–3)–2 (a4b3)3 = a–10b6 a12b9

= a–10 + 12 b6 + 9

= a2b15



(f)

9h5k4

27h8k3 =

13

h5 – 8k4 – 3

=

k

3h3

(g) (2m23n

– 16)6 = 26(m

23)6(n

– 16)6

= 64m4n–1

=

64m4

n

(h) (m–4)12

n (m

23

n)9

= m–2n m6n

= m–2n – 6n

= m–8n

=

1

m8n

(i) 3 p6q–3 = (p6q–3)13

= p2q–1

=

p2

q

6 (a) ( m)3 (3 m)4 (6 m)5

= m32 +

43 –

56

= m2 (b) 3p + 1 9p 27 = 3p + 1 32p 33

= 3p + 1 + 2p – 3

= 33p – 2

(c) 23a – 1 8a + 1 16 = 23a – 1 23(a + 1) 24

= 23a – 1 – 3(a + 1) + 4

= 20

= 1 (d) 32k 9k – 2 812k – 1

= 32k 32(k – 2) 34(2k – 1)

= 32k + 2(k – 2) – 4(2k – 1)

= 3–4k

=

1

34k

(e)

4b + 1 4b – 1

16–b =

4b + 1 + (b – 1)

4–2b

=

42b

4–2b

= 42b – (–2b)

= 44b

(f)

81y + 1 92y + 1

3y – 4 27y + 2

=

34(y + 1) 32(2y + 1)

3y – 4 33(y + 2)

=

34(y + 1) – 2(2y + 1)

3y – 4 + 3(y + 2)

=

32

34y + 2

= 32 – (4y + 2)

= 3–4y

=

1

34y

7 (a) 3x – 1 – 3x + 1 = 3x

31 – 3x 31

=

y3

– 3y

= – 83

y

(b) 4(31 – x) = 4( 3

3x) =

12y

(c) 9x – 2713

x + 23

= 32x – 33(1

3 x + 23)

= 32x – 3x + 2

= 32x – 3x 9 = y2 – 9y (d) 3x + 3x + 1 + 3x + 2

= 3x + 3x 3 + 3x 9 = y + 3y + 9y = 13y

(e) 32x + 1 + 33x – 1 = 32x 3 +

33x

3

= 3y2 + 13

y3

(f) 3x – 912

x + 1 = 3x – 3

2(12

x + 1) = 3x – 3x + 2

= 3x – 3x 9 = y – 9y = –8y

8 (27k2)3 – h( 13k )h

= (33k2)3 – h(3k)–h

= 39 – 3hk6 – 2h(3–hk–h) = 39 – 3h –hk6 – 2h – h

= 39 – 4hk6 – 3h

9 (a) 4n – 22n + 1 + 5(4n + 1) = 22n – 22n 21 + 5[22(n + 1)] = 22n – 2 22n + 5(22n 22) = 22n – 2 22n + 20 22n

= 19 22n

\ 19 22n is divisible by 19 for all positive integers of n.

(b) 2n – 1 + 2n + 1 + 2n

= 2n 2–1 + 2n 21 + 2n

=

12

2n + 2 2n + 2n

=

72

2n

\

72

2n is divisible by 7 for

all positive integers of n. (c) 4(2n + 2) + 2n + 1 – 3(2n) = 4(2n 22) + 2n 2 1 – 3 2n

= 16 2n + 2 2n – 3 2n

= 15 2n


(d) 3n + 3n + 1 + 3n + 2

= 3n + 3n 31 + 3n 32

= 3n + 3 3n + 9 3n

= 13 3n


10 (a) Let 3a = 2b = 6c = k, then 3 2 = 6 k

1a k

1b = k

1c

k1a +

1b = k

1c

a + b

ab =

1c

c =

aba + b

(b) Let 3x = 4y = 12z = k then 3 4 = 12

k1x k

1y = k

1z

k1x +

1y = k

1z

x + yxy =

1z

xz + yz = xy xy – xz = yz x(y – z) = yz

x = yz

y – z

11 (a) 53 = 125 ⇔ log5125 = 3 (b) 73 = 343 ⇔ log7343 = 3 (c) 100 = 1 ⇔ log101 = 0

(d) 4–2 = 1

16 ⇔ log4( 1

16) = –2

(e) 3–1 = 1

3 ⇔

log3(1

3) = –1

(f) 63 = 216 ⇔ log6216 = 3

12 (a) log39 = 2 ⇔ 9 = 32 (b) log81 = 0 ⇔ 1 = 80

(c) log101000 = 3 ⇔ 1000 = 103

(d) log2(12) = –1 ⇔

12

= 2–1

(e) log6( 1

36) = –2 ⇔ 136

= 6–2

(f) log216 = 4 ⇔ 16 = 24

13 (a) log40.25 = x 0.25 = 4x

4x =

14

4x = 4–1

\ x = –1 (b) log5 5 = x 5

12 = 5x

\ x =

12

(c) log9x = – 1

2

x = 9–

12

=

1

912

\ x = 13

(d) logx27 = 3 27 = x3

x3 = 33

\ x = 3 (e) log2(3x + 1) = 4 3x + 1 = 24

3x + 1 = 16 3x = 15 \ x = 5 (f) logx + 181 = 2 81 = (x + 1)2

92 = (x + 1)2

x + 1 = 9 \ x = 8 (g) logx(5x – 6) = 2 5x – 6 = x2

x2 – 5x + 6 = 0 (x – 2)(x – 3) = 0 x = 2 or x = 3


(h) log335 = x

35 = 3x

\ x = 5

14 (a) log264 = x 64 = 2x

26 = 2x

x = 6 \ log264 = 6 (b) log41 = x 1 = 4x

40 = 4x

x = 0 \ log41 = 0 (c) log366 = x 6 = 36x

61 = 62x

2x = 1

x =

12

\ log366 = 12

(d) log124 = x

4 =

(12)x

22 = 2–x

x = –2 \ log1

24 = –2

(e) log80.25 = x 0.25 = 8x

14

= 8x

2–2 = 23x

x = – 2

3

\ log80.25 = – 23

(f) log 2 = x 2 = 2

x

212 = 2

x2

x = 1 \ log 2 = 1 (g) log100.001 = x 0.001 = 10x

10–3 = 10x

x = –3 \ log100.01 = –3 (h) log24

–1 = x 4–1 = 2x

2–2 = 2x

x = –2 \ log24

–1 = –2

(i) log5

125

= x

125

= 5x

5–2 = 5x

x = –2

\ log5 125

= –2

15 (a) log245 = log2(9 5) = log29 + log25 = 2 log23 + log25 = 2p + q

(b) log210 = log2(2 5) = log22 + log25 = 1 + q (c) log275 = log2(3 25) = log23 + 2 log25 = p + 2q

16 (a) log10x2y = 2 log10x + log10y

= 2m + n (b) log10 10xy3

=

12

(log1010 + log10x + 3 log10y)

=

12

(1 + m + 3n)

=

1 + m + 3n

2

(c) log10(100 x

y2 ) = log10100 + log10 x – log10y

2

= 2 log1010 + 12

log10x – 2 log10y

= 2 + 12

m – 2n

(d) log10(10y

x ) = log1010 + log10y – log10x = 1 + n – m

17 (a) log320 = log3(4 5) = 2 log32 + log35 = 2(0.631) + 1.465 = 2.727 (b) log315 = log3(3 5) = log33 + log35 = 1 + 1.465 = 2.465

(c) log3 2 =

12

log32

=

12

(0.631)

= 0.3155

(d) log32.5 = log3(5

2) = log35 – log32 = 1.465 – 0.631 = 0.834

(e) log33 1

3 = log3

103

= log310 – log33 = log3(2 5) – log33 = log32 + log35 – log33 = 0.631 + 1.465 – 1 = 1.096

(f) log3

18

= log31 – log38

= log31 – log323

= log31 – 3 log32 = 0 – 3(0.631) = –1.893

18 (a) loga9loga5a =

2 loga3loga5 + logaa

=

2(0.254)0.721 + 1

= 0.2952

(b) loga(3a2) = loga3 + 2 logaa = 0.254 + 2 = 2.254

(c) loga(25

3a) = 2 loga5 – (loga3 + logaa) = 2(0.721) – (0.254 + 1) = 0.188

19 (a) log2(818 )

+ 2 log2(23)

– 2 log2(34)

= log2[(818 )(4

9)( 916) ]

= log28 = log22

3

= 3 log22 = 3

(b) log48 – log42 = log4(82)

= log44 = 1

(c) log636 + log7(1

7) = log66

2 + log77–1

= 2 log66 + (–log77) = 2 – 1 = 1 (d) log3(3 3) + log3 3

= log33 +

12

log33 + 12

log33

= 1 +

12

+ 12

= 2

(e) 2 log105 – log102 + log10( 4

35) +

log1070

= log10[(25)( 4

35)(70)

2 ] = log10100 = 2 log1010 = 2 (f) log32 + log34 – log372

= log3[(2)(4)

72 ]

= log3

19

= log33–2

= –2

20 (a) 2 logm5 – 3 logm2 = logm25 – logm8

= logm 258

(b) loga7 + loga 7 = loga(7 7) = loga7

1 + 12

=

32

loga7

(c) 2 logp5 + logp4 – 2 logp3 = logp25 + logp4 – logp9


= logp((25)(4)9 )

= logp(1009 )

(d) 4 log10(3

2) + log10( 8

75) – 2 log10(3

5)

= log10[(8116)( 8

75)( 925) ]

= log10(32)

(e) 2 + log25 = 2 log22 + log25 = log2(4 5) = log220 (f) 3 – 3 log102 = 3 log1010 – 3 log102 = log10103 – log102

3

= log101000 – log108

= log10(1000

8 ) = log10125

21 (a) log47 = log107log104

= 1.4037 (b) log312 =

log1012log103

= 2.2619

(c) log128 =

log108

log10 12

= –3 (d) log4(5.2) =

log105.2log104

= 1.1893

(e) log2.56.5 = log106.5log102.5

= 2.0428 (f) log5p =

log10p

log105 = 0.7113

22 (a) log85 log57 log78 =

log105log108

log107log105

log108log107

= 1

(b)

1

logaabc +

1

logbabc +

1

logcabc = logabca + logabcb + logabcc = logabcabc = 1 (c) 4 log35 2 log53 =

4 log55log53 2 log53

= 8 (d) logba logcb logac

= log10alog10b

log10blog10c

log10clog10a

= 1

23 (a) log224 = log2(8 3) = log28 + log23

= log223 + log23

= 3 log22 + log23 = 3 + 1.585 = 4.585

(b) log824 = log224log28

= log224log22

3

=

4.5853

= 1.528

24 (a) log3a3 = 3 log3a

= 3m

(b) log3(1

a) = log3a

–1

= –log3a = –m (c) log9a =

log3alog39

= log3alog33

2

=

m2

25 (a) log2mn = log2m + log2n = p + q

(b) log4(mn ) =

log2(mn )

log24

=

log2m – log2n

2 log22

= p – q

2

(c) logm4n =

log24nlog2m

=

2 log22 + log2n

log2m

= 2 + q

p

26 (a) log4(1a)

= log41 – log4a

= 0 – b = –b

(b) log28a =

log48alog42

=

log48 + log4a

log42

=

log4(4 2) + log4a

log42

=

log44 + log42 + log4a

log42

=

1 + 12

+ b

12

=

32

+ b

12

= 3 + 2b

27 (a) log464h = log464 + log4h = log44

3 + log4h = 3 log44 + log4h = 3 + k

(b) log8h =

log4hlog48

=

log4h

log4(4 2)

=

log4h

log44 + log42

=

k

1 + 12

=

23

k

(c) log28h3 =

log48h3

log42

=

log48 + 3 log4h

log42

=

log4(4 2) + 3 log4h

log42

=

log44 + log42 + 3 log4h

log42

=

1 + 12

+ 3k

12

= 3 + 6k

28 logy81 = log981log9y

=

log99

2

log9y

= 2x

29 m = 2a ⇒ log2m = a n = 2b ⇒ log2n = b

(a) log2(m3n

8 ) = log2m

3 + log2n – log28 = 3 log2m + log2n – 3 log22 = 3a + b – 3

(b) log8m + log4n =

log2mlog28

+ log2nlog24

=

log2mlog22

3 +

log2nlog22

2

= log2m

3 +

log2n

2

=

a3

+ b2

=

2a + 3b

6

30 log1pk =

logpk

logp1p

=

logpklogp p

–1

= –h

31 (a) 4x = 8 22x = 23


\ 2x = 3 x =

32

(b) 22x – 1 = 32 22x – 1 = 25

\ 2x – 1 = 5 2x = 6 x = 3 (c) 4x + 1 = 0.25

4x + 1 =

14

4x + 1 = 4–1

\ x + 1 = –1 x = –2

(d) ( 2 )3x =

18

232

x = 2–3

\

32

x = –3

x = –2 (e) 9x = ( 3 )x + 2

32x = 3x + 2

2

\ 2x =

x + 2

2 4x = x + 2 3x = 2

x =

23

(f) 16x =

12

24x = 2–1

\ 4x = –1

x = – 14

(g) 54 + x = (0.2)x

54 + x =

(15)x

54 + x = 5–x

\ 4 + x = –x 2x = –4 x = –2 (h) 34x = 27x + 3

34x = 33(x + 3)

\ 4x = 3x + 9 x = 9

32 (a) 9x – 1 = (13)4x – 1

32(x – 1) = 3–1(4x – 1)

\ 2x – 2 = –4x + 1 6x = 3 x =

12

(b) 32x + 1 = 9x – 2

32x + 1

2 = 32(x – 2)

\

2x + 1

2 = 2x – 4

2x + 1 = 4x – 8 2x = 9

x =

92

(c)

(14)x – 1

= 3 23x + 1

2–2(x – 1) = 23x + 1

3

\ –2x + 2 =

3x + 1

3 –6x + 6 = 3x + 1 9x = 5

x = 59

(d) 34x – 5 = 3 34x + 5

34x – 5

2 = 34x + 5

3

\

4x – 5

2 =

4x + 5

3 12x – 15 = 8x + 10 4x = 25

x = 254

(e)

53x

25x + 1 =

1

125

53x

52(x + 1) = 5–3

53x – 2(x + 1) = 5–3

\ 3x – 2x – 2 = –3 x = –1

(f)

(12)2x + 1

=

24x – 1

128

2–(2x + 1) =

24x – 1

27

2–2x – 1 = 24x – 8

2

\ –2x – 1 =

4x – 8

2 – 4x – 2 = 4x – 8 8x = 6

x =

34

33 (a) 22x + 1 + 2x – 3 = 0 21 22x + 2x – 3 = 0 Let y = 2x

2y2 + y – 3 = 0 (2y + 3)(y – 1) = 0

y = –

32

or y = 1

When y = –

32

2x = –

32

(inadmissible)

When y = 1 2x = 1 2x = 20

\ x = 0 (b) 22x + 3 + 2x + 3 = 1 + 2x

23 22x + 23 2x = 1 + 2x

8 22x + 8 2x = 1 + 2x

Let y = 2x

8y2 + 8y = 1 + y 8y2 + 7y – 1 = 0 (8y – 1)(y + 1) = 0

y =

18

or y = –1

When y =

18

,

When y = –1

2x = –1

2x =

18

(inadmissible)

2x = 2–3

\ x = –3

(c) 41 – x + 23 – x = 12

22(1 – x) +

82x

= 12

22 – 2x + 82x

= 12

4

22x +

82x

= 12

Let y = 2x

4y2

+ 8y

= 12

4y + 8y2 = 12y3

12y3 – 8y2 – 4y = 0 4y(3y2 – 2y – 1) = 0 4y(3y + 1)(y – 1) = 0

y = 0 or y = –

13

or y = 1

When y = 0 2x = 0 (inadmissible)

When y = –

13

2x = –

13

(inadmissible)

When y = 1 2x = 1 2x = 20

\ x = 0 (d) 6(9x) + 3x = 2 6 32x + 3x = 2 Let y = 3x

6y2 + y – 2 = 0 (3y + 2)(2y – 1) = 0

y = –

23

or y = 12

When y = – 23

3x = –

23

(inadmissible)

When y =

12

3x =

12

x log103 = log10

12

\ x =

log10 12

log103 x = – 0.631 (e) 52x + 5x + 1 = 6 52x + 5 5x = 6 Let y = 5x

y2 + 5y – 6 = 0 (y + 6)(y – 1) = 0 y = –6 or y = 1 When y = –6 5x = –6 (inadmissible) When y = 1 5x = 1 5x = 50

\ x = 0 (f) 9x = 4(3x) – 3 32x = 4(3x) – 3 Let y = 3x

y2 – 4y + 3 = 0 (y – 1)(y – 3) = 0 y = 1 or y = 3


When y = 1 When y = 3 3x = 1 3x = 3 3x = 30 3x = 31

x = 0 x = 1 \ x = 0 or x = 1 (g) 32x + 1 + 9 = 3x + 3 + 3x

3 32x + 9 = 33 3x + 3x

3 32x + 9 = 27 3x + 3x

Let y = 3x

3y2 + 9 = 27y + y 3y2 – 28y + 9 = 0 (3y – 1)(y – 9) = 0

y =

13

or y = 9

When y =

13

When y = 9

3x = 9

3x =

13

3x = 32

3x = 3–1 \ x = 2

\ x = –1 \ x = –1 or x = 2 (h) 5(2x) = 2(4x) + 2 5 2x = 2 22x + 2 Let y = 2x

5y = 2y2 + 2 2y2 – 5y + 2 = 0 (2y – 1)(y – 2) = 0

y = 12

or y = 2

When y =

12

When y = 2

2x = 2–1 2x = 2

\ x = –1 \ x = 1

\ x = –1 or x = 1

34 (a) 8x – 1 = 4y

23(x – 1) = 22y

3x – 3 = 2y 3x – 2y = 3 … 1 27x = 3y + 3

33x = 3y + 3

3x = y + 3 3x – y = 3 … 2 2 – 1 : y = 0 Substitute y = 0 into 1 : 3x – 2(0) = 3 3x = 3 x = 1 \ x = 1, y = 0 (b) 3x 92y = 27 3x 34y = 33

3x + 4y = 33

x + 4y = 3 … 1

2x 4–y =

18

2x 2–2y = 2–3

2x – 2y = 2–3

x – 2y = –3 … 2 1 – 2 : 6y = 6 y = 1 Substitute y = 1 into 1 : x + 4(1) = 3 x = –1 \ x = –1, y = 1

(c) 7x – y = 49 7x – y = 72

x – y = 2 … 1 7x + y = 343 7x + y = 73

x + y = 3 … 2 1 + 2 : 2x = 5

x =

52

Substitute x =

52

into 1 :

52

– y = 2

y =

52

– 2

y =

12

\ x = 52

, y = 12

(d) 3x 81y = 27 3x 34y = 33

3x + 4y = 33

x + 4y = 3 … 1

2x 8y =

116

2x 23y = 2– 4

2x + 3y = 2– 4

x + 3y = – 4 … 2 1 – 2 : y = 7 Substitute y = 7 into 1 : x + 4(7) = 3 x + 28 = 3 x = –25 \ x = –25, y = 7 (e) 52x + y = 625 52x + y = 54

2x + y = 4 … 1

24x – 2y =

116

24x – 2y = 2–4

4x – 2y = –4 2x – y = –2 … 2 1 + 2 : 4x = 2

x =

12

Substitute x =

12

into 1 :

2(1

2) + y = 4

1 + y = 4 y = 3

\ x = 12

, y = 3

(f) 8x = 2y + 1

23x = 2y + 1

3x – y = 1 … 1 5y = 25x + 1

5y = 52(x + 1)

y = 2x + 2 2x – y = –2 … 2 1 – 2 : x = 3

Substitute x = 3 into 1 : 3(3) – y = 1 y = 9 – 1 = 8 \ x = 3, y = 8

35 y = mxn – 5 7 = m(2)n – 5 m(2)n = 12 … 1 22 = m(3)n – 5 m(3)n = 27 … 2

1 2 :

(23)n

=

1227

(23)n

=

49

(23)n

=

(23)2

\ n = 2 Substitute n = 2 into 1 : m(2)2 = 12 4m = 12 \ m = 3 \ m = 3, n = 2

36 (a) 0.1x = 0.25

x log100.1 = 5 log100.2 x = 3.4949 (b) 2x 3x = 5x + 1

6x = 5x + 1

x log106 = (x + 1) log105 x log106 = x log105 + log105 x (log106 – log105) = log105 x = 8.8275 (c) 3x + 1 = 7 (x + 1) log103 = log107 x + 1 = 1.7712 x = 0.7712 (d) 4x = 9(5x)

(45)x

= 9

x log10(4

5) = log109

x = –9.8467 (e) 2x = 3x – 2

x log102 = (x – 2) log103 x log102 = x log103 – 2 log103 x (log103 – log102) = 2 log103 x = 5.419 (f) 3x + 1 = 4x

(x + 1) log103 = x log104 x log103 + log103 = x log104 x (log104 – log103) = log103 x = 3.819

37 (a) 5 logm6 – logm96 = 4

logm( 65

96) = 4

m4 = 81 m4 = 34

\ m = 3 (b) log1025 + log10x – log10(x – 1) = 2

log10( 25xx – 1)

= 2


25xx – 1

= 100

25x = 100x – 100 75x = 100 x =

43

(c) log104 + 2 log10p = 2 log104p2 = 2 4p2 = 100 p2 = 25 p = 25 = 5 \ p = 5 (–5 is rejected) (d) 2 log103 + log102x = log10(3x + 1) log10(3

2 2x) = log10(3x + 1) log1018x = log10(3x + 1) 18x = 3x + 1 15x = 1

x =

115

(e) log2(2x + 5) – 2 log22x = 2

log2

2x + 5(2x)2

= 2

2x + 5

4x2 = 4

2x + 5 = 16x2

16x2 – 2x – 5 = 0 (8x – 5)(2x + 1) = 0

x = 58

or x = – 12

(f) log10(x + 6) = log10(3x – 2) x + 6 = 3x – 2 2x = 8 x = 4

38 (a) log5x = 4 logx5 + 3

log5x = 4(log55

log5x ) + 3

log5x =

4

log5x + 3

Let y = log5x

y = 4y + 3

y2 = 4 + 3y y2 – 3y – 4 = 0 (y + 1)(y – 4) = 0 y = –1 or y = 4 When y = –1 When y = 4 log5x = –1 log5x = 4 x = 5–1 x = 54

= 625

= 15

\ x = 15

or x = 625

(b) log3x + log9x = 6

log3x +

log3x

2 log33 = 6

log3x +

12

log3x = 6

log3x x = 6 x

32 = 36

x 32 = 729

x = 72923

= (36)23

= 34

= 81 (c) log4(6 – x) – log28 = log93 log4(6 – x) – log22

3 = log9912

log4(6 – x) – 3 =

12

log4(6 – x) =

72

6 – x = 472

6 – x = (22)72

6 – x = 128 x = –122 (d) log5x – log25(x + 10) =

12

log5x – log5(x + 10)

log525 =

12

log5x –

12

log5(x + 10) = 12

log5 x

x + 10 =

12

x

x + 10 = 5

12

x = ( 5)( x + 10) x2 = 5(x + 10) x2 – 5x – 50 = 0 (x + 5)(x – 10) = 0 x = –5 or x = 10 \ x = 10 (e) log3x + 2 = 3 logx3

log3x + 2 =

3

log3x Let y = log3x

y + 2 =

3y

y2 + 2y – 3 = 0 (y + 3)(y – 1) = 0 y = –3 or y = 1 When y = –3 When y = 1 log3x = –3 log3x = 1 x = 3–3 x = 3

=

127

\ x =

127

or x = 3

(f) 4 log4x = 9 logx4

4 log4x =

9

log4x Let y = log4x

4y =

9y

y2 =

94

y = 3

2

When y =

32

When y = – 32

log4x =

32

log4x = – 32

x = 432 x = 4

– 32

= (2 2 )32 = (22)

– 32

= 8 = 18

\ x = 8 or x = 18

39 (a) log10y = 2 – log10x log10xy = 2 xy = 100

y = 100

x (b) 2 log10xy = 2 + log10(x + 1)

+ log10y

log10

x2y2

y(x + 1) = 2

x2y

x + 1 = 100

y = 100(x + 1)

x2

(c) 3 + log2(x + y) = log2(x – 2y)

3 = log2(x – 2yx + y )

8 =

x – 2yx + y

8x + 8y = x – 2y 10y = –7x

y = – 710

x

(d) 3 + log10x = 2 log10y

log10

y2

x = 3

y2

x = 1000

y2 = 1000x y = 1000x

40 (a) 2x(4y) = 128 2x 22y = 27

x + 2y = 7 … 1 log10(4x – y) = log102 + log105 log10(4x – y) = log10(2 5) log10(4x – y) = log1010 4x – y = 10 … 2 2 2: 8x – 2y = 20 … 3 1 + 3 : 9x = 27 x = 3 Substitute x = 3 into 1 : 3 + 2y = 7 2y = 4 y = 2 \ x = 3 , y = 2 (b) log5(3x – y) + log56 = log524 log56(3x – y) = log524 6(3x – y) = 24 3x – y = 4 … 1 75x 73y = 1 75x – 3y = 70

5x – 3y = 0 … 2 1 3: 9x – 3y = 12 … 3 3 – 2 : 4x = 12 x = 3 Substitute x = 3 into 1 : 3(3) – y = 4 y = 9 – 4 = 5 \ x = 3, y = 5


1 8x – 1 = 322x + 5

(23)x – 1 = (25)2x + 5

3x – 3 = 10x + 25 7x = –28 x = –4

2 (1

3)435x + 1 = 81

3– 4(35x + 1

2 ) = 34

3– 4 +

5x + 12 = 34

– 4 + 5x + 1

2 = 4

5x + 12

= 8

5x + 1 = 16 5x = 15 x = 3

3 2n + 1 2n – 2

41 – n =

22n – 1

22(1 – n)

= 22n – 1 – 2(1 – n)

= 24n – 3

4

5a + 1 – 5a – 1

6 53a =

5 5a – 5a

51

6 53a

=

(245 5a)

6 53a

=

45

5–2a

\ Compare with m 5an

we have m =

45

and n = –2.

5

(6423x3)5

(1654x

53)3

=

[(26)23x3]5

[(24)54x

53]3

=

(24x3)5

(25x53)3

=

220x15

215x5

= 25x10

= (2x2)5

\ Compare with (2xm)n we have m = 2 and n = 5.

6 92x + 4 = (1

3)–(3x + 3)

32(2x + 4)

2 = 3–1(–3x – 3)

2x + 4 = 3x + 3 x = 1

7 23(x – 1) =

12

32

23(x – 1) = 2–1 252

23(x – 1) = 232

3x – 3 =

32

3x =

92

x =

32

8 Let 2x = 3y = 6z = k 2 3 = 6

k1x k

1y = k

1z

k1x +

1y = k

1z

1x +

1y =

1z

x + y

xy = 1z

z = xy

x + y (shown)

9 4x + 1 + 7(2x) = 2 22(x + 1) + 7(2x) = 2 22x + 2 + 7(2x) = 2 4 22x + 7(2x) = 2 Let y = 2x

4y2 + 7y = 2 4y2 + 7y – 2 = 0 (4y – 1)(y + 2) = 0

y =

14

or y = –2

When y =

14

2x =

14

2x = 2–2

x = –2 When y = –2 2x = –2 (inadmissible) \ x = –2

10 3235 16

14 = (25)

35 (24)

14

= 23 21

= 22

= 4

11

a32b

23

c34

=

43227

23

1634

=

(22)32(33)

23

(24)34

=

2332

23

= 32

= 9

12 2n + 1 + 4(2n + 2) – 3(2n) = 2 2n + 4(4 2n) – 3(2n) = 2 2n + 16 2n – 3 2n

= 15 2n


13 52x + y = 625 52x + y = 54

2x + y = 4 … 1 22(2x – y) = 2–4

4x – 2y = –4 2x – y = –2 … 2 1 + 2 : 4x = 2

x =

12

Substitute x = 12

into 1 :

2(1

2) + y = 4

1 + y = 4 y = 3

\ x = 12

, y = 3

14 2x 4x – 1 = 82x – 1

2x 22(x – 1) = 23(2x – 1)

23x – 2 = 26x – 3

3x – 2 = 6x – 3 3x = 1

x =

13

15 4x + 2 32x = 576 (16 4x)(9x) = 576 4x9x = 36 36x = 36 x = 1

16

( 964)1

2

(38)–1

(2764)2

3

=

38

83

916

=

169

17 y = nxm – 4 8 = n(2)m – 4 12 = n(2)m … 1 71 = n(5)m – 4 75 = n(5)m … 2

1 2 :

425

= (2

5)m

(25)2

=

(25)m

m = 2 Substitute m = 2 into 1 : 12 = n(2)2

n = 3 \ m = 2, n = 3

18

(–2k)3(2k)– 23

(16k4)13

=

–(2k)3(2k)– 23

(2k)43

= –(2k)3 –

23 –

43

= –2k

19 log7x = 9 logx7

log7x =

9 log77log7x

Let y = log7x

y =

9y

y2 = 9 y = 3


When y = 3 When y = –3 log7x = 3 log7x = –3 x = 343 x = 7–3

=

1343

\ x = 343 or x =

1343

20 3 log7x – 2 = log7y 3 log7x – log7y = 2

log7(x3

y ) = 2

x3

y = 49

y =

x3

49

21 2 logx4 + log4x = 3

2 logx4 +

1

logx4 = 3

Let y = logx4

2y + 1y = 3

2y2 – 3y + 1 = 0 (2y – 1)(y – 1) = 0

y =

12

or y = 1

When y =

12

When y = 1

logx4 = 1

logx4 =

12

x = 4

x12 = 4

x = 16 \ x = 16 or x = 4

22 logmp2q = 8 2 logmp + logmq = 8 … 1

logm q2

p = 6

–logmp + 2 logmq = 6 … 2 2 2: –2 logmp + 4 logmq = 12 … 3 1 + 3 : 5 logmq = 20 logmq = 4 Substitute logmq = 4 into 1 : 2 logmp + 4 = 8 2 logmp = 4 logmp = 2 \ logmq = 4, logmp = 2

23 log4mn = 10 log4m + log4n = 10 … 1

2 log8m = 3 log8n

2 log4mlog48 =

3 log4nlog48

2 log4m – 3 log4n = 0 … 2 1 2: 2 log4m + 2 log4n = 20 … 3 3 – 2 : 5 log4n = 20 log4n = 4 n = 44

= 256 Substitute n = 256 into 1 : log4m + log4256 = 10 log4m + 4 log44 = 10 log4m = 6

m = 46

= 4096 \ m = 4096, n = 256

24 log82x + log2x =

53

log22x3 log22

+ log2x =

53

log22x + 3 log2x = 5 log2(2x)(x3) = 5 2x4 = 32 x4 = 16 x = 2

25 logp 27 p125

= logp(33 p12

53 )

= 3 logp 3 + 12

logp p –

3 logp 5

= 3x +

12

– 3y

26 log3y +

12

log3x = 2 log3z

log3y x = log3z2

y x = z2

x =

z2

y

x = ( z2

y )2

= z4

y2

27 log45 log56 log67 log78

= log25log24

log26log25

log27log26

log28log27

=

log28log24

=

3 log222 log22

=

32

28 log927 log381 =

log327log39

log381

=

32

4

= 6

29 log5x = 4 logx5

log5x =

4log5x

Let y = log5x

y =

4y

y2 = 4 y = 2 When y = 2 When y = –2 log5x = 2 log5x = –2 x = 25

x =

125

\ x = 25 or x =

125

30 log2x – 2 = log4(x – 4)

log2x – 2 = log2(x – 4)

2 log22 2 log2x – 4 = log2(x – 4)

log2( x2

x – 4) = 4

x2

x – 4 = 16

x2 = 16x – 64 x2 – 16x + 64 = 0 (x – 8)(x – 8) = 0 x = 8

31 log10(2x + 7) = 1 + log10x

log10 2x + 7

x = 1

2x + 7

x = 10

2x + 7 = 10x 8x = 7

x = 78

32 (a) log4(1n)

= log41 – log4n

= –m (b) log28n = log28 + log2n = 3 log22 +

log4nlog42

= 3 + 2m

33 log28 + log3 19

+ log2 12

= 3 log22 – 2 log33 – log22 = 3 – 2 – 1 = 0

34 log1045 = log10(9 5) = log109 + log105 = 2 log103 + log105 = 2(0.477) + 0.699 = 1.653

35 log264 p = 6 log22

6p = 6 6p log22 = 6 6p = 6 p = 1

36 log38 = 3 log22log23

= 3p

37 logh10h = logh10 + loghh

=

1

log10h + 1

=

1k

+ 1

=

1 + k

k

38 logm(5m2

27 ) = logm5 + 2 logmm – 3 logm3

= q + 2 – 3p


39 log20.3 = log2

310

= log2[3 (5 2)] = log23 – (log25 + log22) = 1.585 – 2.322 – 1 = –1.737

40 2 log2(23)

+ log2(818 )

– 2 log2(34)

= log2

(49)(81

8 )916

= log28 = 3 log22 = 3

41 log98 = a

3 log22log29 = a

log29 = 3a

log43 = log93log94

= 1

2 log94

=

1

(2 log24log29 )

=

1

(4 log22log29 )

=

1

(4a3 )

=

34a

42 uv = 25 v = logu25

v = 2 log55log5u

log5u = 2v

�

BC = (1–1)2+[2–(–1)]2

= 9

= 3units

CA = (1–5)2+(2–2)2

= 16 = 4units Perimeter = 5+3+4 = 12units

7 (a) P(x,0),M(5,2)andN(–1,4) MP =NP

22+(5–x)2 = 42+(–1–x)2

4+25–10x+x2 = 16–8y+y2+1 29–10x = 17+2x 12x = 12 x = 1 ∴P(1,0) (b) Q(0,y),M(5,2)andN(–1,4) MQ=NQ

(2–y)2+52 = (4–y)2+(–1–0)2

4–4y+y2+25 = 16–8y+y2+1 29–4y = 17–8y 4y = –12 y = –3 ∴Q(0,–3)

8 (–3+h2

,4+k

2)=(5,7)

–3+h

2 = 5 and

4+k2

= 7

–3+h = 10 4+k = 14 h = 13 k = 10 ∴h=13,k=10

9m2+n2

2 = 5

m2+n2 = 10… 1

m+n

2 = 1

m+n = 2 m = 2–n… 2 Substitute 2 into 1 : (2–n)2+n2 = 10 4–4n+n2+n2 = 10 2n2–4n–6 = 0 n2–2n–3 = 0 (n+1)(n–3) = 0 n=–1orn=3 Substituten=–1into 2 : m = 2–(–1) = 3 Substituten=3into 2 : m = 2–3 = –1 ∴m=3,n=–1;m=–1,n=3

1 (–4h)2+(–3h)2 = 10 25h2 = 100 h2 = 4 h = ±2

2 AB = (–1–3)2+(–3)2

= 25 = 5units

BC = [–1–(–1)]2+(–5)2

= 25 = 5units

CD = [(3–(–1)]2+[–2–(–5)]2

= 25 = 5units

DA = (3–3)2+(–2–3)2

= 25 = 5units AB=BC=CD=DA ∴ABCDisarhombus.

3 RP=SP (2–y)2+(3–x)2

= (–2–y)2+(4–x)2

4–4y+y2+9–6x+x2

= 4+4y+y2+16–8x+x2

13–6x–4y=20–8x+4y 2x–8y–7=0

4 AB = (6–2)2+(6–1)2

= 41units

BC = (1–6)2+(10–6)2

= 41units

AC = (1–2)2+(10–1)2

= 82 AB=BC=2AC ∴A,BandCaretheverticesofa

right-angledtriangle.

5 (a) PQ = [3–(–4)]2+(3–2)2

= 50units

QR = (3–4)2+[3–(–4)]2

= 50units

(b) PR = (–4–4)2+[2–(–4)]2

= 100 = 10units PQ=QR ≠ PR ∴PQRisanisoscelestriangle.

6 AB = (1–5)2+(–1–2)2

= 25 = 5units

10p+8

2 = 4 and

5+q2

= 3

p+8 = 8 5+q = 6 p = 0 q = 1 ∴p=0,q=1

11 (a) MidpointofAC

= [6+(–4)2

, 4+62 ]

= (1,5) (b) MidpointofAC=midpointofBD LetDbe(x,y),

( x+82

,y+7

2 )=(1,5)

x+8

2 = 1 and

y+72

= 5

x+8 = 2 y+7 = 10 x = –6 y = 3 ∴D(–6,3)

12 MidpointofAC

= (–1+22

,4+5

2 ) = (1

2,

92 )

MidpointofBD

= (–4+52

,6+3

2 ) = (1

2,

92 )

Thesamepoint( 12

,92 )isthe

midpointofACandBD,sothelinesACandBDmustbisectoneanother.

13 BD=AC

(p+(–7)2

,6+q

2 )=(4+(–2)2

,–4+4

2 )

p–72

= 1 and6+q

2=0

p–7 = 2 q =–6 p = 9 ∴p=9,q=–6

14 y=1–2x… 1 y2+(2x+3)2=10… 2 Substitute 1 into 2 : (1–2x)2+(2x+3)2=10 1–4x+4x2+4x2+12x+9 =10 8x2+8x+10 =10 x2+x =0 x(x+1) =0 x=0orx=–1 Whenx=0,y = 1–2(0) = 1 Whenx=–1,y = 1–2(–1) = 3 A(0,1);B(–1,3)



∴MidpointofAB

= (0+(–1)2

,1+3

2 ) = (– 1

2,2)

15 (a) A = ( –2+42

,7+9

2 ) = (1,8)

B = (6+42

,1+9

2 ) = (5,5)

(b) AB = (5–8)2+(5–1)2

= 25 = 5units

PQ = (1–7)2+[6–(–2)]2

= 100 = 10units

∴AB=12

PQ

16 (a) P = (2(3)+3(–2)2+3

,2(3)+3(–2)

2+3 ) = (0,0)

(b) P=(2(4)+3(–1)2+3

,2(–2)+3(3)

2+3 ) =(1,1)

17 (a) M= ( 2+02

,–3+1

2 ) = (1,–1)

(b) (1,–1)=( 3+x2

,0+4

2 )

3+x2

=10+y

2 = –1

x =–1 y = –2 S(–1,–2)

18 (a) P = (3(4)+1(–4)3+1

,3(3)+1(–1)

3+1 ) = (2,2)

(b) P=(1(1)+2(–2)1+2

,1(–2)+2(4)

1+2 ) =(–1,2)

19 (a)2x+1(2)

3= 0

2x+2 = 0 2x = –2 x = –1 and

2y+1(4)

3= –2

2y+4 = –6 2y = –10 y = –5 ∴B(–1,–5)

(b)2x+5(3)

7= 1

2x+15 = 7 2x = –8 x = –4 and

2y+5(–4)

7 = –2

2y–20 = –14 2y = 6

y = 3 ∴B(–4,3)

20 (a)7m+(–3n)

m+n = 1

7m–3n = m+n 6m = 4n

mn

=46

=23

∴m:n=2:3

(b) k =2(3)+3(–2)

2+3

=6–6

5

= 0

21 (a)3m+8nm+n

= 5

3m+8n = 5m+5n 3n = 2m

mn

=32

∴m:n=3:2

(b) k =3(4)+2(–1)

3+2

=105

=2

22 (a) Area =12 |

5 1 –6 5| 2 6 –7 2

=12

(30–7–12–2+ 36+35)

= 40unit2

(b) Area =12 |

0 4 6 –50| 0 1 5 3 0

=12

(20+18–6+25)

= 28.5unit2

23 (a) A =12 |

t 2t –2 t |–3 3 –1 –3

=12

(3t–2t+6+6t+6+t)

=12

(8t+12)

= 4t+6 (b) 4t+6 = 14 4t = 8 t = 2

2412 |

5 2 8 5|10 1 r 10=±24

12

(5+2r+80–20–8–5r)=±24

12

(57–3r)=±24

12

(57–3r) = 24

57–3r = 48 3r = 9 r = 3 or

12

(57–3r) = –24

57–3r = –48 3r = 105 r = 35 ∴r=3orr=35

2512 |

1 m 4 1| 2 3 5 2= ±6

12

(3+5m+8–2m–12–5)= ±6

12

(3m–6)= ±6

12

(3m–6) = 6

3m–6 = 12 3m = 18 m = 6 or

12

(3m–6) = –6

3m–6 = –12 3m = –6 m = –2 ∴m=6orm=–2

26 (a) AreaofABC

=12 |

4 3 –4 4| 1 5 3 1

=12

(20+9–4–3+20–12)

=12

(30)

= 15unit2

AreaofACD

=12 |

4 –4–2 4| 1 3 –3 1

=12

(12+12–2+4+6+12)

=12

(44)

= 22unit2

(b) AreaofABCD = AreaofABC+Areaof ACD = 15+22 = 37unit2

2712 |

–1 0 2 –1|–5 –2 k –5 = 0

2–10+4+k = 0 k–4 = 0 k = 4

28 (a) M= ( –3+112

,1+(–3)

2 ) = (4,–1) LetSbe(x,y),

x+4

2 =4 and

y+92

= –1

x+4 =8 y+9 = –2 x =4 x = –11 ∴S(4,–11) (b) AreaofPQRS

=12 |

–3 4 11 4 –3| 1 –11–3 9 1

=12

(33–12+99+4–4+121 +12+27)

=12

(280)

= 140unit2

29m2+4–2m

6–m = 1

m2–2m+4 = 6–m


m2–m–2 = 0 (m+1)(m–2) = 0 m=–1orm=2

30–6–(–k)

k–2 = –3

–6+k = –3k+6 4k = 12 k = 3

31h–24–2

=2–(–1)

2–1

h–2

2 = 3

h–2 = 6 h = 8

32 (a) y–(–3)=3(x–2) y+3=3x–6 y =3x–9 (b) y–4= –2[x–(–1)] y–4= –2(x+1) y–4= –2x–2 y+2x = 2

(c) y–6 =13

(x–5)

3y–18 = x–5 3y–x = 13

(d) y–7 = –25

[x–(–5)]

y–7 = –25

x–2

y = –25

x+5

33k–2

–3–2 = –4

k–2 = 20 k = 22

34 (a) m =3–15–1–5

= 2 y–15 = 2(x–5) y–15 = 2x–10 y = 2x+5 (b) Atx-axis, y = 0 2x = –5

x = –52

∴A(– 52

,0) Aty-axis, x = 0 y = 5 ∴B(0,5)

AB = 52+(52)2

= 1254

= 5.59units

35 m =5–26–3

= 1 y–2= 1(x–3) y–2= x–3 x–y = 1 … 1 x+2y = –5… 2 1 – 2 : –3y = 6 y = –2

Substitutey=–2into 1 : x–(–2)= 1 x = –1 Thepointofintersectionis(–1,–2).

36 2x+y=2… 1 3x+6y=3… 2 1 6: 12x+6y = 12… 3 3 – 2 : 9x = 9 x = 1 Substitutex=1into 1 : 2(1)+y = 2 y = 0 Thepointofintersectionis(1,0). x+2y+h=0 At(1,0), 1+2(0)+h = 0 h = –1

37 x+y=–4… 1 3x+y=–2… 2 2 – 1 : 2x = 2 x = 1 Substitutex=1into 1 : 1+y = –4 y = –5 Twopointsare(0,0)and(1,–5). Theequationis:

yx

=–5–01–0

y = –5x

38 3x–y=–1… 1 2x+3y=–8… 2 1 3: 9x–3y = –3… 3 2 + 3 : 11x = –11 x = –1 Substitutex=–1into 1 : –3–y = –1 y = –2 Theequationis: y–(–2)= –2[x–(–1)] y+2= –2(x+1) y+2= –2x–2 y+2x = –4

39 (a) A(–1,–1)andC(4,3) Theequationis:

y–(–1)x–(–1)

=3–(–1)4–(–1)

y+1x+1

=45

5y+5= 4x+4 5y–4x = –1 B(3,–2)andD(0,2) Theequationis:

y–(–2)

x–3 =

2–(–2)0–3

y+2x–3

= –43

3y+6 = –4x+12 3y+4x = 6 (b) 5y–4x=–1 … 1 3y+4x=6 … 2 1 + 2 : 8y = 5

y =58

Substitutey=58

into 1 :

5(58 )–4x = –1

4x =258

+1

=338

x =3332

∴Thepointofintersectionis

(3332

,58 ).

40 (a) y–x=1… 1 3y–x=–1… 2 2 – 1 :2y = –2 y = –1 Substitutey=–1into 1 : –1–x = 1 x = –2 ∴P(–2,–1) TheequationofABis: y+1 = 2(x+2) y+1 = 2x+4 y = 2x+3

∴A(0,3)andB(– 32

,0) (b)

–2m+n(0)m+n

= –32

–4m = –3m–3n m = 3n

mn

= 3

∴m:n=3:1

ABBP

=31

41 (a) C= ( 2(6)+1(–3)2+1

,2(6)+1(0)

2+1 ) = (3,4)

(b)y

x+3 =

6–06–(–3)

9y = 6x+18 9y–6x = 18 3y–2x = 6 (c) Q(0,6)andC(3,4) Theequationis:

y–6

x =

4–63

3y–18 = –2x 3y+2x = 18

42 (a)y–2

x–(–1) =

–3–24–(–1)

y–2x+1

= –55

5y–10 = –5x–5 5y+5x = 5 y+x = 1

(b) y+x=1… 1 y–2x=–5… 2 1 – 2 : 3x = 6 x = 2 Substitutex=2into 1 : y+2 = 1 y = –1 ∴P(2,–1)


(c)4m+n(–1)

m+n = 2

4m–n = 2m+2n 2m = 3n

mn

=32

∴AP:PB=3:2

43 (a) 2x = y+6 y = 2x–6 m

1 = 2

y = 2x+4 m

2 = 2

∴Parallel (b) x+y = 3 y = –x+3 m

1 = –1

2x+2y = 5 2y = –2x+5

y = –x+52

m2 = –1

∴Parallel

(c)x2

+y3

= 1

3x+2y = 6 2y = –3x+6

y = –32

x+3

m1 = –

32

2y+3x = 8 2y = –3x+8

y = –32

x+4

m2 = –

32

∴Parallel (d) 2x–5y+4 = 0 5y = 2x+4

y =25

x+45

m1 =

25

2y = 5x+6

y =52

x+3

m2 =

52

∴Notparallel

44 (a) m=4,(–1,2) y–2 = 4(x+1) y–2 = 4x+4 y = 4x+6

(b) m=–23

,(–1,–1)

y+1 = –23

(x+1)

3y+3 = –2x–2 3y+2x = –5

45 (a) m=–12

,(2,3)

y–3 = –12

(x–2)

2y–6 = –x+2 2y+x = 8

(b) m=12

,(–6,1)

y–1 =12

(x+6)

y–1 =12

x+3

y =12

x+4

(c) m = –43

,(4,–1)

y+1 = –43

(x–4)

3y+3 = –4x+16 3y+4x = 13

(d) m=–23

,(–3,–5)

y+5 = –23

(x+3)

y+5 = –23

x–2

y = –23

x–7

46 (a)5–(–3)

p+8 =

23

24 = 2p+16 2p = 8 p = 4

(b) m=23

,(4,5)

y–5 =23

(x–4)

3y–15 = 2x–8 3y–2x = 7

47 (a) y–x=–2… 1 y+2x=1… 2 2 – 1 : 3x = 3 x = 1 Substitutex=1into 1 : y–1 = –2 y = –1 ∴A(1,–1) (b) m=3,A(1,–1) Theequationis: y+1 = 3(x–1) y+1 = 3x–3 y = 3x–4

48 (a) x+y = 5 x–y = 3 y = –x+5 y = x–3 m

1 = –1 m

2 = 1

∴Perpendicular (b) 2x+y = 1 y = –2x+1 m

1 = –2

x+2y = 4 2y = –x+4

y = –12

x+2

m2 = –

12

∴Notperpendicular (c) y = 3x+6 x = 3y+7 m

1= 3 3y = x–7

y =13

x–73

m2 =

13

∴Notperpendicular

(d)x3

–y5

= 1

5x–3y = 15 3y = 5x–15

y =53

x–5

m1 =

53

5y+3x = 10 5y = –3x+10

y = –35

x+2

m2 = –

35

∴Perpendicular

49 (a) m=–1,(0,0) Theequationis: y=–x (b) m=2,(3,–2) Theequationis: y+2= 2(x–3) y+2= 2x–6 y = 2x–8 (c) m=–2,(–4,–5) Theequationis: y+5 = –2(x+4) y+5 = –2x–8 y+2x+13 = 0

(d) m=–12

,(5,6)

Theequationis:

y–6 = –12

(x–5)

2y–12 = –x+5 2y+x = 17

50 (a) m=–12

y=–12

x+n

At(4,–5), –5 = –12

(4)+n

n = –5+2 = –3

∴m=–12

,n=–3

(b)k3

= –23

k = –2

51 (a) m1 =

3–5–2–4

=13

m2 = –3

Midpoint = (4+(–2)2

,5+3

2 ) = (1,4) Theequationofperpendicular

bisectoris: y–4 = –3(x–1) y–4 = –3x+3 y+3x = 7

(b) m1

=4–(–8)–2–3

= –125


m2 =

512


,–8+4

2 ) = ( 1

2,–2)

Theequationofperpendicularbisectoris:

y+2 =512 (x–

12 )

12y+24 = 5x–52

24y+48 = 10x–5 24y–10x+53 = 0

(c) m1 =

8–74–3

= 1 m

2 = –1

Midpoint = (3+42

,7+8

2 ) = ( 7

2,

152 )


y–152

= –1(x–72 )

y–152

= –x+72

2y–15 = –2x+7 2y+2x = 22 y+x = 11

(d) m1 =

6–3–1–2

= –1 m

2 = 1


,3+6

2 ) = ( 1

2,

92 )


y–92

= 1(x–12 )

2y–9 = 2x–1 2y–2x = 8 y–x = 4

52 (a) mAB

=4–21–3

= –1 m

CD = 1

MidpointofAB

= (1+32

,4+2

2 ) = (2,3) Theequationis: y–3 = 1(x–2) y = x+1 (b) At(4,t), t = 4+1 = 5 (c) LetDbe(0,y),

(0+42

,y+5

2 ) = (2,3)

y+5

2 = 3

y+5 = 6 y = 1 ∴D(0,1)

53 (a) m=–13

,A(2,–3)

Theequationis:

y+3 = –13

(x–2)

3y+9 = –x+2 3y+x = –7 (b) 3y+x=–7… 1 y–3x=1 … 2 1 3 : 9y+3x = –21… 3 2 + 3 : 10y = –20 y = –2 Substitutey=–2into 1 : –6+x = –7 x = –1 ∴P(–1,–2)

54 (a) m =2–4

2–(–1)

= –23

Theequationis:

y–9 = –23

(x+2)

3y–27 = –2x–4 3y+2x = 23

(b) m=32

,A(–1,4)

Theequationis:

y–4 =32

(x+1)

2y–8 = 3x+3 2y–3x = 11 (c) 3y+2x= 23… 1 2y–3x = 11… 2 1 3: 9y+6x = 69… 3 2 2: 4y–6x = 22… 4 3 + 4 : 13y = 91 y = 7 Substitutey=7into 1 : 3(7)+2x = 23 21+2x = 23 2x = 2 x = 1 ∴D(1,7)

55 (a) (x+1)2+(y+3)2 = 3 x2+2x+1+y2+6y+9 = 9 x2+y2+2x+6y+1 = 0

(b) x2+y2 = 3 x2+y2 = 9 x2+y2–9 = 0

(c) (x–2)2+(y–5)2 = 3 x2–4x+4+y2–10y+25 = 9 x2+y2–4x–10y+20 = 0

(d) (x–1)2+(y+2)2 = 3 x2–2x+1+y2+4y+4 = 9 x2+y2–2x+4y–4 = 0

56 (a) (x+1)2+y2= (x–1)2+y2

x2+2x+1+y2= x2–2x+1+y2

2x+1= –2x+1 4x = 0 x = 0

(b) (x+1)2+(y–1)2

= (x–1)2+(y+1)2

x2+2x+1+y2–2y+1 = x2–2x+1+y2+2y+1

2x–2y+2 = –2x+2y+2 4x–4y = 0 x–y = 0

(c) (x–2)2+(y–4)2

= (x–8)2+(y–6)2 x2–4x+4+y2–8y+16 = x2–16x+64+y2–12y+36 20–4x–8y = 100–16x–12y 12x+4y–80 = 0 3x+y–20 = 0

(d) x2+(y–2)2 = (x–2)2+y2

x2+y2–4y+4 = x2–4x+4+y2

4–4y = 4–4x 4x–4y = 0 x–y = 0

57 (a) 2PA = PB

2 (x+4)2+y2 = (x–4)2+y2

4(x2+8x+16+y2)= x2–8x+16+y2

4x2+4y2+32x+64= x2+y2–8x+16 3x2+3y2+40x+48= 0

(b) 3PA=2PB 3 (x+4)2+y2 =2 (x–4)2+y2

9(x2+8x+16+y2) =2(x2–8x+16+y2)9x2+9y2+72x+144 =2x2+2y2–16x+327x2+7y2+88x+112 =0

(c) PA2+PB2 = 10

( (x+4)2+y2 )2+( (x–4)2+y2 )2

= 10

x2+8x+16+y2+x2–8x+16+y2 = 10 2x2+2y2+22 = 0 x2+y2+11 = 0

(d) 3PA= PB

3 (x+4)2+y2 = (x–4)2+y2

9(x2+8x+16+y2)= x2–8x+16+y2

9x2+72x+144+9y2= x2–8x+16+y2

8x2+8y2+80x+128= 0 x2+y2+10x+16= 0

58 PA = x (x–3)2+(y–2)2 = x x2–6x+9+y2–4y+4 = x2

y2–6x–4y+13 = 0

59 PA=3PB

(x–2)2+y2 =3 (x+4)2+y2

x2–4x+4+y2=9(x2+8x+16+y2) x2–4x+4+y2=9x2+72x+144+9y2

8x2+8y2+76x+140=0 2x2+2y2+19x+35=0

60 PA=x+3

(x–3)2+y2 = x+3 x2–6x+9+y2= (x+3)2

x2–6x+9+y2= x2+6x+9 y2= 12x


61 3OP =2PA 3 x2+y2 = 2 (x–2)2+y2

9(x2+y2)= 4(x2–4x+4+y2) 9x2+9y2= 4x2–16x+16+4y2

5x2+5y2+16x–16= 0

62 (x–4)2+y2 = x x2–8x+16+y2 = x2

y2–8x+16 = 0

1h2+k2

2 = 5

h2+k2 = 10… 1

h+k

2 = 1

h+k = 2 h = 2–k… 2 Substitute 2 into 1 : (2–k)2+k2 = 10 4–4k+k2+k2 = 10 2k2–4k–6 = 0 k2–2k–3 = 0 (k+1)(k–3) = 0 k=–1ork=3 Substitutek=–1into 2 : h = 2–(–1) = 3 Substitutek=3into 2 : h= 2–3 = –1 ∴h=3,k=–1;h=–1,k=3

2 (p+(–7)2

,6+q

2 ) =(4+(–2)

2,

–4+42 )

p–7

2 = 1 and

6+q2

= 0

p–7 = 2 q = –6 p = 9 ∴p=9,q=–6

3 Area=6unit2

12

2 5 –1 2 | 3 r 4 3 | = ±6

12

(2r+20–3–15+r–8)= ±6

12

(3r–6)= ±6

3r–6 = 12 or 3r–6 = –12 3r = 18 3r = –6 r = 6 r = –2

4 y=5–2x… 1 x2+y2=10… 2 x2+(5–2x)2= 10 x2+25–20x+4x2 = 10 5x2–20x+15 = 0 x2–4x+3 = 0 (x–1)(x–3) = 0 x=1orx=3 Substitutex=1into 1 : y = 5–2(1) = 3

Substitutex=3into 1 : y = 5–2(3) = –1 P(1,3)andQ(3,–1) ∴MidpointofPQ

= (1+32

,3+(–1)

2 ) = (2,1)

512

0 2 4m 0 | 0 3m 6 0 | = 0

12–12m2 = 0 12m2 = 12 m2 = 1 m = ±1

6 mBC

=–12

B(0,3)

3–00–

= –12

= 6

7 (a)3t–(t+2)

t2–t =

t–32t

2t(2t–2) = (t–3)(t2–t) 4t2–4t = t3–t2–3t2+3t t3–8t2+7t = 0 t2–8t+7 = 0 (t–1)(t–7) = 0 t=1ort=7

(b) ( 3tt2 )( t–3–(t+2)

2t–t )=–1

( 3t )(–5

t )= –1

t2= 15 t = ± 15

8 x=y+6… 1 y2=8x … 2 Substitute 1 into 2 : y2= 8(y+6) y2–8y–48= 0 (y+4)(y–12)= 0 y=–4ory=12 Substitutey=–4into 1 : x = –4+6 = 2 Substitutey=12into 1 : x = 12+6 = 18 A(2,–4)andB(18,12)

∴AB = [12–(–4)]2+(18–2)2

= 512 = 22.627units

9 Midpoint = (–1+72

,8+(–2)

2 ) = (3,3) Theequationofthelinethrough(3,3) andparalleltotheline2x+5y=9:

y–3 = –25

(x–3)

5y–15 = –2x+6 5y+2x = 21

10 x–2y=1… 1 x+3y=6… 2 2 – 1 : 5y = 5 y = 1 Substitutey=1into 1 : x–2(1) = 1 x = 3 ∴Thepointofintersectionis(3,1). 3x+4y = 8 4y = –3x+8

y = –34

x+2

∴m=–34

Therequiredequationis:

y–1 = –34

(x–3)

4y–4 = –3x+9 4y+3x = 13

11 (a) MidpointofAB

= ( –6+(–2)2

,1+7

2 ) = (–4,4)

mAB

=7–1

–2–(–6)

=32

mPQ

=–23

TheequationofthelinePQis:

y–4 = –23

(x+4)

3y–12 = –2x–8 3y+2x = 4 (b) 3y+2x=4 Aty-axis,x=0 3y+2(0)= 4 3y = 4

y =43

∴P(0,43 )

Atx-axis,y=0 3(0)+2x = 4 2x = 4 x = 2 ∴Q(2,0)

12 ( –3–32–k )(–3–1

2–10 ) = –1

( –62–k )( 1

2 ) = –1

–3 = –1(2–k) –3 = k–2 k = –1

13 (a) mAB

mAC

= (5–(–3)3–(–1) )(5–2

3–9 ) = 2(– 1

2 ) = –1 Sincem

ABm

AC=–1,theangle

BACistherightangleandhenceABCisaright-angledtriangle.


(b) AreaofABC

=12

3 –1 9 3 | 5 –3 2 5 |

=12

(–9–2+45+5+27–6)

=12

(60)

= 30unit2

14 Midpoint = ( –4+82

,6+(–2)

2 ) = (2,2)

Gradient =–2–6

8–(–4)

= –23

Sogradientoftheperpendicular

bisector=32

.

Equationoftheperpendicularbisectoris:

y–2 =32

(x–2)

2y–4 = 3x–6 2y–3x = –2

15 (a) 2x+3y =7 … 1 3x–4y =2 … 2 1 4: 8x+12y = 28… 3 2 3: 9x–12y = 6 … 4 3 + 4 : 17x = 34 x = 2 Substitutex=2into 1 : 2(2)+3y = 7 3y = 3 y = 1 ∴A(2,1) Hence,theequationoftheline

throughA(2,1)andpassesthroughB(5,7)is

y–1x–2

=7–15–2

y–1= 2(x–2) y–1= 2x–4 y = 2x–3

(b) GradientofAB =7–15–2

= 2 Sothegradientofthe

perpendicularline=–12

.

TheequationoftheperpendicularlinethroughA(2,1)is:

y–1 = –12

(x–2)

2y–2 = –x+2 2y+x = 4

16 (a) AreaofABC

=12

1 7 1 1 |–2 6 2 –2|

=12

(6+14–2+14–6–2)

=12

(24)

= 12unit2

(b) TheequationofthelineAC:

y+2x–1

=6–(–2)

7–1 3y+6 = 4x–4 3y–4x = –10… 1

mBD

=–34

andB(1,2)

TheequationofthelineBD:

y–2= –34

(x–1)

4y–8= –3x+3 4y+3x=11… 2 1 3: 9y–12x = –30… 3 2 4: 16y+12x = 44… 4 3 + 4 : 25y = 14

y = 1425

Substitutey=1425

into 1 :

3(1425 )–4x = –10

4x =4225

+10

=29225

=7325

∴D(7325

,1425 )

17 C = (3+92

,5+1

2 ) = (6,3)

mAB

=5–13–9

so mCD

=32

= –23

TheequationoftheperpendicularbisectorofABis:

y–3 =32

(x–6)

2y–6 = 3x–18 2y–3x = –12… 1 TheequationofthelineODis:

y–0= –23

(x–0)

3y= –2x 3y+2x = 0… 2 1 2: 4y–6x = –24… 3 2 3: 9y+6x = 0… 4 3 + 4 : 13y = –24

y = –2413

Substitutey=–2413

into 1 :

2(– 2413 )–3x = –12

3x =10813

x =3613

∴D(3613

,–2413 )

18 (2k)2+k2 = (2k+1)2+(k–3)2 4k2+k2 = 4k2+4k+1+k2–6k+9 –2k+10 = 0 –2k = –10 k = 5

19 (a)x

2a+

y3a

=1

(b)142a

+(–93a ) = 1

7a

–3a

= 1

4a

= 1

a = 4

20 MidpointofAB = (0+82

,5+7

2 ) = (4,6)

mAB

=7–58–0

=14

Sogradientofthebisector=–4 EquationofthebisectorofABis: y–6 = –4(x–4) y–6 = –4x+16 y+4x = 22… 1 MidpointofBC=(6,4)

mBC

=7–18–4

=32

Sogradientofthebisector=–23

EquationofthebisectorBCis:

y–4 =–23

(x–6)

3y–12 = –2x+12 3y+2x= 24… 2 2 2: 6y+4x=48… 3 3 – 1 : 5y = 26

y =265

Substitutey=265

into 1 :

265

+4x = 22

4x =845

x =215

∴Thepointofintersectionis(215

,265 ).

21 LetRbe(x,y), PR=QR

(x+3)2+(y–1)2

= (x+1)2+(y–7)2 x2+6x+9+y2–2y+1 = x2+2x+1+y2–14y+49 6x–2y+10 = 2x–14y+50 4x+12y = 40 x+3y = 10… 1 y–2x = –6… 2 1 2: 6y+2x = 20… 3 2 + 3 : 7y = 14 y = 2 Substitutey=2into 1 : x+3(2) = 10 x+6 = 10 x = 4 ∴R(4,2)


22 mPQ

=–1

10–3–

= –1

10– = –3+ + = 13… 1

MidpointofPQ=(+32

, +102 )

+102

= +32

+5

–+72

= 5

–+7 = 10 – = 3… 2 1 + 2 : 2 = 16 = 8 Substitute=8into 1 : +8 = 13 = 5 ∴=5,=8

23 (x–2)2+(y–3)2 = (x–1)2+y2 x2–4x+4+y2–6y+9 =x2–2x+1+y2

13–4x–6y = 1–2x 2x+6y–12 = 0 x+3y–6 = 0

24 (a) mPQ

=3andR(–2,3) TheequationofthelinePQ: y–3 = 3(x+2) y–3 = 3x+6 y = 3x+9 (b) y=3x+9 Atx-axis, y = 0 0 = 3x+9 –3x = 9 x = –3 ∴P(–3,0) Aty-axis, x = 0 y = 3(0)+9 = 9 ∴Q(0,9)

(c)m(0)+n(–3)

m+n = –2

–3n = –2m–2n 2m = n

mn

=12

∴PR:RQ=1:2

25 (a) LetCbe(0,y), m

BC = –5

y–10–1

= –5

y–1 = 5 y = 6 ∴C(0,6)

(b) mBD

=7–1–8–1

= –69

= –23

mAC

=32

TheequationofAC:

y–6=32

(x–0)

2y–12= 3x 2y–3x = 12 (c) 2y–3x=12 Atx-axis,y=0 2(0)–3x = 12 x = –4 ∴A(–4,0)

M = (–4+02

,0+6

2 ) = (–2,3)

(d) Area =12

–4 1 0 –8 –4 | 0 1 6 7 0 |

=12

(–4+6+48+28)

= 39unit2

26 (a) mAB

=4–(–4)–2–(–6)

=84

= 2

mBC

= –12

TheequationofthelineBC:

y–4 = –12

(x+2)

2y–8 = –x–2 2y+x = 6 (b) 2y+x=6 Aty-axis, x=0 2y = 6 y = 3 ∴D(0,3) TheequationofthelineAB: y+4 = 2(x+6) y+4 = 2x+12 y = 2x+8 Atx-axis,y=0 0 = 2x+8 2x = –8 x = –4 ∴E(–4,0)

mAC

=mED

=0–3–4–0

=34

TheequationofAC:

y+4 =34

(x+6)

4y+16 = 3x+18 4y–3x = 2 2y+x = 6… 1 4y–3x = 2… 2 1 3: 6y+3x =18… 3 2 + 3 : 10y =20 y =2 Substitutey=2into 1 : 2(2)+x = 6 x = 2 ∴C(2,2)

(c) Area =12

–6 2 0 –4 –6 |–4 2 3 0 –4|

=12

(–12+6+16+8+12)

=12

(30)

= 15unit2

27 (a) mAB

=12

andA(–3,–1)

TheequationofAB:

y+1 =12

(x+3)

2y+2 = x+3 2y–x = 1 m

AD=–2andA(–3,–1)

TheequationofAD: y+1 = –2(x+3) y+1 = –2x–6 y+2x = –7 (b) 2y–x= 6… 1 y+2x= –7… 2 1 2: 4y–2x =12… 3 2 + 3 : 5y = 5 y = 1 Substitutey=1into 1 : 2(1)–x = 6 x = –4 ∴D(–4,1) AreaofACD = 7.5

12

–3 x –4 –3 |–1 y 1 –1| = 7.5

12

(–3y+x+4+x+4y+3) = 7.5

2x+y+7 = 15 2x+y=8… 1 2y–x=6… 2 2 2: –2x+4y = 12… 3 1 + 3 : 5y = 20 y = 4 Substitutey=4into 1 : 2x+4 = 8 2x = 4 x = 2 ∴C(2,4) LetBbe(x,y),

(–4+x2

,1+y

2 )=(–3+22

,–1+4

2 )

–4+x2

= –12

and1+y

2 =

32

–4+x = –1 1+y =3 x = 3 y =2 ∴B(3,2)

28 (a) M = (–3+52

,–3+(–1)

2 ) = (1,–2)

mCD

=–1–(–3)5–(–3)

=28

=14

mAM

= –4


TheequationofperpendicularbisectorofAM:

y+2 = –4(x–1) y+2 = –4x+4 y+4x = 2 Aty-axis,x=0 y+4(0) = 2 y = 2 ∴A(0,2)

(b) mAB

=14

andA(0,2)

TheequationofAB:

y–2 =14

(x–0)

4y–8 = x 4y–x = 8 m

BC=–4andC(5,–1)

TheequationofBC: y+1 = –4(x–5) y+1 = –4x+20 y+4x = 19 (c) 4y–x=8… 1 y+4x=19… 2 1 4: 16y–4x = 32… 3 2 + 3 : 17y = 51 y = 3 Substitutey=3into 1 : 4(3)–x = 8 x = 12–8 = 4 ∴B(4,3) (d) Area

=12

0 –3 5 4 0 | 2 –3 –1 3 2 |

=12

(3+15+8+6+15+4)

= 2512

unit2

29 (a) mAB

=5–3

–1–(–5)

=24

=12

mDE

=–2 MidpointofAB,

D = (–5+(–1)2

,3+5

2 ) = (–3,4) TheequationofDE: y–4 = –2(x+3) y–4 = –2x–6 y+2x = –2

mAC

=–3–3

3–(–5)

= –68

= –34

∴mBF

=43

TheequationofBF:

y–5 =43

(x+1)

3y–15 = 4x+4 3y–4x = 19

(b) y+2x =–2… 1 3y–4x =19… 2 1 2: 2y+4x =–4… 3 2 + 3 : 5y =15 y =3 Substitutey=3into 1 : 3+2x = –2 2x = –5

x = –52

∴G(– 52

,3)30 (a) m

AB = –3m

4–1–1–1

= –3m

–32

= –3m

m =12

(b) mAC

= 3m

= 3(12 )

=32

TheequationofAC:

y–1 =32

(x–1)

2y–2 = 3x–3 2y–3x = –1… 1 m

BC = m

=12

TheequationofBC:

y–4 =12

(x+1)

2y–8 = x+1 2y–x = 9… 2 2 – 1 : 2x = 10 x = 5 Substitutex=5into 1 : 2y–3(5) = –1 2y–15 = –1 2y = 14 y = 7 ∴C(5,7)

(c) AB = (4–1)2+(–1–1)2

= 13

AC = (5–1)2+(7–1)2

= 52

= 413

= 2 13 ∴AC=2AB

31 (a) mPQ

=7–34–6

= –42

= –2 (b) m

AB=–2andR(3,1)

TheequationofAB: y–1 = –2(x–3) y–1 = –2x+6 y+2x = 7 (c) m

AC = m

RP

=3–16–3

=23

Thegradientoftheperpendicularbisectoris–

32

.

TheequationoftheperpendicularbisectorofAC:

y–7 = –32

(x–4)

2y–14 = –3x+12 2y+3x = 26

32 (a) y–x= 2… 1 y+3x = 10… 2 2 – 1 : 4x = 8 x = 2 Substitutex=2into 1 : y–2 = 2 y = 4 ∴A(2,4) LetCbe(x,y),

( 2+x2

,4+y

2 )=(312

,712 )

2+x

2=

72

and4+y

2=

152

x = 5 y = 11 ∴C(5,11) (b) m

BC=1andC(5,11)

TheequationofBC: y–11= x–5 y–x = 6

33 (a) LetDbe(x,y),

x+9

2= –1 and

y+112

=7

x+9= –2 y+11= 14 x = –11 y = 3 ∴D(–11,3)

(b) mAC

=12–7

–3–(–1)

= –52

TheequationofAC:

y–7= –52

(x+1)

2y–14= –5x–5 2y+5x = 9 (c) LetCbe(x,y),

–3+x

2 = –1

–3+x = –2 x = 1 and

12+y

2 = 7

12+y = 14 y = 2 ∴C(1,2)

AC = (1+3)2+(2–12)2

= 116 = 10.77units

(d) Area =12

–3 –11 1 9 –3 |12 3 2 11 12|

�0© Cerdik Publications Sdn. Bhd. (203370-D) 2010 ISBN: 978-983-70-3258-3

=12

(–9–22+11+108

+132–3–18+33)

=12

(232)

= 116unit2

34 (a) mAB

=8–22–4

= –62

= –3

mBC

=13

andB(2,8)

TheequationofBC:

y–8 =13

(x–2)

3y–24 = x–2 3y–x = 22 (b) 3y–x=22… 1 y–x=–2… 2 1 – 2 : 2y = 24 y = 12 Substitutey=12into 1 : 3(12)–x = 22 36–x = 22 x = 14 ∴C(14,12) MidpointofAC

= ( 4+142

,2+12

2 ) = (9,7) LetDbe(x,y),

2+x

2 = 9 and

8+y2

= 7

2+x = 18 8+y = 14 x = 16 y = 6 ∴D(16,6) (c) TheareaofABCD

=12

4 16 14 2 4 | 2 6 12 8 2|

=12

(24+192+112+4–32–

84–24–32)

=12

(160)

= 80unit2

35 (a) 3y–5x=–34…1 4y–x=0… 2 2 5: 20y–5x = 0… 3 3 – 1 : 17y = 34 y = 2 Substitutey=2into 1 : 3(2)–5x = –34 6–5x = –34 5x = 40 x = 8 ∴C(8,2) LetBbethepoint(x,y),

( x2

,y2 )=(3+8

2,

5+22 )

x2

=112

andy2

=72

x = 11 y = 7 ∴B(11,7) m

AD=–4andA(3,5)

TheequationofAD: y–5=–4(x–3) y–5=–4x+12 y+4x=17… 1 4y–x=0… 2 2 4:16y–4x = 0… 3 1 + 3 : 17y = 17 y = 1 Substitutey=1into 1 : 1+4x = 17 4x = 16 x = 4 ∴D(4,1) (b) TheareaofOABC

=12

0 8 11 3 0 | 0 2 7 5 0 |

=12

(56+55–22–21)

=12

(68)

= 34unit2

36 (a) LetAbe(x,y),

x2+y2 = 20 x2+y2=20… 1 y=2x… 2 Substitute 2 into 1 : x2+4x2 = 20 5x2 = 20 x2 = 4 x = ±2 Substitutex=2into 2 : y = 2(2) = 4 ∴A(2,4)

mAB

=–12

andA(2,4)

TheequationofAB:

y–4 = –12

(x–2)

2y–8 = –x+2 2y+x = 10 Aty-axis, x = 0 2y = 10 y = 5 ∴B(0,5) m

BC=2andB(0,5)

TheequationofBC: y–5 = 2(x–0) y–2x = 5… 1 y+3x = 0… 2 2 – 1 : 5x = –5 x = –1 Substitutex=–1into 1 : y–2(–1) = 5 y+2 = 5 y = 3 ∴C(–1,3)

(b) Area =12

0 2 0 –1 0 | 0 4 5 3 0|

=12

(10+5)

= 7.5unit2

37 (a) AB = (6–2)2+(6–3)2

= 25

= 5units

BC = (6–6)2+(6–1)2

= 25 = 5units SinceAB=BC,ABCisan

isoscelestriangle.

(b) m = (2+62

,3+1

2 ) = (4,2)

mCD

=mBA

=6–36–2

=34

TheequationofCD:

y–1 =34

(x–6)

4y–4 = 3x–18 4y–3x = –14… 1

mAC

=1–36–2

= –12

SomBD

=2 TheequationofBD: y–6=2(x–6) y–6=2x–12 y–2x =–6… 2 2 4: 4y–8x = –24… 3 1 – 3 : 5x = 10 x = 2 Substitutex=2into 1 : 4y–3(2)= –14 4y = –8 y = –2 ∴D(2,–2) (c) DP=2MP

(x2–2)2+(y+2)2 =2 (x–4)2+(y–2)2 x2–4x+4+y2+4y+4= 4(x2–8x+16+ y2–4y+4)3x2+3y2–28x–20y+72=0

38 (a) 3x–y=–9… 1 2x+y=–1… 2 1 + 2 : 5x = –10 x = –2 Substitutex=–2into 1 : 3(–2)–y = –9 –6–y = –9 y = 3 ∴P(–2,3)

mPO

=–32

andO(0,0)

TheequationofPO:

y=–32

x

(b) mPC

=–13

andP(–2,3)

TheequationofPC:

y–3 = –13

(x+2)

3y–9 = –x–2 3y+x = 7

1

Statistics

1 (a) Mean = 34812

= 29 Mode = 25

Median =

25 + 31

2 = 28

(b) Mean =

4215

= 2.8 Mode = 3 Median = 3

(c) Mean =

11011

= 10 Mode = 12 Median = 10

(d) Mean =

1418

= 17.625 Mode = 22

Median =

17 + 18

2 = 17.5

2 (a) Mean =

3120

= 1.55 Mode = 0

Median =

1 + 2

2 = 1.5

(b) Mean =

5025

= 2 Mode = 1 Median = 2

(c) Mean =

7828

= 2.786 Mode = 1

Median =

2 + 3

2 = 2.5

3 (a) x– = 8

55 + x

8 = 8

x = 64 – 55 = 9 (b) x– = 6

36 + y

7 = 6

36 + y = 42 y = 42 – 36 = 6

4 x– = 6

800 + 8x140 + x

= 6

800 + 8x = 840 + 6x 2x = 40 x = 20

5 (a) Modal class = 15 – 19

Mean =

68540

= 17.125

Median = 14.5 +

[20 – 1412 ]5

= 17 (b) Modal class = 161 – 165

Mean =

16 024

98 = 163.51

Median = 160.5 +

[49 – 2540 ]5

= 163.5 (c) Modal class = 50 – 59

Mean =

266550

= 53.3

Median = 49.5 + [25 – 19

16 ]10

= 53.25 (d) Modal class = 53 – 55

Mean =

164130

= 54.7

Median =

52.5 +

[15 – 98 ]3

= 54.75

6 x– = 68.25

1077 + 72x

x + 16 = 68.25

1077 + 72x = 68.25x + 1092 3.75x = 15 x = 4

7 a + b + 18 = 50 a + b = 32 b = 32 – a … 1 Median = 58

55.5 +

[25 – (b + 8)a ]10 = 58 … 2


25 – (32 – a + 8)a = 0.25

a – 15 = 0.25a 0.75a = 15 a = 20 Substitute a = 20 into 1 : b = 32 – 20 = 12

8 (a)Number of apples

Mass (g)54.5 60.5 66.5 72.563.8

14

12

10

8

6

4

2

0

Mode = 63.8 (b)

Number of people

14

12

10

8

6

4

2

0 42.5 48.5 54.5 60.550

Age(years)

Mode = 50 (c)Number of pupils

16

14

12

10

8

6

4

2

0

14.25

Arrival time(minutes)6.5 11.5 16.5 21.5

Mode = 14.25



(d)Number of workers

12

10

8

6

4

2

038.5 43.5 48.5 53.5

45.75

Amount spent (RM)

Mode = 45.75 9 (a) Cumulative frequency

40

35

30

25

20

15

10

5

024.5 44.529.5 49.534.5 54.539.5 59.5

43.75

Length (mm)

m

Median = 43.75 (b)

Cumulative frequency

m

100

90

80

70

60

50

40

30

20

10

0.5 40.510.5 50.520.5 60.530.5 70.5 80.533.5

Mark

Median = 33.5

10 (a) Mean = 5510

= 5.5 Mode = 6 Median = 6 (b) (i) New mean = 5.5 + 1 = 6.5 New mode = 6 + 1 = 7 New median = 6 + 1 = 7 (ii) New mean = 5.5 – 2 = 3.5 New mode = 6 – 2 = 4 New median = 6 – 2 = 4

(iii) New mean = 5.5(3) = 16.5 New mode = 6(3) = 18 New median = 6(3) = 18

(iv) New mean = 5.5

12

= 11

New mode = 6

12

= 12

New median = 6

12

= 12 11 (a) Range = 32 – 15 = 17 Interquartile range = 29 – 16 = 13 (b) Range = 9 – 2 = 7 Interquartile range = 7 – 3.5 = 3.5 (c) Range = 16 – 7 = 9 Interquartile range = 15 – 9 = 6 (d) Range = 8 – 2 = 6 Interquartile range = 7 – 3 = 4

12 (a) Range = 5 – 1 = 4 Interquartile range = 4 – 1 = 3 (b) Range = 6 – 1 = 5 Interquartile range = 5 – 2 = 3 (c) Range = 6 – 0 = 6 Interquartile range = 4.5 – 2 = 2.5 (d) Range = 8 – 4 = 4 Interquartile range = 8 – 4 = 4

13 (a) Range = 28 – 3 = 25

Q1 = 0.5 +

[10 – 010 ]5

= 5.5

Q3 = 10.5 +

[30 – 246 ]5

= 15.5 Interquartile range = 15.5 – 5.5 = 10 (b) Range = 75.5 – 5.5 = 70

Q1 = 30.5 +

[40 – 1968 ]10

= 30.5 + 3.088 = 33.5880


Q3 = 40.5 +

[120 – 8751 ]10

= 40.5 + 6.471 = 46.971 Interquartile range = 46.971 – 33.588 = 13.383 (c) Range = 70.5 – 42.5 = 28

Q1 = 48.5 +

[25 – 817 ]4

= 52.5

Q3 = 60.5 +

[75 – 6916 ]4

= 62 Interquartile range = 62 – 52.5 = 9.5 (d) Range = 65.5 – 10.5 = 55

Q1 = 40.5 +

[14 – 1315 ]20

= 41.83

Q3 = 50.5 +

[42 – 2820 ]10

= 57.5 Interquartile range = 57.5 – 41.83 = 15.67

14 (a) x– = 8010

= 8

σ2 =

66810

– (8)2

= 2.8 σ = 2.8 = 1.673

(b) x– =

255

= 5

σ2 =

1355

– (5)2

= 2 σ = 2 = 1.414

(c) x– =

7212

= 6

σ2 =

49412

– (6)2

= 5.167 σ = 5.167 = 2.273

(d) x– =

488

= 6

σ2 =

3008

– (6)2

= 1.5 σ = 1.5 = 1.225

15 (a) x– = 14020

= 7

σ2 =

42010

– (6)2

= 6 σ = 6 = 2.449 (a) x– = 6 + 2 = 8 σ2 = 6 σ = 2.449 (b) x– = 6(3) = 18 σ2 = 32 6 = 54 σ = 3 2.449 = 7.347

(c) x– =

62

= 3

σ2 =

622

= 1.5

σ =

2.449

2 = 1.2245

18 Range = 35 – 23 = 12 Interquartile range = 31 – 25.5 = 5.5 (a) New range = 12 New interquartile range = 5.5 (b) New range = 12 New interquartile range = 5.5

(c) New range =

12

12

= 6 New interquartile range

=

12

5.5

= 2.75

(d) New range =

124

= 3

New interquartile range =

5.54

= 1.375 19 (a) x– = 4

Σx5

= 4

Σx = 20

20 + x6

= 4

20 + x = 24 x = 4 (b) σ2 = 2

Σx2

5 – (4)2 = 2

Σx2 = 90 The new variance,

=

90 + 42

6 – (4)2

= 1.667

σ2 =

101620

– (7)2

= 1.8 σ = 1.8 = 1.342

(b) x– =

26440

= 6.6

σ2 =

181440

– (6.6)2

= 1.79 σ = 1.79 = 1.338

(c) x– =

24040

= 6

σ2 =

1512

40 – (6)2

= 1.8 σ = 1.8 = 1.342

(d) x– =

3015

= 2

σ2 =

8015

– (2)2

= 1.333 σ = 1.333 = 1.155

16 (a) x– =

274040

= 68.5

σ2 =

188 980

40 – (68.5)2

= 32.25 σ = 32.25 = 5.679

(b) x– =

1560

30 = 52

σ2 =

81 750

30 – (52)2

= 21 σ = 21 = 4.583

(c) x– =

43020

= 21.5

σ2 =

10 340

20 – (21.5)2

= 54.75 σ = 54.75 = 7.399

(d) x– =

85550

= 17.1

σ2 =

17 155

50 – (17.1)2

= 50.69 σ = 50.69 = 7.12

17 x– =

6010

= 6


20 (a)


Q1

Q3

200

180

160

140

120

100

80

60

40

20

0.5 12.53.5 15.5 24.56.5 18.5 27.59.5 21.5 30.518.2 24.94

Life span(weeks)

Interquartile range = 24.94 – 18.2 = 6.74 (b)


Q1

Q3

100

90

80

70

60

50

40

30

20

10

19.5 39.524.5 44.5 59.529.5 49.534.5 54.531 42.5

Age (years)

Interquartile range = 42.5 – 31 = 11.5

(b) m =

3 + 4

2 = 3.5

4 m = 14.5 + (15 – 10

10 )5

= 14.5 + 2.5 = 17

5 σ = 3 12

49 + y2

4 – (11 + y

4 )2

= 3

12

49 + y2

4 – (121 + 22y + y2

16 ) = 12.25

196 + 4y2 – 121 – 22y – y2 = 196 3y2 – 22y – 121 = 0 (3y + 11)(y – 11) = 0

y = –

113

or y = 11

\ y = 11

6 (a) x– = 8

7x + 44

= 8

7x + 4 = 32 7x = 28 x = 4 (b) When x = 4 the set of data is 2, 7,

7 and 16.

σ2 =

3584

– (8)2

= 25.5

7 (a) x– = 9

2m + 184

= 9

2m + 18 = 36 2m = 18 m = 9 σ = 5

164 + (9 – n)2 + (9 + n)2

4 – (9)2

= 5

164 + 81 – 18n + n2 + 81 + 18n + n2

4 – 81

= 25

2n2 + 326 = 424 2n2 = 98 n2 = 49 n = 7 \ m = 9, n = 7

8 σ2 = Σ(x – x–)2

N

=

15020

= 7.5

9 (a) Median mark = 34.5 (b) Q1 = 27.5 Q3 = 42.5 \ Interquartile range = 42.5 – 27.5 = 15 (c) 400 – 275 = 125 students

1

28 + x4

= 10

28 + x = 40 x = 12 \ The fourth number is 12.

2 (a) RM250

(b)

1270 + x

6 = 240

1270 + x = 1440 x = 170 \ Mukhriz’s weekly wage is

RM170.

3 (a)

Number of goals scored 0 1 2 3 4 5

Number of matches 2 1 3 4 7 3

0

0


10 x– = 3.5

6 + 3k + 8 + 20 + 6k + 10

= 3.5

3k + 40 = 3.5k + 35 0.5k = 5 k = 10

11 (a) x– = 357

= 5

σ2 =

2877

– (5)2

= 16 (b) 16m2

12 Q1 = 39.5 + (7 – 3

5 )5

= 43.5

Q3 = 49.5 +

(21 – 186 )5

= 52 \ Interquartile range = 52 – 43.5 = 8.5

13 (a) x– = 1086

= 18

(b) σ2 =

21846

– (18)2

= 40

14 New mean = 4(2) – 3 = 5 New variance = 22(1.5) = 6

15 (a) x– = 20m

5 = 4m σ2 = 8

90m2

5 – (4m)2 = 8

90m2 – 80m2

5 = 8

10m2 = 40 m2 = 4 m = 2 \ m = 2

16 σ2 = 4n

505

– ( m )2 = 4n

10 – m = 4n m = 10 – 4n

17 (a) Student A : x– = 205

= 4

σ =

905

– (4)2

= 2 = 1.414

Student B : x– = 205

= 4

σ =

985

– (4)2

= 3.6 = 1.897

22 (a) Number of goals scored 0 1 2 3 4 5 6

Number of matches 2 3 4 5 2 1 3

(b) (i) m = 3 (ii) Interquartile range = 3

(iii) x– =

3 + 8 + 15 + 8 + 5 + 18

20

=

5720

= 2.85

σ =

22920

– (2.85)2

= 3.3275 = 1.824

23 (a) (i) m = 8 (ii) Interquartile range = 13 – 3.5 = 9.5

(iii)

x– =

729

= 8

σ2 =

7929

– (8)2

= 24

(b)

72 + x10

= 8

72 + x = 80 x = 8

24 (a) Mode = 2

(b) m =

2 + 32

= 2.5 (c) Range = 6 – 0 = 6 (d) x– =

5220

= 2.6

σ =

19220

– (2.6)2

= 2.84 = 1.685

(b) Student A

18 (a) m = 4

(b) x– = 7820

= 3.9

σ =

35620

– (3.9)2

= 2.59 = 1.609

19 (a) k = 9, l = 5 (b) Estimated mode = 12.5

20 (a) x = 5 (b)

x + 112

x

x + 11 2x –x –11 x 11 \ x = 10 (c) x– = 2

x + 12 + 6 + 12x + 11

= 2

x + 30 = 2x + 22 x = 8

21 (a)Number of cars

6

5

4

3

2

1

060.5 70.5 80.5 90.5 100.5 110.5

Speed (km h–1)92.5

(b) Estimated mode = 92.5

25 (a) Frequency

10

8

6

4

2

050.5 60.5 70.5 80.5 90.5 100.5

76.5110.5

Mass (g)

Mode = 76.5

(b) m = 70.5 +

(18 – 1110 )10

= 77.5


(c) x– =

281836

= 78.278

26 Number of books Number of students

1 – 3 5

4 – 6 7

7 – 9 8

10 – 12 10

13 – 15 6

16 – 18 4

(a) Q1 = 3.5 +

(10 – 57 )3

= 5.643

Q3 = 9.5 +

(30 – 2010 )3

= 12.5 \ Interquartile range = 12.5 – 5.643 = 6.857

(b) x– = 37140

= 9.275

σ =

424940

– (9.275)2

= 20.1994 = 4.494

27 (a) Frequency

12

10

8

6

4

2

00.5 5.5 10.5 15.5 20.5 25.5

9.530.5

Time(minutes)

Mode = 9.5

(b) x– =

32530

= 10.833

σ2 =

445530

– (10.833)2

= 31.146

(c) m = 5.5 +

(15 – 412 )5

= 10.083

28 (a) Range = 52.5 – 36.5 = 16 Modal class = 47 – 50

(b)Number of workers

14

12

10

8

6

4

2

034.5 38.5 42.5 46.5 50.5 54.5

48.7

Number ofhours

Mode = 48.7

(c) x– =

186040

= 46.5

σ =

87 418

40 – (46.5)2

= 23.2 = 4.817

29 (a) (i) Median mark = 29.5 (ii) Interquartile range = 37.5 – 21.5 = 16 (b) 40 marks

30 (a)Cumulative frequency

Q3

Q1

m

40

3532.5

30

25

20

15

10

5

25.5 30.5 35.5 40.5 45.5 50.5 55.5 60.536 46 4841.75

Length (mm)

(i) m = 41.75 (ii) Interquartile range = 46 – 36 = 10 (iii) The number of leaves which are longer than 48 mm = 40 – 33 = 7 leaves (b) (i) p = 9, q = 3

0


(ii)

Frequency

12

10

8

6

4

2

025.5 30.5 35.5 40.5 45.5 50.5 55.5 60.5

42.5Length (mm)

Mode = 42.5

31 (a) x– =

168080

= 21

(b) Score 5 10 15 20 25 30 35

Number of students 0 4 14 34 64 76 80

(c)

Number of students

m

8076

70

60

50

40

30

20

10

5.5 10.5 15.5 20.5 25.5 30.5 35.521.5

Score

(i) Median score = 21.5 (ii) The percentage of students whose scores are 31 and above

=

80 – 7680

100

= 5% 32 (a) (i) Σx = 3m + 112

(ii) x– = 50

3m + 112

5 = 50

3m + 112 = 250 3m = 138 m = 46

(b) σ2 =

12 652

5 – (50)2

= 30.4

New variance =

30.422

= 7.6

33 (a) (i) x– = 6

Σx5

= 6

Σx = 30

(ii) σ = 2

12

Σx2

5 – (6)2

= 2

12

Σx2

5 – 36 = 6.25

Σx2 = 211.25 (b) (i) New mean = (6 – 1)(2) = 5(2) = 10

(ii) New standard deviation = 2

12

(2)

= 5

34 (a) (i) x– = 8

Σx50

= 8

Σx = 400 (ii) σ = 3

Σx2

50 – (8)2

= 3

Σx2

50 – 64 = 9

Σx2 = 3650

(b) σ =

3650 + 82

51 – (8)2

= 8.824 = 2.97

35 (a) x– = 12010

= 12

σ2 =

165010

– (12)2

= 21

(b) (i)

120 – k

9 = 11

120 – k = 99 k = 21

(ii) σ2 =

1650 – 212

9 – (11)2

= 13.33

0


1 (a) 18° =

18°180°

p

=

110

p radian

(b) 45° =

45°180°

p

=

14

p radian

(c) 80° =

80°180°

p

=

49

p radians

(d) 150° = 150°180°

p

=

56

p radians

(e) 225° = 225°180°

p

=

54

p radians

(f) 300° = 300°180°

p

=

53

p radians

2 (a) 20° =

20°180°

p

= 0.349 radian

(b) 42.5° = 42.5°180°

p

= 0.742 radian

(c) 75° =

75°180°

p

= 1.309 radians (d) 142° 25 =

142° 25

180° p

= 2.486 radians

(e) 250° = 250°180°

p

= 4.363 radians

(f) 320° 15 = 320° 15

180° p

= 5.589 radians

3 (a) 0.72 rad = 0.72

p 180°

= 41° 15 (b) 1.5 rad =

1.5p 180°

= 85° 57

(c) 2.425 rad = 2.425

p 180°

= 138° 57

(d) 56 p rad =

56

p

p 180°

= 150°

(e) 45 p rad =

45

p

p 180°

= 144°

(f) 74 p rad =

74

p

p 180°

= 315°

4 (a) s = 8(1.6) = 12.8 cm

(b) s = 20(290

180 p)

= 20(5.061) = 101.22 cm (c) s = 10(0.65) = 6.5 cm

(d) s = 50(3

4 p)

= 117.81 cm

5 (a) r =

120.8

= 15 cm

(b) r =

222.2

= 10 cm

(c) r =

91.003

= 8.97 cm

(d) r =

353.491

= 10.03 cm

6 (a) q = 154

= 3.75 radians

(b) q =

87

= 1.143 radians

(c) q = 1512

= 1.25 radians

(d) q =

918

= 0.5 radian

7 p = rq + 2r sin q2

(a) p = 9(1.047) + 2(9)(sin 30°) = 9.423 + 9 = 18.423 cm (b) p = 10(2.5) + 2(10)(sin 1.25 rad) = 25 + 18.98 = 43.98 cm

(c) p = 15(106° 16

180 p)

+ 2(15)

(sin 53° 8) = 27.821 + 24.001 = 51.822 cm

(d) p = 8(1.5) + 2(8)(sin 0.75 rad) = 12 + 10.906 = 22.906 cm

8 (a) ∠POQ =

710

= 0.7 rad

PR10

= sin 0.7 rad

PR = 10(sin 0.7 rad) = 6.442 cm

OR10

= cos 0.7 rad

OR = 10(cos 0.7 rad) = 7.648 cm QR = 10 – 7.648 = 2.352 cm \ Perimeter of the shaded region = 7 + 6.442 + 2.352 = 15.794 cm

(b) sPQ = 6(p

2) = 9.425 cm PQ = 62 + 62

= 72 = 8.485 cm \ Perimeter of the shaded region = 9.425 + 8.485 = 17.91 cm (c)

3 cm

P

SO q

1.25 cm

sin q = 1.25

3 q = 24° 37 \ Perimeter of the shaded region

= 3(49° 14

180° p)

+ 2.5

= 3(0.859) + 2.5 = 5.077 cm (d) sPQ = 8(0.6) = 4.8 cm

PR8

= tan 0.6 rad

PR = 8(tan 0.6 rad) = 5.473 cm OR = 5.4732 + 82

= 9.693 cm RQ = 9.693 – 8 = 1.693 cm \ Perimeter of the shaded region = 4.8 + 5.473 + 1.693 = 11.966 cm

Circular Measures



9

8 cm

R

C

O30°

8

OR = tan 30°

OR =

8

tan 30° = 13.856 cm BR = AP = 24 – 13.856 = 10.144 cm

sPQR = 8(240°

180° p)

= 8(4.189) = 33.512 cm

sAQB = 24(p

3) = 25.133 cm \ Perimeter of the shaded region = 2(10.144) + 33.512 + 25.133 = 78.933 cm

10

OBBD

=

21

2BD = 8 BD = 4 cm sAB = 8(0.5) = 4 cm sCD = 12(0.5) = 6 cm \ Perimeter of the shaded region = 2(4) + 4 + 6 = 18 cm

11

20AC

= tan 0.5 rad

AC =

20

tan 0.5 rad = 36.61 cm

20OC

= sin 0.5 rad

OC =

20

sin 0.5 rad = 41.717 cm BC = 41.717 – 20 = 21.717 cm

∠AOC = p –

p2

– 0.5

= 1.071 rad sAB = 20(1.071) = 21.42 cm \ Perimeter of the shaded region = 36.61 + 21.717 + 21.42 = 79.747 cm

12

4 cm

5 cm

9 cm

A

E

D C

B

13 cm

CD = 132 – 52

= 144 = 12 cm

cos ∠BAE =

513

∠BAE = 67° 23 = 1.176 rad ∠ABC = 2p – p – 1.176 = 1.966 rad \ Perimeter of the shaded region = 12 + 9(1.176) + 4(1.966) = 30.448 cm

13 (a) ∠AOB = 1020

=

12

rad

= 0.5 radian (b) sBC = 20(p – 0.5) = 52.832 cm

14 (a) A =

12

(5)2(2.4)

= 30 cm2

(b) A =

12

(8)2(p3)

= 33.51 cm2

(c) A =

12

(20)2(300180

p) = 1047.198 cm2

(d) A =

12

(10)2(109

p) = 174.533 cm2

15 (a)

12

r2(2) = 64

r2 = 64 r = 8 cm (b)

12

r2(4.4) = 55

r2 = 25 r = 5 cm

(c)

12

r2(2p3 )

= 12p

r2 = 36 r = 6 cm

(d)

12

r2(3.5) = 175

r2 = 100 r = 10 cm

16 (a)

12

(5)2(q) = 10

25q = 20

q =

2025

= 45

radian

(b)

12

(8)2(q) = 24p

32q = 24p

q =

34

p

= 2.356 radians

(c)

12

(10)2(q) = 30

50q = 30

q =

35

radian

(d)

12

(6)2(q) = 27

18q = 27 q = 1.5 radians

17 (a) q = 1210

= 1.2 rad

Area =

12

(102)(1.2) – 12

(102)

(sin 1.2 rad) = 60 – 46.602 = 13.398 cm2

(b) q =

512

p rad

Area =

12

(12)2( 512

p) – 12

(12)2

(sin

512

p rad)

= 94.248 – 69.547 = 24.701 cm2

(c)

Area =

12

(10)2( 25

p) – 12

(10)2

(sin 25

p rad) = 62.832 – 47.553 = 15.279 cm2

(d)

12 cm 13 cm

D

C

Oq

sin q =

1213

q = 67° 23 = 1.176 rad ∠AOC = 2 1.176 = 2.352 rad

Area =

12

(13)2(2.352) – 12

(13)2(sin 2.352 rad) = 198.744 – 60.001 = 138.743 cm2

18 Area of the shaded segment

=

12

r2(1.8) – 12

r2(sin 1.8 rad)

= 0.9 r2 – 0.4869 r2

= 0.4131 r2

Area of circle = pr2

The ratio of the shaded segment to

the circle

=

0.4131r2

pr2

=

0.41313.1416

=

4131

31 416 = 81 : 616

19 ∠AOB = 26

180°

= 60°

= 13

p rad


sAB = 4p

r(13

p) = 4p

r = 12 cm

∠BOC =

36

180°

= 90°

=

12

p rad

∠COD =

16

180°

= 30°

=

16

p rad

Area of shaded segment BC

=

12

(12)2(12

p) – 12

(12)2 sin 90°

= 36p – 72 Area of shaded segment CD

=

12

(12)2(16

p) – 12

(12)2 sin 30°

= 12p – 36 \ Area of the shaded region = 36p – 72 + (12p – 36) = (48p – 108) cm2

20 (a) Area of DOPR = 12

(5)(12)

= 30 cm2

Area of sector OPQ

=

12

(5)2(1.176)

= 14.7 cm2

Area of the shaded region = 30 – 14.7 = 15.3 cm2

(b) Area of sector OPQ

=

12

(14.142)2(14

p) = 78.538 cm2

Area of DORQ =

12

(10)(10)

= 50 cm2

Area of the shaded region = 78.538 – 50 = 28.538 cm2

21 AP10

= tan 0.6 rad

AP = 6.841 cm

Area of OAPB = 2[1

2 (10)(6.841)]

= 68.41 cm2

Area of sector OAB =

12

(10)2(1.2)

= 60 cm2

\ Area of the shaded region = 68.41 – 60 = 8.41 cm2

22

Area of semicircle = 12

(6.5)2(p)

= 66.366 cm2

Area of DPQR = 12

(5)(12)

= 30 cm2


23 (a) sPQ = 12 8q = 12

q = 32

radians

(b) Area of minor sector TOR

=

12

(4)2(32)

= 12 cm2

Area of circle = p(4)2

= 16p Area of the shaded region = (16p – 12) cm2

1 (a) tan ∠AOX =

86

=

43

∠AOX = 53° 8 = 0.927 rad

(b) Area of DAOX =

12

(6)(8)

= 24 cm2

Area of sector AOB

=

12

(6)2(0.927)

= 16.686 cm2

\ Area of the shaded region = 24 – 16.686 = 7.314 cm2

2

10 cm6 cm

qO

A

B

sin q =

610

=

35

q = 36° 52 ∠AOB = 73° 44 = 1.287 rad Area of the shaded segment

=

12

(10)2(1.287) – 12

(10)2 sin 73° 44

= 64.35 – 47.998 = 16.352 cm2

\ Area of the shaded region = p(10)2 – 16.352 = 314.159 – 16.352 = 297.81 cm2

3 (a) sACB = 15(53

p) = 78.54 cm

(b) Area of the shaded region

=

12

(15)2(p3 )

– 12

(15)2 sin p3

= 117.81 – 97.428 = 20.38 cm2

4 (a) sAB = 2p 12q = 2p

q =

p6

radian

\ ∠AOB =

p6

radian


=

12

(12)2(p6 )

– 12

(12)2 sin p6

= 37.699 – 36 = 1.699 cm2

5 Area of DOAT = 60

12

(10)(AT) = 60

AT = 12 cm

tan ∠AOT =

1210

= 1.2 ∠AOT = 50° 12 = 0.876 rad \ Area of the sector AOB

=

12

(10)2(0.876)

= 43.8 cm2

6 (a) sAB = 10(0.8) = 8 cm (b) Area of the shaded region

=

12

(10)2(0.8) – 12

(5)(10)

sin 0.8 rad = 40 – 17.934 = 22.07 cm 2

7 (a) sAB = 6(0.5) = 3 cm

(b)

BC6

= tan 0.5 rad

BC = 6 tan 0.5 rad = 3.278 cm \ Area of the shaded region

=

12

(6)(3.278) – 12

(6)2(0.5)

= 9.834 – 9 = 0.834 cm2

8 (a)

12 cm

9 cm

Q

O

T

tan ∠OTQ =

912

=

34

∠OTQ = 36° 52 ∠PTQ = 2(36° 52) = 73° 44 = 1.287 radians



= 2[1

2 (9)(12)]

– 12

(9)2(1.855)

= 108 – 75.128 = 32.872 cm2

9 2r + rq = 15 r(2 + q) = 15

r =

15

2 + q … 1

12

r2q = 9 … 2


12 ( 15

2 + q)2q = 9

225q = 18(4 + 4q + q 2) 225q = 72 + 72q + 18q 2

18q 2 – 153q + 72 = 0 2q 2 – 17q + 8 = 0 (2q – 1)(q – 8) = 0

q =

12

or q = 8 (rejected)

Substitute q =

12

into 1 :

r =

15

2 + 12

= 6

r = 6, q =

12

radian

10 sAB = 10(0.6) = 6 cm

BP10

= tan 0.3 rad

BP = 10 tan 0.3 rad = 3.093 cm \ Perimeter of the shaded region = 2(3.093) + 6 = 12.186 cm

11 (a) sACB = 24 6q = 24 q = 4 radians (b) Area of minor sector AOB

=

12

(6)2(2p – 4)

= 41.097 cm2

12 (a) ∠CAO = 45°

=

p4

rad

sCD = 8.485(p

4) = 6.664 cm (b) Area of the shaded region

=

12

(8.485)2(p4)

– 12

(6)(6)

= 28.27 – 18 = 10.27 cm2

13 Let r be the radians of sector COD

12

r2(0.5) – 12

(6)2(0.5) = 16

12

r2(0.5) = 25

r2 = 100 r = 10 0

sAB = 6(0.5) = 3 cm sCD = 10(0.5) = 5 cm \ Perimeter of the shaded region = 2(4) + 3 + 5 = 16 cm

14 (a)

5 cm4 cm

B

OT

sin ∠BOT =

45

∠BOT = 53° 8 \ q = 53° 8 = 0.927 radian (b) Area of the minor segment BCD

=

12

(5)2(1.854) – 12

(5)2

sin 1.854 rad = 23.175 – 12.002 = 11.173 cm2

15 (a) sAB = 6(0.9) = 5.4 cm (b) Area of the shaded region

=

12

(6)2(0.9) – 12

(3)2sin 0.9 rad

= 16.2 – 3.525 = 12.675 cm2

16 (a) 2q + 6q + 8 = 12 8q = 4

q = 12

radian


=

12

(6)2(12) –

12

(2)2(12)

= 9 – 1 = 8 cm2

17 (a)

13 cm 12 cm

q

sin q =

1213

q = 67° 23 \ ∠AOC = 2(67° 23) = 134° 46 = 2.352 radians (b) sABC = 13(2.352) = 30.576 cm

(c) Area of the shaded region

=

p(13)2 –

12

(13)2 sin 134° 46

= 530.93 – 59.99 = 470.94 cm2

18 (a) Let r be the length of OQ

13

r + 13

(r + 2) + 4 = 10

23

r + 23

= 6

2r + 2 = 18 2r = 16 r = 8 \ OQ = 8 cm (b) Area of the shaded region

=

12

(10)2(13) –

12

(8)2(13)

= 16.67 – 10.67 = 6 cm2

19 (a) q = 45°

=

p4

radian

(b) sCD = 14.142 (1

4 p)

= 11.107 cm sACB = 10 p = 31.416 cm

\ Perimeter of the shaded region = 31.416 + 11.107 + 14.142

+ 5.858 = 62.52 cm

(c) Area of semicircle =

12

p(10)2

= 157.08 cm2

Area of sector CBD

=

12

(14.142)2(14

p) = 78.54 cm2

\ Area of the shaded region = 157.08 – 78.54 = 78.54 cm2

20 (a)

12

(3k)2(34)

– 12

(2k)2(34)

= 30

278

k2 – 128

k2 = 30

158

k2 = 30

k2 = 16 k = 4 0

(b) sCD = 12 (3

4) = 9 cm

sAB = 8

(34)

= 6 cm \ The difference in length = 9 – 6 = 3 cm

21 (a)

12

(8)2q = p (2)2

32q = 4p

q =

18

p rad \ ∠AOB = p

8 radian


(b) sAB = 2 (p

8) = 0.785 cm

sCD =

8 (p

8) = 3.412 cm \ Perimeter of the shaded

region ABDC = 0.785 + 3.142 + 12 = 15.927 cm

22 (a)

sABsBC =

45

10q

10 (p2

– q)

= 45

5q = 4(p

2 – q)

9q = 2 p

q =

29

p rad

(b) sAB = 10(29

p) = 6.981 cm

10 cmB

C

O

T

CT10

= sin 0.436 rad

CT = 10 sin 0.436 rad = 4.223 cm BC = 8.446 \ Perimeter of the shaded region = 20 + 6.981 + 8.446 = 35.427 cm

23 (a) sBC = 10(1.2) = 12 cm (b) (i) Area of sector COB

=

12

(10)2(1.2)

= 60 cm2

(ii) ∠CAO =

1.22

= 0.6 radC

T

O

10 cm

0.9708 rad

CT10

= sin 0.9708 rad

CT = 10 sin 0.9708 = 8.2534 cm CA = 2(8.2534) = 16.507 cm

Area of sector CAD

=

12

(16.507)2(0.6)

= 81.744 cm2

(c) Area of DAOC

=

12

(10)2(sin 1.9416)

= 46.602 cm2

Area of COD = 81.744 – 46.602 = 35.142 \ Area of the shaded region = 60 – 35.142 = 24.86 cm2

24 (a) Area of sector LOM

=

12

(8)2(1.25)

= 40 cm2

(b)

LN8

= tan 1.25 rad

LN = 8 tan 1.25 = 24.077 cm

Area of DLON =

12

(8)(24.077)

= 96.308 cm2


25 (a) ∠PRQ = p – 2(0.9) = 1.3416 rad Area of the shaded region

=

12

(10)(10) sin 1.3416 rad – 12

(10)2(0.9) = 48.692 – 45 = 3.69 cm2

(b) sRS = 10(0.9) = 9 cm

P T

R

10 cm

0.9 rad

PT10

= cos 0.9 rad

PT = 10 cos 0.9 = 6.216 cm PQ = 2(6.216) = 12.432 cm Perimeter of the shaded region = 9 + 10 + (12.432 – 10) = 21.432 cm

26 (a) 12q = 18

q = 1812

=

32

radians

(b) S

U

6 cm32

rad

O

SU6

= sin 32

rad

SU = 6 sin

32

= 5.985 cm \ RT = 5.985 cm (c)

OT

R

5.98512

sin ∠ROT =

5.985

12 ∠ROT = 29° 55 = 0.522 radian (d) Area of sector OQR

=

12

(12)2(0.522)

= 37.584 cm2

Area of DORT

=

12

(10.401)(5.985)

= 31.125 cm2

\ Area of the shaded region = 37.584 – 31.125 = 6.459 cm2

27 (a)

cos ∠POQ =

610

=

35

∠POQ = 53° 8 = 0.927 radian (b) sAB = 3(0.927) = 2.781 cm PQ = 102 – 62

= 8 cm \ Perimeter of the shaded region = 7 + 8 + 3 + 2.781 = 20.781 cm (c) Area of the shaded region

=

12

(6)(8) – 12

(3)2(0.927)

= 24 – 4.172 = 19.828 cm2

28 (a)

15 cm

1.1 rad

A

F O

AF15

= sin 1.1 rad

AF = 15 sin 1.1 rad = 13.368 cm


AC = 2(13.368) = 26.736 cm (b)

F

A

D

20 cm13.368

sin ∠ADF =

13.368

20 ∠ADF = 41° 57 ∠ADC = 2(41° 57) = 83° 54 = 1.464 radians (c) sABC = 15(2.2) = 33 cm sAEC = 20(1.464) = 29.28 cm \ Perimeter of the shaded region = 33 + 29.28 = 62.28 cm Area of segment AEC

=

12

(20)2(1.464) – 12

(20)2

sin 1.464 rad = 292.80 – 198.86 = 93.94 cm2

Area of segment ABC

=

12

(15)2(2.2) – 12

(15)2 sin 2.2

= 247.50 – 90.956 = 156.544 cm2

Area of the shaded region = 156.544 – 93.94 = 62.604 cm2

29 (a)

12

(20)2(0.8)

12

(20 + r)2(0.8) – 12

(20)2(0.8)

= 45

5(160) = 4(0.4r2 + 16r) 1.6r2 + 64r – 800 = 0 r2 + 40r – 500 = 0 (r + 50)(r – 10) = 0 r = 10 0 (b) sPQ = 20(0.8) = 16 cm sRS = 30(0.8) = 24 \ Perimeter of region A = 16 + 24 + 20 = 60 cm

30 (a)

OC16

= cos 1.5 rad

OC = 16(cos 1.5 rad) = 1.132 cm \ BC = 16 – 1.132 = 14.87 cm (b) AC = 162 – 1.1322

= 15.96 cm sAB = 16(1.5) = 24 cm \ Perimeter of the shaded region = 24 + 15.96 + 14.87 = 54.83 cm Area of sector OAB

=

12

(16)2(1.5)

= 192 cm Area of DAOC

=

12

(1.132)(15.96)

= 9.033 cm2

\ Area of the shaded region = 192 – 9.033 = 182.97 cm2

Differentiation

1 (a) limx→–1(x2–1

x+1 )

= lim

x→–1

(x+1)(x–1)(x+1)

= –1–1 = –2

(b) lim

x→3(x2–9

x–3 )

= limx→3

(x+3)(x–3)(x–3)

= limx→3

(x+3)

= 3+3 = 6

(c) lim

x→4( x2–x–12x2–11x+28)

= lim

x→4(x+3)(x–4)(x–4)(x–7)

= lim

x→4(x+3)(x–7)

=

4+34–7

= – 7

3

(d) lim

x→2(x2–2xx2–4 )

= lim

x→2

x(x–2)(x+2)(x–2)

= lim

x→2( xx+2)

=

24

=

12

2 (a) limx→∞(x+1

x–1)

= limx→∞

( x

x +

1x

xx

–1x)

= limx→∞

( 1+

1x

1–1x

)

=

1+01–0

= 1

(b) lim

x→∞(2x+1x )

= limx→∞

( 2x

x +1x

xx

)

= limx→∞(2+

1x)

= 2

(c) lim

x→∞( xx+5 )

= limx→∞

( x

xxx

+5x)

= limx→∞

( 1

1+5x )

=

11+0

= 1

(d) lim

x→∞( 3–4x6–7x )

= limx→∞

( 3

x –4

6x

–7)

=

47

(e) lim

x→∞(x2+7xx2–4 )

= limx→∞

( x2

x2+7x

x2

x2

x2–4x2

)

= limx→∞

(1+7x

1– 4x2

) = 1

(f) lim

x → ∞(2x3–5x2+4x–63–8x3 )

=

( 2x3

x3–5x2

x3 +4xx3 – 6

x3

3x3

–8x3

x3)

= limx→∞

(2–5x

+ 4x2

– 6

x3

3x3 –8 )

=

– 2

8

= – 1

4

3 (a) limx→ 0( x

x–1)=

0

0–1 = 0

(b) lim

x→ 0(x+1x–1)

=

0+10–1

= –1

(c) lim

x→ 0(x+31–x)

=

0+31–0

= 3

(d) lim

x→ 0( 23+x)

=

23+0

=

23

4 (a) limx→2

(x2–3x+1) = 22–3(2)+1 = –1

(b) lim

x→–3( 3x2)

=

3(–3)2

=

39

=

13

(c) lim

x→∞(4+5x2)

= 4+0

=4

(d) limx→1

3x(x–2) = 3(–1) = –3

(e) lim

x→3( 4–xx+2 )

=

4–33+2

= 1

5 (f) lim

x→4(x–2)(x–3)

= (4–2)(4–3) = 2

5 (a) mAB =

9–43–2

= 5

(b)

dydx

= 4

6 (a) y = 3x–4 y+δy = 3(x+δx)–4 δy = 3x+3δx–4–(3x–4) δy = 3δx

δyδx

= 3

∴

dydx

=limδx→0

δyδx

= 3

(b) y = 2x2+3 y+δy = 2(x+δx)2+3 = 2[x2+2xδx+(δx)2]+3 δy = 2x2+4xδx+2(δx)2+ 3–(2x2+3) δy = 4xδx+2(δx)2

δyδx

= 4x+2δx

∴

dydx

=limδx→0

δyδx

= 4x

(c) y =

1x

y+δy =

1

x+δx

δy =

1

x+δx–1x

δy =

x–(x+δx)x(x+δx)

δy =

–

δxx(x+δx)

δyδx

=

–

1x(x+δx)

∴

δyδx

=

limδx→0

δyδx

= – 1x2

�©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3


(d) y =

12

x2

y+δy =

12

(x+δx)2

=

12

[x2+2xδx+(δx)2]

=

12

x2+xδx+12

(δx)2

–12

x2

δy = xδx+

12

(δx)2

δyδx

=x+δx2

∴

δyδx

= limδx→0

δyδx

= x (e) y = 2x3

y+δy = 2(x+δx)3

= 2(x+δx)(x+δx)2

= 2(x+δx)[x2+2xδx+ (δx)2]

= 2[x3+2x2δx+x(δx)2+ x2δx+2x(δx)2+(δx)3] δy = 2x3+4x2δx+2x(δx)2

+2x2δx+4x(δx)2+ 2(δx)3–2x3

δyδx

= 4x2+2xδx+2x2+

4xδx+2(δx)2

= 6x2+6xδx+2(δx)2

∴

δyδx

= limδx→0

δyδx

= 6x2

(f) y =

3x

–2

y+δy =

3

x+δx –2

δy =

3

x+δx –2–

(3x

–2)

=

3x+δx

–3x

=

3x–3(x+δx)

x(x+δx)

=

–

3δxx(x+δx)

δyδx

=

–

3x(x+δx)

δyδx

=

limδx→0

δyδx

= – 3x2

(g) y = 2x2–3x+2 y+δy = 2(x+δx)2–3(x+δx)

+2 = 2[x2+2xδx+(δx)2]

–3x–3δx+2 δy = 4xδx+2(δx)2–3δx

δyδx

= 4x+2δx–3

∴

δyδx

= limδx→0

δyδx

= 4x–3

(h) y =

x

x+2

y+δy =

x+δx

x+δx+2

δy =

x+δx

x+δx+2 –

xx+2

=

(x+2)(x+δx)–x(x+δx+2)

(x+2)(x+δx+2)

=

x2+xδx+2x+2δx–x2–xδx–2x

(x+2)(x+δx+2)

=

2δx

(x+2)(x+δx+2)

δyδx

=

2(x+2)(x+δx+2)

∴

δyδx

=

limδx→0

δyδx

= 2

(x+2)2

7 (a)

dydx

=6x+5

(b)

dydx

=5x4–21x2+6

(c) y = 9x2–4x–2

dydx

= 18x+8x–3

=

18x+

8x3

(d) y = 4x+3x–1

dydx

= 4–3x–2

= 4–

3x2

(e) y = 3x+4x12–3

dydx

= 3+2x– 1

2

= 3+

2

x

(f) y = 7x2+3x–x12

dydx

= 14x+3–12

x– 1

2

= 14x+3–

1

2 x

(g) y = 4x2+

23

x– 1

2

+1

dydx

= 8x–13

x– 3

2

= 8x–

1

3x32

(h) y =

x3

–3x–1–5x–2

dydx

=

13

+3x–2+10x–3

= 1

3 +

3x2

+10x3

(i) y =

32

x12

+6x

– 12

dydx

=

34

x– 1

2

–3x

– 32

=

3

4 x–

3

x32

(j) y = 2x4+

12

x2–1x

+9

= 2x4+

12

x2–x–1+9

dydx

= 8x3+x+x–2

= 8x3+x+1x2

8 (a) y =

2x2+6x

x = 2x+6

dydx

= 2

(b) y =

4x3–2x–3

8x

=

12

x2–14

–38

x–1

dydx

= x+38

x–2

= x+

38x2

(c) y =

3x2–4xx

= 3x32–4x

12

dydx

=

92

x12

–2x

– 12

=

92

x–

2

x

(d) y =

x3+x

x2

= x+x–1

dydx

= 1–x–2

= 1–1x2

(e) y = x +

1

x

= x12+x

– 12

=

12

x– 1

2

–12

x– 3

2

=

1

2 x–

1

2x32

(f) y =

6x2– x+3

2x

= 3x–

12

x– 1

2

+

32

x–1

dydx

=

3+14

x– 3

2

+

–32

x–2

= 3+

1

4x32

–

32x2

(g) y =

x+3 xx

= x12 +x

– 16

dydx

=

12

x– 1

2–16

x– 7

6

=

1

2 x–

1

6x76

(h) y =

2+ x

x2

= 2x–2+x– 3

2

dydx

= –4x–3–

32

x– 5

2

= –

4x3

–

3

2x52


(i) y =

x2+2x

= x32+2x

– 12

dydx

=

32

x12

–x

– 32

=

32

x–

1

x32

(j) y =

x2+3

x2

= 1+3x–2

dydx

= –6x–3

= – 6

x3

9 (a) y = (3x+1)2

= 9x2+6x+1

dydx

= 18x+6

(b) y =

(x2+2x)2

= x4+4x+4x–2

dydx

= 4x3+4–8x–3

= 4x3+4–8x3

(c) y = (x+1)(2x–3) = 2x2–x–3

dydx

= 4x–1

(d) y = x( x –4)

= x32–4x

dydx

=

32

x12

–4

= 3

2x

–4

(e) y = (1+ x )(1– x ) = 1–x

dydx

= –1

(f) y =

( x +

3x )2

= x+6+9x–1

dydx

=

1–9x–2

= 1–9x2

(g) y = (x2+3)2

= x4+6x2+9

dydx

= 4x3+12x

(h) y =

(2+1x2)2

= 4+4x–2+x–4

dydx

=

–8x–3–4x–5

=

– 8

x3 –4x5

(i) y = (5x+

13x)2

= 25x2+

103

+19

x–2

dydx

= 50x–29

x–3

= 50x–

29x3

(j) y = (2–5 x )2

= 4–20x12+25x

dydx

= –10x– 1

2+25

= 25–

10

x

10 (a)

dydx

=4x–5

Whenx=2,

dydx

= 4(2)–5

=3

(b) y = 2x12–x

32

dydx

= x– 1

2

–32

x12

=

1

x –32

x

Whenx=9,

dydx

=

13

–32

(3)

= – 256

(c) y = 3x2–2x–1+4

dydx

= 6x+2x–2

= 6x+

2x2

Whenx=1,

dydx

= 6(1)+

212

= 8 (d) y = 6x2–11x+4

dydx

= 12x–11

Whenx=3,

dydx

= 12(3)–11 =25

(e) y = 1–9x –1

dydx

= 9x–2

=

9x2

Whenx=–3,

dydx

=

9(–3)2

= 1

(f) y = 4x+

3x

+

1x2

dydx

=4–

3x2

–

2x3

Whenx=–1,

dydx

=

4–

3(–1)2

–

2(–1)3

= 4–3+2 = 3

(g) y =

2x2–x–3

x = 2x–1–3x–1

dydx

= 2+3x–2

= 2+

3x2

Whenx=

12

,dydx

=2+

3

(12 )2

=2+12 =14

(h) y = x32+x

12+4x

– 12

dydx

=

32

x12

+

12

x– 1

2 –2x

– 32

=

32

x +

1

2 x

–

2

x32

Whenx=4,

dydx

= 3+14

–28

=3

(i) y = 2x12+3

dydx

=

x– 1

2

=

1

x

Whenx=9,

dydx

=13

(j) y = 4x2+4x+1

dydx

=8x+4

Whenx=

14

,dydx

= 8(14)

+4

=6

11 (a) y = (x3+2)(x2–1)

dydx

= 3x2(x2–1)+2x(x3+2)

= 3x4–3x2+2x4+4x = 5x4–3x2+4x (b) y = (3+2 x)(3–2 x)

dydx

= x– 1

2(3–2 x)–x– 1

2(3+2 x)

= 3x– 1

2–2x– 1

2 + 12–3x

– 12

–2x– 1

2 + 12

= –4 (c) y = (x+2)(x2+3x)

dydx

= x2+3x+(2x+3)(x+2)

= x2+3x+2x2+7x+6 = 3x2+10x+6 (d) y = (x2+4)(2x2–x+6) = 2x(2x2–x+6)+(4x–1) (x2+4) = 4x3–2x2+12x+4x3+16x

–x2–4 = 8x3–3x2+28x–4

(e) y = (3x3+2x)(x2+

1x

+4)

dydx

= (9x2+2)(x2+1x

+4)+

(2x–

1x2 )(3x3+2x)

= 9x4+9x+36x2+2x2+

2x

+8+6x4+4x2–3x–

2x

= 15x4+42x2+6x+8 (f) y = 5x2(4– x)

dydx

=

10x(4– x)–

1

2 x (5x2)

= 40x–10x

32

–52

x32

= 40x–252

x32

(g) y = x( x –6)

dydx

= x –6+

1

2 x (x)


= x –6+

x2

=

12

x–6

(h) y =

(x+1x)(1–

1x)

dydx

= (1– 1

x2)(1–1x)

+1x2

(x+1x)

= 1–

1x

–1x2

+1x3

+1x

+1x3

= 1–1x2

+

2x3

(i) y = x (x+2)

dydx

=

12 x

(x+2)+

x

=

12

x+

1x

+ x

=

32

x+

1

x

(j) y =

( x+

1x )(x+

1x )

dydx

= ( 1

2 x –

1

2x32 )(x+

1x )

+

(1–1x2 )( x

+

1x )

=

12

x+

1

2x32

–

12 x

–

1

2x52

+

x +1x

–

1

x32

–

1

x52

=

32

x–

1

2x32

+

12 x

–

3

2x52

12 (a) y =

2x+53x–4

dydx

=

2(3x–4)–3(2x+5)(3x–4)2

=

6x–8–6x–15

(3x–4)2

=

– 23

(3x–4)2

(b) y =

3

5–x

dydx

=

0(5–x)–(–1)(3)(5–x)2

=

3(5–x)2

(c) y =

x2

2x–1

dydx

=

2x(2x–1)–2(x2)(2x–1)2

=

4x2–2x–2x2

(2x–1)2

=

2x2–2x(2x–1)2

=

2x(x–1)(2x–1)2

(d) y =

3x2+5xx

dydx

=

(6x+5) x –

12 x

(3x2+5x)x

=

6x32+5 x –3

2x

32–

52

x

x

=

92

x32

+

52

x

x

=

92

x +

5

2 x

(e) y =

2x–x4

x2

dydx

=

(2–4x3)(x2)–2x(2x–x4)x4

=

2x2–4x5–4x2+2x5

x4

=

–2x2–2x5

x4

=

– 2

x2 –2x

(f) y =

x2–x+1

2x+3

dydx

=

(2x–1)(2x+3)–2(x2–x+1)(2x+3)2

=

4x2+4x–3–2x2+2x–2

(2x+3)2

=

2x2+6x–5

(2x+3)2

(g) y =

xx+1

dydx

=

12 x

(x+1)– x

(x+1)2

=

12

x +1

2 x – x

(x+1)2

=

12 x

–12

x

(x+1)2

=

1–x2 x (x+1)2

(h) y =

x2–1x2+1

dydx

=

2x(x2+1)–2x(x2–1)(x2+1)2

=

2x3+2x–2x3+2x

(x2+1)2

=

4x

(x2+1)2

(i) y =

12

x3+2x

dydx

=

0–(3x2+2)(12)(x3+2x)2

=

– 12(3x2+2)

(x3+2x)2

(j) y =

4x+1x2+3

dydx

=

4(x2+3)–2x(4x+1)(x2+3)2

=

4x2+12–8x2–2x

(x2+3)2

=

12–2x–4x2

(x2+3)2

=

2(6–x–2x2)

(x2+3)2

13 (a) y = (1+3x)12

dydx

=

12

(1+3x)– 1

2(3)

=

32 1+3x

(b) y = (5x2+2)13

dydx

=

13

(5x2+2)– 2

3(10x)

=

10x

3(5x2+2)23

(c) y = (x+2)4

dydx

= 4(x+2)3

(d) y = (2–3x2)5

dydx

= 5(2–3x2)4(–6x)

= –30x(2–3x2)4

(e) y =

(14

x+3)8

dydx

= 8(14

x+3)7(14)

= 2(14

x+3)7

(f) y = (x2+1)–1

dydx

= –(x2+1)–2(2x)

=

–2x

(x2+1)2

(g) y = 3(3–4x)–3

dydx

= –9(3–4x)–4(–4)

=

36

(3–4x)4

(h) y = 4(2–x)–2

dydx

= –8(2–x)–3(–1)

=

8

(2–x)3

(i) y = (2– x )6

dydx

= 6(2– x )5(–1

2 x )

=

–

3(2– x )5

x

(j) y = 2(x2–3)12

dydx

=(x2–3)– 1

2(2x)

=

2xx2–3

14 (a) y = x 1+x2

dydx

=

1+x2+12

(1+x2)– 1

2(2x)(x)

= 1+x2 +

x2

1+x2

=

1+2x2

1+x2

(b) y = x(2x–1)3

dydx

= (2x–1)3+3(2x–1)2(2)(x)

= (2x–1)3+6x(2x–1)2


= (2x–1)2[(2x–1)+6x] = (2x–1)2(8x–1)

(c) y = x (1–x)2

dydx

=

12 x (1–x)2+2(1–x)

(–1)( x )

=

(1–x)2

2 x –2 x (1–x)

=

(1–x)2–4x(1–x)2 x

=

1–2x +x2–4x+4x2

2 x

=

5x2–6x+12 x

(d) y = x2 1–2x2

dydx

= 2x 1–2x2

+

12

(1–2x2)– 1

2

(–4x)(x2)

= 2x 1–2x2

–

2x3

1–2x2

=

2x(1–2x2)–2x3

1–2x2

=

2x–6x3

1–2x2

=

2x(1–3x2)1–2x2

(e) y =

x1–x

dydx

=

1–x –12

(1–x)– 1

2

(–1)(x)( 1–x)2

=

1–x +

x2 1–x

1–x =

2(1–x)+x2 1–x (1–x)

=2–x

2(1–x)32

(f) y =

2x2x+1

dydx

=

2 2x+1–12

(2x+1)– 1

2

(2)(2x)

2x+1

=

2 2x+1–

2x2x+1

(2x+1)

=

2(2x+1)–2x(2x+1)( 2x+1)

=

2x+2

(2x+1)32

=

2(x+1)

(2x+1)32

(g) y = (x–9) x–3

dydx

=

x–3+

12

(x–3)– 1

2(x–9)

= x–3 +

x–92 x–3

=

2(x–3)+x–92 x–3

=

3x–152 x–3

=

3(x–5)2 x–3

(h) y = x 9+3x

dydx

=

9+3x +12

(9+3x)– 1

2(3)(x)

= 9+3x +

3x2 9+3x

=

2(9+3x)+3x2 9+3x

=

18+9x2 9+3x

=

9(2+x)2 9+3x

15 (a) y = (2x–1)4

dydx

= 4(2x–1)3(2)

= 8(2x–1)3

Whenx=1,

dydx

=8

(b) y = 5–2x

dydx

=12

(5–2x)– 1

2(–2)

=

–

15–2x

Whenx=

12

,dydx

=–1

2

(c) y =

1

2x–3

dydx

=

–2(2x–3)2

Whenx=12

, dydx

=

–2

[2(12 )–3]2

=

–2

4=

–1

2

(d) y =

3x2–85–2x

dydx

=

6x(5–2x)–(–2)(3x2–8)(5–2x)2

=

30x–12x2+6x2–16

(5–2x)2

=

30x–6x2–16

(5–2x)2

Whenx=2,

dydx

=

30(2)–6(2)2–16[5–2(2)]2

=20

(e) y =

3x2

1–4x2

dydx

=

6x(1–4x2)–(–8x)(3x2)(1–4x2)2

=

6x–24x3+24x3

(1–4x2)2

=

6x

(1–4x2)2

Whenx=1,

dydx

=

69

=

23

(f) y =

x2+4

x2

dydx

=

2x(x2)–2x(x2+4)x4

=

2x3–2x3–8x

x4

= – 8

x3

Wheny=5, 5=x2+4

x2

4x2= 4 x2= 1 x = 1

Whenx=1,

dydx

=–8

Whenx=–1,

dydx

=8

(g) y = x (2–x)

dydx

=

12

x– 1

2(2–x)+(–1)( x )

=

2–x2 x

–

x

=

2–x–2x

2 x

=

2–3x2 x

Whenx=9,

dydx

=

2–3(9)2 9

=

–25

6

(h) y =

5–x2x

dydx

=

–2x–2(5–x)4x2

=

–

104x2

=

– 5

2x2

Wheny=2,2= 5–x

2x 4x = 5–x 5x = 5 x = 1

Whenx=1,

dydx

=–5

2

16 (a) y=3x2–4x+5 Whenx = –1, y =3(–1)2–4(–1)+5 = 12

dydx

=6x–4

Atx=–1,

dydx

=–10

Equationofthetangent: y–12 = –10(x+1) y–12 = –10x–10 y+10x = 2 (b) y = x2–3x+4

dydx

=2x–3

Atx=3,

dydx

=3

Equationofthetangent: y–4 = 3(x–3) y–4 = 3x–9 y = 3x–5

(c) y=x+

2x

Whenx=–2,y = –2+

2

–2 =–3


dydx

=1–

2x2

Atx=–2,

dydx

= 1–

2(–2)2

=

12

Equationofthetangent:

y+3 =

12

(x+2)

y+3 =

12

x+1

y =

12

x–2

(d) y=

x–1x+1

Whenx=1,y =

1–11+1

= 0

dydx

=

x+1–(x–1)(x+1)2

=

2

(x+1)2

Atx=1,

dydx

=12


y =

12

(x–1)

2y = x–1 2y–x = –1 (e) y = 1+4x–x2

dydx

=

12

(1+4x–x2)– 1

2(4–2x)

=

2–x1+4x–x2

Atx=3,

dydx

=

2–31+4(3)–32

=

–12


y–2=

–12

(x–3)

2y–4= –x+3 2y+x =7

(f) y=

x2+5x+1

Whenx=1,y =

62

= 3

dydx

=

2x(x+1)–(x2+5)(x+1)2

=

2x2+2x–x2–5

(x+1)2

=

x2+2x–5

(x+1)2

Atx=1,

dydx

=

–24

=

–12


y–3=

–12

(x–1)

2y–6= –x+1 2y+x = 7

17 (a) y=2(x–1)3

Whenx=

12

,y = 2(12

–1)3

=2(–

18 )

=

–14

dydx

=6(x–1)2

Atx=

12

,dydx

=64

=32

Equationofthenormal:

y+

14

=

–23 (x–

12 )

y+14

= –23

x+13

y =

–23

x+

112

(b) y=x 1–2x Whenx=–4,y = –4 1–2(–4) = –4(3) = –12

dydx

= 1–2x +12

(1–2x)–1

2

(–2)(x)

= 1–2x –

x1–2x

=

(1–2x)–x1–2x

=

1–3x1–2x

Atx=–4,

dydx

=

1–3(–4)1–2(–4)

=

133


y+12=

–

313

(x+4)

13y+156= –3x–12 13y+3x = –168

(c) y=

x

x+1

Whenx=2,y=

23

dydx

=

x+1–x(x+1)2

=

1

(x+1)2

Atx=2,

dydx

=19


y–

23

= –9(x–2)

y–

23

= –9x+18

3y–2= –27x+54 3y+27x = 56

(d) y=2+

1x

Whenx=1,y=3

dydx

=–1x2

Atx=1,

dydx

=–1

Equationofthenormal: y–3= 1(x–1) y–3= x–1

y–x=2

(e) y=7x–

6x

Whenx=3,y = 7(3)–

63

=21–2 =19

dydx

=7+

6x2

Atx=3,

dydx

= 7+23

=

233


y–19=

–

323

(x–3)

23y–437= –3x+9 23y+3x = 446

(f) y=

x–2

2x+1 Whenx=2,y=0

dydx

=

2x+1–2(x–2)(2x+1)2

=

5

(2x+1)2

Atx=2,

dydx

=

525

=

15

Equationofthenormal: y = –5(x–2) y = –5x+10 y+5x = 10

18 (a) y = 8x+

12x2

= 8x+

12

x–2

dydx

= 8–x–3

= 8–

1x3

Forturningpoint,

dydx

= 0

8–

1x3

=0

8=

1x3

x3=

18

x=

12

Whenx=

12

,y = 8(12)

+

1

2(12)2

=4+2 =6

Theturningpointis

(12

,6).

(b) y = 4x+x–1

dydx

=4–x–2

=4–

1x2

4–

1x2

= 0

4=

1x2


x2 =

14

x = ± 1

2

Whenx=

12

,y = 4(12 )

+

1

(12)

= 2+2 = 4

Whenx=

–

12

,

y=4(– 12)

+

1

(–12)

= –2–2 = –4

∴

(12

,4)and

(–12

,–4) (c) y=x3–6x2+9x

dydx

= 0

3x2–12x+9= 0 x2–4x+3= 0 (x–1)(x–3)= 0 x=1orx=3 Whenx=1,y=13–6(1)2+9(1) =4 Whenx=3,y=33–6(3)2+9(3) =0 ∴(1,4)and(3,0) (d) y = x(x–2)2

= x(x2–4x+4) =x3–4x2+4x

dydx

=0

3x2–8x+4=0 (3x–2)(x–2)=0

x=

23

orx=2

Whenx=

23

,y = (2

3 )3

–4(2

3 )2

+4(2

3 )

=

3227

Whenx=2,y=23–4(2)2+4(2) =0

∴

(23 ,

3227 )

and(2,0)

(e) y = 4–x2–16x–2

dydx

=0

–2x+

32x3

= 0

32x3

= 2x

16=x4

x=2

Whenx=2,y = 4–(2)2–

16(2)2

=–4 ∴(2,–4) (f) y = x2–2x4

dydx

= 0

2x–8x3 = 0

2x = 8x3

1x2

= 4

x2 =

14

x = ± 1

2

Whenx=

12

,y

=(1

2)2

–2(1

2)4

=

14

–2( 116)

=

18

Whenx=–

12

,y = (–

12)2

–2(– 1

2)4

=

18

–2( 116)

=

18

∴(1

2 ,18)

and(–

12

,18)

19 (a) y = x(x–6)2

=x(x2–12x+36) =x3–12x2+36x

dydx

= 0

3x2–24x+36= 0 x2–8x+12= 0 (x–2)(x–6)= 0 x=2orx=6 Whenx= 2, y= 23–12(2)2+36(2) = 8–48+72 =32 Whenx= 6, y=63–12(6)2+36(6) =216–432+216 =0

d2ydx2

= 6x–24

Whenx=2,

d2ydx2

= 6(2)–24

=–120 ∴(2,32)isamaximumpoint.

Whenx=6,

d2ydx2

=6(6)–24

=36–24 =120 ∴(6,0)isaminimumpoint. (b) y = x+16x–1

dydx

= 1–16x–2

= 1–

16x2

dydx

=0

1–

16x2

=0

1=

16x2

x2 = 16 x =±4

Whenx=4,y =4+

164

=4+4 =8

Whenx=–4,y= –4+

16

(–4) = –8

d2ydx2

=32x –3

=

32x3

Whenx=4,

d2ydx2

=

3264

=

12

0

∴(4,8)isaminimumpoint.

Whenx=–4,

d2ydx2

=

–3264

=

–12

0

(–4,–8)isamaximumpoint. (c) y = x3–2x2

dydx

= 3x2–4x

dydx

= 0

x(3x–4)= 0

x=0orx=

43

Whenx=0,y=0

Whenx=4

3,y=

(43)3

–2(4

3)2

=

6427

–329

=

–3227

d2ydx2

=6x–4

Whenx=0,

d2ydx2

=–40

∴(0,0)isamaximumpoint.

Whenx=

43

,d2ydx2

= 6(43)

–4

=8–4 =40

(43

,–3227)isaminimumpoint.

(d) y = 4x+9x–1

dydx

=4–

9x2

dydx

= 0

4–

9x2

=0

4 =

9x2

x2 =

94

x =± 3

2

Whenx=

32

,y=4(32)

+

9

(32)

= 6+6 =12

Whenx =

–32

,

y = 4(–

32)

+

9

(– 32)

= –6–6 = –12


d2ydx2

=

18x3

Whenx=

32

,d2ydx2

=

18

(32)3

=

163

0

∴(3

2 ,12)

isaminimumpoint.

Whenx =

–32

,d2ydx2

=

18

(– 32)3

=

–

163

0

∴(– 3

2,–12)

isamaximumpoint.

(e) y =

x2–6x+9

x = x–6+9x–1

dydx

= 1–9x–2

dydx

= 0

1–

9x2

= 0

1=

9x2

x2=9 x=±3 Whenx=3,y=0 Whenx=–3,y=–12

d2ydx2

=18x3

Whenx=3,

d2ydx2

=1827

=

23

0


Whenx=–3,

d2ydx2

=

–1827

=

– 2

30

∴(–3,–12)isamaximumpoint.

(f) y =

x2

x+1

dydx

=

2x(x+1)–x2

(x+1)2

=

x2+2x(x+1)2

dydx

= 0

x2+2x(x+1)2

=0

x(x+2)= 0 x=0orx=–2 Whenx=0,y=0 Whenx=–2,y=–4

d2ydx2

=

(2x+2)(x+1)2–2(x+1)(x2+2x)

(x+1)4

=

(2x+2)(x2+2x+1)–2(x3+2x2+x2+2x)

(x+1)4

=

2x3+4x2+2x+2x2+4x+2–2x3–4x2–2x2–4x

(x+1)4

=

2x+2(x+1)4

Whenx=0,

d2ydx2

=20


Whenx=–2,

d2ydx2

=

2(–2)+2(–2+1)4

=

–2(–1)4

=–20 ∴(–2,–4)isamaximumpoint.

20 y = ax2+bx+c

dydx

= 0

2ax+b= 0 Atx=2, 4a+b=0… 1 Atthepoint(0,10), 10 = a(0)2+b(0)+c c = 10 Atthepoint(2,18), 18=a(2)2+b(2)+10 4a+2b= 8 2a+b = 4… 2

1 – 2 : 2a = –4 a = –2 Substitutea=–2into 1 : 4(–2)+b = 0 b=8 ∴a=–2,b=8,c=10

21 (a)

dydx

= 0

3x2–12= 0 3x2= 12 x2=4 x = ±2

(b)

dydx

=3x2–12

Atx=3,

dydx

=3(3)2–12 =27–12 =15 Equationofthetangent: y+9= 15(x–3) y+9= 15x–45 y=15x–54

22 y=mx+nx2

Atthepoint(3,5),

5= 3m+

n9

27m+n = 45… 1

dydx

=m–2nx3

Atx=3,

dydx

= 0

m–

2n27

=0

27m–2n = 0… 2

1 – 2 : 3n =45 n=15 Substituten=15into 2 : 27m–2(15)= 0

27m = 30

m=

109

∴m=109

,n=15

23 (a) 4(3x)+4(x)+4(h)= 1200 16x+4h =1200 4x+h =300 h= 300–4x Volumeofthebox, V =3x2h =3x2(300–4x) =900x2–12x3

=12x2(75–x)(shown) (b) ForthemaximumvalueofV,

dVdx

= 0

1800x–36x2= 0 36x(50–x)=0 Sincex0,x= 50

24 (a)

(24 – h) cm

R

A

B

24 cm

7 cm

cmx2

Fromthediagram,

724

=

( x2 )

24–h 7(24–h)=12x

24–h =

127

x

h =24–

127

x

TheareaofΔPQR,

A =

12

xh

=

12

x (24–127

x)

= 12x–

67

x2

=

67

x (14–x)(shown)

(b) ForamaximumvalueofA,

dAdx

=0

12–

127

x=0

127

x=12

x=7

25 (a) sAB = 40–2r rθ=40–2r θ=

40–2r

r (b) Areaofsector,

A =

12

r2θ


=

12

r2(40–2rr )

=20r–r2

=r(20–r)(shown) (c) ForamaximumvalueofA,

dAdr

= 0

20–2r=0 2r=20 r=10

andso,

d2Adr2

=–20

Hence,forthearea,Atobemaximum,r=10andthemaximumareais20(10)–(10)2=100cm2.

26 (a) y = 2x2+4x

dydx

=4x–4x2

Whenx=–2,

dydx

=

4(–2)–

4(–2)2

=–8–1 x =–9 Bythechainrule,

dydt

=

dydx

dxdt

=–9(3) =–27

(b) y =

2

(2x–3)3

dydx

=

–

12(2x–3)4

Whenx=2,

dydx

=–12

Bythechainrule,

dydt

=

dydx

dxdt

=–12(3) =–36 (c) y =(3x–5)4

dydt

=12(3x–5)3

Whenx=

43

,

dydx

= 12[3(43 )

–5]3

= –12 Bythechainrule,

dydt

=

dydx

dxdt

=–12(3) =–36 (d) y = 3x2–5

dydx

=6x

Whenx=

– 1

3,dydx

=6(– 13 )

= –2 Bythechainrule,

dydt

=

dydx

dxdt

=–2(3) =–6

27 (a) y = 2x2+5x+2

dydx

= 4x+5

Whenx=–1,

dydx

=1

Bythechainrule,

dydt

=

dydx

dxdt

2=1

dxdt

dxdt

=2

(b) y =x(x–4)

dydx

=2x–4

Whenx=3,

dydx

=2

Bythechainrule,

dydt

=

dydx

dxdt

2 = 2

dxdt

dxdt

= 1

(c) y = 2x +3

dydx

=

12x +3

Whenx=3,

dydx

=13

Bythechainrule,

dydt

=

dydx

dxdt

2 =

13

dxdt

dxdt

=6

(d) y=

x

x+2

Aty=

13

,13

=

xx+2

x+2= 3x 2x = 2 x = 1

dydx

=

2(x+2)2

Whenx=1,

dydx

=29

Bythechainrule,

dydt

=

dydx

dxdt

2 =

29

dxdt

dxdt

= 9

28 LetAbetheareaandrtheradius. ThenA=πr2.

Bythechainrule,dA

dt =

dAdr

drdt

= 2πr(4) =8πr

(a) Whenr=2,

dAdt

= 8π(2)

= 16πcm2s–1

(b) WhenA=9π,πr2= 9π r2= 9

r = 3(r0)

∴ dA

dt = 8π(3)

= 24πcm2s–1

29 (a) y = x2+8x

dydx

= 2x–8x2

(b) Bythechainrule,

dydt

=

dydx

dxdt

dydt

=(2x–

8x2) dx

dt

If

dydt

=6whenx=2,

6= 2

dxdt

dxdt

=3unitss–1

30 LetAbethetotalsurfaceareaandxthelengthofitsside.

ThenA=6x2

Bythechainrule,

dAdt

=

dAdx

dxdt

=12x(3) =36x

(asdxdt

=3cms–1) Whenvolume=64,x3=64 x =4

∴dA

dt =36(4)

= 144cm2s–1

31 LetVbethevolumeandrtheradius,

thenV=

43

πr3.

Bythechainrule,

dVdt

=

dVdr

drdt

8= 4πr2

drdt

drdt

=

84πr2

=

2

πr2(asdVdt

=8cm3s–1)

WhenA=16π, 4πr2= 16π r2= 4 r = 2(r0)

∴

drdt

=

2π(2)2

=

1

2π cms–1

32 LetAbetheareaandxthelengthofitsside.ThenA=x2.

Bythechainrule,

dAdt

=

dAdx

dxdt

12=2x

dxdt

dxdt

=

6x (as

dAdt

=12cm2s–1) WhenA=9,x2= 9 x = 3(x0)

�0©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

∴

dxdt

=

63

= 2cms–1

33 (a) A =

12

xy

=

12

x(4x–x2)

=

12

(4x2–x3)

(b) Ifxincreasesatarateof2units

persecond,thendxdt

=2.

dAdt

=

dAdx

dxdt

=

12

(8x–3x2)(2)

=8x–3x2

=x(8–3x) (i) Whenx = 2,

dAdt

= 2[8–3(2)]

=4unit2s–1

(ii) Whenx=3

dAdt

= 2[8–3(3)]

= 2(–1) =–2unit2s–1

34 (a)

r

x

10

50

Fromthediagram,

r

10=

x

50

r =

x5

Thevolumeof thewater in thecontaineris:

V =

13

πr2x

=

13

π(x5)2

x

=

13

π( x2

25)x

=

175

πx3(shown)

(b)

dVdt

=

dVdx

dxdt

18 =

( 125

πx2)dxdt

dxdt

=

450πx2

Whenx=3,dx

dt =

450

π(3)2

=

50π

cms–1

35 (a) y = 4x3–7x2

dydx

=12x2–14x

Atthepoint(2,4),

dydx

=12(2)2–14(2) =20 (b) Wheny=4,x =2 δy =4.05–4 =0.05

dydx

=20

Thenδy

dydx

δx

0.0520δx δx0.0025

36 (a) y = x3+1

dydx

=

3x2

2 x3+1

(b) Whenx=2,δx = 2.02–2 =0.02

and

dydx

= 2

Thenδy

dydx

δx

2(0.02) 0.04

37 (a) y = 2x3–7x2+15

dydx

= 6x2–14x

(b) Whenx=2,δx = 2.01–2 =0.01

and

dydx

= 6(2)2–14(2)

=24–28 =–4

Thenδy

dydx

δx

–4(0.01) –0.04

38 (a) y =

13x+1

dydx

=

–3(3x+1)2

Whenx=3,

dydx

=– 3

100 (b) Theapproximatechangeiny,

δy

dydx

δx

–

3100

(p)

– 3p

100

39 y =

3x+2x+2

dydx

=

3(x+2)–(3x+2)(x+2)2

=

4

(x+2)2

Whenx=2,δx = 2+p–2 =p

anddy

dx =

14

Thenδy

dydx

δx

14

(p)

p4

40 (a) y = 3x–9x

dydx

=3+9x2

(b) Whenx=3,δx = 3+p–3 =p

and

dydx

= 3+

9(3)2

=4

Thenδy

dydx

δx

4p

41 V =43

πr3

dVdr

= 4πr2

Whenr=4,δr = 3.8–4 =–0.2

and

dVdr

=4π(4)2

=64π

ThenδV

dVdr

δr

64π(–0.2) –12.8π

42 T = 2πl

10

=

2π10

(l)12

dTdl

=

12(

2π10 )l–1

2

=

π10l

Whenl=2.5,δl = 2.6–2.5 = 0.1

and

dTdl

=

π10(2.5)

=

π5

ThenδT

dTdl

δl

π5

(0.1)

0.02π

43 y =

1x

= x– 1

2

dydx

=

–12

x– 3

2

=

–1

2x32

(a) Whenx=100, y =

1

10 δx = 100.5–100 =0.5

��©CerdikPublications Sdn.Bhd. (203370-D) 2010 ISBN:978-983-70-3258-3

dydx

=

– 12000

Then

1

100.5

1

10+

(– 12000)

(0.5) 0.09975

(b) Whenx=49,y =

17

δx = 49.2–49 =0.2

dydx

= –

1686

Then

1

49.2

17

+(– 1

686)(0.2)

0.14257

(c) Whenx=25,y =

15

δx = 24.05–25 = –0.95

dydx

= –

1250

Then

124.05

15

+(– 1

250) (–0.95) 0.2038

(d) Whenx=9,y =

13

δx = 8.96–9 =–0.04

dydx

= –

154

Then

18.96

13

+(– 1

54) (–0.04) 0.33407

44 y = x– 1

3

dydx

= –13

x– 4

3

= –

1

3x43

Whenx=8,y =

12

δx =8.95–8 =0.95

dydx

= –

148

Then

13

8.95

12

+(– 1

48)(0.95)

0.4802

45 (a) y = (x+1)2

= x2+2x+1

dydx

= 2x+2

d2ydx2

= 2

(b) y = x+

1x

= x+x–1

dydx

= 1–x–2

= 1–

1x2

d2ydx2

=

2x3

(c) y = x–5

dydx

=

12

(x–5)– 1

2

=

12 x–5

d2ydx2

= –

1

4(x–5)32

(d) y =

1

x+1

dydx

= –

1(x+1)2

d2ydx2

=

2(x+1)3

(e) y =

1x

dydx

= –

1

2x32

d2ydx2

=

3

4x52

(f) y =

x

x–1

dydx

=

(x–1)–x(x–1)2

= –

1

(x–1)2

d2ydx2

=

2(x–1)3

46 (a) y = x3–6x2+5

dydx

= 3x2–12x

d2ydx2

= 0

6x–12= 0 6x= 12 x = 2 (b) y = x2–27x–2

dydx

= 2x+54x–3

d2ydx2

= 0

2–

162x4

= 0

2=

162x4

x4= 81 x = 3 (c) y = (x–2)3

dydx

= 3(x–2)2

d2ydx2

= 0

6(x–2)= 0 x = 2 (d) y = 3x2–x3

dydx

=

6x–3x2

d2ydx2

= 0

6–6x = 0 x = 1

47 y = x3–6x2+4

dydx

= 3x2–12x

d2ydx2

= 0

6x–12= 0 x = 2

∴

dydx

= 3(2)2–12(2)

= 12–24 = –12

48 y = x4–108x

dydx

=0

4x3–108= 0 4x3= 108 x3=27 x= 3

d2ydx2

=12x2

Whenx=3,

d2ydx2 = 12(3)2

= 108

49 y = x3–2x2+3x+4

dydx

= 3x2–4x+3

d2ydx2

=6x–4

Whenx=1,

dydx

= 3(1)2–4(1)+3

=2

Whenx=1,

d2ydx2

= 6(1)–4

=2

50 y =

1–x2

x

dydx

=

–2x(x)–(1–x2)x2

=

–x2–1

x2

d 2ydx2

=

–2x(x2)–2x(–x2–1)x4

=

2xx4

=

2x3

xd2y

dx2 +2dy

dx+2= x(2

x3)+2(–x2–1x2 )+2

=

2x2

+2(–1–1x2)

+2

=

2x2

–2–2x2

+2

= 0(shown)

51 (a) y =

1x+1

dydx

= –1

–

1(x+1)2

= –1

1= (x+1)2

x2+2x =0 x(x+2)=0 x=0orx=–2

Whenx=0,y =

1

0+1 =1

Whenx=–2,y =

1

–2+1 = –1


∴Thecoordinatesare(0,1)and(–2,–1).

(b)

d2ydx2

=

2(x+1)3

Atthepoint(1,1),

d2ydx2

=

2(1+1)3

=

28

=

14

1 (a) y = x12+x–

12

dydx

=

12

x– 12

–12

x– 32

=

12 x

–

1

2x32

(b) y =

(x+1)2

x

=

x2+2x+1

x = x+2+x–1

dydx

= 1–x–2

= 1–1x2

2 y = 8x–1–6x12

dydx

= –8x–2–3x– 12

= –

8x2

–

3

x

Atx=4,

dydx

= –

8(4)2

–

3

4

= –

12

–32

= –2

3 y=ax

+bx2

Atthepoint(3,6),6=a

3+b(3)2

a+27b = 18… 1

and

dydx

= 7

–ax2

+2bx = 7

Atthepoint(3,6),–

a9

+6b=7

–a+54b=63… 2

1 + 2 : 81b = 81 b = 1 Substituteb=1into 1 : a+27(1)= 18 a = –9 ∴a=–9,b=1

4 y =

x2

x–2

dydx

=

2x(x–2)–x2

(x–2)2

=

2x2–4x–x2

(x–2)2

=

x2–4x(x–2)2

Forturningpoints,

dydx

= 0

x2–4x(x–2)2

=0

x(x–4)= 0 x=0orx = 4 Whenx=0,y=0

Whenx=4,y =

42

4–2 =8 ∴Theturningpointsare(0,0)and(4,8)

5 y =

3x2x–3

dydx

= –94

3(2x–3)–2(3x)

(2x–3)2 =

–94

–9

(2x–3)2 =

–94

(2x–3)2= 4 4x2–12x+9=4 4x2–12x+5=0 (2x–1)(2x–5)=0

x=12

orx=52

6 y=2x+1x2+2

Aty-axis,x =0

y=

2(0)+102+2

=

12

Wheny=

12

,12

=2x+1x2+2

x2+2= 4x+2 x2–4x=0 x(x–4)=0 x=0orx=4

dydx

=

2(x2+2)–2x(2x+1)(x2+2)2

=

2x2+4–4x2–2x

(x2+2)2

=

4–2x –2x2

(x2+2)2

Whenx=0,

dydx

=

4–2(0)–2(0)2

(02+2)2

=1

Whenx=4,

dydx

=

4–2(4)–2(4)2

(42+2)2

= –

36

324

= – 19

7 (a) y = (x+6)7(x–9)8

dydx

= (x+6)6(x–9)7[7(x–9) +8(x+6)] = (x+6)6(x–9)7(7x–63 +8x+48) = (x+6)6(x–9)7(15x–15) = 15(x–1)(x+6)6(x–9)7

(b)

dydx

= 0

15(x–1)(x+6)6(x–9)7= 0 ∴x=1,x=–6,x=9

8 y = (x2+

2x )8

dydx

=

8(x2+2x )7(2x– 2

x2)

Whenx=1,dydx

= 8(3)7(0)

=0

9 y = (2x–3)3

dydx

= 6

3(2x–3)2(2)= 6 6(2x–3)2=6 4x2–12x+9=1 4x2–12x+8=0 x2–3x+2=0 (x–1)(x–2)=0 x=1orx=2 Whenx=1,y=[2(1)–3]3

=–1 Whenx=2,y=[2(2)–3]2

=1 ∴(1,–1)and(2,1)

10 y = (2x+3)3(x–4)

dydx

= 3(2x+3)2(2)(x–4)+(2x+3)3

= 6(x–4)(2x+3)2+(2x+3)3

= (2x+3)2[6(x–4)+(2x+3)] = (2x+3)2(8x–21)

Comparewith

dydx

=(2x+3)2(mx+n)

∴m=8,n=–21

11 (a) y = (1+x)(1+ x )

dydx

= 1+ x + 12 x

(1+x)

=

2 x +2x+1+x

2 x

=1+

3x

2 x +

1

2 x

=1+

32

x+

1

2 x (b) y = x(1+3x)5

dydx

= (1+3x)5+5(1+3x)4(3)(x)

= (1+3x)5+15x(1+3x)4

= (1+3x)4[(1+3x)+15x] = (1+3x)4(18x+1)

12 (a) y = 4x+x–1

dydx

= 4–x–2

= 4–1x2

(b) y=4x+

1x

Atthepointwherex=2,

y=4(2)+

12

=

172


Atx=2,

dydx

= 4–122

=

154


y–

172

=

154

(x–2)

4y–34= 15x–30 4y–15x=4

13 (a) y=3x2–8x+7

Foraminimumpoint,

dydx

= 0

6x–8=0 6x=8

x =

43

(b) Atx=2,y = 3(2)2–8(2)+7 = 3 and dy

dx = 6(2)–8

= 4 Equationofthenormal:

y–3= –

14

(x–2)

4y–12= –x+2 4y+x = 14

14 y = x2–x+3

dydx

= –3

2x–1 = –3 2x = –2 x = –1 Whenx=–1,y =(–1)2–(–1)+3 =5 Atthepoint(–1,5),3(5)–(–1) = c c = 16

15 (a) dydx

= 2

2x–2 = 2 2x =4 x =2 Whenx=2,y= 22–2(2)+2 = 4–4+2 = 2 ∴A(2,2) (b) Atthepoint(2,2), 2= 2(2)+c c = 2–4 c =–2

16 (a) y =

2x–4x+2

dydx

=

2(x+2)–(2x–4)(x+2)2

=

8

(x+2)2

Atx=–6,

dydx

=

8(–6+2)2

=

8

16

=

12

EquationofthenormalatA: y–4= –2(x+6) y–4= –2x–12 y+2x = –8

(b) y=

2x–4x+2

… 1

y=–2x–8… 2


–2x–8=

2x–4x+2

–2x2–12x–16= 2x–4 2x2+14x+12= 0 x2+7x+6= 0 (x+6)(x+1)= 0 x=–6orx = –1

Substitutex=–1into 2 : y= –2(–1)–8 =2–8 =–6 ∴B(–1,–6)

17 (a) y =

12x+3

dydx

= –

2(2x+3)2

Comparewith

dydx

=

k(2x+3)2

∴k = –2 (b) Atthepoint(–1,1),

dydx

=

– 2[2(–1)+3]2

= –2 Equationofthenormal:

y–1=

12

(x+1)

2y–2= x+1 2y–x = 3

18 (a) y =

2x–8x+2

dydx

=

2(x+2)–(2x–8)(x+2)2

=

12

(x+2)2

Comparewith

dydx

=

k(x+2)2

∴ k = 12 (b) Atx-axis,y=0

0 =

2x–8x+2

2x =8 x = 4 ∴P(4,0) Aty-axis,x=0

y =

2(0)–8

0+2 = –4 ∴Q(0,–4) AtpointP(4,0),

dydx

=

12(4+2)2

=

1236

=

13

EquationofthenormalatP: y= –3(x–4) y = –3x+12 y+3x = 12 Aty-axis,x= 0

y+3(0)=12 y = 12 ∴R(0,12) ∴ThelengthofRQ = 12+4 = 16units

19 (a) y=x+6x–2

Atx-axis,y=0

0=

x+6x–2

x= –6 ∴A(–6,0) Aty-axis,x=0

y = –

62

= –3 ∴B(0,–3)

(b)

dydx

=

x–2–(x+6)(x–2)2

= –

8

(x–2)2

AtpointB(0,–3),

dydx

= –

8(–2)2

= –2 EquationofthenormalatB:

y+3 =

12

x

y =

12

x–3

Atx-axis,y=0

12

x = 3

x = 6 ∴C(6,0)

∴MidpointofBC=

(62

,–32 )

=

(3,–32 )

20 y =

13 x

= x– 1

3

dydx

= –13

x– 43

= –

1

3x43

Whenx=8, y =

13 8

=

12

δx = 7.9–8 = –0.1

and

dydx

= –

1

3(8)43

= –

1

48

Then

1

3 7.9

12

+(–

148)(–0.1)

0.5021

21 y = x3

dydx

= 3x2

(a) Whenx=1,δx = 1.05–1 = 0.05

and

dydx

= 3(1)2

= 3


Then δy

dydx

δx

3(0.05) 0.15 (b) Wheny=27,x3= 27 x = 3 δy= 26.8–27 = –0.2

and

dydx

= 3(3)2

= 27

Then δy

dydx

δx

–0.227δx δx–0.00741

22 y =

3x2

x+1

dydx

=

6x(x+1)–3x2

(x+1)2

=

3x2+6x(x+1)2

=

3x(x+2)(x+1)2

Bythechainrule,

dydt

=

dydx

dxdt

=

[3x(x+2)(x+1)2 ]dx

dt

If

dydt

=4whenx=2

4 =

( 83 )dx

dt

dxdt

=

128

=

32

unitspersecond

23 LetAbetheareaandrtheradius. ThenA=πr2

Bythechainrule,

dAdt

=

dAdr

drdt

5=

2πr

drdt

drdt

=

52πr

=

2.5πr

Whenthecircumference= 40cm 2πr =40

r=

402π

=

20π

∴ drdt

=

2.5

π( 20π )

=0.125cms–1

24 LetVbe thevolumeof sphereandrtheradius,then

V =43

πr3

dVdr

= 4πr2

Whenr=10,δr = 9.98–10

= –0.02

and

dVdr

= 4π(10)2

=400π

δV

dVdr

δr

400π(–0.02) –25.133

25 (a) y = x+5x–2

dydx

= 1–10x–3

= 1–

10x3

Whenx=4,

dydx

= 1–

10(4)3

=

2732

(b)

dydt

=

dydx

dxdt

2.7 =

(2732)dx

dt

dxdt

= 3.2unitspersecond

26 (a) y =

51–3x

dydx

=

15(1–3x)2

(b) Whenx=2, δx = 2+p–2 = p

and

dydx

=

1525

=

35

Then δy

dydx

δx

35

p

27 limx→∞( x2+7x

x2–5 )=

limx→∞(1+

7x

1–5x2

)

=

1+01–0

=1

28 limx→2( x2–2x

x2–4 )=lim

x→2[ x(x–2)(x+2)(x–2)]

=lim

x→2( xx+2)

=

2

2+2

=

24

=

12

29 (a) Forturningpoint,dydx

= 0

6x2+2px = 0 Atthepoint(–3,19), 54–6p =0 6p=54 p=9 Atthepoint(–3,19), 19 =2(–3)3+9(–3)2+q

19=27+q q=–8 p=9,q=–8

(b)

d2ydx2

=12x+18

Whenx=–3,

d2ydx2

=12(–3)+18

=–180 (–3,19)isamaximumpoint

30 (a) y = x3–3x2+4

dydx

= 3x2–6x

(b) Forturningpoints,

dydx

= 0

3x2–6x=0 3x(x–2)=0 x=0orx=2 Whenx=0,y=4 ∴B(0,4) Whenx=2,y=23–3(2)2+4 =0 ∴A(2,0)

31 (a) A(2,3)andB(1,0)

mAB =

–3–1

=3 3px2–3= 3 AtthepointA(2,3), 12p–3= 3 12p =6

p =

12

y=

12

x3–3x+q

AtthepointA(2,3),

3=

12

(2)3–3(2)+q

3= 4–6+q q = 5

∴p=12

,q=5

(b) TheequationofthenormalatA:

y–3= –1

3(x–2)

3y–9= –x+2 3y+x = 11 (c) Atx-axis,y=0 3(0)+x = 11 x = 11 ∴C(11,0)

32 (a) mAB = –124

=–3 2x–9=–3 2x=6 x=3 Whenx=3,y = 32–9(3)+18 =0 ∴P(3,0) (b) EquationofthenormalatP:

y =

13

(x–3)

3y = x–3


(c) y=x2–9x+18… 1

x=3y+3… 2


y=(3y+3)2–9(3y+3)+18 y=9y2+18y+9–27y–27+18 9y2–10y = 0 y(9y–10)= 0

y=0ory=

109

Substitutey=

109

into

2 :

x=3(10

9 )+3

=

193

∴Q(193

,109 )

33 (a) y =x2–3x+4

dydx

=2x–3

AtthepointP(1,2),

dydx

= 2(1)–3

=–1 EquationofthetangentatP: y–2 = –1(x–1) y–2 =–x+1 y+x= 3 (b) AtthepointQ(3,4),

dydx

= 2(3)–3

= 3 EquationofthenormalatQ(3,4):

y–4= –

13

(x–3)

3y–12= –x+3 3y+x = 15 (c) y+x =3… 1

3y+x = 15… 2

2 – 1 : 2y = 12 y = 6 Substitutey=6into 1 : 6+x = 3 x=–3 ∴R(–3,6)

34 (a) y = ax+bx

dydx

= a–

bx2

AtP(3,5),

dydx

=

13

a–

b9

=

13

9a–b=3… 1

and 5=3a+

b3

9a+b = 15 … 2

1 + 2 :18a=18 a=1 Substitutea=1into 1 : 9–b = 3 b=6 ∴a=1,b=6

(b) y= x+

6x

xy=x2+6… 1

y=14–3x … 2

Substitute 2 into 1 : x(14–3x)= x2+6 14x–3x2= x2+6 4x2–14x+6=0 2x2–7x+3=0 (2x–1)(x–3)=0

x=

12

orx=3

Substitutex=

12

into

2 :

y=14–3(1

2)

=

252

∴Q(12

,252 )

35 (a) A(1,2)andB(0,–1)

mAB =

–3–1

=3 AtA(1,2),3ax2–6= 3 3a–6=3 3a=9 a=3 and 2= 3(1)3–6(1)+b 2=–3+b b=5 ∴a=3,b=5 (b) LetCbe(x,0),

mAC = –

13

2

1–x =

–13

6= –1+x x=7 ∴C(7,0) (c) AreaofΔABC

=

12 12

0–1

70

12

=

12

(–1+14+7)

=

12

(20)

=10unit2

36 V=

43

πr3

dVdr

= 4πr2

(a) Whenr=16,δr= 15.9–16 =–0.1

and

dVdr

=4π(16)2

=1024π

Then

dV

dVdr

δr

1024π(–0.1) –102.4π (b) Bythechainrule,

dVdt

=dVdr

drdt

1000 = 4πr2

drdt

drdt

=

10004πr2

Whenr=5,

dAdt

=

1000100π

=

10π

cms–1

37 (a) V =

43

πr3

dVdr

= 4πr2

Bythechainrule,

dVdt

=

dVdr

drdt

dVdt

= 4πr2(0.25)

Whenr=4,

dVdt

= 64π(0.25)

= 16πcm3s–1

(b) A = 4πr2

dAdr

= 8πr

Bythechainrule,

dAdt

=

dAdr

drdt

=8πr(0.25)

Whenr=4,

dAdt

= 32π(0.25)

=8πcm2s–1

38 (a)

r

x

5

10

Fromthediagram,

rx

=

510

r =

12

x

Thevolumeofwaterinthecone,

V=

13

πr2x

=

13

π(12

x)2

x

=

112

πx3(shown)

(b) V =

1

12 πx3

dVdx

=

14

πx2

Whenx=4,δx = 4.05–4 =0.05

and

dVdx

=

14

π(4)2

=4π

Then δV=

dVdx

δx

=4π(0.05) =0.2πcm3


39 (a) V =

13

πx3

dVdx

= πx2

Bythechainrule,

dVdt

=

dVdx

dxdt

25=πx2

dxdt

dxdt

=

25πx2

Whenx=5,

dxdt

=

2525π

=

1π

cms–1

(b) A = πx2

dAdx

=2πx

Bythechainrule,

dAdt

=

dAdx

dxdt

= 2πx(1

π ) = 2x

Whenx=5,

dAdt

= 2(5)

=10cm2s–1

40 (a) V =2x2+16x

dVdx

=4x+16

Bythechainrule,

dVdt

=

dVdx

dxdt

12=(4x+16)

dxdt

Whenx=2,

dxdt

=

1224

=

12

cms–1

(b) 12=(4x+16)(0.2) 4x+16= 60 4x =44 x=11cm

41 (a) y =

c(x2+1)3

y =c(x2+1)–3

dydx

=–3c(x2+1)–4(2x)

= –

6cx

(x2+1)4

(b) Whenx=1,δx = 1+p–1 = p

and

dydx

= –6c16

=

–3c

8

Then δy

dydx

δx

– 3

4p

–3c

8(p)

c2

42 (a) 50x+2y = 480 25x+y=240 y=240–25x

A =24xy+

12

(24x)(5x)

=24x(240–25x)+60x2

=5760x–600x2+60x2

=5760x–540x2(shown) (b) FormaximumvalueofA,

dAdx

= 0

5760–1080x =0

x= 16

3

andso,

d2Adx2

=–1080

∴Aismaximum.

Whenx=

163

,

y=240–25(16

3 )

=106 23

Hence,themaximumarea,

A=5760(16

3 )–540(16

3 )2

=15360cm2

43 (a)

C

y

R

Q10 – x

45°

Fromthediagram,

y10–x

= tan45°

y =10–x AreaofPQRS, A=2xy =2x(10–x)(shown) (b) ForamaximumvalueofA,

dAdx

= 0

20–4x = 0 4x =20 x=5

andso

d2Adx2

=–40

Hence,themaximumareais2(5)(10–5)=50cm2.

44 (a) 2y+2x+πx = 200 2y=200–2x–πx

y=100–x–

12

πx

A= 2xy+

12

πx2

=2x(100–x–

12

πx)+12

πx2

=200x–2x2–πx2+

12

πx2

=200x–2x2–12

πx2(shown)

(b) ForamaximumvalueofA,

dAdx

= 0

200–4x–πx = 0 x(4+π)=200

x=

200

4+π =28

andso

d2Adx2

=–7.1420

Hencethemaximumarea,

A= 200(28)–2(28)2–

12

π(28)2

=5600–1568–1232 =2800cm2

45 (a) 2r+rθ = 16 rθ=16–2r

θ =

16–2r

r

A =

12

r2θ

=

12

r2(16–2rr )

=

12

r(16–2r)

= 8r–r2

=r(8–r)(shown)

(b)

dAdr

= 8–2r

dAdt

=

dAdr

drdt

=(8–2r)(0.05)

Whenr=3,

dAdt

= 2(0.05)

=0.1cm2s–1

46 (a) s =rθ

dsdθ

=r

dsdt

=

dsdθ

dθdt

2=r dθ

dt

Whenr=16,2=16 dθ

dt

dθdt

=18

radianpersecond

(b) A =

12

r2θ

dAdθ

=

12

r2

dAdt

=

dAdθ

dθdt

=

12

r2dθdt

Whenr=16,

dAdt

=

12

(16)2( 18 )

=16cm2s–1

10 Solution of Triangles

1

Booster Zone

1 (a) xsin 40°

= 6sin 15°

x = 6 sin 40°sin 15°

= 14.9 cm

(b) xsin 40°

= 12

sin 120°

x = 12 sin 40°sin 120°

= 8.91 cm

(c) 8.5sin 123°

= 5.5sin θ

sin θ = 5.5 sin 123°

8.5 = 0.5427 θ = 32° 52'

(d) 7sin 110°

= 4sin ∠QPR

sin ∠QPR = 4 sin 110°

7 = 0.5370 ∠QPR = 32° 29' θ = 180° – (110° + 32° 29') = 37° 31'

(e) 10sin 160°

= 5

sin ∠PRQ

sin ∠PRQ = 5 sin 160°

10 = 0.1710 ∠PRQ = 9° 51' θ = 180° – (160° + 9° 51') = 10° 9'

(f) xsin 75°

= 9

sin 70°

x = 9 sin 75°sin 70°

= 9.25 cm

2 (a) 5sin 25°

= 8sin θ

sin θ = 8 sin 25°

5 = 0.6762 θ = 42° 33', 137° 27' ∴ θ = 137° 27'

(b) 8

sin 10° = 15

sin θ

sin θ = 15 sin 10°8

= 0.3256 θ = 19°, 161° ∴ θ = 161°

(c) 16sin 20°

= 20sin θ

sin θ = 20 sin 20°

16 = 0.4275 θ = 25° 19', 154° 41'

∴ θ = 154° 41'

(d) 5sin 8°

= 9

sin θ

sin θ = 9 sin 8°5

= 0.2505 θ = 14° 30', 165° 30'

∴ θ = 165° 30'

3 (a)

5

sin 25° =

9sin ∠ABC

sin ∠ABC = 9 sin 25°

5 = 0.7607 ∠ABC = 49° 32' ∴ ∠AB'C = 180° – 49° 32' = 130° 28'

(b)

12sin ∠ABC

= 6sin 15°

sin ∠ABC = 12 sin 15°6

= 0.5176 ∠ABC = 31° 10', 148° 50' ∴ ∠AB'C = 148° 50'

(c)

7sin 10°

= 16sin ∠ABC

sin ∠ABC = 16 sin 10°7

= 0.3969 ∠ABC = 23° 23', 156° 37' ∴ ∠AB'C = 156° 37'

(d)

8

sin 20° =

14sin ∠ABC

sin ∠ABC = 14 sin 20°

8 = 0.5985 ∠ABC = 36° 46', 143° 14' ∴ ∠AB'C = 143° 14'

4 (a) 10sin 130°

= 6

sin ∠RQS

sin ∠RQS = 6 sin 130°10

= 0.4596 ∠RQS = 27° 22'

(b) PRsin 42° 22'

= 10

sin 115°

PR = 10 sin 42° 22'

sin 115° = 7.44 ∴PS = 7.44 – 6 = 1.44 cm

5 (a) 12sin 85°

= 6

sin ∠QRS

sin ∠QRS = 6 sin 85°

12 = 0.4981 ∠QRS = 29° 52' ∴ ∠RQS = 180° – (85° + 29° 52') = 65° 8'

(b) PQsin 95°

= 6sin 50°

PQ = 6 sin 95°sin 50°

= 7.8 cm

6 (a) x2 = 82 + 102 – 2(8)(10) cos 160° = 314.351

x = 314.351 = 17.73 cm

(b) x2 = 42 + 92 – 2(4)(9) cos 110° = 121.625

x = 121.625 = 11.03 cm

(c) x2 = 72 + 92 – 2(7)(9) cos 115° = 183.25

x = 183.25 = 13.54 cm

(d) x2 = 52 + 122 – 2(5)(12) cos 135° = 253.853 x = 253.853 = 15.93 cm

A

9 cm 5 cm

C B' B25°

A

16 cm

7 cm

C

B'

B

10°

A14 cm 8 cm

CB' B

20°

A

12 cm

6 cm

C

B'

B

15°


2 ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010

7 (a) 172 = 82 + 122 – 2(8)(12) cos θ

cos θ = 82 + 122 – 172

2(8)(12) = –0.4219 θ = 114° 57'

(b) 162 = 142 + 202 – 2(14)(20) cos θ cos θ = 142 + 202 – 162

2(14)(20) = 0.6071 θ = 52° 37'

(c) 72 = 62 + 92 – 2(6)(9) cos θ

cos θ = 62 + 92 – 72

2(6)(9) = 0.6296 θ = 50° 59'

(d) 202 = 72 + 182 – 2(7)(18) cos θ

cos θ = 72 + 182 – 202

2(7)(18) = –0.1071 θ = 96° 9'

8 (a) BD2 = 62 + 102 – 2(6)(10) cos 50° = 58.865

BD = 58.865 = 7.67 cm

(b) 92 = 52 + 7.672 – 2(5)(7.67) cos ∠BDC

cos ∠BDC = 52 + 7.672 – 92

2(5)(7.67) = 0.0369 ∠BDC = 87° 54', 92° 6' ∴ ∠BDC = 92° 6'

9 (a) QSsin 145°

= 4sin 15°

QS = 4 sin 145°

sin 15° = 8.87 cm

(b) 8.872 = 82 + 122 – 2(8)(12) cos ∠QRS

cos ∠QRS = 82 + 122 – 8.872

2(8)(12) = 0.6736 ∠QRS = 47° 39'

10 (a) AC2 = 22 + 52 – 2(2)(5) cos 110° = 35.84

AC = 35.84 = 5.99 cm

(b) 5.99sin ∠ADC

= 8

sin 80°

sin ∠ADC = 5.99 sin 80°

8 = 0.7374 ∠ADC = 47° 30' ∴ ∠ACD = 180° – (80° + 47° 30') = 52° 30'

11 (a) A = 12

(4)(9) sin 130°

= 13.79 cm2

(b) A = 12

(3)(4.5) sin 70°

= 6.34 cm2

(c) A = 12

(5)(6.2) sin 33°

= 8.44 cm2

(d) 9sin 54°

= 10sin ∠PQR

sin ∠PQR = 10 sin 54°9

= 0.8989 ∠PQR = 64° 1' ∠QPR = 180° – (54° + 64° 1') = 61° 59'

A = 12

(9)(10) sin 61° 59'

= 39.73 cm2

(e) 22 = 42 + 52 – 2(4)(5) cos ∠PRQ

cos ∠PRQ = 42 + 52 – 22

2(4)(5) = 0.925 ∠PRQ = 22° 20'

A = 12

(4)(5) sin 22° 20'

= 3.8 cm2

(f) 122 = 62 + 92 – 2(6)(9) cos ∠ABC

cos ∠ABC = 62 + 92 – 122

2(6)(9) = –0.25 ∠ABC = 104° 29'

A = 12

(6)(9) sin 104° 29'

= 26.14 cm2

12 (a)

16.162 = 102 + 172 – 2(10)(17) cos θ

cos θ = 102 + 172 – 16.162

2(10)(17) = 0.3760 θ = 67° 55'

A = 12

(10)(17) sin 67° 55'

= 78.76 cm2

(b)

132 = 52 + 13.422 – 2(5)(13.42) cos θ

cos θ = 52 + 13.422 – 132

2(5)(13.42) = 0.2690 θ = 74° 23'

A = 12

(5)(13.42) sin 74° 23'

= 32.31 cm2

13 (a)

9.612 = 12.52 + 172 – 2(12.5)(17) cos θ

cos θ = 12.52 + 172 – 9.612

2(12.5)(17) = 0.8303 θ = 33° 52'

(b) Area of ΔACT

= 12

(12.5)(17) sin 33° 52'

= 59.18 cm2

14 (a)

tan ∠VCW = 125

= 2.4 ∠VCW = 67° 23'

(b) Area of ΔAVC = 12

(10)(13)

sin 67° 23' = 60 cm2

SPM Appraisal Zone

1 (a) 6sin 85°

= 4sin ∠ADC

sin ∠ADC = 4 sin 85°6

= 0.6641 ∠ADC = 41° 37' ∴ ∠CAD = 180° – (85° + 41° 37') = 53° 23'

(b) 42 = 22 + 32 – 2(2)(3) cos ∠ABC

cos ∠ABC = 22 + 32 – 42

2(2)(3) = – 0.25 ∠ABC = 104° 29'

(c) Area of ΔABC

= 12

(2)(3) sin 104° 29'

= 2.91 cm2

Area of ΔCAD

= 12

(4)(6) sin 53° 23'

= 9.63 cm2

∴ Area of ABCD = 2.91 + 9.63 = 12.54 cm2

16.16 cm

17 cm θ

10 cm

5 cm

13 cm

θ

13.42 cm

9.61 cm

17 cmθ

12.5 cm

A

T

C

13

5

12

W

V

C


2 (a) QS2 = 72 + 92 – 2(7)(9) cos 65° = 76.75 QS = 8.76 cm

(b) 8.76sin 110°

= 5sin ∠QSR

sin ∠QSR = 5 sin 110°

8.76 = 0.5364 ∠QSR = 32° 26'

(c) Area of ΔQPS = 12

(7)(9) sin 65°

= 28.55 cm2

Area of ΔRQS

= 12

(8.761)(5) sin 37° 34'

= 13.35 cm2

∴ Area of PQRS = 28.55 + 13.35 = 41.9 cm2

3 (a) PR2 = 62 + 82 – 2(6)(8) cos 55° = 44.937 PR = 6.7 cm

(b) 6.7sin 60°

= 5

sin ∠RPS

sin ∠RPS = 5 sin 60°6.7

= 0.6463 ∠RPS = 40° 15' ∴ ∠PRS = 180° – (60° + 40° 15') = 79° 45'

(c) Area of Δ PQR = 12

(6)(8) sin 55°

= 19.66 cm2

Area of ΔPRS

= 12

(5)(6.7) sin 79° 45'

= 16.48 cm2

∴ Area of PQRS = 19.66 + 16.48 = 36.14 cm2

(d)

6.7sin 55°

= 8sin ∠PRQ

sin ∠PRQ = 8 sin 55°

6.7 = 0.9781 ∠PRQ = 77° 59' ∴ ∠PR'Q = 180° – 77° 59' = 102° 1'

4 (a) (i)

tan θ = 8

17.09 = 0.4681 θ = 25° 5'

(ii)

tan θ = 86

= 1.3333 θ = 53° 8'

(b)

162 = 12.812 + 12.812 – 2(12.81)(12.81) cos ∠ATB

cos ∠ATB

= 12.812 + 12.812 – 162

2(12.81)(12.81) = 0.22 ∠ATB = 77° 17'

(c) Area of ΔATB

= 12

(12.81)(12.81) sin 77° 17'

= 80 cm2

5 (a) (i) BD2 = 42 + 72 – 2(4)(7) cos 60° = 37 BD = 6.08 cm

(ii) 6.08

sin 30° =

8sin ∠ADB

sin ∠ADB = 8 sin 30°

6.08 = 0.6579 ∠ADB = 41° 8'

(iii) Area of ΔABD

= 12

(6.08)(8) sin 108° 52'

= 23.01 cm2

Area of ΔBCD

= 12

(4)(7) sin 60°

= 12.12 cm2

∴ Area of ABCD = 23.01 + 12.12 = 35.13 cm2

(b) (i)

(ii) ∠AD'B = 180° – 41° 8' = 138° 52'

6 (a) 12

(12)(20) sin ∠BAD = 60

sin ∠BAD = 60120

= 0.5 ∠BAD = 30°

(b) BD2 = 122 + 202 – 2(12)(20) cos 30° = 128.308 BD = 11.33 cm

(c) 11.33sin 20°

= 28

sin ∠BDC

sin ∠BDC = 28 sin 20°

11.33 = 0.8452 ∠BDC = 57° 41' ∴ ∠BDC = 180° – 57° 41' = 122° 19'

(d) Area of ΔCBD

= 12

(11.33)(28) sin 37° 41'

= 96.96 cm2

∴ Area of ABCD = 60 + 96.96 = 156.96 cm2

7 (a) (i) QSsin 30°

= 10sin 105°

QS = 10 sin 30°sin 105°

= 5.18 cm

(ii) 5.182 = 922 + 112 – 2(9)(11) cos ∠QPS

cos ∠QPS = 92 + 112 – 5.182

2(9)(11) = 0.8847 ∠QPS = 27° 48'

(iii) Area of ΔQPS

= 12

(9)(11) sin 27° 48'

= 23.09 cm2

Area of ΔRQS

= 12

(5.18)(10) sin 45°

= 18.3 cm2

∴ Area of PQRS = 23.09 + 18.3 = 41.39 cm2

(b) (i)

6 cm

R'8 cm

R

Q

P

17.09 cm

8 cm

C

F

Aθ

6 cm

8 cm

θ

12.81 cm

16 cm B

T

A

8 cm

6.08 cm

A

D

B30°

D'

41° 8'

9 cm

5.18 cm

Q

P

SQ'

27° 48'


(ii) 5.18sin 27° 48'

= 9

sin ∠PQS

sin ∠PQS = 9 sin 27° 48' 5.18

= 0.8103 ∠PQS = 54° 7' ∠PQ'S = 180° – 54° 7' = 125° 53' ∠PSQ = 180° – (27° 48'

+ 125° 53') = 26° 19'

Area of ΔPQ'S

= 12

(5.18)(9) sin 26° 19'

= 10.33 cm2

8 (a) CD2 = 182 + 242 – 2(18)(24) cos 30° = 151.754 CD = 12.32 cm

(b) 12.322 = 19.312 + 252 – 2(19.31)(25) cos ∠CAD

cos ∠CAD = 19.312 + 252 – 12.322

2(19.31)(25) = 0.8763 ∠CAD = 28° 48'

(c) Area of ΔCAD

= 12

(19.31)(25) sin 28° 48'

= 116.28 cm2

9 (a) (i) PR2 = 92 + 182 – 2(9)(18) cos 120°

= 567 PR = 23.81 cm

(ii) 6

sin 12° = 23.81

sin ∠RSP

sin ∠RSP = 23.81 sin 12°6

= 0.8251 ∠RSP = 55° 36'

(iii) Area of ΔPQR

= 12

(9)(18) sin 120°

= 70.15 cm2

Area of ΔPRS

= 12

(6)(23.81) sin 112° 24'

= 66.04 cm2

∴ Area of PQRS = 70.15 + 66.04 = 136.19 cm2

(b) (i)

(ii) PS'

sin 43° 36' =

6sin 12°

PS' = 6 sin 43° 36'sin 12°

= 19.9 cm

10 (a) 12

(8)(12) sin ∠PQS = 24

sin ∠PQS = 0.5 ∠PQS = 30°

(b) PS2 = 82 + 122 – 2(8)(12) cos 30° = 41.723 PS = 6.46 cm

(c) 13sin 20°

= 8sin ∠QRS

sin ∠QRS = 8 sin 20° 13

= 0.2105 ∠QRS = 12° 9'

(d) Area of ΔQSR

= 12

(8)(13) sin 147° 51'

= 27.67 cm2

∴ Area of PQRS = 24 + 27.67 = 51.67 cm2

6 cm

23.81 cm

R

P

S

12°

124° 24'

S'55° 36'

11 Index Number

1

Booster Zone

1 I10/08

= 13 23012 600

× 100 = 105

2 I09/07

= 45643260

× 100 = 140

3 (a) I06/05

= 8401200

× 100

= 70

(b) I10/08

= 1.801.20

× 100

= 150

(c) I04/08

= 26002000

× 100

= 130

(d) I09/06

= 2.101.50

× 100

= 140

4 (a) 420P

0

× 100 = 105

P0 = 420 × 100

105

= RM400

P

1

1200 × 100 = 87.5

P1 = 87.5 × 1200

100

= RM1050

(b) 560P

0

× 100 = 112

P0 = 560 × 100

112

= RM500

P

1

60 × 100 = 125

P1 = 125 × 60

100

= RM75

(c) 29P

0

× 100 = 145

P0 = 29 × 100

145

= RM20

P

1

80 × 100 = 90

P1 = 90 × 80

100

= RM72

5 p = 1110

× 100 = 110

7.8q

× 100 = 104

q = 7.80 × 100104

= 7.50

r

5.00 × 100 = 112

r = 112 × 5.00100

= 5.60

(s + 1)

s × 100 = 105

s + 1 = 1.05s 0.05s = 1 s = 20 ∴ p = 110, q = 7.50, r = 5.60, s = 20

6 P

08

P00

÷ P

05

P00

= 1.261.20

P

08

P05

= 1.05

p = 1.05 × 100 = 105

1.32 ÷ P

05

P00

= 1.10

P

05

P00

= 1.321.10

= 1.2 q = 1.2 × 100 = 120

P

08

P00

÷ 1.25 = 1.16

P

08

P00

= 1.16 × 1.25

= 1.45 r = 1.45 × 100 = 145 ∴ p = 105, q = 120, r = 145

7 x = 2424

× 100 = 100

y24

× 100 = 125

y = 125 × 24100

= 30

z = 3624

× 100

= 150 ∴ x = 100, y = 30, z = 150

8 (a) –I =

115(6) + 120(5) + 150(4)15

= 189015

= 126

(b) –I =

125(2) + 110(5) + 105(3)10

= 111510

= 111.5

9 (a) –I

09/07 =

120(4) + 110(3) + 105(2) + 125(1)

10

= 114510

= 114.5

(b) –I

09/07 =

115(4) + 104(6) + 105(8) + 110(2)

20

= 214420

= 107.2

(c) –I

09/07 =

105(2) + 115.5(4) + 108(3) + 115(6)

15 =

168615

= 112.4

10 (a)

128(2) + 110(6) + m + 85(4)

13 = 105

m + 1256 = 1365 m = 109

(b)

122(4) + 120m + 132(5) + 86(2)

m + 11 = 110

1320 = 110m + 1210 110m = 110 m = 1

(c)

120(m – 2) + 112m + 115(2)

2m = 115

232m – 10 = 230m 2m = 10 m = 5

(d)

150(m + 5) + 130(2) + 90m

+ 110(7)2m + 14

= 125

1780 + 240m = 250m + 1750 10m = 30 m = 3

11 (a)

115(2) + 8m + 125(4) + 105(6)

20 = 116

1360 + 8m = 2320 8m = 960 m = 120

(b) –I

10/05 = 116 × 1.05

= 121.8



12 (a) –I

07/04 =

110(3) + 105(4) + 115(2) + 140(1)

10

= 112010

= 112

(b) –I

09/04 =

120(3) + 125(4) + 115(2) + 130(1)

10

= 122010

= 122

(c) –I

09/07 =

122112

× 100

= 108.93

13 (a) p = 18.0012.00

× 100

= 150

(b) 28q

× 100 = 140

q = 28 × 100140

= 20

(c) r

7.20 × 100 = 125

r = 125 × 7.20100

= 9

(d)

150(7) + 140(3) + 125(6) + 110s

s + 16 = 133

2220 + 110s = 133s + 2128 23s = 92 s = 4

14 (a) A : 110 × 1.05 = 115.5 B : 112 × 1.1 = 123.2 C : 124 × 0.95 = 117.8 D : 115

(b) –I

10/06 =

115.5(2) + 123.2(5) + 117.8(5) + 115(2)

14

= 166614

= 119

(c) P

10

2500 × 100 = 119

P10

= 119 × 2500100

= RM2975

SPM Appraisal Zone

1 (a) x

1.20 × 100 = 125

x = 125 × 1.20100

= 1.50

(b) y + 0.30

y × 100 = 120

y + 0.3 = 1.2y 0.2y = 0.3 y = 1.50

∴ y = 1.50 and z = y + 0.30 = 1.50 + 0.30 = 1.80

(c) (i) P

09

5 × 100 = 133

P09

= 133 × 5100

= RM6.65

(ii)

125(4) + 150(7) + 120(6) + 130m

m + 17 = 133

2270 + 130m = 133m + 2261 3m = 9 m = 3

2 (a) x = 0.900.60

× 100

= 150

y

1.20 × 100 = 125

y = 125 × 1.20100

= 1.50

0.90

z × 100 = 112.5

z = 0.90 × 100112.5

= 0.80 ∴ x = 150, y = 1.50, z = 0.80

(b) (i) –I

10/07 =

150(3) + 125(6) + 120(4) + 112.5(2)

15

= 190515

= 127

(ii) 4.70P

07

× 100 = 127

P07

= 4.70 × 100127

= RM3.70

(c) –I

11/07 = 127 × 1.1

= 139.7

3 (a) (i) 0.50P

06

× 100 = 125

P06

= 0.50 × 100125

= RM0.40

(ii) P

08

P06

= 1.4

P

06

P04

= 1.1

P

08

P04

= 1.4 × 1.1

= 1.54 I

08/04 = 1.54 × 100

= 154

(b) (i) –I

08/06 = 124

140(2) + 4x + 105(3) + 125(1)

10 = 124

4x + 720 = 1240 4x = 520 x = 130

(ii) P

08

2.50 × 100 = 124

P08

= 124 × 2.50100

= RM3.10

4 (a) 1.80

x × 100 = 150

x = 1.80 × 100150

= 1.20

y

0.80 × 100 = 175

y = 175 × 0.80100

= 1.40

z = 1.201.00

× 100

= 120 ∴ x = 1.20, y = 1.40, z = 120

(b) –I

07/05 =

130(5) + 150(3) + 175(2) + 125(4) + 120(1)

15

= 207015

= 138

(c) Let a = new weightage of banana b = new weightage of

pineapple 6 + 5 + 2 + a + b = 20 13 + a + b = 20 a + b = 7 ... 1

–I = 137

130(6) + 150(5) + 175(2) + 125a + 120b

20 = 137

1880 + 125a + 120b = 2740 125a + 120b = 860 25a + 24b = 172 ... 2 1 × 24 : 24a + 24b = 168 ... 3 2 – 3 : a = 4

Substitute a = 4 into 1 : 4 + b = 7 b = 3 ∴ a = 4, b = 3

5 (a) p = 4.804.00

× 100

= 120

q

2.50 × 100 = 112

q = 112 × 2.50100

= 2.80

2.75

r × 100 = 125

r = 2.20 ∴ p = 120, q = 2.80, r = 2.20


(b) –I

08/06 =

120(4) + 112(3) + 125(1) + 150(2)

10

= 124110

= 124.1

(c) P

08

2000 × 100 = 124.1

P08

= 124.1 × 2000100

= RM2482

(d) I10/06

= 124.1 × 1.2 = 148.92

6 (a) (i) P

04

8.50 × 100 = 120

P04

= 120 × 8.50100

= RM10.20

(ii) 16.50

P00

× 100 = 132

P00

= 16.50 × 100132

= RM12.50

(b) (i) U : 1.501.20

× 100 = 125

V : 1.321.10

× 100 = 120

W : 1.471.05

× 100 = 140

X : 1.041.30

× 100 = 80

∴ U : 125, V : 120, W : 140, X : 80

(ii)

125(2) + 120(3) + 140(4) + 80m

m + 9 = 125

1170 + 80m = 125m + 1125 45m = 45 m = 1

7 (a) x = 6050

× 100

= 120

y

42 × 100 = 140

y = 140 × 42100

= 58.8 ∴ x = 120, y = 58.8

(b) –I

10/08 = 137

120(200) + 140(700) + 150zz + 900

= 137

122 000 + 150z = 137z + 123 300 13z = 1300 z = 100

(c) P

10

4100 × 100 = 137

P10

= 137 × 4100100

= RM5617

8 (a) –I

09/07 =

120(4) + 200(2) + 150(1) + 128(3)

10

= 141410

= 141.4

(b) P

09

1000 × 100 = 141.4

P09

= 141.4 × 1000100

= RM1414

(c) (i) –I

10/07 = 141.4 × 1.1

= 155.54

(ii) –I

10/09 = 110

9 (a)

Ingredient

Price index in

2010 (2008 = 100)

Weightage

Noodles 120 5

Sweet potatoes

105 3

Bean curds 102 1

Bean sprouts

104 2

Onions 110 4

–I

10/08 =

120(5) + 105(3) + 102(1) + 104(2) + 110(4)

15

= 166515

= 111

(b) (i) P

10

1.50 × 100 = 120

P10

= RM1.80

(ii) 2.20P

08

× 100 = 105

P08

= RM2.10

(c) P

10

4.50 × 100 = 111

P10

= 111 × 4.50100

= RM5

10 (a) (i)Item I08/07

Coffee 105

Milk 150

Sugar 125

Cream 150

(ii) –I

08/07 =

105(4) + 150(3) + 125(2) + 150(1)

10

= 127010

= 127

(b) (i) –I

10/08 = 127

(ii) P

10

2500 × 100 = 127

P10

= 127 × 2500100

= RM3175


SPM-Cloned Questions

Chapter 1: Functions 1 (a) 7

(b) 2, 3 2 h(x) = 1—

2x – m

Let y = 1—

2x – m

1—

2x = y + m

x = 2y + 2m∴h–1(x) = 2x + 2mCompare with h–1(x) = nx + 6∴ 2m = 6 and n = 2 m = 3∴ m = 3, n = 2

3 h(x) = 3x – 5Let y = 3x – 5 3x = y + 5

x = y + 5——–

3

∴ h–1(x) = x + 5——–

3 ∴ gh–1(x) = g� x + 5

——–3 �

= 2� x + 5

——–3 � – 1

= 2x + 10 – 3—————

3 =

2x + 7———

3 4 (a) k(x) =

3——–4 – x

Let y = 3

——–4 – x

4y – xy = 3 xy = 4y – 3

x = 4y – 3———

y

∴ k –1(x) = 4x – 3———

x, x ≠ 0

(b) k –1� 1—2 � =

4� 1—2 � – 3

————1—2

=

–1—–1—2

= –2

5 (a) Many to one relation.(b) f : x x2 – 1

6 (a) Let y = m(x) Then y = 2x – 5 2x = y + 5 x =

y + 5——–

2

∴ m–1(x) = x + 5——–2

m–1(–3) = –3 + 5———

2 = 1(b) nm(x) = n(2x – 5) = (2x – 5)2 + 3(2x – 5) – 4 = 4x2 – 20x + 25 + 6x – 15 – 4 = 4x2 – 14x + 6

7 (a) 3 and 7(b) Domain = {3, 5, 7}

8 (a) Let y = f(x) Then y = 2x – 3 2x = y + 3

x = y + 3——–

2

∴ f –1(x) = x + 3——–

2 (b) f –1g(x) = f –1� x—

4 + 1�

=

x—4

+ 1 + 3————–

2 =

� x + 16———4 �

————2

= x + 16———8

(c) hg(x) = 1—2

x + 5

h� x—4

+ 1� = 1—2

x + 5

Let k = x—4

+ 1 x—

4 = k – 1

x = 4k – 4 h(k) = 1—

2(4k – 4) + 5

= 2k – 2 + 5 = 2k + 3 ∴ h(x) = 2x + 3

Chapter 2: Quadratic Equations

1 Sum of roots: p + q + q – 3 = 2 p + 2q = 5 p = 5 – 2q … 1Product of roots: (p + q)(q – 3) = –24 pq – 3p + q2 – 3q = –24 … 2

Substitute 1 into 2 : q(5 – 2q) – 3(5 – 2q) + q2 – 3q = –24 5q – 2q2 – 15 + 6q + q2 – 3q = –24 q2 – 8q – 9 = 0 (q + 1)(q – 9) = 0 q = –1 or q = 9Substitute q = –1 into 1 : p = 5 – 2(–1) = 7

Substitute q = 9 into 1 : p = 5 – 2(9) = –13∴ p = 7, q = –1; p = –13, q = 9

2 2x2 + 4x + p = 0Sum of roots: α + 3α = –2 4α = –2 α = – 1—

2 … 1

Product of roots: α(3α) =

p—2

3α2 = p—2

… 2


3�– 1—2 �

2

= p—2

3—4

= p—2

p = 6—4

= 3—2

∴ The roots are – 1—2

and 3�– 1—2 � = – 3—

2and p = 3—

2.

3 x2 + 2k + 10 = x – 3kx x2 + (3k – 1)x + 2k + 10 = 0For two distinct roots, b2 – 4ac � 0 (3k – 1)2 – 4(1)(2k + 10) � 0 9k2 – 6k + 1 – 8k – 40 � 0 9k2 – 14k – 39 � 0 (9k + 13)(k – 3) � 0

∴ k � – 13—–

9 or k � 3

4 x2 + 6x + a = 2x + 1 x2 + 4x + a – 1 = 0For the line not to intersect the curve, b2 – 4ac � 0 42 – 4(1)(a – 1) � 0 16 – 4a + 4 � 0 20 – 4a � 0 4a � 20 a � 5

5 (a) 4x2 – 11x + 6 = 0 (4x – 3)(x – 2) = 0

k3 13– —–

9

ISBN: 978-983-70-3258-3© Cerdik Publications Sdn. Bhd. (203370-D) 2010


x = 3—4

or x = 2 (b) kx2 + px = 9 kx2 + px – 9 = 0 For two equal roots, b2 – 4ac = 0 p2 – 4(k)(–9) = 0 p2 + 36k = 0

6 x2 + x – k = 0Since –2 is one root of the equation, (–2)2 + (–2) – k = 0 k = 4 – 2 = 2

7 x2 – 2x = p – 2 x2 – 2x + 2 – p = 0For two different roots, b2 – 4ac � 0 (–2)2 – 4(1)(2 – p) � 0 4 – 8 + 4p � 0 4p � 4 p � 1

8 px2 – px + 2q = 2 + 4x px2 + (–p – 4)x + 2q – 2 = 0(a) Sum of roots:

1—p + q = p + 4——–

p 1 + pq = p + 4 pq = p + 3 q =

p + 3——–

p … 1

Product of roots:

1—p (q) = 2q – 2———p

q = 2q – 2 q = 2 Substitute q = 2 into 1 :

2 = p + 3——–p

2p = p + 3 p = 3 ∴ p = 3, q = 2(b) Sum of roots: p + (–3q) = 3 – 3(2) = –3 Product of roots: p(–3q) = 3[–3(2)] = –18 The quadratic equation with roots

p and –3q is x2 – (–3)x + (–18) = 0 x2 + 3x – 18 = 0

Chapter 3: Quadratic Functions

1 f(x) = –3x2 – 6x + 8

= –3�x2 + 2x – 8—3 �

= –3�x2 + 2x + � 2—2 �

2

– 8—3

– � 2—2 �

2

� = –3�(x + 1)2 – 11—–

3 � = 11 – 3(x + 1)2

∴ The maximum value is 11 and the axis of symmetry is x = –1.

2 (5x + 4)(x – 1) � 2(x – 1) 5x2 – x – 4 � 2x – 2 5x2 – 3x – 2 � 0 (5x + 2)(x – 1) � 0

∴ x � – 2—

5 or x � 1

3 (a) k = 1

(b) x = 1(c) (1, –4)

4 (a) p = –2(b) q = 4(c) f(x) = a(x – 2)2 + 4 At point (0, 2), 2 = a(–2)2 + 4 4a = –2

a = – 1—2

5 (a) a = 4(b) x = –4

6 f(x) = x2 + 2x – a2

= x2 + 2x + (1)2 – a2 – (1)2

= (x + 1)2 – a2 – 1∴ –a2 – 1 = –10 a2 = 9 a = ±3

7 (a) f(x) = x2 – 2px + 2p2 + 9 = x2 – 2px + (–p)2 + 2p2 + 9 – (–p)2

= (x – p)2 + p2 + 9 ∴ p2 + 9 = h2 + 6p h2 = p2 – 6p + 9 = (p – 3)2

h = p – 3 (shown)(b) p = h2 + 1 … 1 h = p – 3 … 2 Substitute 2 into 1 : p = (p – 3)2 + 1 p = p2 – 6p + 10 p2 – 7p + 10 = 0 (p – 2)(p – 5) = 0 p = 2 or p = 5 Substitute p = 2 into 2 : h = 2 – 3 = –1 Substitute p = 5 into 2 : h = 5 – 3 = 2 ∴ p = 2, h = –1; p = 5, h = 2

8 (a) At f(x)-axis, x = 0 f(x) = 02 – k(0) + 13 = 13 ∴ A(0, 13)

(b) f(x) = x2 – kx + 13

= x2 – kx + �– k—2 �

2

+ 13 – �– k—2 �

2

= �x – k—

2 �2

+ 13 – k2

—4

∴ k—2

= 3 and p = 13 – k2

—4

k = 6 = 13 – 36—–4

= 13 – 9 = 4 ∴ k = 6, p = 4

Chapter 4: Simultaneous Equations

1 1—2

y = 1 – x y = 2 – 2x … 1 y2 = 2x + 10 … 2

Substitute 1 into 2 : (2 – 2x)2 = 2x + 10 4 – 8x + 4x2 = 2x + 10 4x2 – 10x – 6 = 0 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 x = – 1—

2 or x = 3

Substitute x = – 1—

2 into 1 :

y = 2 – 2�– 1—2 �

= 3Substitute x = 3 into 1 : y = 2 – 2(3) = –4

∴ �– 1—2

, 3�; (3, –4)

2 y – 3x = 7 y = 3x + 7 … 1

x2 + y2 – xy = 7 … 2

Substitute 1 into 2 : x2 + (3x + 7)2 – x(3x + 7) = 7 x2 + 9x2 + 42x + 49 – 3x2 – 7x = 7 7x2 + 35x + 42 = 0 x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = –3 or x = –2Substitute x = –3 into 1 : y = 3(–3) + 7 = –9 + 7 = –2Substitute x = –2 into 1 : y = 3(–2) + 7 = –6 + 7 = 1∴ x = –3, y = –2; x = –2, y = 1

3 e – f = 3 f = e – 3 … 1 e2 – 4f = 10 … 2Substitute 1 into 2 : e2 – 4(e – 3) = 10 e2 – 4e + 2 = 0

e = ————————–2(1)

4 ± (–4)2 – 4(1)(2)

x1 2– —

5


= ———2

4 ± 8

= 3.414 or 0.586Substitute e = 3.414 into 1 : f = 3.414 – 3 = 0.414Substitute e = 0.586 into 1 : f = 0.586 – 3 = –2.414∴ e = 3.414, f = 0.414; e = 0.586,

f = –2.414 4 x + 1 —

2y = 2

x = 2 – 1 —

2y … 1

y2 = 4x … 2


y2 = 4�2 – 1 —2

y� y2 = 8 – 2y y2 + 2y – 8 = 0 (y + 4)(y – 2) = 0y = –4 or y = 2

Substitute y = –4 into 1 : x = 2 – 1 —

2(–4)

= 2 + 2 = 4

Substitute y = 2 into 1 : x = 2 – 1 —

2(2)

= 2 – 1 = 1

∴ x = 4, y = –4; x = 1, y = 2

5 2x + y = 3 y = 3 – 2x … 1 2x2 + y2 + xy = 6 … 2

Substitute 1 into 2 : 2x2 + (3 – 2x)2 + x(3 – 2x) = 6 2x2 + 9 – 12x + 4x2 + 3x – 2x2 = 6 4x2 – 9x + 3 = 0 x = ————————–

2(4)9 ± (–9)2 – 4(4)(3)

= ———

89 ± 33

= 1.843 or 0.407Substitute x = 1.843 into 1 : y = 3 – 2(1.843) = –0.686Substitute x = 0.407 into 1 : y = 3 – 2(0.407) = 2.186∴ x = 1.843, y = –0.686; x = 0.407,

y = 2.186

6 x + 2y = 10 x = 10 – 2y … 1 2y2 – 7y + x – 1 = 0 … 2

Substitute 1 into 2 : 2y2 – 7y + 10 – 2y – 1 = 0 2y2 – 9y + 9 = 0 (2y – 3)(y – 3) = 0

y = 3 —2

or y = 3 Substitute y = 3 —

2 into 1 :

x = 10 – 2� 3 —

2 � = 7Substitute y = 3 into 1 : x = 10 – 2(3) = 4 ∴ x = 7, y = 3 —

2; x = 4, y = 3

7 2y – x = 1

x = 2y – 1 … 1 x2 + xy = 26 … 2 (2y – 1)2 + y(2y – 1) = 26 4y2 – 4y + 1 + 2y2 – y = 26 6y2 – 5y – 25 = 0 (3y + 5)(2y – 5) = 0

y = – 5 —3

or y = 5 —2

Substitute y = – 5 —3

into 1 : x = 2�– 5 —

3 � – 1

= – 13 —–3

Substitute y = 5 —2

into 1 : x = 2� 5 —

2 � –1 = 4

∴ x = – 13 —–3

, y = – 5 —3

; x = 4, y = 5 —2

8 k – 2p = –3 k = 2p – 3 … 1 p + pk – 4k = 0 … 2

Substitute 1 into 2 : p + p(2p – 3) – 4(2p – 3) = 0 p + 2p2 – 3p – 8p + 12 = 0 2p2 – 10p + 12 = 0 p2 – 5p + 6 = 0 (p – 2)(p – 3) = 0 p = 2 or p = 3Substitute p = 2 into 1 : k = 2(2) – 3 = 1Substitute p = 3 into 1 : k = 2(3) – 3 = 3∴ k = 1, p = 2; k = 3, p = 3

Chapter 5: Indices and Logarithms

1 log3R – log

9T = 2

log3R –

log3T

——–log

39

= 2 log

3R – 1 —

2 log

3T = 2

log

3

R —–T

= 2

R —–T

= 9 R = 9 T

2 9x = � 3 � x + 6

32x = 3x + 6 ——–

2

∴ 2x = x + 6 ——–2

4x = x + 6 3x = 6 x = 2

3 log

36.25 = log

3

25 —–4

= log325 – log

34

= 2 log35 – 2 log

32

= 2n – 2m 4 log

h � 81h ——16 � = log

h81 + log

hh – log

h16

= 4 logh3 + 1 – 2 log

h4

= 4p + 1 – 2q

5 log8 �4m —–n � =

log2�4m —–n �

————log

28

= log

24 + log

2m – log

2n

—————————log

28

= 2 log

22 + log

2m – log

2n

—————————–3 log

22

= 2 + x – y

————–3

6 4x

——2x – 3

= 82–x

22x

——2x – 3

= (23)2–x

22x – (x – 3) = 26 – 3x

∴ 2x – (x – 3) = 6 – 3x x + 3 = 6 – 3x 4x = 3

x = 3—4

7 2n–7 × 8n = 512

2n–7 × 23n = 29

2n–7 + 3n = 29

∴ n – 7 + 3n = 9 4n = 16 n = 4

8 log81

p – log3q = 0

log3p

——–log

381

= log3q

log3p

———4 log

33

= log3q

1—4

log3p = log

3q

log

3p

1—4 = log

3q

∴ p1—4 = q

p = q4

Chapter 6: Coordinate Geometry

1

A(h, 3h)

1

B(p, 2t)

C(4p, 5t)

2


2h + 4p———–

3 = p

2h + 4p = 3p p = –2h … 1 6h + 5t

———–3

= 2t 6h + 5t = 6t 6h = t

h = t—6

… 2 Substitute 2 into 1 : p = –2� t—

6 � = – 1—

3t

2 x

—2

+ y

—3

= 1 3x + 2y = 6 2y = –3x + 6 y = – 3—

2x + 3

The gradient of this line is – 3—

2.

3y = 2x + 6 y = 2—

3x + 2

The gradient is 2—

3.

Since �– 3—

2 �� 2—3 � = –1 so these two lines

are perpendicular to each other.

3 x—4

– y—2

= 1

The gradient of this line,

m1 = – �– 2—

4 � = 1—

2 and the coordinates of K is (4, 0).∴ Equation of the perpendicular line

through K is y – 0 = –2(x – 4) y = –2x + 8

4 Equation of the perpendicular line CD through C is y – 2 = 1—

2(x – 0)

2y – 4 = x 2y – x = 4 … 1 y + 2x = 7 … 21 × 2 : 4y – 2x = 8 … 32 + 3 : 5y = 15 y = 3Substitute y = 3 into 1 : 2(3) – x = 4 x = 6 – 4 = 2∴ D(2, 3)

5 a =

2(–2) + 1(13)——————

3 = 9—

3 = 3

b = 2(0) + 1(15)—————–

3

= 15—–3

= 5∴ Q(3, 5)

6 (a) The equation of AB: y – 2 = 1—

2(x + 2)

2y – 4 = x + 2 2y – x = 6 At x-axis, y = 0 2(0) – x = 6 x = –6 ∴ A(–6, 0)(b)

3(–6) + 2x—————5

= –2

–18 + 2x = –10 2x = 8 x = 4 3(0) + 2y————–

5 = 2

2y = 10 y = 5 ∴ C(4, 5)(c) Equation of the perpendicular line

through C: y – 5 = –2(x – 4) y – 5 = –2x + 8 y + 2x = 13

7 (a) QP = 5 units y2 + (x – 5)2 = 5 y2 + x2 – 10x + 25 = 25 x2 + y2 – 10x = 0(b) (i) x2 + y2 – 10x = 0

At M(2, k), 4 + k2 – 10(2) = 0 k2 – 16 = 0 k2 = 16 k = ±4 ∴ k = 4 (ii) Let N be (x, y),

� 2 + x——–2

, k + y——–

2 � = (5, 0) 2 + x——–

2 = 5 and k + y——–

2 = 0

2 + x = 10 4 + y = 0 x = 8 y = –4 ∴ N(8, –4)

8 (a) (i) � 1—k �(–2) = –1

k = 2 (ii) 2y – x = 10 … 1 y + 2x = 10 … 2

1 × 2: 4y – 2x = 20 … 3 2 + 3 : 5y = 30 y = 6 Substitute y = 6 into 1 : 2(6) – x = 10 x = 2 ∴ Q(2, 6)

(b) (i) 2y = x + 10 At y-axis, x = 0 2y = 10 y = 5 ∴ D(0, 5)

1(x) + 3(2)————–

4 = 0

x + 6 = 0 x = –6 and

1(y) + 3(6)————–

4 = 5

y + 18 = 20 y = 2 ∴ R(–6, 2) (ii) m = 2 – 0——–

–6 – 0 = – 1—

3

y – 0 = – 1—3

(x – 0)

y = – 1—3

x

(c) 2EQ = ER

2 (y – 6)2 + (x – 2)2

= (y – 2)2 + (x + 6)2

4(y2 – 12y + 36 + x2 – 4x + 4) = y2 – 4y + 4 + x2 + 12x + 36 4y2 – 48y + 4x2 – 16x + 160 = y2 – 4y + x2 + 12x + 40 3x2 + 3y2 – 28x – 44y + 120 = 0

Chapter 7: Statistics 1

Score f x fx x2 fx2

21–25 1 23 23 529 529

26–30 2 28 56 784 1568

31–35 9 33 297 1089 9801

36–40 11 38 418 1444 15 884

41–45 14 43 602 1849 25 886

46–50 3 48 144 2304 6912

Σf=40 Σfx=1540 Σfx2=60 580

–(a) x = Σfx

——Σf

= 1540——

40 = 38.5(b) Median class = 36 – 40

m = L + �

1—2

N – F————

fm

�C

2

A(–6, 0)

3

B(–2, 2)

C(x, y)

3

1

R(x, y)

D(0, 5)

Q(2, 6)


= 35.5 + �

1—2

(40) – 12—————–

11�5

= 39.14 (c) σ =

–Σfx2

——Σf

– (x)2

= 60 580———

40 – (38.5)2

= 5.679

2 81—–

3 – � k �2 = 2h

27 – k = 4h2

k = 27 – 4h2

– 3 (a) x = 8

Σx—–

6 = 8

Σx = 48 (b) 48 + m———

7 = 9

48 + m = 63 m = 63 – 48 = 15

4 (a) σ2 = 21, Σx2 = 1275, N = 15

1275——15

– (–x)2 = 21 (–x)2 = 85 – 21 = 64 –x = 64 = 8

(b) Σx—–15

= 8

Σx = 120

5 (a) N = 8, Σx = 120, Σx2 = 2448 –x = Σx—–

N –x = 120—–

8 = 15 σ =

Σx2

—–N

– (–x)2

=

2448——

8 – 152

= 81 = 9

(b) (i)

120 + m———–

9 = 14

120 + m = 126 m = 6 ∴ The number that being

added to the set is 6. (ii) σ =

2448 + 36————–

9 – 142

= 80 = 8.944

6 (a) Age (years)

Number of patients

1 – 10 8

11 – 20 6

21 – 30 5

31 – 40 7

41 – 50 10

Median class = 21 – 30 Median = L + �

N—2

– F———

fm

�C

= 20.5 + �18 – 14———

5 �10 = 20.5 + 8 = 28.5

(b) –x =

8(5.5) + 6(15.5) + 5(25.5) + 7(35.5) + 10(45.5)

———————————36

= 968——36

= 26.89

x f x2 fx2

5.5 8 30.25 242

15.5 6 240.25 1441.5

25.5 5 650.25 3251.25

35.5 7 1260.25 8821.75

45.5 10 2070.25 20 702.5

Σf=36 Σfx2=34 459

σ =

34 459———

36– 26.892

= 234.122 = 15.3

7 (a) Median class = 21–30 Median = 24.5

20.5 +

k + 24�———�2 – 12�——————�

1010 = 24.5

k + 24�———�2 – 12 = 4

k + 24———

2 = 16

k + 24 = 32 k = 8(b)

Mode time = 26.5

(c) From the histogram the mode time is 26.5. If the time of each athlete is decreased by 2, so the new mode is 26.5 – 2 = 24.5

8 (a) Mark Number of students

0 – 9 6

10 – 19 8

20 – 29 12

30 – 39 9

40 – 49 5

(b) Q1 = 9.5 + � 10 – 6———

8 �10

= 14.5 Q

3 = 29.5 + � 30 – 26———–

9 �10

= 33.94 ∴ Interquartile range = 33.94 – 14.5 = 19.44

Chapter 8: Circular Measures

1 Let the radius of a sector COD be r cm, 2r + 2 = 10 2r = 8 r = 4 cm s

CD = 2

4θ = 2

θ = 1—2

radian 2 (a) θ = 360° – 280°

= 80°

= 80° × π——180°

= 1.396 radians(b) s

XY = 20

1.396r = 20 r = 20——–

1.396 = 14.33 cm

3 (a) sQT

= 3 4θ = 3 θ = 0.75 radian ∴ ∠QST = 0.75 radian(b) Area of semicircle PQR = 1—

2(3)2(π)

= 14.14 cm2

Area of sector QST

= 1—2

(4)2(0.75) = 6 cm2

∴ Area of the shaded region = 14.14 – 6 = 8.14 cm2

4 (a) sQR

= 4(2.1) = 8.4 cm

Number of athletes

10

8

6

4

2

0 0.5 10.5 20.5 30.5 40.5 50.5

Time (minutes)26.5


(b) Area of sector QOR =

1—2

(4)2(2.1)

= 16.8 cm2

Area of ∆POS

= 1—2

(2)(2) sin �2.1 × 180°——π � = 1.73 cm2

Area of the shaded region = 16.8 – 1.73 = 15.07 cm2

5 (a) sQR

= 12(1.18) = 14.16 cm(b) Area of sector OQR = 1—

2(12)2(1.18)

= 84.96 cm2

Area of ∆OPS

= 1—2

(6)(6) sin �1.18 × 180°——π � = 16.64 Area of the shaded region = 84.96 – 16.64 = 68.32 cm2

6 (a) OP = 3—4

(12) = 9 cm

cos θ = 9—–12

= 0.75 θ = 41° 24' = 0.723 radian

(b) RP = 122 – 92

= 7.937 cm

Area of the shaded region = 1—

2 (12)2(0.723) – 1—

2(9)(7.937)

= 52.056 – 35.717 = 16.34 cm2

7 (a) Area of sector POQ = 1—

2(5)2(1.8)

= 22.5 cm2

(b) sPQ

= 5(1.8) = 9 cm s

PS = 8(3.142 – 1.8)

= 10.736 cm ∴ Perimeter of the shaded region = 10.736 + 9 + (8 – 5) = 22.74 cm(c) Area of sector PTS

= 1—2

(8)2(1.342)

= 42.94 cm2

Area of ∆QOT

= 1—2

(5)(3) sin �1.342 × 180°——π � = 7.31 cm2

∴ Area of the shaded region = 42.94 – 22.5 – 7.31 = 13.13 cm2

8 (a)

sin θ = 3—5

= 0.6 θ = 36° 52' ∠AOB = 0.644 radian(b) s

AB = 10(0.644)

= 6.44 cm

sDE

= 3� π—2 �

= 4.71 cm ∴ Perimeter of the shaded region = 6.44 + 5 + 4.71 + (10 – 7) = 19.15 cm(c) Area of sector OAB =

1—2

(10)2(0.644) = 32.2 cm2

Area of ∆OCD = 1—

2(4)(3)

= 6 cm2

Area of quadrant DCE

= 1—2

(3)2� π—2 �

= 7.07 cm2

Area of the shaded region = 32.2 – 6 – 7.07 = 19.13 cm2

Chapter 9: Differentiation 1 y = (3x – 4)3

dy

—–dx

= 3(3x – 4)2(3)

= 9(3x – 4)2

When x = 2,

dy—–dx

= 9[3(2) – 4]2 = 36The rate of change of y is given by

dy

—–dt

= dy

—–dx

× dx

—–dt

18 = 36 � dx—–dt �

dx

—–dt

= 18—–36

= 0.5 unit s–1

∴ x is increasing at a rate of 0.5 unit s–1.

2 h(x) = 2� 1—

2x – 4�

–2

h'(x) = –4� 1—2

x – 4�–3

� 1—2 �

= –2� 1—2

x – 4�–3

h"(x) = 6� 1—2

x – 4�–4

� 1—2 �

= 3————

� 1—2

x – 4�4

h"(4) =

3—————

� 1—2

(4) – 4�4

=

3——(–2)4

= 3—–

16 3 y = 3—

5u4

= 3—

5 (5x – 2)4

dy

—–dx

= 12—–5

(5x – 2)3(5)

= 12(5x – 2)3

4 (a) y = 2x2 + 3x – 2

dy—–dx

= 4x + 3

When x =

1—2

, dy

—–dx

= 4� 1—2 � + 3

= 5(b) When x = 2, δx = 2 + p – 2 = p

dy—–dx

= 4(2) + 3

= 11 Then, δy ≈

dy—–dx

× δx ≈ 11(p) ≈ 11p ∴ The approximate change in y is

11p.

5 Let V be the volume of a packet of butter. Then, V = x3

By the chain rule, dV—–dt

= dV—–dx

× dx—–dt

= 3x2 × dx—–dt

When x = 4, –12 = 3(4)2 dx—–dt

dx—–dt

= – 12—–48

= –0.25 cm s–1

∴ Rate of change of x is –0.25 cm s–1. 6 y =

1—2

x + 3 The gradient of this line is

1—2

.

So the gradient of the tangent at A(1, 4) is 2x – kx2 = –2 2(1) – k(1)2 = –2 2 – k = –2 k = 4

R

12

θP O9

3 5

θ

D

C O


7 (a) At the turning point (k, 2),

dy

—–dx

= 0

1 – 4—x2

= 0 1 – 4—

k2 = 0

1 =

4—k2

k2 = 4 k = ±2 ∴ k = 2

(b) dy

—–dx

= 1 – 4x–2

d 2y—–dx2

= 8x–3

=

8—x3

At (2, 2),

d 2y——dx2 � 0.

∴ (2, 2) is a minimum point.

8 y = 3 – 2x –1

dy

—–dx

= 2x –2

= 2—

x2 When x = 2, dy—–

dx = 2—–

(2)2

= 1—

2 So the gradient of the tangent at

A(2, 2) is 1—2

and its equation is y – 2 = 1—

2 (x – 2)

2y – 4 = x – 2 2y – x = 2

Chapter 10: Solution of Triangles

1 (a) BD2 = 32 + 42 – 2(3)(4) cos α = 25 – 24 cos α(b) BD2 = 12 + 22 – 2(1)(2) cos β = 5 – 4 cos β α + β = 180° β = 180° – αSubstitute β = 180° – α into 5 – 4 cos β:5 – 4 cos (180° – α)= 5 – 4 [cos 180° cos α + sin 180° sin α]= 5 – 4 [–1(cos α) + 0(sin α)]= 5 – 4 (– cos α)= 5 + 4 cos α 25 – 24 cos α = 5 + 4 cos α 28 cos α = 20

cos α = 20—–28

= 5—7

2

QS———sin 40°

= 4———sin 55°

QS = 4 sin 40°————sin 55°

= 3.14 cm

PS2 = 32 + 3.142 – 2(3)(3.14) cos 95° = 20.5

PS = 20.5 = 4.53 cm

3 (a) PV = 62 + 82

= 100 = 10 cm

QR = 172 – 82

= 225 = 15 cm

PR = 62 + 152

= 261 = 16.16 cm16.162 = 102 + 172 – 2(10)(17) cos ∠PVR 340 cos ∠PVR = 127.85 cos ∠PVR = 0.3760 ∠PVR = 67° 55'(b) Area of PVR

= 1—2

(10)(17) sin 67° 55'

= 78.76 cm2

4 (a) (i) AC 2 = 62 + 102 – 2(6)(10) cos 130°

= 213.135

AC = 213.135 = 14.6 cm (ii) 16———

sin 30° = 14.6————–

sin ∠ABC sin ∠ABC =

14.6 sin 30°—————16

= 0.4563 ∠ABC = 27° 9' (iii) Area of quadrilateral ABCD = 1—

2(6)(10) sin 130° +

1—2

(14.6)(16) sin 122° 51'

= 22.981 + 98.123 = 121.104 cm2

(b) (i)

(ii) 14.6———–sin 130°

= 10————–sin ∠ACD

sin ∠ACD = 10 sin 130°—————14.6

= 0.5247 ∠ACD = 31° 39' ∴AC'D = 180° – 31° 39' = 148° 21'

5 (a) AB2 = 52 + 72 – 2(5)(7) cos 130° = 118.995 AB = 118.995 = 10.9 cm

(b)

10.9————–

sin ∠AD2B

= 5.5———sin 30°

sin ∠AD

2B =

10.9 sin 30°—————

5.5

∠AD2B = 82° 17'

∠AD1B = 180° – 82° 17'

= 97° 43'(c) (i) ∠AD

2B = 82° 17'

∠ABD2 = 180° – 30° – 82° 17'

= 67° 43'

AD2————–

sin 67° 43' = 10.9————–

sin 82° 17' AD

2 = 10.18 cm

(ii) Area of ∆ACB

= 1—2

(5)(7) sin 130°

= 13.41 cm2

Area of ∆ABD2

= 1—2

(10.9)(5.5) sin 67° 43'

= 27.74 cm2

Area of quadrilateral ACBD = 13.41 + 27.74 = 41.15 cm2

6 (a) Area = 10 cm2

1—2

(5)(6) sin ∠BAD = 10 sin ∠BAD = 10—–

15 = 0.6667 ∠BAD = 41° 48'(b) BD2 = 52 + 62 – 2(5)(6) cos 41° 48' = 16.271

BD = 16.271 = 4.03 cm (c)

BC———sin 80°

= 4.03———

sin 65° BC = 4.03 sin 80°—————–

sin 65° = 4.38 cm(d) Area of quadrilateral ABCD = 1—

2(5)(6) sin 41° 48' +

1—2

(4.03)(4.38) sin 35° = 9.998 + 5.062 = 15.06 cm2

7 (a) (i) 72 = 52 + 82 – 2(5)(8) cos ∠BDC 80 cos ∠BDC = 40

cos ∠BDC = 1—2

∠BDC = 60°

D2

5.5 cm30°

5.5 cm

7 cm130°

D1

A

C

B

5 cm

A C'

10 cm6 cm

C

D


(ii) AD———sin 40°

= 12———sin 60°

AD = 12 sin 40°————–

sin 60° = 8.907 cm (iii) Area of ∆ABD = 1—

2(5)(8.907) sin 120°

= 19.28 cm2

(b) (i)

(ii) Area of ∆AD'E

= 1—2

(8.907)(12) sin 20°

= 18.28 cm2

8 (a)

QS———sin 61°

= 6.8

———sin 80°

QS =

6.8 sin 61°————–

sin 80° = 6.04 cm(b) ∠PSQ = 180° – (61° + 80°) = 39° RS2 = (2.2)2 + (6.04)2 – 2(2.2)(6.04)

cos 39° = 20.668

RS = 20.668 = 4.55 cm

(c) sin ∠QRS————

6.04 =

sin 39°———

4.55 sin ∠QRS =

6.04 sin 39°—————

4.55 = 0.8354 ∠QRS = 56° 39', 123° 21' ∴ ∠QRS = 123° 21' (d) Area of ∆QRS

= 1—2

(2.2)(4.55) sin 123° 21'

= 4.18 cm2

Chapter 11: Index Number 1 I

– = 115

120(m – 2) + 112m + 115(2)———————————–

2m = 115

232m – 10 = 230m 2m = 10 m = 5

2 (a) I–

= 113

110(6) + 120p + 80(3)

+ 140(5) + 90(2)—————————

p + 16 = 113

120p + 1780 = 113p + 1808 7p = 28 p = 4

(b) I09/07

= 90

P

09——6.50

× 100 = 90 P

09 = 90 × 6.50————

100 = RM5.85

3 (a) x——

2.50 × 100 = 112

x = 112 × 2.50————–

100 = 2.80

(b) I–

10/08 = 113

105(2) + 112(5) + 120y——————————–

y + 7 = 113

770 + 120y = 113y + 791 7y = 21 y = 3

4 (a) (i) x——0.90

× 100 = 150 x = 150 × 0.90————–

100 = 1.35 (ii) y = 1.80——

1.50 × 100

= 120 (iii) 3.85——z × 100 = 110 z = 3.85 × 100—————

110 = 3.50

(b) I–

05/00 =

150(3) + 125(6) + 120(4) + 160(10) + 110(7)

——————————–30

= 4050——30

= 135

(c) P

05——420

× 100 = 135

P05

= 135 × 420————–

100 = 567 ∴ The total monthly cost for the

year 2005 is RM567.

(d) I–

10/00 = 135 × 1.1

= 148.5 5 (a) I

07/04 = 140

4.20——

m × 100 = 140

m = 4.20 × 100————–140

= 3.00(b) I

07/04 = 150

n + 1——–n

× 100 = 150

100n + 100 = 150n 50n = 100 n = 2 ∴ n = 2, p = 3

(c) (i) I–

07/04 = 130

P

07——6.50

× 100 = 130

P07

= 130 × 6.50————–100

= RM8.45

(ii)

140(2) + 125(4) + 150(3) + 110q———————

q + 9 = 130

1230 + 110q = 130q + 1170 20q = 60 q = 3

6 (a) 1.40——x × 100 = 175 x = 1.40 × 100————–

175 = 0.80 y = 3.00——

2.50 × 100

= 120

(b) I

–10/09

=

150(90) + 175(45) + 140(108) + 120(36) + 125(81)

————————————360

= 50 940———360

= 141.5

(c) I–

11/09 = 141.5 × 1.2

= 169.8

P11—–

25 × 100 = 169.8

P

11 = 169.8 × 25————–

100 = 42.45 ∴ The production cost for the year

2011 is RM42.45.

7 (a) h = 1.65——1.20

× 100 = 137.5 0.80——

k × 100 = 160

k = 0.80 × 100————–160

= 0.50

(b) I

–10/08

=

120(20) + 137.5(40) + 160(30) + 150(10)

—————————100

= 14 200———100

= 142(c) (i) I

–12/08

= 142 × 1.1 = 156.2 (ii)

P12——

1.60 × 100 = 156.2

P

12 = 156.2 × 1.60—————

100 = 2.50 ∴ The price of a muffi n in the

year 2012 is RM2.50. 8 (a) (i) x = 1.80——

2.40 × 100

= 75

(ii) y

——3.20

× 100 = 150

y = 150 × 3.20————–100

= 4.80

A

60°

8.907 cm

D'40°

D E

12 cm


(iii) 7.15——

z × 100 = 110

z =

7.15 × 100————–

110 = 6.50

(b) I–

10/08 = 117.5

75(4) + 112(5) + 150(7) + 110m

———————m + 16

= 117.5

1910 + 110m = 117.5m + 1880 7.5m = 30 m = 4 (c)

P10——

5240 × 100 = 117.5

P

10 = 117.5 × 5240—————–

100 = 6157 ∴ The total expenditure in the year

2010 is RM6157.

(d) I–

11/08 = 140

I–

10/08 = 112

∴ I

11/10 =

140—–112

× 100 = 125

11

Paper 1 1 (a) f : x → x2

(b) q = 16

2 (a)

(b) –4 � f –1(x) � 6

3 f 2(x) = gf(1)

4(4x – 5) – 5 = 5

2(1)–3 16x – 25 = –1 16x = 24 x = 1.5

4 (a) f(–4) = 6

–4–2 = –1

(b) y = 6

x – 2 xy – 2y = 6

xy = 2y + 6

x = 2y + 6

y

f –1(x) = 2x + 6

x ∴ a = 2, b = 6

5 3x(2x – 4) = x – 5 6x2 – 12x = x – 5

6x2 – 13x + 5 = 0

x = 13± (–13)2 – 4(6)(5) 2(6)

= 13± 49

12

= 13 + 712

or 13 – 7

12

= 53

or 12

6 x2 – [3 + �– 12 �]x + (3)�– 1

2 � = 0

x2 – 52

x – 32

= 0

2x2 = 5x + 3∴ p = 5, q = 3

7 y = 2k+ x

8 … 1

2y2 = k + x … 2


2�2k + x8 �

2

= k + x

2� 4k2 + 4kx + x2

64 � = k + x

x2 + 4kx + 4k2 = 32k + 32x x2 + (4k – 32)x + 4k2 – 32k = 0

b2 – 4ac = 0 (4k – 32)2 – 4(1)(4k2 – 32k) = 0 16k2 – 256k + 1024 – 16k2 + 128k = 0 1024 – 128k = 0 128k = 1024 k = 8

8 (a) f(x) = –x2 + 4x + 6 = –(x2 – 4x – 6) = –[x2 – 4x + (–2)2 – (–2)2 – 6] = –[(x – 2)2 – 10] = 10 – (x – 2)2

∴ h = –2, k = 10

(b) The maximum value is 10.

9 (a) f(x) = a(x + 1)2 + 6 At the point (0, 4),

4 = a + 6 a = –2 f(x) = –2(x + 1)2 + 6 ∴ a = –2, p = 1, q = 6

(b) f(x) = –2(x – 1)2 + 6

10 2x2 – 4 � 7x 2x2 – 7x – 4 � 0

(2x + 1)(x – 4) � 0

∴ – 12

� x � 4

11

�64

2––3 x

3––2 �5

�165––4 x

2––3 �3

= [�26� 2––

3 x3––2 ]5

[�24� 5––4 x

2––3 ]3

= 220 x15––2

215 x2

= 25 x11––2

= �2x11––10�5

∴ m = 1110

, n = 5

12 8x

2y = 64

23x–y = 26

3x – y = 6 … 1

34x × �19 �

y–1

= 81

34x × (3–2)y–1 = 34

34x – 2y + 2 = 34

4x – 2y + 2 = 4 4x – 2y = 2 2x – y = 1 … 2 1 – 2 : x = 5

Substitute x = 5 into 1 : 3(5) – y = 6 y = 15 – 6 = 9 ∴ x = 5, y = 9

13

log4� x

y � = log

2 � x

y �log

24

= log

2x – log

2y

2

= h – k2

14 (a) AC = (10 – 2)2 + (10 – 4)2

= 64 + 36

= 100

= 10 units

(b) Area of ΔABC

= 12

⎪ 4 10 2 4 ⎪⎪ 2 10 6 2 ⎪

= 12

(40 + 60 + 4 – 20 – 20 – 24)

= 12

(40)

= 20 unit2

15 2y = 4x + 7

y = 2x + 72

The gradient of the line is 2.

Midpoint = �–2 + 62

, 7 + (–9)2 �

= (2, –1)

Then the required equation is y + 1 = 2(x – 2) y + 1 = 2x – 4 y = 2x – 5

16 (a) Let P(x, y) be the moving point, AP = BP

(y–0)2 + (x–1)2 = (y – 7)2 + (x – 6)2

y2 + x2 – 2x + 1 = y2 – 14y + 49 + x2 – 12x + 36

y

O

y = f –1(x)

x

y = x(3, 6)

(6, 3)2

2–4

–4

x4

1—2

–



1 – 2x = 85 – 12x – 14y 10x + 14y – 84 = 0 5x + 7y – 42 = 0

(b) At y-axis, x = 0 5(0) + 7y – 42 = 0 7y = 42 y = 6

∴ C(0,6)

17 M = 40.5 + �15 – 139 �20

= 40.5 + 4.44 = 44.94

18 (a) Range = 15 – 3 = 12

(b) x = 405

= 8

σ = 4045

–(8)2

= 16.8 = 4.099

19 (a) The new mean = 4(9) + 2 = 38

(b) The new variance = 42�3 12 �

= 56

20 (a) tan ∠AOC = 106

= 53

∠AOC = 59° 2' = 1.03 radians

(b) sAB

= 6(1.03) = 6.18 cm

OC = 62 + 102

= 11.662

∴ Perimeter of the shaded region = 6.18 + 10 + (11.662 – 6) = 21.842 cm

21 (a) sAB

= 10�8π5 �

= 16π cm/50.265 cm

(b) Area of sector AOB

= 12

(10)2 �2π5 �

= 20π = 62.832 cm2

Area of ΔAOB

= 12

(10)2 sin �2π5 �

= 47.553 cm2

∴ Area of the shaded region = 62.832 – 47.553 = 15.28 cm2

22 lim x(x + 10) =

lim x + 10 x→0 x(x2 + 2) x→0 x2 + 2

= 102

= 5

23 (a) y = 116

(3x–4)5

dydx

= 516

(3x – 4)4 (3)

= 1516

(3x – 4)4

(b) When x = 2,

dydx

= 1516

[3(2) – 4]4

= 15

and y = 116

[3(2) – 4]5

= 2

∴ Equation of the normal:

y – 2 = – 115

(x – 2)

15y – 30 = –x + 2 15y + x = 32

24 y = x– 1–

3

dydx

= – 13

x– 4–

3

= –1

3x4–3

When x = 8, δx = 8.2 – 8 = 0.2

y = 12

and dydx

= – 148

∴ 13 8.2

= 12

+ �– 148 � (0.2)

= 0.4958

25 y = x + 5x–2

dydx

= 1 – 10x3

By the chain rule,dydt

= dydx

× dxdt

1.5 = �1 – 10x3 � dx

dt

When x = 2, 1.5 = –0.25 dxdt

dxdt

= –6

∴ x changes at a rate of –6 units s–1.

Paper 2Section A/Bahagian A

1 3y + 2x = 14 3y = 14 – 2x

y = 14 – 2x

3 … 1

y = 4x

… 2


14 – 2x3

= 4x

14x – 2x2 = 12

2x2 – 14x + 12 = 0 x2 – 7x + 6 = 0 (x – 1)(x – 6) = 0 x = 1 or x = 6

Substitute x = 1 into 2 : y = 4

Substitute x = 6 into 2 :

y = 46

= 23

∴ x = 1, y = 4; x = 6, y = 23

∴ P(1, 4); Q�6, 23 �

2 (a) f(–1) = 2 g(2) = 3

8(–1) + m = 2 n

2 + 6 = 3

m = 10

n = 24 ∴ m = 10, n = 24

(b) f(x) = 8x + 10 y = 8x + 10

x = y – 108

f –1(y) = y – 10

8

(c) gf(x) = g(8x + 10)

= 248x + 10 + 6

= 248x + 16

= 248(x + 2)

= 3x + 2

, x ≠ – 2

3 (a) (i) a = 1 (ii) (1, –4) (iii) x = 1

(b)

4 (a) LN10

= tan 1.2

LN = 10 tan 1.2 = 25.72 cm

ON = 102 + (25.72)2

= 27.6 cm s

LM = 10(1.2)

= 12 cm

∴ Perimeter of the shaded region = 25.72 + 12 + (27.6 – 10)= 55.32 cm

(b) Area of ΔLON

= 12

(10)(25.72)

= 128.6 cm2

x

y

y = –f(x)

3

3

O–1


Area of section LOM

= 12

(10)2(1.2)

= 60 cm2

∴ Area of the shaded region = 128.6 – 60 = 68.6 cm2

5 (a) (i) x = ΣxN

= 12010

= 12

(ii) σ2 = Σx2

N – (x)2

= 160010

– (12)2

= 16

(b) (i) 120 – x9

= 12

x = 12

(ii) σ = 1600 – (12)2

9 – (12)2

= 17.778 = 4.216

6 (a) y = x2 – 3x + 4

dydx

= 2x – 3

At point A(1, 2), dydx

= 2(1) – 3

= –1

∴ Equation of the tangenty – 2 = –1(x – 1)y – 2 = –x + 1y + x = 3

(b) At point B(3, 4), dydx

= 2(3) – 3

= 3

∴ Equation of the normal:

y – 4 = – 13

(x – 3)

3y – 12 = –x + 3 3y + x = 15

(c) y + x = 3 … 1 3y + x = 15 … 2 2 – 1 : 2y = 12

y = 6

Substitute y = 6 into 1 : 6 + x = 3 x = –3

∴ C(–3, 6)

Section B/Bahagian B

7 (a) 12

(x – 1)[2(x – 1) + x + 3] = 112

(x – 1)(3x + 1) = 224 3x2 – 2x – 1 = 224 3x2 – 2x – 225 = 0 (shown)

(b) x = 2 + (–2)2 – 4(3)(–225)2(3)

= 2 + 27046

= 2 + 526

or 2 – 52

6

= 9 or –253

(c) (i) AB = 2(9 – 1) = 16 cm

(ii)

tan ∠DAE = 84

= 2 ∠DAE = 63° 26' ∴ ∠BAD = 63° 26'

8 (a) 4x

2y = 2

22x–y = 21

2x – y = 1 … 1

log10

(2x + 2y) = 1 2x + 2y = 10 x + y = 5 … 2

1 + 2 : 3x = 6 x = 2

Substitute x = 2 into 1 : 2(2) – y = 1 y = 3 ∴ x = 2, y = 3

(b) (i) log4 � 1

m� = log4 1 – log

4 m

= –n

(ii) log28m =

log4 8m

log42

= log

48 + log

4 m

log42

= 3 log

42 + log

4 m

log42

= 3�1

2� + n

12

= 3 + 2n

(c) (i) By January 2012 ⇒ 4 years RM50 000(1.0425)4

= RM59 057.39

(ii) 50 000(1.0425)t � 80 000 t log

10 1.0425 � log

10 1.6

t � 11.29 ∴ 12 years. It will be in the

year 2020.

9 (a) At y-axis, x = 0 y – 2(0) = 2 y = 2

∴ P(0, 2)

D

A E

8 cm

4 cm

Let S be (x, 0),

mPS

= – 12

0 – 2x – 0

= –12

x = 4 ∴ S(4, 0) ∴ P(0, 2); S(4, 0)

(b) Equation of QR:

y – 5 = –12

(x – 9)

2y – 10 = –x + 9 2y + x = 19

(c) y – 2x = 2 … 1 2y + x = 19 … 2 2 × 2: 4y + 2x = 38 … 3 1 + 3 : 5y = 40

y = 8 Substitute y = 8 into 1 :

8 – 2x = 2 2x = 6 x = 3

∴ Q(3, 8)

(d) Area of PQRS

= 12

⎪ 0 4 9 3 0 ⎪⎪ 2 0 5 8 2 ⎪

= 12

(20 + 72 + 6 – 8 – 15)

= 12

(75)

= 37.5 unit2

10 (a)

(i) Median time = 13.75 (ii) Interquartile range = 17.5 – 10.5 = 7 (iii) The number of teenagers who

spent more than 20 hours = 100 – 87 = 13 teenagers

(b) x =

3(6) + 8(18) + 13(39) + 18(25) + 23(9) + 28(3)

100

= 1410

100 = 14.1

100

90

80

70

60

50

40

30

20

10

00.5 5.5 10.5 15.5 20.5 25.5 30.5 13.75 17.5


Time (hours)

Q3

Q1

Median

87


11 (a) (i) Volume of water in he hemispherical bowl

= 23

π(6)3

= 144π ∴ dV

dt = 144π

4 = 36π cm3 s–1

(ii)

rh

= 816

r = 12

h

Volume of water in the cone,

V = 13

πr2h

= 13

π�12

h�2h

= 112

πh3

dVdh

= 14

πh2

By the chain rule,

dVdt

= dVdh

× dhdt

36π = 14

πh2 × dhdt

When h = 8,

36π = 16π dhdt

∴ dhdt

= 2.25 cm s–1

(b) (i) 36x + 2y = 120 2y = 120 – 36x y = 60 – 18x

Area = 16xy + 12

(16x)(6x)

= 16x(60 – 18x) + 48x2

= 960x – 288x2 + 48x2

= 960x – 240x2

= 240x(4 – x) (shown)

(ii) For a maximum value of A,

dAdx

= 0

960 – 480x = 0 x = 2

and so d2Adx2

= –480 � 0

Hence, for the area, A to be maximum, x = 2 and

y = 60 – 18(2) = 24 ∴ x = 2, y = 24

Section C/Bahagian C

12 (a) (i) 15sin 82°

= 8sin ∠BAD

sin ∠BAD = 8 sin 82°15

= 0.5281 ∠BAD = 31° 53'

(ii) 82 = 52 + 72 – 2(5)(7) cos ∠ BCD

cos ∠BCD = 52 + 72 – 82

2(5)(7) = 0.1429 ∠BCD = 81° 47'

(iii) Area of ΔABD

= 12

(8)(15) sin 66° 7'

= 54.86 cm2

Area of ΔBCD

= 12

(5)(7) sin 81° 47'

= 17.32 cm2

∴ Area of ABCD = 54.86 + 17.32 = 72.18 cm2

(b) (i)

(ii) 8sin 81° 47'

= 7sin ∠BDC

sin ∠BDC = 7 sin 81° 47'8

= 0.8660 ∠BDC = 60° ∴ ∠BD'C = 180° – 60° = 120°

13 (a) (i) BC2 = 72 + 12.22 – 2(7)(12.2) cos 60°

= 112.44

BC = 112.44 = 10.604 cm

(ii) 10.604sin 120°

= 3.5

sin ∠CBD

sin ∠CBD = 3.5 sin 120°

10.604 = 0.2858 ∠CBD = 16° 37'

(b) (i)

(ii) ∠ BCD = 180° – 120° – 16° 37' = 43° 23'

Area of ΔBCD

= 12

(3.5)(10.604) sin 43° 23'

= 12.75 cm2

∠CDC' = 180° – 2(43° 23') = 93° 14'

Area of ΔCDC'

= 12

(3.5)(3.5) sin 93° 14'

= 6.12 cm2

∴ Area of ΔBC'D = 12.75 – 6.12 = 6.63 cm2

14 (a) x

1.50 × 100 = 120

x = 120 × 1.50

100 = 1.80

4.20y

× 100 = 105

y = 4.20 × 100105

= 4

z = 7.005.00

× 100

= 140 ∴ x = 1.80, y = 4, z = 140

(b) –I

10/08 =

125(4) + 120(8) + 112(6) + 105(2) + 140(9)

29

= 360229

= 124.21

(c) P

10

520 × 100 = 124.21

P10

= 124.21 × 520100

= RM645.89

(d) –I

12/08 = 124.21 × 0.9

= 111.79

15 (a) x = 0.500.40

× 100

= 125

2.20

y × 100 = 110

y = 2.20 × 100110

= 2.00

z

4.00 × 100 = 90

z = 90 × 4.00100

= 3.60 ∴ x = 125, y = 2.00, z = 3.60

(b) (i) –I

11/09 =

125(9) + 110(2) + 150(10) + 90(15)

36

= 419536

= 116.53

(ii) P

11

1.20 × 100 = 116.53

P11

= 116.53 × 1.20100

= RM1.40

(iii) –I

12/09 = 116.53 × 1.2

= 139.84

h

r

8

16

B

C'

C

D

3.5 cm

B

CD

D'

5 cm

7 cm

12 Progressions

1

Booster Zone

1 (a) T1 = 1—

2 [4(1) – 3]

= 1—2

T2 = 1—

2[4(2) – 3]

= 5—2

(b) d = 5—2

– 1—2

= 2

2 T4 = 27

a + 3d = 27 6 + 3d = 27 3d = 21 d = 7∴ p = 6 + 7 = 13and q = 13 + 7 = 20∴p = 13, q = 20

3 (a) a + 4d = 28 … 1 a + 19d = 103 … 2 2 – 1 : 15d = 75 d = 5 Substitute d = 5 into 1 : a + 4(5) = 28 a = 28 – 20 = 8 ∴ a = 8, d = 5(b) T

10 = a + 9d

= 8 + 9(5) = 53

4 (a) k + 3 – (k–1) = 2k + 1 – (k + 3) 4 = k – 2 k = 6(b) The fi rst three terms are 5, 9 and

13 So a = 5 and d = 4 ∴ T

12 = a + 11d

= 5 + 11(4) = 49

5 (a) d = 12 – 7 = 5(b) T

8 = a + 7d

= 7 + 7(5) = 42

6 4(2a + 7d) = 24 2a + 7d = 6 … 1

9(2a + 7d) = 90 2a + 17d = 10 … 2

2 – 1 : 10d = 4

d = 2—5

Substitute d = 2—5

into 1 :

2a + 7� 2—5 � = 6

2a = 16—–5

a = 8—5

∴ T15

= a + 14d

= 8—5

+ 14� 2—5 �

= 36—–5

7 a + 5d = 27 … 1

a + 13d = 59 … 2

2 – 1 : 8d = 32 d = 4Substitute d = 4 into 1 : a + 5(4) = 27 a = 7∴ S

10 = 5[2(7) + 9(4)]

= 5(50) = 250

8 (a) n—2

(4 + 104) = 1134 54n = 1134 n = 21

(b) 21—–2

[2(4) + 20d] = 1134

8 + 20d = 108 20d = 100 d = 5

9 S10

= S15

– S10

2S10

= S15

2[5(24 + 9d)] = 15—–2

(24 + 14d)

240 + 90d = 180 + 105d 15d = 60 d = 4

∴ S15

= 15—–2

[24 + 14(4)]

= 15—–2

(80) = 600

10 3—2

(2a + 2d) = 21 2a + 2d = 14 … 1

S6 – S

3 = 57

3(2a + 5d) = 57 + 21 2a + 5d = 26 … 2

2 – 1 : 3d = 12 d = 4Substitute d = 4 into 1 : 2a + 2(4) = 14 2a = 6 a = 3∴ a = 3, d = 4

11 2(2a + 3d) = 62 2a + 3d = 31 … 1 5(2a + 9d) = 365 2a + 9d = 73 … 22 – 1 :

6d = 42 d = 7Substitute d = 7 into 1 : 2a + 3(7) = 31 2a = 10 a = 5∴ T

6 = a + 5d

= 5 + 5(7) = 40

12 (a) T1 = S

1 = 2(1)2 + 3(1)

= 5(b) T

2 = S

2 – T

1

= 2(2)2 + 3(2) – 5 = 14 – 5 = 9 ∴ d = 9 – 5 = 4

13 3(2a + 5d) = 96 2a + 5d = 32 … 1

S10

= 1—3

S20

5(2a + 9d) = 1—3

[10(2a + 19d)] 15(2a + 9d) = 10(2a + 19d) 30a + 135d = 20a + 190d 10a – 55d = 0 2a – 11d = 0 … 21 – 2 :

16d = 32 d = 2Substitute d = 2 into 1 : 2a + 5(2) = 32 2a = 22 a = 11∴ T

10 = a + 9d

= 11 + 9(2) = 29



14 100, 104 , 117 , ..., 494 , 500

So 104 + (n–1) 13 = 494 13(n–1) = 30 n – 1 = 30 n = 31

15 (a) S500

= 500——2

(1 + 500)

= 125 250

(b) 1, 5 , 10 , ..., 495 , 500

So, 5 + (n – 1) 5 = 500 5(n – 1) = 495 n – 1 = 99 n = 100

S100

= 100—–2

[2(5) + 99(5)]

= 100—–2

(505) = 25 250∴ The sum of all integers between

1 and 500 which are not the multiples of 5.

= 125 250 – 25 250= 100 000

16 a + 3d = 9 … 1 2(2a + 3d) = 21 2a + 3d = 10.5 … 22 – 1 : a = 1.5

Substitute a = 1.5 into 1 : 1.5 + 3d = 9 3d = 7.5 d = 2.5∴ S

8 = 4 [2(1.5) + 7(2.5)]

= 4(20.5) = 82

17 Sn � 230

n—2

[2(5) + (n–1) 4] � 230

n—2

(4n + 6) � 230

2n2 + 3n – 230 � 0 (2n + 23)(n – 10) � 0

So, n � 10∴ The least value of n is 11

18 (a) n—2

(8 + 52) = 360

30n = 360 n = 12

(b) 12—–2

[2(8) + 11d] = 360 16 + 11d = 60 11d = 44 d = 4 ∴ Angles of the other sides = 12°, 16°, 20°, 24°, 28°, 32°,

36°, 40°, 44°, 48°

19 (a) n—2

(1.2 + 3.6) = 12

2.4 n = 12 n = 5

(b) 5—2

[2(1.2) + 4d] = 12

5—2

(2.4 + 4d) = 12

2.4 + 4d = 4.8 4d = 2.4 d = 0.6 ∴ Length of the other sides = 1.8 cm, 2.4 cm, 3 cm

20 (a) a = 100 and d = –5 ∴ T

10 = 100 + 9(–5)

= 55

(b) S10

= 10—–2

[2(100) + 9(–5)]

= 5(155) = 775

21 T

2—–T

1

= 52x

—–5x

= 5x

T3—–

T2

= 53x

—–52x

= 5x

So, the sequence is a geometric progression with common ratio 5x.

22 27 + 27r + 27r2 = 21 27r2 + 27r + 6 = 0 9r2 + 9r + 2 = 0 (3r + 1)(3r + 2) = 0

r = – 1—3

or r = – 2—3

When r = – 1—3

,

x = 27�– 1—3 �

= –9

y = –9�– 1—3 �

= 3

When r = – 2—3

,

x = 27�– 2—3 �

= –18

y = –18�– 2—3 �

= 12∴ x = –9, y = 3 or x = –18, y = 12

23 (a) x——–x – 4

= 5x – 12———–x

x2 = (5x – 12)(x – 4) x2 = 5x2 – 32x + 48 4x2 – 32x + 48 = 0 x2 – 8x + 12 = 0 (x – 2)(x – 6) = 0 x = 2 or x = 6

(b) x – 2——–x + 1

=

1—2

x——–x – 2

x2 – 4x + 4 = 1—2

x2 + 1—2

x

2x2 – 8x + 8 = x2 + x

x2 – 9x + 8 = 0 (x – 1)(x – 8) = 0 x = 1 or x = 8

(c) x + 3——–x + 1

= x + 8——–x + 3

x2 + 6x + 9 = x2 + 9x + 8 3x = 1

x = 1—3

(d) x – 1——–x – 2

= 3x – 5——–x – 1

x2 – 2x + 1 = 3x2 – 11x + 10 2x2 – 9x + 9 = 0 (2x – 3)(x – 3) = 0 x = 3—

2 or x = 3

24 (a) T

1 = 32(1)

= 9 T

2 = 32(2)

= 81

(b) r = 81—–9

= 9

25 –2(–2)n–1 = 1024 (–2)n–1 = –512 (–2)n–1 = (–2)9

n–1 = 9 n = 10

26 3(2)n–1 � 600 (n – 1)log

102 � log

10 200

n – 1 � 7.644 n � 8.644∴ n = 9

27 ar3 = 24 … 1 ar 6 = 192 … 22 ÷ 1 :

r3 = 8 r = 2Substitute r = 2 into 1 : a(2)3 = 24 a = 3∴ T

13 = ar12

= 3(2)12

= 12 288 28 (a) 6h + k———

2h + k = 14h + k———

6h + k 36h2 + 12hk + k2 = 28h2 + 16hk + k2

8h2 = 4hk k = 2h(b) T

1 = 2h + 2h

= 4h T

2 = 6h + 2h

= 8h

r = 8h—–4h

= 2

29 (a) x + 4——–x

= 2x + 2——–x + 4

x2 + 8x + 16 = 2x2 + 2x x2 – 6x – 16 = 0

23– —– 2

10n


(x + 2)(x – 8) = 0 x = –2 or x = 8 ∴ x = 8 (�0)

(b) r = 12—–8

= 3—2

ar2 = 8 a� 3—

2 �2

= 8 a = 8� 4—

9 � = 32—–9

T6 = ar5

= 32—–

9 � 3—

2 �5

= � 32—–

9 ��243—–32 �

= 27

30 –2(–2)n–1 = 1024 (–2)n–1 = –512 (–2)n–1 = (–2)9

n–1 = 9 n = 10

31 ar – a = 4 a(r – 1) = 4 … 1 ar2 – ar = 16 ar(r – 1) = 16 … 2

2 ÷ 1 : r = 4

Substitute r = 4 into 1 : 3a = 4 a = 4—

3 ∴ T

5 = ar4

= 4—

3 (4)4

= 4—

3 (256)

= 341 1—

3

32 a + ar = 17 1—2

a(1 + r) = 35—–2

… 1 ar2 = 14—–

3 … 2

2 ÷ 1 : r 2

——1 + r

= � 14—–3 �� 2—–

35 � 105r2 = 28 + 28r 105r2 – 28r – 28 = 0 15r2 – 4r – 4 = 0 (5r + 2)(3r – 2) = 0

r = – 2—5

or r = 2—3

∴ r = 2—3

(�0)

33 a = 8 and r = 24—–8

= 3The sum from 8th term to 10th term= S

10 – S

7

= 8(310 – 1)———–

3 – 1 –

8(37– 1)———–

3 – 1= 236 192 – 8 744= 227 448

34 (a) T

1 = S

1 = 4 – 4—

21 = 2 T

2 = S

2 – T

1

= �4 – 4—–22 � – 2

= 1 r =

T2—–

T1

= 1—2

(b) T5 = ar4

= 2� 1—

2 �4

= 2� 1—–16 �

= 1—8

35 16�—–�p

q——– = ——–

4

16�—–�p

4q = 256——p2

q = 64—–p2

36 (a) ar4 = 32

2r4 = 32 r4 = 16 r4 = 24

r = 2 (b) S

8 = 2(28 – 1)————

2 – 1 = 510

37 (a) r = 12—–4

= 3(b) S

n = 13 120

4(3n – 1)

————3 – 1

= 13 120 3n – 1 = 6 560 3n = 6 561 3n = 38

n = 8

38 (a) T4 = 24

81r3 = 24

r3 = 8—–27

r3 = � 2—3 �3

r = 2—3

(b) S∞ =

81———1 –

2—3

= 243

39

a(1 – r3)———–

1 – r = 14 … 1

a——–1 – r

= 16 … 2 Substitute 2 into 1 : 16(1 – r3) = 14 1 – r3 = 7—

8

r3 = 1—8

r3 = � 1—2 �3

r = 1—

2 Substitute r = 1—

2 into 2 :

a———

1 – 1—2

= 16 a = 16� 1—

2 � = 8

40 (a) S∞ = 1———1 – 1—

3 = 3—

2

(b) S∞ =

1�—�2————–1 – 1�– —� 2

= 1—

2 ÷ 3—

2

= 1—3

(c) S∞ = 0.2———1 – 0.1

= 2—9

(d) S∞ = 24———1 – 1—

2 = 48

••41 (a) 0.15 = 0.15 + 0.0015 + 0.000015

+ ... S∞ = 0.15———–

1 – 0.01 = 0.15——

0.99

= 5—–33

••(b) 0.06 = 0.06 + 0.0006 + 0.000006

+ ... S∞ = 0.06———–

1 – 0.01

= 0.06——0.99

= 2—–33 •

(c) 2.4 = 2 + 0.4 + 0.04 + 0.004 + ...

= 2 + 0.4———1 – 0.1

= 2 + 4—9

= 2 4—9


••(d) 1.45 = 1 + 0.45 + 0.0045 + 0.000045 + ...

= 1 + 0.45———–1 – 0.01

= 1 + 5—–11

= 1 5—–11

42 ar3 = 8a r3 = 8 r = 2 S

4 = 360

a(24 – 1)————

2 – 1 = 360

15a = 360 a = 24∴ The area of the largest sector = 24°, 48°, 96°, 192°

SPM Appraisal Zone

1 (a) T6 = 2(6) + 7 = 19

(b) T5 = 2(5) + 7 = 17

d = T6 – T

5

= 19 – 17 = 2

2 T15

= 2T5

–4 + 14d = 2(–4 + 4d) –4 + 14d = –8 + 8d 6d = –4

d = – 2—3

T13

= a + 12d

= –4 + 12�– 2—3 �

= –12

3 (a) For an arithmetic progression a, a + 4, a + 8, ...

d = (a + 4) – a = 4 T

8 = 33

a + 7(4) = 33 a = 5(b) T

10 = a + 9d

= 5 + 9(4) = 41

4 T16

= 3T5

a + 15d = 3(a + 4d) a + 15d = 3a + 12d 2a – 3d = 0 … 1 T

12 – T

7 = 20

a + 11d – (a + 6d) = 20 5d = 20 d = 4Substitute d = 4 into 1 : 2a – 3(4) = 0 2a = 12 a = 6∴ a = 6, d = 4

5 (a) a + 4d = –4 … 1 a + 9d = 16 … 2

2 – 1 : 5d = 20 d = 5 Substitute d = 5 into 1 : a + 4(5) = –4 a = –24 ∴ a = –24, d = 5(b) T

20 = –24 + 19(5)

= 71

6 x + 3 – (x + 2) = 2x2 + 1 – (x + 3) 1 = 2x2 – x – 2 2x2 – x – 3 = 0 (2x – 3)(x + 1) = 0

x = 3—2

or x = –1

7 (a) a + 9d = 21 … 1

S30

– S20

= 675

30—–2

(2a + 29d) – 20—–2

(2a + 19d)

= 675 10a + 245d = 675 2a + 49d = 135 … 2

1 × 2: 2a + 18d = 42 … 3

2 – 3 : 31d = 93 d = 3 Substitute d = 3 into 1 : a + 9(3) = 21 a = –6 ∴ a = –6, d = 3

(b) S10

= 10—–2

[2(–6) + 9(3)]

= 5(–12 + 27) = 75

8 S10

= 80 2a + 9d = 16 … 1 S

22 – S

10 = 624

22—–2

(2a + 21d) – 80 = 624

22a + 231d = 704 2a + 21d = 64 … 22 – 1 : 12d = 48

d = 4

Substitute d = 4 into 1 : 2a + 9(4) = 16 2a = –20 a = –10∴ T

1 = –10

T2 = –10 + 4 = –6

T3 = –6 + 4 = –2

∴ The fi rst three terms is –10, –6 and –2.

9 (a) 2x + 6 – (x + 3) = 8 – (2x + 6) x + 3 = 2 – 2x 3x = –1

x = – 1—3

(b) The fi rst three terms of an arithmetic

progression are 8—3

, 16—–3

and 8 with

a = 8—3

and d = 8—3

∴ S8 = 8—

2 �2� 8—3 � + 7� 8—

3 �� = 4(24) = 96

10 (a) S6 = 6

3[2(–9) + 5d] = 6 –18 + 5d = 2 5d = 20 d = 4(b) S

n = 90

n—2

[2(–9) + (n – 1)(4)] = 90

n—2

(4n – 22) = 90

4n2 – 22n = 180 2n2 – 11n – 90 = 0 (2n + 9)(n – 10) = 0

n = – 9—2

or n = 10 ∴ The number of terms in the

arithmetric progression is 10.(c) T

10 = a + 9d

= –9 + 9(4) = 27

11 S20

= 140

20—–2

(3 + l) = 140

3 + l = 14 l = 11

12 (a) x + 3——–x + 1

= x + 8——–x + 3

x2 + 6x + 9 = x2 + 9x + 8 3x = 1

x = 1—3

(b) r =

1—3

+ 3———1—3

+ 1

= 21—2

13 ar3 = 3 … 1

ar7 = 1—–27

… 2

2 ÷ 1 : r 4 = 1—–81

r 4 = �± 1—3 �

4

r = ±

1—3

Substitute r = 1—3

into 1 :

a� 1—3 �

3

= 3

1—–27

a = 3 a = 81 Substitute r = – 1—

3 into 1 :

a�– 1—3 �3 = 3

– 1—–27

a = 3 a = –81

∴ a = –81, r = – 1—3

and a = 81, r = 1—3


14 (a) ar2 = 21—4

… 1

ar5 = – 2—3

… 2

2 ÷ 1 : r3 = – 8—–27

r3 = �– 2—3 �

3

r = – 2—3

Substitute r = – 2—3

into 1 :

a�– 2—3 �

2

= 9—4

4—9

a = 9—4

a = 81—–16

∴ a = 81—–16

, r = – 2—3

(b) T2 = ar

= � 81—–16 ��– 2—

3 � = – 27—–

8

15 a = 3—4

and r = 1—2

÷ 3—4

= 2—3

3—4 � 2—

3 �n – 1

= 4—–27

� 2—3 �

n–1

= 16—–81

� 2—3 �

n–1

= � 2—3 �

4

n – 1 = 4 n = 5

16 ar + ar2 = 30 ar (1 + r) = 30 … 1 ar3 = 27 … 2

2 ÷ 1 : r2

——–1 + r

= 9—–10

10r2 = 9 + 9r 10r2 – 9r – 9 = 0 (5r + 3)(2r – 3) = 0

r = – 3—5

or r = 3—2

17 For a geometric progression 2, 3, 9—2

, ...

a = 2 and r = 3—2

Sn � 30

2�� 3—

2 �n

– 1�—————–

3—2

– 1 � 30

� 3—2 �

n

– 1 � 71—2

n log10

� 3—2 � � log

10 �8 1—

2 � n � 5.278 n = 6

18 (a) a(ar6) = ar3

ar3 = 1 … 1 a + ar3 = 9 … 2 Substitute 1 into 2 :

a + 1 = 9 a = 8 Substitute a = 8 into 1 : 8r3 = 1

r3 = 1—8

r = 1—2

∴ a = 8, r = 1—2

(b) S4 =

8�1 – � 1—2 �

4

�—————

1 – 1—2

= 15

19 (a) T2 = S

2 – S

1

= 15�1 – 1—9 � – 15�1 –

1—3 �

= 131—3

– 10

= 31—3

(b) S∞ = 15 �1 – 1—3

∞� = 15

20

� 1—12�

——–1 – r

= 2—3

1—–12

= 2—3

(1 – r)

1—8

= 1 – r

r = 7—8

21 a——1 – r

= 9

a = 9(1 – r) … 1 ar = 2

a = 2—r

… 2

From 1 and 2 : 2—r = 9(1 – r)

2 = 9r – 9r2

9r2 – 9r + 2 = 0 (3r – 1)(3r – 2) = 0

r = 1—3

or r = 2—3


into 2 :

a = 2—–

� 1—3 �

= 6


into 2 :

a = 2—–

� 2—3 �

= 3

∴ a = 3, r = 2—3

and a = 6, r = 1—3

22 (a) 1 + 3h———1 + h

= 1 + 4h———1 + 3h

1 + 6h + 9h2 = 1 + 5h + 4h2

5h2 + h = 0

h(5h + 1) = 0

h = 0 or h = – 1—5

∴ h = – 1—5

(b) The fi rst 3 terms are 4—5

, 2—5

and 1—5

with a = 4—5

and r = 1—2

S∞ =

� 4—5 �

———1 – 1—

2 = 1

3—5

23 (a) ar2 = 22—3

… 1

ar5 = 8—–81

… 2

2 ÷ 1 r3 = 1—–27

r = 1—3


into 1 :

a� 1—3 �

2

= 8—3

a = � 8—3 �(9)

= 24

∴ a = 24, r = 1—3

(b) S∞ = 24———1 – 1—

3 = 36

24 2.555 ... = 2 + 0.5 + 0.05 + 0.005 + ... = 2 +

0.5———1 – 0.1

= 2 +

5—9

= 25—9

25 (a) Let PQ and ∠QPR be x cm and θ respectively.

A

1 =

1—2

x2 sin θ A

2 =

1—2

� x—2 �� x—

2 � sin θ

= 1—8

x2 sin θ A

3 =

1—2

� x—4 �� x—

4 � sin θ

= 1—–32

x2 sin θ

A2—–

A1

= A

3—–A

2

= 1—4

Since A

2—–A

1

= A

3—–A

2

, thus the areas of

triangles form a geometric

progression with common ratio 1—4

.


(b) S∞ =

1—2

x2 sinθ————

1 – 1—4

= 2—3

x2 sin θ

∴

2—3

x2 sinθ————1—2

x2 sinθ =

2—3

÷ 1—2

= 4—3

26 (a) a = 80 and r =

3—4

T4 = ar3

= 80� 3—

4 �3

= 33.75° (b) S∞ = 80———

1 – 3—4

= 320°

27 (a) h1 = 32� 3—

4 � = 24 cm h

2 = 24� 3—

4 � = 18 cm h

3 = 18� 3—

4 � = 13.5 cm The respective height of the ball

are as follows: 24 cm, 18 cm, 13.5 cm

h

2—h

1

= 18—–24

= 3—4

h

3—h

2

= 13.5——18

= 3—4

Since the ratios are equal, the height of the ball form a geometric progression with the common ratio

of 3—4

.

(b) The height of 6th bounce, T

6 = ar 5

= 24� 3—4 �

5

= 5.695 cm(c) The total distance traveled = 32 + 2(24) + 2(18) + 2(13.5) + ...

= 32 + 48———1 – 3—

4 = 32 + 192 = 224 cm

28 (a) A1 = x2

A2 = � x—

2 �� x—2 �

= 1—

4x2

A3 = � x—

4 �� x—4 �

= 1—–16

x2

A

3—–A

2

= A

2—–A

1

= 1—4

Since A

3—–A

2

= A

2—–A

1

, so the areas of

the squares form a geometric


.

(b) (i) A1 = 162

= 256 cm

T3 = 256� 1—

4 �2

= 16 cm2

∴ The area of the 3rd square is 16 cm2.

(ii) S∞ = 256———

1 – 1—4

= 341

1—3

cm

29 (a) P

1 = 3x

P

2 = 3—

2x

P

3 = 3—

4x

P

3—–P

2

= P

2—–P

1

= 1—2

Since

P3—–

P2

= P

2—–P

1

so the perimeters

of the triangles form a geometric


.

(b) (i) P1 = 3(64)

= 192 cm T

9 = 192 � 1—

2 �8

= 0.75 cm So, the length of the side of

the 9th triangle

= 0.75——

3 = 0.25 cm

(ii) S∞ = 192———1 – 1—

2 = 384 cm

30 (a) x——–

x + 5 =

x – 4——–x

x2 = x2 + x – 20 x = 20 (b) r =

20—–25

=

4—5

(c) S∞ =

25———1 – 4—

5 = 125 S

3 = 25 + 20 + 16

= 61 ∴ The difference is 125 – 61 = 64

31 (a) Sn = 126

n—2

[2(21) + (n – 1)(–1)] = 126

n—2

(43 – n) = 126 n(43 – n) = 252 n2 – 43n + 252 = 0 (n – 7)(n – 36) = 0 n = 7 or n = 36 ∴ n = 7(b) T

7 = a + 6d

= 21 + 6(–1) = 15

32 (a) S10

= 310 5(2a + 9d) = 310 2a + 9d = 62 … 1 S

3 = 114

3—2

(2a + 2d) = 114 2a + 2d = 76 … 2 1 – 2 : 7d = –14 d = –2 Substitute d = –2 into 1 : 2a + 9(–2) = 62 2a = 80 a = 40 ∴ a = 40, d = –2(b) T

n = 30

40 + (n – 1)(–2) = 30 2(n – 1) = 10 n – 1 = 5 n = 6 ∴ The 6th part has a length of

30 cm. (c) S

7 =

7—2

[2(40) + 6(–2)] =

7—2

(68) = 238 ∴ The sum of the last three parts = 310 – 238 = 72 cm

33 (a) S2 = 22 + 3(2)

= 10(b) T

2 = S

2 – S

1

= 10 – (1 + 3) = 6(c) d = T

2 – T

1

= 6 – 4 = 2

34 (a) Sn = 40

n—2

(2 + 14) = 40

n—2

(16) = 40 8n = 40 n = 5

(b) 5—2

[2(2)] + 4d] = 40 4 + 4d = 16 4d = 12 d = 3 ∴ The lengths of the other sides

= 5 cm, 8 cm, 11 cm

13 Linear Law

1

Booster Zone

1 (a) y = ax + b—x

y—x

= a + b—x2

= b � 1—x2 � + a

where Y = y—

x and X = 1—

x2 ,

and the gradient, m = bthe Y-intercept = a

(b) y = a—x + b x

y

—–x

= a——x x + b

= a� 1——x x � + b

where Y = y

—–x

and X = 1——x x

,

and the gradient, m = a the Y-intercept = b

(c) y = a—–x

+ b x

y—–

x = a—x + b

= a� 1—x � + b

where Y = y—x and X = 1—x and the gradient, m = a the Y-intercept = b (d) y = a—x + b—

x2

x2y = ax + b where Y = x2y and X = x, and the gradient, m = a the Y-intercept = b(e) y = abx

log10

y = log10

abx

= log10

a + log10

bx

= x log10

b + log10

a = (log

10 b)x + log

10 a

where Y = log10

y and X = x and the gradient, m = log

10 b

the Y-intercept = log10

a(f) y = axb

log10

y = log10

axb

= log10

a + log10

x b

= b log10

x + log10

a where Y = log

10y and X = log

10x

and the gradient, m = b the Y-intercept = log

10 a

(g) xy = a(x + b) xy = ax + ab where Y = xy and X = x, and the gradient, m = a the Y-intercept = ab

(h) y = ax2 + x + b y – x = ax2 + b where Y = y – x and X = x2, and the gradient, m = a the Y-intercept = b (i) y = x———

ax + b y(ax + b) = x

ax + b = x—y

x—y

= ax + b

where Y = x—y

and X = x,

and the gradient, m = a the Y-intercept = b

(j) y = a——–

x + b y(x + b) = a

x + b———a

= 1—y

1—y

= � 1—a �x + b—

a

where Y = 1—y

and X = x,

and the gradient, m = 1—a

the Y-intercept = b—a

(k) ax + by = xy

ax—–y

+ b = x

ax—–y

= x – b

x—y

= x—a

– b—a

= � 1—a �x – � b—

a � where Y = x—

y and X = x,

and the gradient, m = 1—a

the Y-intercept = – b—a

(l) y = ba–x

log10

y = log10

ba–x

= log10

b + log10

a–x

= –x log10

a + log10

b = (–log

10 a)x + log

10 b

where Y = log10

y and X = x, and the gradient, m = –log

10 a

the Y-intercept = log10

b

2 (a) m = 5 – 1——–3 – 5

= – 4—2

= –2 Y = –2X + c

The line passes through point (5, 1).

1 = –2(5) + c c = 11 The equation of the straight line

is Y = –2X + 11

y

—–x

= –2x2 + 11

y = –2x5—2 + 11 x

(b) m =

1 – (–3)———–

1 – 5 = – 4—

4 = –1 Y = – X + c The line passes through point (1, 1). 1 = –1 + c c = 2 Y = –X + 2

y—x

= –x + 2

y = –x2 + 2x

(c) m = 5 – 2——–2 – 1

= 3 Y = 3X + c The line passes through point (1, 2). 2 = 3(1) + c c = –1 The equation of the straight line

is Y = 3X – 1 log

10y = 3log

10(x + 1) – 1

= 3log10

(x + 1) – log10

10

= log10

(x + 1)3

———10

y = (x + 1)3

———10

(d) m = 3 – 0——–0 – 2

= – 3—2

Y-intercept, c = 3

Y = – 3—2

X + 3

1—y

= – 3—2

� 1—x � + 3

= –3 + 6x———–2x

y = 2x———

6x – 3

3 (a) m = 4 – 1——–4 – 2

= 3—2



The line passes through point (2, 1).

Y = 3—2

X + c

1 = 3—2

(2) + c

c = –2

The equation of the straight line is

Y = 3—2

X – 2 xy = 3—

2x – 2

y = – 2—

x + 3—

2 (b) y = – 2—–

(4) + 3—

2 = 1

4 m = 14 – (–1)————4 – 1

= 15—–3

= 5The line passes through point (1, –1). Y = 5X + c –1 = 5(1) + c c = –6∴ y = 5X – 6

y—x

= 5 x – 6

y—–(1)

= 5 (1) – 6 y = – 1

5 (a) m = 4 – 1——–6 – 0

= 3—6

= 1—2

Y-intercept, c = 1

∴ Y = 1—2

X + 1

log10

y = 1—2

log10

x + 1 2 =

1—2

log10

x + 1

1—2

log10

x = 1 log

10 x = 2

x = 102

= 100

(b) log10

y = 1—2

log10

x + 1

= log10

x1—2 + log

1010

= log10

10x1—2

y = 10 x

6 y = Abx

log10

y = log10

Abx

= log10

A + log10

bx

= (log10

b)x + log10

A

m = 4 – 2——–3 – 2

= 2Y = 2X + log

10A

The line passes through point (2, 2). 2 = 2(2) + log

10A

log10

A = –2 A = 10–2

= 0.01Compare log

10Y = (log

10 b)x + log

10 A

with Y = 2X + log10

A log

10 b = 2

b = 102

= 100

7 m =

1—5——

1– — 2

= – 2—5

Y = – 2—5

X + 1—5

1—y

= – 2—5

� 1—x � + 1—

5

= – 2—–5x

+ x—–5x

5x = –2y + xyxy = 5x + 2yCompare to xy = px + qy∴ p = 5, q = 2

8 m = 2 – 0———

12 – 6

= 2—6

= 1—3

Y = 1—3

X + c

The line passes through point (6, 0). 0 = 1—

3(6) + c

c = –2The equation of the straight line is Y = 1—

3 X – 2

xy = 1—

3x2 – 2

y = 1—

3x – 2—

x 3y = x – 6—x Compare to ky = x + h—

x ∴ k = 3, h = –6

9 2y2 + 6y = x y + 3 = x—–

2y y = 1—

2 � x—y � – 3

where c = –3∴ p = –3

The line passes through point (q, –1).

Y = 1—2

X – 3

–1 = 1—2

q – 3

1—2

q = 2

q = 4∴ p = –3, q = 4

10 y = ax b

log10

y = log10

axb

= log10

a + log10

xb

= b log10

x + log10

a

b = –(–1)——

3 = 1—

3 log

10a = –1

a = 10–1

= 0.1 ∴ a = 0.1, b = 1—

3

11 y (2x + 1) = 3x

2x + 1———

3x = 1—

y

1—y

= 1—3

� 1—x � + 2—

3

The equation of the straight line is

Y = 1—3

X + 2—3

q = 1—3

(7) + 2—3

= 7 + 2——–3

= 3

2 = 1—3

p + 2—3

1—3

p = 2 – 2—3

= 4—3

p = 4∴ p = 4, q = 3

12 xy = – 1—4

x2 + 4 Y = – 1—

4 X + 4

where c = 4 ∴ p = 4The line passes through point (q, 2).

2 = – 1—4

q + 4 1—4

q = 4 – 2 q = 8∴ p = 4, q = 8

13 (a)

x 0.5 1.5 2.5 3.5 4.5 5.5

y—x 20.5 17.5 14.5 11.5 8.5 5.5

(b) y = ax + bx2

y—x = bx + a


(i) a = 22 (intercept on y-axis)

(ii) b = 19 – 7——–1 – 5

= 12—––4

= –3 (iii) When x = 3

y

—3

= –3(3) + 22

y = 13(3) = 39

14 (a) x 1 2 3 4 5

y x 1.50 4.95 8.49 12.20 15.43

(b) y x = ax + b From the graph,

a = 12.2 – 2———–4 – 1.2

= 10.2——2.8

= 3.64 b = –2.2 ∴ a = 3.64, b = –2.2

15 (a) x 1 3 4 7 9

log10 y 1.15 1.41 1.56 1.99 2.24

(b) y = ab x

log10

y = log10

ab x

= log10

a + log10

b x

= (log10

b)x + log10

a log

10 a = 1 (intercept on log

10 y-axis)

a = 10

log10

b = 2.20 – 1.15—————8.4 – 1

= 1.05——7.4

= 0.142 b = 100.142

= 1.3868 ∴ a = 10, b = 1.3868

SPM Appraisal Zone

Paper 1

1 m = 2 – 4——–6 – 0

= – 2—6

= – 1—3

Y = – 1—3

X + 4 xy = – 1—

3x2 + 4

y = – 1—3

x + 4—x

2 (a) m = 11 – 3———5 – 1

= 8—4

= 2 Y = 2X + c 3 = 2(1) + c c = 1 Y = 2X + 1 y = 2� 1—

x � + 1 (b) 5 = 2� 1—

x � + 1 5 – 1 =

2—x

x = 2—4

= 1—2

3 (a) qy2 = –px3 + 1 y2 = –

p—q x3 + 1—q

Y = 2—

3X + c

6 = 2—3

(3) + c c = 6 – 2 = 4 Y = 2—

3 X + 4

y2 = –

p—q x3 + 1—q

1—q = 4

q = 1—4

–p

—–

� 1—4 �

= 2—3

–p = 1—

6 p = – 1—

6 ∴ p = – 1—

6, q = 1—

4 (b) Y = 2—

3X + 4

2 = 2—3

k + 4 2—

3k = –2

k = –3

4 (a) y(2x + k) = h

1—y

= 2x + k———h

1—y

= � 2—h �x +

k—h

2—h

= 6 – 2——–8 – 0

= 4—8

= 1—2

h = 4

k—h

= 2

k—4

= 2

k = 8 ∴ h = 4, k = 8

(b) 4 – 2——–k – 0

= 1—2

2—k

= 1—2

k = 4

5 y = –x2 + 2x

y

—x = –x + 2

Y = –X + 2The line passes through point (1, p). p = –(1) + 2 = 1The line passes through point(q, –3). –3 = –(q) + 2 q = 5∴ p = 1, q = 5

y—x

25

20

15

10

5

01 2 3 4 5 6

x

(5, 7)

(1, 19)22

13

y x

16

14

12

10

8

6

4

2

0

–2

–4

1 2 3 4 5 6x

(1.2, 2)

–2.2

(4, 12.2)

log10

y

2.5

2.0

1.5

1.0

0.5

02 4 6 8 10 12

x

(1, 1.15)

(8.4, 2.20)


6 4x + 3y = 9xy

4x—y

+ 3 = 9x

4x—y

= 9x – 3

1—y

= – 3—4

� 1—x � + 9—4

Y = – 3—4

X + 9—4

The line passes through point �p, 3—2 �.

3—2

= – 3—4

(p) + 9—4

3—4

p = 9—4

– 3—2

p = 1

The line passes through point (2, q).

q = – 3—4

(2) + 9—4

= 3—4

∴ p = 1, q = 3—4

7 log10

y = log10

axb

= log10

a + log10

xb

= b log10

x + log10

a

b = 1 – 3——–2 – 1

= –2Y = –2X + cThe line passes through point (2, 1). 1 = –2(2) + c c = 5log

10 a = c

= 5 a = 105

= 100 000∴ a = 100 000, b = –2

8 xy = h� x—y � – k

h = 9 – 1——–3 – 1

= 8—2

= 4Y = 4X + cThe line passes through point (1, 1). 1 = 4(1) + c c = –3∴ –k = c = –3 k = 3∴ h = 4, k = 3

9 y x = –px + q

–p = 2 – 8——–4 – 1

= – 6—3

p = 2Y = 2X + cThe line passes through point (1, 8). 8 = 2(1) + c c = 6 q = c = 6∴ p = 2, q = 6

10 xy = –x + h

k – 0——–1 – 3

= –1

k = 2Y = – X + cThe line passes through point (3, 0). 0 = –(3) + c c = 3 h = c = 3∴ h = 3, k = 2

Paper 211 (a) x 1 2 3 4 5

y x 3.1 11.5 19.1 26.4 34.2 (b) y = a x + b—–

x y x = ax + b

(i) a = 26.5 – (– 3.5)——————4 –0

= 7.5 (ii) b = –3.5

12 (a) 1—x 1.00 0.50 0.33 0.25 0.20 0.17

1—y 1.85 1.18 0.89 0.75 0.67 0.61

(b) b—y

= 1 – a—x

1—y

= �– a—b �� 1—

x � + 1—b

1—b

= 0.38 b =

1——0.38

= 2.6316

–a = 1.85 – 0.38—————1 – 0

a = –3.8684

13 (a) x 1.0 1.5 2.0 2.5 3.0

y—x2 2.00 3.56 5.00 6.56 8.11

(b) y

—x2

= ax + b

(i) a = 6.56 – (–1.15)——————2.5 – 0

= 3.084 (ii) b = –1.15 (iii) From the graph, when x = 0.5

y—x2

= 0.4

y = (0.5)2(0.4) = 0.1

14 (a) x x 0.59 1.84 3.04 4.94 7.41

y x 8.62 7.10 5.65 3.41 0.39

y—x2

8

7

6

5

4

3

2

1

0

–1

0.5 1.0 1.5 2.0 2.5 3.0 3.5x

–1.15

y x

10

9

8

7

6

5

4

3

2

1

01 2 3 4 5 6 7 8

x x

y x

40

35

30

25

20

15

10

5

0

–5

1 2 3 4 5 6–3.5

x

(4, 26.5)

1—y

2.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

00.2 0.4 0.6 0.8 1.0 1.2

1—x

0.38

1.1

(1, 1.85)

(2.5, 6.56)

9.35

7.75


(b) From the graph, y x = ax x + b

(i) a = 0 – 9.35———–

7.75 – 0 = –1.2065 (ii) b = 9.35

15 (a)

x 0.5 1.0 1.5 2.0 2.5 3.0

log10 y 0.792 0.672 0.568 0.462 0.342 0.230

(b) log10

y = log10

ab x

= log10

a + log10

b x

= (log10

b)x + log10

a

From the graph, (i) log

10 a = 0.9

a = 100.9

= 7.9433 (ii) log

10b = 0.9 – 0.23————–

0 – 3 = –0.2233 b = 10–0.2233

= 0.598 (iii) When x = 2.7 log

10 y = 0.295

y = 100.295

= 1.9724

log10

y

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00.5 1.0 1.5 2.0 2.5 3.0 3.5

x

(3, 0.23)

14 Integration

Booster Zone

1 (a) �2x5 dx

= 2x6

—–6

+ c

= 1—3

x6 + c

(b) � 1—3

x dx

= x3—2

——–3� 3—

2 � + c

= 2—9

x3—2 + c

(c) � – 1—x2

dx

= �–x–2 dx

= –x–1

——–1

+ c

= 1—x + c

(d) � 1—–5x6

dx

= � 1—5

x–6 dx

= x–5

——–5(–5)

+ c

= – 1——25x5

+ c

(e) � 4—x

dx

= �4x 1– — 2 dx

= 4x1—2

——1—2

+ c

= 8 x + c

(f) �4x4 dx

= 4x5

——5

+ c

2 (a) �(2 – 1—x

) dx

= �(2 – x 1– — 2 ) dx

= 2x – x1—2

——1—2

+ c

= 2x – 2 x + c

(b) �(3x5 – 4 x ) dx

= (3x5 – 4x1—2 ) dx

= 3x6

——6

– 4x3—2

——3—2

+ c

= 1—2

x6 – 8—3

x3—2 + c

(c) �(6x2 + 4—x2

) dx

= 6x3

—–3

+ 4x–1

—–––1

+ c

= 2x3 – 4—x

+ c

(d) �(1 – 2x + 3x2) dx

= x – 2x2

——2

+ 3x3

——3

+ c

= x – x2 + x3 + c

(e) �(x3 + 1—–x3

) dx

= x4

—–4

+ x–2

—––2

+ c

= 1—4

x4 – 1——2x2

+ c

(f) �(3 – x ) dx

= 3x – x3—2

——3—2

+ c

= 3x – 2—3

x3—2 + c

3 (a) �x2 – 1——–x2

dx

= �(1 – x–2) dx

= x – x–1

——–1

+ c

= x + 1—x + c

(b) � x4 + 7x——–—x3

dx

= �(x + 7x–2) dx

= x2

—–2

+ 7x–1

——–1

+ c = x2

—–2

– 7—x + c

(c) �(2x – 1—2

)2 dx

= �(4x2 – 2x + 1—4

) dx

= 4x3

——3

– 2x2

——2

+ 1—4

x + c

= 4—3

x3 – x2 + 1—4

x + c

(d) �(2 – 3x)2 dx

= �(4 – 12x + 9x2) dx

= 4x – 12x2

——2

+ 9x3

——3

+ c

= 4x – 6x2 + 3x3 + c

(e) �(1 + x)(3 + 2x) dx

= �(3 + 5x + 2x2) dx

= 3x + 5x2

——2

+ 2x3

——3

+ c

(f) �(1 – x)(1 – x ) dx

= �(1 – x – x + x3—2 ) dx

= x – x3—2

——3—2

– x2

—–2

+ x5—2

——5—2

+ c

= x – 2—3

x3—2 – 1—

2x2 + 2—

5x

5—2 + c

(g) � x ( x + 5) dx

= �(x + 5 x ) dx

= x2

—–2

+ 5x3—2

——3—2

+ c

= 1—2

x2 + 10—–3

x3—2 + c

(h) � x (x + 2—x

) dx

= �(x3—2 + 2x

1– — 2 ) dx

= x5—2

——5—2

+ 2x1—2

——1—2

+ c

= 2—5

x5—2 + 4 x + c

(i) �(2 – x)(2 + x)—————–x2

dx

= ��4 – x2

——–x2 � dx

= �(4x–2 – 1) dx

= 4x–1

—–––1

– x + c

= – 4—x – x + c

(j) �� x – 2——–x3 � dx

= �(x –2 – 2x–3) dx

= x –1

—––1

– 2x –2

—–––2

+ c

= – 1—x + 1—–x2 + c



(k) �(3x – x )2 dx

= �(9x2 – 6x3—2 + x) dx

= 9x3

—–3

– 6x5—2

——5—2

+ x2

—2

+ c

= 3x3 – 12—–5

x5—2 + 1—

2x2 + c

(l) �(x – 2 x )2 dx

= �(x2 – 4x3—2 + 4x) dx

= x3

—3

– 4x5—2

——5—2

+ 4x2

—–2

+ c

= 1—3

x3 – 8—5

x5—2 + 2x2 + c

4 (a) �(3x – 5)4 dx

= (3x – 5)5

———–5(3)

+ c

= (3x – 5)5

———–15

+ c (b) �(1 – x)5 dx

= (1 – x)6

———–6(–1)

+ c

= – 1—6

(1 – x)6 + c

(c) �(2 – 3x)2 dx

= (2 – 3x)3

———–3(–3)

+ c

= – 1—9

(2 – 3x)3 + c

(d) � 1 – 2x dx

= �(1 – 2x)1—2 dx

= (1 – 2x)3—2

————–3—2

(–2) + c

= – 1—3

(1 – 2x)3—2 + c

(e) � 4x + 5 dx

= (4x + 5)3—2

————–3—2

(4) + c

= 1—6

(4x + 5)3—2 + c

(f) � 4———–(x + 2)3

dx

= �4(x + 2)–3 dx

= 4(x + 2)–2

———–—–2

+ c

= –2———–(x + 2)2

+ c

(g) � 1———–(2x + 1)2

dx

= �(2x + 1)–2 dx

= (2x + 1)–1

———–––1(2)

+ c

= –1———––2(2x + 1)

+ c

(h) �� 3——––2x – 3 �3

dx

= �27(2x – 3)–3 dx

= 27(2x – 3)–2

———––—–2(2)

+ c

= –27———––4(2x – 3)2

+ c

(i) � –6——––(x – 2)5

dx

= �–6(x – 2)–5 dx

= –6(x – 2)–4

———––—–4

+ c

= 3———–—2(x – 2)4

+ c

(j) � 5———2 – 6x

dx

= �5(2 – 6x) 1– — 2 dx

= 5(2 – 6x)1—2

————–1—2

(–6) + c

= – 5—3

2 – 6x + c

5 (a) dy—–dx

= 2x – 5

y = �(2x – 5) dx = x2 – 5x + c At (4, –2) –2 = (4)2 – 5(4) + c –2 = –4 + c c = 2 ∴ y = x2 – 5x + 2

(b) dy—–dx

= 15x2 – 12

y = �(15x2 – 12) dx

= 5x3 – 12x + c At (1, 3), 3 = 5(1)3 – 12(1) + c 3 = –7 + c c = 10 ∴ y = 5x3 – 12x + 10

(c) dy—–dx

= x2(2x + 1)

y = �(2x3 + x2) dx

= 1—2

x 4 + 1—3

x3 + c

At (1, –1),

–1 = 1—2

+ 1—3

+ c

c = – 11—–6

∴ y = 1—2

x 4 + 1—3

x3 – 11—–6

(d) dy—–dx

= x2(x – 3)

y = �(x3 – 3x2) dx

= x4

—–4

– x3 + c

At (2, –6), –6 = 4 – 8 + c c = –2

∴ y = 1—4

x4 – x3 – 2

(e) dy—–dx

= 2x(x – 3)

y = �(2x2 – 6x) dx

= 2—3

x3 – 3x2 + c

At (3, 6), 6 = 18 – 27 + c c = 15

∴ y = 2—3

x3 – 3x2 + 15

(f) dy—–dx

= (3x – 2)2

y = (9x2 – 12x + 4) dx = 3x3 – 6x2 + 4x + c At (1, 2), 2 = 3 – 6 + 4 + c c = 1 ∴ y = 3x3 – 6x2 + 4x + 1

(g) dy—–dx

= 4———–(x + 2)2

y = �4(x + 2)–2 dx

= 4(x + 2)–1

———–—–1(1)

+ c

= –4——–x + 2

+ c

At (2, 7), 7 = –1 + c c = 8 ∴ y = –4——–

x + 2 + 8

(h) dy—–dx

= 2x + 3

y = (2x + 3)1—2 dx

= (2x + 3)3—2

————–3—2

(2) + c

= (2x + 3)3—2

————–3

+ c

At (3, 5),

5 = 93—2

—–3

+ c

5 = 9 + c c = –4

∴ y = (2x + 3)3—2

————–3

– 4

6 dy—–dx

= 3x2 + 2—–x2

y = �(3x2 + 2x–2) dx

= x3 + 2x–1

—–––1

+ c

= x3 – 2—x

+ c


Since (1, 3) lies on the curve,

3 = (1)3 – 2—–(1)

+ c

3 = –1 + cc = 4∴ y = x3 – 2—x + 4

7 dy—–dx

= 1 + 2x

y = � 1 + 2x dx

= (1 + 2x)3—2

————–3—2

(2) + c

= 1—3

(1 + 2x)3—2 + c

When x = 4 and y = 30, we have

30 = 1—3

(9)3—2 + c

30 = 9 + c c = 21

y = 1—3

(1 + 2x)3—2 + 21

Thus, when x = 0,

y = 1—3

(1) + 21

= 64—–3

8 (a) dy—–dx

= kx – 6

At (2, 1), dy—–dx

= 4

2k – 6 = 4 2k = 10 k = 5

(b) dy—–dx

= 5x – 6

y = �(5x – 6) dx

= 5x2

—–2

– 6x + c

At (2, 1),

1 = 5(2)2

—–—2

– 6(2) + c

1 = 10 – 12 + c c = 3 ∴ y = 5—

2x2 – 6x + 3

9 (a) dy—–dx

= p – x

At (2, 3), dy—–dx

= 0

p – 2 = 0 p = 2

(b) dy—–dx

= 2 – x

y = ��(2 – x) dx

= 2x – x2

—2

+ c

At (2, 3),

3 = 2(2) – (2)2

——2

+ c

3 = 4 – 2 + c c = 1

∴ y = 2x – 1—2

x2 + 1

10 (a) dy—–dx

= 3x2 + px + q

At (1, 0), dy—–dx

= 0

3(1)2 + p(1) + q = 0 p + q = –3 ... 1

At (–3, 32), dy—–dx

= 0

32 = 3(–3)2 + p(–3) + q

3p – q = –5 ... 2

1 + 2 : 4p = –8 p = –2 Substitute p = –2 into 1 : –2 + q = –3 q = –1

(b) dy—–dx

= 3x2 – 2x – 1

y = �(3x2 – 2x – 1) dx

= x3 – x2 – x + c At (1, 0) 0 = 1 –1 – 1 + c c = 1 ∴ y = x3 – x2 – x + 1

11 (a) 3

�� 1

�2 – 3—–x2 � dx

3

= �2x + 3—x � 1

= (6 + 1) – (2 + 3) = 2

(b) 4

�� 1

�x2 – 4 + 4—–x2 � dx

4

= � x3

—3

– 4x – 4—x � 1

= � 64—–3

– 16 – 1� – � 1—3

– 4 – 4� = 12

(c) 4

�� 1

(6x – 3 x ) dx

4

= �3x2 – 2x3—2 �

1

= (48 – 16) – (3 – 2) = 31

(d) 4

�� 2

�x3 – 10—–x2 � dx

4

= � 1—4

x4 + 10—–x �

2

= (64 + 5—2

) – (4 + 5)

= 57 1—2

(e)

4—3��

1(3x – 2) dx

= � 3—

2x2 – 2x�

4—3

1

= 3—2 � 16—–

9 � – 2� 4—3 � – � 3—

2 – 2�

= 1—2

(f) 9

�� 4

1—x

dx

9

= �2 x � 4

= 6 – 4 = 2

(g) 27

�� 1

3 x dx

27

= � 3—4

x4—3 �

1

= 243——4

– 3—4

= 60

(h) 2

�� 1

(3 – 2x2 + 4—–x3

) dx

2

= �3x – 2x3

—–3

– 2—–x2 �

1

= �6 – 16—–3

– 1—2 � – �3 – 2—

3 – 2�

= – 1—6

12 (a) 2

�� 1

2x – 1———x3

dx

= 2

�� 1

(2x–2 – x–3) dx

2

= � –2—–x + 1—–2x2 �

1

= �–1 + 1—8 � – �–2 + 1—

2 � = 5—

8

(b) 3

�� 2

(x – 2)(x – 3) dx

= 3

�� 2

(x2 – 5x + 6) dx

3

= � x3

—–3

– 5x2

—–2

+ 6x� 2

= �9 – 45—–2

+ 18� – � 8—3

– 10 + 12� = – 1—

6

(c) 2

�� 1

x2 + 1—–—–x2

dx

= 2

�� 1

(1 + x–2) dx

2

= �x – 1—x � 1

= �2 – 1—2 � – (1 – 1)

= 3—2


(d) 4

�� 1

2 + x—–—–

x2 dx

= 4

�� 1

(2x–2 + x–3—2 ) dx

4

= � –2—–x

– 2—–x �

1

= �– 1—2

– 1� – (–2 – 2)

= 2 1—2

(e) 3

�� 1

1—x2

(4x2 + 9) dx

= 3

�� 1

(4 + 9x–2) dx

3

= �4x – 9—x � 1

= (12 – 3) – (4 – 9) = 14

(f) 4

�� 1

�3x2 + 2 x—–—–—–

x2� dx

= 4

�� 1

(3 + 2x–3—2 ) dx

4

= �3x – 4—–x �

1

= (12 – 2) – (3 – 4) = 11

(g) –1

�� –2

�x2 + 2—x �2

dx

= –1

�� –2

(x4 + 4x + 4x–2) dx

–1

= � x5

—–5

+ 2x2 – 4—x � –2

= �– 1—5

+ 2 + 4� – �– 32—–5

+ 8 + 2� = 2 1—

5

(h) 4

�� 0

x (1 – x ) dx

= 4

�� 0

( x – x) dx

4

= � 2—3

x3—2 – 1—

2x2�

0

= 16—–3

– 8

= – 8—3

13 (a) 1

�� –1

1——–—(x + 3)2

dx

1

= � –1——–x + 3 �

–1

= – 1—4

– �– 1—2 �

= 1—4

(b)

4—3��

1(3x – 2)5 dx

= � (3x – 2)6

————18 �

4—3

1

= 32—–9

– 1—–18

= 3 1—2

(c) 2

�� 1

1———3x – 2

dx

2

= � 2—3

3x – 2 � 1

= 4—3

– 2—3

= 2—3

(d) 1

�� 0

1———–(3x + 2)2

dx

1

= � –1——–—–3(3x + 2)� 0

= – 1—–15

– �– 1—6 �

= 1—–10

(e) 5

�� 1

(3x + 1) 1–— 2 dx

5

= � 2—3

3x + 1� 1

= 8—3

– 4—3

= 4—3

(f) 4

�� 3

(5 – x)5 dx

4

= � –1—–6

(5 – x)6� 3

= – 1—6

– �– 32—–3 �

= 10 1—2

(g) 4

�� 0

2x + 1 dx

4

= � 1—3

(2x + 1)3—2 �

0

= 9 – 1—3

= 26—–3

(h) 3

�� 2

1——––(4 –x)4

dx

3

= � 1————3(4 – x)3 �

2

= 1—3

– 1—–24

= 7—–24

14 (a) 4

�� 0

f(x) dx = 4

�� 1

f(x) dx + 1

�� 0

f(x) dx

1

�� 0

f(x) dx = 5 – 8

= –3

(b) 4

�� 0

[ x + f(x)] dx

= 4

�� 1

x dx + 4

�� 1

f(x) dx

4

= � 2—3

x3—2 � + 8

1

= � 16—–3

– 2—3 � + 8

= 12 2—3

(c) 4

�� 0

[5f(x) – 2x] dx

= 5 4

�� 0

f(x) dx – 4

�� 0

2x dx

4 = 5(5) – �x2� 0

= 25 – 16 = 9

15 (a) 1

�� 3

f(x) dx = – 3

�� 1

f(x) dx

= –2

(b) 4

�� 1

f(x) dx = 4

�� 3

f(x) dx + 3

�� 1

f(x) dx

= 3 + 2 = 5

(c) 3

�� 1

f(x) dx + 3

�� 4

f(x) dx

= 2 – 3 = –1

16 (a) 3

�� 1

1—3

f(x) dx = 1—3

3

�� 1

f(x) dx

= 1—3

(3)

= 1

(b) 3

�� 1

[f(x) + x] dx

= 3

�� 1

f(x) dx + 3

�� 1

x dx

3

= 3 + � x2

—–2 �

1

= 3 + � 9—2

– 1—2 �

= 3 + 4 = 7

(c) 6

�� 1

[2 f(x) + x] dx

= 2 6

�� 1

f(x) dx + 6

�� 1

x dx

= 2 � 6

�� 3

f(x) dx + 3

�� 1

f(x) dx� +

6

� x2

—–2 �

1

= 2(4 + 3) + �18 – 1—2 �

= 14 + 35—–2

= 31 1—2

17 5

�� 1

[f(x) + kx] dx = 28

5

�� 1

f(x) dx + 5

�� 1

kx dx = 28

5

4 + � kx2

—–2 � = 28

1

25—–2

k – 1—2

k = 24

12k = 24 k = 2


18 (a) 5

�� 1

3—5

f(x) dx

= 3—5

5

�� 1

f(x) dx

= 3—5

(10)

= 6

(b) 1

�� 5

–2 f(x) dx

= –2 1

�� 5

f(x) dx

= –2(–10) = 20

(c) 3

�� 1

f(x) dx + 5

�� 3

[f(x) + 2] dx

= 3

�� 1

f(x) dx + 5

�� 3

f(x) dx + 5

�� 3

2 dx

5 = 10 + �2x� 3

= 10 + (10 – 6) = 14

19 y = x———

1 + 5x

dy—–dx

= 1 + 5x – 5x—————

(1 + 5x)2

= 1————

(1 + 5x)2 (shown)

3

�� 1

� 4———1 + 5x �2

dx = 16 3

�� 1

1————

(1 + 5x)2 dx

3

= 16 � x———1 + 5x �

1

= 16� 3—–16

– 1—6 �

= 1—3

20 y = x 6 + 3x2

dy—–dx

= 6 + 3x2 + 3x2

———–6 + 3x2

= 6 + 3x2 + 3x2

——————6 + 3x2

= 6 + 6x2

———–6 + 3x2

(shown)

6 5

�� 1

1 + x2

———–6 + 3x2

dx = �x 6 + 3x2 �5

1

5

�� 1

1 + x2

———–6 + 3x2

dx = 1—6

(45 – 3)

= 7

21 (a) Area = 3

�� 0

(x2 – 2x + 2) dx

3

= � x3

—–3

– x2 + 2x� 0

= 9 – 9 + 6 = 6 unit2

(b) Area = 3

�� 1

(x2 + 1) dx

3

= � x3

—–3

+ x� 1

= 9 + 3 – � 1—3

+ 1� = 10 2—

3 unit2

(c) Area = 3

�� 2

4—x2

dx

3

= � –4—–x �

2

= – 4—3

– (–2)

= 2 – 4—3

= 2—3

unit2

(d) Area = 3

�� 0

(x – 3)2 dx

3

= �(x – 3)3

———3 �

0

= –(–9) = 9 unit2

22 (a) Area

= 4

�� 1

(y – 3)2 + 2 dy

= 4

�� 1

(y2 – 6y + 11) dy

4

= � y3

—–3

– 3y2 + 11y� 1

= 64—–3

– 48 + 44 – � 1—3

– 3 + 11� = 9 unit2

(b) Area = 2

�� 0

y(y – 2) dy

2

= � y3

—–3

– y2� 0

= 8—3

– 4

= – 4—3

= 4—3

unit2

(c) Area = 27

�� 8

y1—3 dy

27

= � 3—4

y4—3 �

8

= 243——4

– 12

= 48 3—4

unit2

(d) Area = 3

�� 0

y2

—–16

dy

3

= � y3

—–48 �

0

= 9—–16

unit2

23 (a) y = x ... 1 y = 8x – x2 ... 2

Substitute 1 into 2 : x = 8x – x2

x2 – 7x = 0 x(x – 7) = 0 x = 0 or x = 7 Substitute x = 7 into 1 : y = 7 Area of A = 1—

2(7)(7)

= 24 1—2

unit2

At x-axis, y = 0 8x – x2 = 0 x2 – 8x = 0 x(x – 8) = 0 x = 0 or x = 8 Area of B

= 8

�� 7

(8x – x2) dx

8

= �4x2 – x3

—–3 �

7

= �256 – 512——3 � – �196 – 343——

3 � = 3 2—

3 unit2

∴ Area of combined region

= 24 1—2

+ 3 2—3

= 28 1—6

unit2

(b) y = 8x ... 1 y = 9 – x2 ... 2

Substitute 1 into 2 : 8x = 9 – x2

x2 + 8x – 9 = 0 (x + 9)(x – 1) = 0 x = –9 or x = 1 Substitute x = 1 into 1 , y = 8(1) = 8

Area of A = 1—2

(1)(8)

= 4 unit2

At x-axis, y = 0 9 – x2 = 0 x = ±3 Area of B

= 3

�� 1

(9 – x2) dx

3

= �9x – x3

—–3 �

1

= (27 – 9) – �9 – 1—3 �

= 9 1—3

unit2

∴ Area of combined region

= 4 + 9 1—3

= 13 1—3

unit2


(c) Area of A = 2 × 2 = 4 unit2

Area of B = 4

�� 2

8—x2

dx

4

= � –8—–x �

2

= –2 – (–4) = 2 unit2

∴ Area of combined region = 4 + 2 = 6 unit2

(d) At x-axis, y = 0 x(x – 1)(x – 4) = 0 x = 0 or x = 1 or x = 4 Area of A

= 1

�� 0

(x3 – 5x2 + 4x) dx

= � x4

—–4

– 5x3

—–3

+ 2x2�1

0

= 1—4

– 5—3

+ 2

= 7—–12

unit2

Area of B

= 4

�� 1

(x3 – 5x2 + 4x) dx

4

= � x4

—–4

– 5x3

—–3

+ 2x2� 1

= 64 – 320——3

+ 32 – � 1—4

– 5—3

+ 2� = –11 1—

4

= 11 1—4

unit2

Area of combined region

= 7—–12

+ 11 1—4

= 11 5—6

unit2

24 (a) y = 9 – x ... 1 y = x2 – 4x + 5 ... 2 Substitute 1 into 2 : 9 – x = x2 – 4x + 5 x2 – 3x – 4 = 0 (x + 1)(x – 4) = 0 x = –1 or x = 4 Substitute x = –1 into 1 : y = 9 – (–1) = 10 Substitute x = 4 into 1 : y = 9 – 4 = 5 Area of trapezium

= 1—2

(5)(10 + 5)

= 37 1—2

unit2

Area under the curve

= 4

�� –1

(x2 – 4x + 5) dx

4

= � x3

—–3

– 2x2 + 5x� –1

= 64—–3

– 32 + 20 – �– 1—3

– 2 – 5� = 16 2—

3 unit2

∴ Area of the shaded region

= 37 1—2

– 16 2—3

= 20 5—6

unit2

(b) y = 7 ... 1 y = 8x – x2 ... 2 Substitute 1 into 2 : 7 = 8x – x2

x2 – 8x + 7 = 0 (x – 1)(x – 7) = 0 x = 1 or x = 7 Area of square = 6 × 7 = 42 unit2


= 7

�� 1

(8x – x2) dx

7

= �4x2 – x3

—–3 �

1

= 196 – 343——3

– �4 – 1—3 �

= 78 unit2

∴ Area of the shaded region = 78 – 42 = 36 unit2

(c) Area of trapezium

= 1—2

(1)(1 + 2)

= 3—2

unit2


= 2

�� 1

1—x2

dx

2

= �– 1—x �

1

= – 1—2

– (–1)

= 1—2

unit2


= 3—2

– 1—2

= 1 unit2

(d) y = 8 ... 1 y = x2 – 8x + 20 ... 2

Substitute 1 into 2 : 8 = x2 – 8x + 20 x2 – 8x + 12 = 0 (x – 2)(x – 6) = 0 x = 2 or x = 6 Area of square = 4 × 8 = 32 unit2


= 6

�� 2

(x2 – 8x + 20) dx

6

= � x3

—–3

– 4x2 + 20x� 2

= 72 – 144 + 120 – � 8—3

– 16 + 40� = 21 1—

3 unit2


= 32 – 21 1—3

= 10 2—3

unit2

25 (a) k

�� 1

4—x2

dx = 2

k

�– 4—x � = 2 1

– 4—k

– (–4) = 2

4—k

= 2

k = 2

(b) k

� 2

3x(x – 2) dx = 4 k �x3 – 3x2� = 4 2

k3 – 3k2 – (8 – 12) = 4 k3 – 3k2 = 0 k2(k – 3) = 0 ∴ k = 3

(c) 3

�� k

3x2 dx = 26 3 �x3� = 26 k 27 – k3 = 26 k3 = 1 k = 1

(d) k

�� 1

[(y – 2)2 + 1] dy = 6

k

� y3

—–3

– 2y2 + 5y� = 6 1

k3

—3

– 2k2 + 5k – ( 1—3

– 2 + 5) = 6

k3

—3

– 2k2 + 5k – 28—–3

= 0

k3 – 6k2 + 15k – 28 = 0 (k2 – 2k + 7)(k – 4) = 0 ∴ k = 4

26 (a) Volume = π 4

�� 2

144———(x + 2)2

dx

4

= π� –144———x + 2 �

2

= π [–24 – (–36)] = 12π unit3

(b) Volume = π 1

�� –1

1———(x + 2)2

dx

1

= π� 1– ——– x + 2 �

–1

= π �– 1—3

– (–1)� = 2—

3 π unit2


(c) Volume = π 4

�� 1

(5 + 4x – x2) dx

4

= π�5x + 2x2 – x3

—–3 �

1

= π �20 + 32 – 64—–3

–

�5 + 2 – 1—3 ��

= 24π unit3

(d) Volume = π 4

�� 0

16x dx 4 = π�8x2� 0

= 128π unit3

27 (a) Volume

= π 4

� 3

(4 – y) dy

4

= π�4y – y2

—–2 �

3

= π �16 – 8 – �12 – 9—2 ��

= 1—2

π unit3

(b) Volume = π 4

�� 0

y—2

dy

4

= π � y2

—–4 �

0

= 4π unit3

(c) Volume = π 4

�� 2

(y – 2) dy

4

= π� y2

—–2

– 2y� 2

= π [8 – 8 – (2 – 4)] = 2π unit3

(d) Volume

= π 4

�� 0

(4y – y2)2 dy = π

4

�� 0

(16y2 – 8y3 + y4)dy

4

= π� 16y3

—––3

– 2y4 + y5

—5 �

0

= π �1024——

3 – 512 + 1024——

5 � = 34 2—–

15 π unit3

28 (a) π k

�� 1

8x dx = 32π k �4x2� = 32 1

4k2 – 4 = 32 4k2 = 36 k2 = 9 k = ±3 ∴ k = 3

(b) π 2

�� k

16———(3 – x)2

= 8π

2

� 16——–3 – x � = 8

k

16 – 16——–3 – k

= 8

16——–3 – k

= 8

2 = 3 – k k = 1

(c) π k

�� 1

4—–y2

dy = 3π

k

�– 4—y � = 3 1

– 4—k

– (–4) = 3

4—k

= 1

k = 4

(d) π 6

�� k

(6 – y) dy = 8π

6

�6y – y2

—–2 � = 8

k

36 – 18 – �6k – k2

—–2 � = 8

k2

—–2

– 6k + 10 = 0

k2 – 12k + 20 = 0 (k – 2)(k – 10) = 0 k = 2 or k = 10 ∴ k = 2

SPM Appraisal Zone

1 (a) �(2 – 3x)2 dx

= �(4 – 12x + 9x2) dx

= 4x – 6x2 + 3x3 + c

(b) 9

�� 4

1—x

dx

9 = �2 x �

4

= 6 – 4 = 2

2 (a) �x4 + 7x—–—–x3

dx

= �(x + 7x–2) dx

= x2

—2

+ 7x–1

——–1

+ c

= 1—2

x2 – 7—x

+ c

(b) 3

�� 0

1———1 + 5x

dx

3

= � 2—5

1 + 5x � 0

= 8—5

– 2—5

= 6—5

3 dy—–dx

= 4——–—(x + 2)2

y = �4(x + 2)–2 dx

= 4(x + 2)–1

——–——–1

+ c

= –4——–x + 2

+ c

Since (2, 7) lies on the curve

7 = –4—–4

+ c

c = 8

∴ y = –4——–x + 2

+ 8

4 y = x———1 + 5x

dy—–dx

= 1 + 5x – 5x—————(1 + 5x)2

= 1———–(1 + 5x)2

(shown)

3

�� 1

� 4———1 + 5x �

2

dx = 16 3

�� 1

1————(1 + 5x)2

dx

3

= 16 � x———1 + 5x �

1

= 16 � 3—–16

– 1—6 �

= 1—3

5 (a) (i) 1

�� 5

f(x) dx = – 5

�� 1

f(x) dx

= –4

(ii) 5

�� 1

2 f(x) dx = 2 5

�� 1

f(x) dx

= 2(4) = 8

(b) 5

�� 1

[f(x) + kx] dx = 28

5

�� 1

f(x) dx + 5

�� 1

kx dx = 28

5

4 + � kx2

—–2 � = 28

1

25—–2

k – 1—2

k = 24

12k = 24 k = 2

6 (a) y = x2

———2x – 1

dy—–dx

= 2x(2x – 1) – 2x2

——————–(2x – 1)2

= 2x2 – 2x————–(2x – 1)2

= 2x(x – 1)————–(2x – 1)2

(b) � 2x(x – 1)————–(2x – 1)2

dx = x2

———2x – 1

2

2 2

�� 1

x(x – 1)———––(2x – 1)2

= � x2

———2x – 1 �

1

2

�� 1

x(x – 1)———––(2x – 1)2

= 1—2 � 4—

3 – 1�

= 1—2 � 1—

3 �

= 1—6


7 (a) (i) 3

�� 0

1—6

f(x) dx = 1—6

3

�� 0

f(x) dx

= 1—6

(12)

= 2

(ii) 0

�� 3

[f(x) – x] dx

= 0

�� 3

f(x) dx – 0

�� 3

x dx

0

= –12 – � x2

—–2 �

3

= –12 – � –9—–2 �

= –7 1—2

(b) 3

�� 0

[f(x) + mx] dx = 39

3

�� 0

f(x) dx + 3

�� 0

mx dx = 39

3

12 + � mx2

—––2 � = 39

0

9—2

m = 27

m = 27� 2—9 �

= 6

8 dy—–dx

= x2 – 4—––—x2

y = �(1 – 4x–2) dx

= x + 4—x

+ c

Since (2, 7) lies on the curve,7 = 2 + 2 + cc = 3

y = x + 4—x

+ 3

When y = 8

8 = x + 4—x

+ 3

8x = x2 + 4 + 3x x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0x = 1 or x = 4

∴ The coordinates are (1, 8) and (4, 8).

9 (a) At point P, dy—–dx

= 5

4x – 3 = 5 4x = 8 x = 2 When x = 2, y = 2(2) = 4 ∴ P(2, 4)

(b) dy—–dx

= 4x – 3

y = �(4x – 3) dx

= 2x2 – 3x + c When x = 2 and y = 4, we have 4 = 2(2)2 – 3(2) + c c = 2 ∴ y = 2x2 – 3x + 2

10 dy—–dx

= x(2 + 3x)

y = �(2x + 3x2) dx

= x2 + x3 + cAt point (1, –3),–3 = 1 + 1 + cc = –5y = x2 + x3 – 5At point (2, k),k = (2)2 + (2)3 – 5 = 4 + 8 – 5 = 7

11 (a) 4

�� 2

�x3 – 8—x2 � dx

4

= � x4

—4

+ 8—x �

2

= 64 + 2 – (4 + 4) = 58

(b) 4

�� 1

�3 x + 2—x � dx

4

= �2x3—2 + 4 x �

1

= 16 + 8 – (2 + 4) = 18

12 ds—–dt

= 3(t – 1)2 + 2

s = �[3(t – 1)2 + 2] dt

= �(3t2 – 6t + 5) dt

= t3 – 3t2 + 5t + cWhen t = 1 and s = 5, we have5 = 1 – 3 + 5 + cc = 2∴ s = t3 – 3t2 + 5t + 2

13 2

�� 1

(2x + k) dx = 5

2 �x2 + kx� = 5 1

4 + 2k – (1 + k) = 5 3 + k = 5 k = 2

14 5

�� 3

[2f(x) – kx] dx = 12

2 5

�� 3

f(x) dx – 5

�� 3

kx dx = 12

5

2(8) – � kx2

—–2 � = 12

3

25—–2

k – 9—2

k = 4

8k = 4

k = 1—2

15 (a) �x2(3 + 1—x4 ) dx

= �(3x2 + x–2) dx

= x3 – 1—x

+ c

(b) �(2x + 5)3 dx

= (2x + 5)4

———––4(2)

+ c

= 1—8

(2x + 5)4 + c

16 4

�� 0

y dx + 8

�� 0

x dy

= 4 × 8= 32

17 Area of the shaded region

= 3

�� 0

(4x – x2) dx – 1—2

(3)(3)

3

= �2x2 – x3

—3 � – 9—

2 0

= 18 – 9 – 9—2

= 4 1—2

unit2

18 (a) x + 1 = (x – 1)2

x + 1 = x2 – 2x + 1 x2 – 3x = 0 x(x – 3) = 0

x = 0 or x = 3 When x = 0, y = 0 + 1 = 1 When x = 3, y = 3 + 1 = 4 ∴ A(0, 1) and B(3, 4)


= 1—2

(3)(1 + 4) – 3

� 0

(x – 1)2 dx

3

= 15—–2

– �(x – 1)3

—–—–3 �

0

= 15—–2

– � 8—3

– �– 1—3 ��

= 4 1—2

unit2

19 x = 8—x2

x3 = 8x = 2When x = 2, y = 2Area of the shaded region

= 1—2

(2)(2) + 4

� 2

� 8—x2 � dx

4

= 2 + � –8—–x �

2

= 2 + [–2 – (–4)]= 4 unit2

20 k

�� 1

8—x2 dx =

2

�� k

8—x2 dx

k 2

�– 8—x � = �– 8—

x � 1 k

– 8—k

– (–8) = –4 – �– 8—k �

8 – 8—k

= 8—k

– 4

16—–k

= 12

k = 11— 3


21 Area of region M

= 1

�� 0

(x3 – 3x2 + 2x) dx

1

= � x4

—4

– x3 + x2� 0

= 1—4

– 1 + 1

= 1—4

unit2

Area of region N

= 2

�� 1

(x3 – 3x2 + 2x) dx

2

= � x4

—4

– x3 + x2� 1

= 4 – 8 + 4 – � 1—4

– 1 + 1�= – 1—

4

= 1—4

unit2

∴ Areas of regions A and B are equal.

22 Area of region A

= 2

�� 1

� 4—x2 � dx

2

= � –4—–x �

1

= –2 – (–4)= 2 unit2

Area of region B

3

= � –4—–x �

2

= – 4—3

– (–2)

= 2—3

unit2

The ratio of the areas of region A to

region B is 2 : 2—3

= 3 : 1.

23 a

�� 0

2x2 dx = 18

a

� 2x3

—–3 � = 18

0

2a3

—–3

= 18

a3 = 18� 3—2 �

= 27 a = 3

24 k

�� 1

5x4 dx = 31

k

�x5� = 31 1 k5 – 1 = 31 k5 = 32 k5 = 25

k = 2

25 Volume formed = π 4

�� 1

1—x2

dx

4

= π� –1—–x �

1

= π�– 1—4

– (–1)� = 3—

4π unit3

26 Volume generated = π 2

�� 1

y dy

2

= π� y2

—–2 �

1

= π�2 – 1—2 �

= 3—2

π unit3

27 Volume formed

= π 4

�� 2

9 � x2

—4

– 1� dx

4

= π� 3—4

x3 – 9x� 2

= π[48 – 36 – (6 – 18)]= 24π unit3

28 π p

�� 1

4x dx = 6π

p

�2x2� = 6 1

2p2 – 2 = 6 2p2 = 8 p2 = 4 p = ±2∴ p = 2

29 (a) x2 = 8 – x2

2x2 = 8 x2 = 4 x = ±2 When x = 2, y = (2)2

= 4 ∴ P(2, 4)

(b) Area of region A = 2

�� 0

x2 dx

2

= � x3

—–3 �

0

= 8—3

unit2

(c) Volume of region B generated

= π 8

�� 4

(8 – y) dy

8

= π�8y – y2

—–2 �

4

= π[64 – 32 – (32 – 8)] = 8π unit3

30 (a) 8 – x = 12—–x

8x – x2 = 12 x2 – 8x + 12 = 0 (x – 2)(x – 6) = 0 x = 2 or x = 6

When x = 2, y = 12—–2

= 6

When x = 6, y = 12—–6

= 2

∴ A(2, 6) and B(6, 2)

(b) The total area of regions P and Q = Area of trapezium

= 1—2

(6 – 2)(6 + 2)

= 16 unit2

(c) Volume of region Q generated

= π 6

�� 2

144—––x2

dx

6

= π� –144——–x �

2

= π[–24 – (–72)] = 48π unit3

31 (a) 2x = 6x – x2

x2 – 4x = 0 x(x – 4) = 0 x = 0 or x = 4 When x = 4, y = 2(4) = 8 ∴ A(4, 8) At x-axis, y = 0 6x – x2 = 0 x2 – 6x = 0 x(x – 6) = 0 x = 6 ∴ B(6, 0)

Hence, A(4, 8) and B(6, 0)

(b) Area of region P

= 4

�� 0

(6x – x2) dx – 1—2

(4)(8)

4

= �3x2 – x3

—3 � – 16

0

= 48 – 64—–3

– 16

= 10 2—3

unit2

(c) Volume generated

= π 6

�� 4

(6x – x2)2 dx

= π 6

�� 4

(36x2 – 12x3 + x4) dx

6

= π�12x3 – 3x4 + x5

—–5 �

4

= π�2592 – 3888 + 7776——–5

–

�768 – 768 + 1024——–5 ��

= 54 2—5

π unit3


32 (a) At y-axis, x = 0 2y = 3(0) + 1

y = 1—2

∴ A�0, 1—2 �

y = 2—x2

When y = 1—2

, 1—2

= 2—x2

x2 = 4 x = ±2

∴ C�2, 1—2 �

Hence, A�0, 1—2 � and C�2, 1—

2 �(b) Area of the shaded region ABC

= 1—2

(1)�2 1—2 � +

2

�� 1

2—x2

dx – 2� 1—2 �

2

= 5—4

+ � –2—–x � – 1

1

= 1—4

+ [– 1 – (–2)]

= 5—4

unit2


= π 2

�� 1

� 4—–x4 � dx

2

= π� –4—–3x3 �

1

= π�– 1—6

– �– 4—3 ��

= 7—6

π unit3

33 (a) y = 4x + 1 At y-axis, x = 0

y = 4(0) + 1 = 1 ∴ A(0, 1)

(b) dy—–dx

= 1—2

(4x + 1) 1– — 2 (4)

= 2———4x + 1

At P(2, 3), dy—–dx

= 2———–4(2) + 1

= 2—3

∴ Equation of the normal PB:

y – 3 = –3—–2

(x – 2)

2y – 6 = –3x + 6 2y + 3x = 12(c) 2y + 3x = 12 At x-axis, y = 0 2(0) + 3x = 12 x = 4 ∴ B(4, 0)

(d) Area of shaded region X

= 2

�� 0

4x + 1 dx

2

= � 1—6

(4x + 1)3—2 �

0

= 9—2

– 1—6

= 4 1—3

unit2

Area of shaded region Y

= 1—2

(4 – 2)(3)

= 3 unit2

(e) Volume generated

= π 2

�� 0

(4x + 1) dx

2

= π�2x2 + x� 0

= 10π unit3

34 (a) 3x = 4 – x2

x2 + 3x – 4 = 0 (x + 4)(x – 1) = 0 x = –4 or x = 1 When x = 1, y = 3(1) = 3 ∴ P(1, 3)

(b) Area of region R

= 1—2

(1)(3) + 2

�� 1

(4 – x2) dx

2

= 3—2

+ �4x – 1—3

x3� 1

= 3—2

+ �8 – 8—3

– �4 – 1—3 ��

= 3—2

+ 5—3

= 3 1—6

unit2

(c) Volume of region S, generated

= 1—3

π(1)2(3) + π 4

�� 3

(4 – y) dy

4

= π + π �4y – y2

—2 �

3

= π + π �16 – 8 – �12 – 9—2 ��

= 1 1—2

π unit3

35 (a) 2x2 + 3x = 27 2x2 + 3x – 27 = 0 (2x + 9)(x – 3) = 0

x = – 9—2

or x = 3

When x = 3, y = 32

= 9 ∴ A(3, 9) At x-axis, y = 0 3x = 27 x = 9 ∴ B(9, 0) Hence, A(3, 9) and B(9, 0).


= 3

�� 0

x2 dx + 1—2

(9 – 3)(9)

3

= � x3

—–3 � + 27

0

= 9 + 27 = 36 unit2

(c)

Volume generated = π 9

� 0

y dy

9

= π� y2

—2 �

0

= 40 1—2

π unit3

36 (a) (i) At x-axis, y = 0 4(0) = x – 1 x = 1 ∴ B(1, 0)(ii) Area of the shaded region

= 1—2

(2 – 1)� 1—4 � +

4

� 2

� 1—x2 � dx

4

= 1—8

+ �– 1—x �

2

= 1—8

+ �– 1—4

– �– 1—2 ��

= 3—8

unit2

(b) π k

�� 2

� 12——–x + 2 �

2

dx = 12π

k

� –144——–x + 2 � = 12

2

–144——–k + 2

– (–36) = 12

144——–k + 2

= 24

k + 2 = 6 k = 4

37 (a) 7 – 3x = 1—4

x2

x2 + 12x – 28 = 0 (x + 14)(x – 2) = 0 x = –14 or x = 2 When x = 2, y + 3(2) = 7 y = 1 ∴ P(2, 1)

(b) Area of region A

= 2

�� 0

1—4

x2 dx

2

= � 1—–12

x3� 0

= 2—3

unit2

y

x0

9 y = x2


(c) Volume of region B, generated

= 1—3

π (2)2(6) + π 1

�� 0

4y dy 1 = 8π + π �2y2�

0

= 8π + 2π = 10π unit3

38 (a) (i) x = x2 – 5x + 8 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x = 2 or x = 4 When x = 2, y = 2 When x = 4, y = 4 ∴ P(2, 2) and Q(4, 4)(ii) Area of the shaded region

= 1—2

(4 – 2)(2 + 4) –

4

�� 2

(x2 – 5x + 8) dx

4

= 6 – � x3

—3

– 5x2

—–2

+ 8x� 2

= 6 – � 64—–3

– 40 + 32 –

� 8—3

– 10 + 16�� = 6 – 4 2—

3

= 1 1—3

unit2

(b) π 4

�� k – 1

4—–x2

dx = 3π

4

� –4—–x � = 3

k – 1

–1 – � –4——–k – 1 � = 3

4——–k – 1

= 4

k – 1 = 1 k = 2

39 (a) y = 6x2 – x3

dy—–dx

= 0

12x – 3x2 = 0 3x2 – 12x = 0 3x(x – 4) = 0 x = 0 or x = 4 When x = 4, y = 6(4)2 – 43

= 32 ∴ P(4, 32) At x-axis, y = 0 0 = 6x2 – x3

x3 – 6x2 = 0 x2(x – 6) = 0 x = 0 or x = 6 ∴ Q(6, 0)


= 6

�� 0

(6x2 – x3) dx – 1—2

(6 – 4)(32)

6

= �2x3 – x4

—4 � – 32

0

= 432 – 324 – 32 = 76 unit2

40 (a) 3 = x2 + 2 x2 = 1 x = ±1 ∴ A(–1, 3) and B(1, 3)(b) The area of the shaded region

= (2 × 3) – 1

�� –1

(x2 + 2) dx

1

= 6 – � x3

—3

+ 2x� –1

= 6 – � 1—3

+ 2 – �– 1—3

– 2�� = 1 1—

3 unit2

(c) Volume of the shaded region generated

= π 3

� 2

(y – 2) dy

3

= π� y2

—2

– 2y� 2

= π� 9—2

– 6 – (2 – 4)� = 1—

2π unit3

15 Vectors

Booster Zone

1 (a) a~ = 2p~

, b~ = –3p~

(b) p~

= 1—2

a~ and p~

= 1–— 3

b~

So, 1—2

a~ = 1–— 3

b~

b~ = 3–— 2

a~

(c) |a~| = |2p~

|

= 2(2) = 4 units |b~| = |–3p

~|

= |–3| × |p~

|

= 3 × 2 = 6 units

2 (a) (i) ⎯→FC = 2a~

(ii) ⎯→EB = –2c~

(iii) ⎯→OF = –a~

(iv) ⎯→AF = c~

(b) (i) |⎯→ED| = 3 units

(ii) ⎯→CF = –2a~

|⎯→CF | = |–2| × |a~|

= 2(3) = 6 units

PQ 3 3 —— = — SR 5

PQ = 3—5

SR

⎯→PQ = 3—

5 (10a~)

= 6a~

4 (a) ⎯→AB = 2u~,

⎯→AC = 2v~

(b) ⎯→AD = 4—

5

⎯→AE

= 4—5

w~

⎯→DE = 1—

5

⎯→AE

= 1—5

w~

⎯→ED = 1–—

5w~

5 u~ = k v~3a~ + 6b~ = k(2a~ + pb~)

3a~ + 6b~ = 2ka~ + kpb~

By comparison,

2k = 3

k = 3—2

and kp = 6

3—2

p = 6

p = 6( 2—3

)

= 4

6 ⎯→AB = p

⎯→PQ

2u~ – 5v~ = p[4u + (k – 6)v~]

2u~ – 5v~ = 4pu~ + p(k – 6)v~

By comparison, 4p = 2

p = 1—2

and p(k – 6) = –5

1—2

(k – 6) = –5

k – 6 = –10 k = –10 + 6 = –4

7 ⎯→PQ = 3a~

a~ = 1—3

⎯→PQ

⎯→QR = 6a~

a = 1—6

⎯→QR

∴ 1—3

⎯→PQ = 1—

6

⎯→QR

⎯→PQ = 1—

2

⎯→QR

Since ⎯→PQ = 1—

2

⎯→QR and Q is the

common point, therefore, P, Q and R are collinear.

8 h – 4 = 0 h = 4

2k – 1 = 0 2k = 1

k = 1—2

9 p + 2q – 5 = 0 p + 2q = 5 1… 2q – p + 3 = 0 p – 2q = 3 2… 1 + 2 : 2p = 8 p = 4Substitute p = 4 into 1 : 4 + 2q = 5 2q = 1

q = 1—2

10 4m – 9n = 5 1… m – 6n = 0 2…

2 × 4 : 4m – 24n = 0 3…1 – 3 : 15n = 5

n = 1—3

Substitute n = 1—3

into 2 :

m – 6( 1—3

) = 0

m – 2 = 0 m = 2

11 (a) ⎯→MD =

⎯→MC +

⎯→CD

= 3b~ – a~

(b) ⎯→ND = 2—

3

⎯→BC

= 2—3

(6b~)

= 4b~

∴ ⎯→DN = –4b~

(c) ⎯→NM =

⎯→ND +

⎯→DM

= 4b~ + a~ – 3b~ = a~ + b~

12 (a) ⎯→PR =

⎯→PS +

⎯→SR

= 2b~ + a~ = a~ + 2b~

⎯→PX = 1—

2

⎯→PR

= 1—2

(a~ + 2b~)

= 1—2

a~ + b~

(b) ⎯→PY = 1—

2

⎯→PX

= 1—2

( 1—2

a~ + b~)

= 1—4

a~ + 1—2

b~

⎯→QY =

⎯→QP +

⎯→PY

= –a~ + 1—4

a~ + 1—2

b~

= – 3—4

a~ + 1—2

b~

(c) ⎯→PZ = 2—

3

⎯→PS

= 2—3

(2b~)

= 4—3

b~



∴ ⎯→XZ =

⎯→XP +

⎯→PZ

= 1–— 2

a~ – b~ + 4—3

b~

= 1–— 2

a~ + 1—3

b~

13 (a) ⎯→QP =

⎯→QR +

⎯→RP

= 4a~ + 2b~

∴ ⎯→ZP =

⎯→ZQ +

⎯→QP

= –a~ – 4a~ – 2b~ = –5a~ – 2b~

(b) ⎯→ZP = 5

⎯→ZY

⎯→ZY = 1—

5

⎯→ZP

= 1—5

(–5a~ – 2b~)

= –a~ – 2—5

b~

∴ ⎯→QY =

⎯→QZ +

⎯→ZY

= a~ + (–a~ – 2—5

b~)

= – 2—5

b~

(c) ⎯→YX =

⎯→YZ +

⎯→ZR +

⎯→RX

= (a~ + 2—5

b~) + 3 a~ + b~

= 4a~ + 7—5

b~

14 (a) (i) ⎯→MQ =

⎯→MP +

⎯→PQ

= –2b~ + 3a~ = 3a~ – 2b~

∴ ⎯→MN = 1—

5

⎯→MQ

= 1—5

(3a~ – 2b~)

= 3—5

a~ – 2—5

b~

(ii) ⎯→SN =

⎯→SP +

⎯→PM +

⎯→MN

= –a~ + 2b~ + ( 3—5

a~ – 2—5

b~) = – 2—

5a~ + 8—

5b~

(iii)

⎯→SR =

⎯→SP +

⎯→PR

= – a~ + 4b~

(b) ⎯→SN = – 2—

5 a~ + 8—

5b~

= 2—5

(–a~ + 4b~)

= 2—5

⎯→SR

∴ The points S, N and R are collinear.

(c) ⎯→SN = 2—

5

⎯→SR

⎯→SN

—— = 2—5

⎯→SR

SN : SR = 2 : 5

∴ SN : NR = 2 : 3

15 (a) ⎯→OM =

⎯→OP +

⎯→PM

= 2a~ + b~

(b) ⎯→ON =

⎯→OR +

⎯→RN

= 2b~ + a~

= a~ + 2b~

(c) ⎯→PN =

⎯→PQ +

⎯→QN

= 2b~ + (–a~)

= –a~ + 2b~

16 (a) ⎯→MC = 3—

2

⎯→ND

= 3—2

(2b~)

= 3b~

(b) ⎯→AC =

⎯→AM +

⎯→MC

= 3—2

(2a~) + 3b~

= 3a~ + 3b~

(c) ⎯→ME =

⎯→MN +

⎯→NE

= –a~ – 2b~

17 (a) ⎯→AB =

⎯→AO +

⎯→OB

= –a~ + b~

(b) (i) ⎯→OR =

⎯→OA +

⎯→AR

= a~ + h(–a~ + b~)

= a~ – ha~ + hb~

= (1 – h)a~ + hb~

(ii) ⎯→OR = k

⎯→OC

= k(a~ + 2b~)

= ka~ + 2kb~ (1 – h)a~ + hb~ = ka~ + 2kb~(c) By comparison, 1 – h = k h + k = 1 1… h = 2k h – 2k = 0 2… 1 – 2 : 3k = 1

k = 1—3

Substitute k = 1—3

into 1 :

h + 1—3

= 1

h = 1 – 1—3

= 2—3

18 (a) ⎯→PS =

⎯→PR +

⎯→RS

= 2q~

+ 2p~

= 2p~

+ 2q~

⎯→QR =

⎯→QP +

⎯→PR

= –p~

+ 2q~

(b) (i) ⎯→PT = λ

⎯→PS

= λ(2p~

+ 2q~

)

= 2λp~

+ 2λq~

(ii) ⎯→QT = �

⎯→QR

= �(–p~

+ 2q~

)

= –�p~

+ 2�q~

(c) ⎯→PT =

⎯→PQ +

⎯→QT

2λp~

+ 2λq~

= p~

– �p~

+ 2�q~

2λp~

+ 2λq~

= (1 – �)p~

+ 2�q~

By comparison, 2λ = 2� λ = �(shown)

19 (a) (i) ⎯→OR = 2—

3

⎯→OQ

= 2—3

(9q~

)

= 6q~

⎯→PR =

⎯→PO +

⎯→OR

= –4p~

+ 6q~

(ii) ⎯→SQ =

⎯→SO +

⎯→OQ

= –2p~

+ 9q~

(b) (i) ⎯→TR = h

⎯→PR

= h(–4p~

+ 6q~

)

= –4hp~

+ 6hq~

(ii) ⎯→TQ = k

⎯→SQ

= k(–2p~

+ 9q~

)

= –2kp~

+ 9kq~

(c) ⎯→RQ =

⎯→RT +

⎯→TQ

3q~

= 4hp~

– 6hq~

– 2kp~

+ 9kq~

3q~

= (4h – 2k)p~

+ (9k – 6h)q~

By comparison, 4h – 2k = 0 1… –6h + 9k = 3 –2h + 3k = 1 2… 2 × 2 : –4h + 6k = 2 3… 1 + 3 : 4k = 2

k = 1—2

1 Substitute k = — into 1 : 2 4h – 2( 1—

2) = 0

4h – 1 = 0

h = 1—4

20 (a) ⎯→AB =

⎯→AO +

⎯→OB

= –a~ + 2b~

(b) ⎯→OD =

⎯→OC +

⎯→CD

= h⎯→OA + k

⎯→CE

= ha~ + k(–ha~ + b~ )

= ha~ – hka~ + kb~


= (h – hk)a~ + kb~

(c) ⎯→AD =

⎯→AO +

⎯→OD

= –a~ + (h – hk)a~ + kb~

= (h – hk – 1)a~ + kb~ Since A, D and B are collinear,

⎯→AB = p

⎯→AD

–a~ + 2b~ = p[(h – hk – 1)a~ + kb~]

–a~ + 2b~ = (h – hk – 1)pa~ + kpb~

By comparison, kp = 2

p = 2—k

and (h – hk – 1)p = –1

(h – hk – 1)( 2—k

) = –1

h – hk = 1 – 1—2

k

h(1 – k) = 2 – k———2

h = 2 – k————2(1 – k)

21 (a) (i) 3 i~ + 4 j~

(ii) 3� �4

(b) |⎯→O~

A| = 5 units

1(c)

⎯→OA = — (3 i~ + 4 j

~)

5

22 (a) |a~

| = 17 units

82 + k2 = 17 k2 + 64 = 289 k2 = 225

k = 225 = ±15 k = 15, –15

(b) |⎯→PQ | = 13 units

122 + (–k)2 = 13

k2 + 144 = 169 k2 = 25 k = ±5 k = 5, –5

23 (a) 2a~ + b~ = 2

2 � � –6 +

2� �4

= 4� � –12

+ 2� �4

= 6 � � –8

(b) |2a~ + b~| = 62 + (–8)2

= 100 = 10 units(c) The unit vector in the direction of

2a~ + b~

1 = —— (6 i~ – 8 j

~)

10

1 = — (3 i~ – 4 j

~)

5

24 (a) ⎯→AB =

⎯→AO +

⎯→OB

= – i~ + 4 j~

+ 10 i~ + 8 j~

= 9 i~ + 12 j~

|⎯→AB | = 92 + 122

= 225

= 15 units

∴ The unit vector parallel to ⎯→AB

1 = —— (9 i~ + 12 j

~)

15 1 = — (3 i~ + 4 j

~)

5 2(b)

⎯→AP = —

⎯→AB

3

= 2—3

(9 i~ + 12 j~

)

= 6 i~ + 8 j~

⎯→OP =

⎯→OA +

⎯→AP

= i~ – 4 j~

+ 6 i~ + 8 j~

= 7 i~ + 4 j~

39 25 (a)

⎯→OP = ——

–5� � –12 13

= 3 –5� �12

= –15� � 36

25

⎯→OQ = ——

3� �4

5 = 5

3� �4

=

15 � � 20

(b) ⎯→PQ =

⎯→PO +

⎯→OQ

= 15� � –36

+

15 � � 20

= 30� � –16

|⎯→PQ | = 302 + (–16)2

= 1156

= 34 units

26 (a) ⎯→BT =

⎯→BA +

⎯→AT

=

–3 � � –4 +

6� �3

=

3 � � –1

⎯→BC = 2

3 � � –1

=

6 � � –2

⎯→AC =

⎯→AB +

⎯→BC

= 3� �4

+

6 � � –2

= 9� �2

(b) ⎯→CD = –

⎯→AB

=

–3 � � –4

|⎯→CD | = (–3)2 + (–4)2

= 25 = 5 units

27 (a) ⎯→OA =

4� �3

(b) ⎯→AB =

⎯→AO +

⎯→OB

= –4 i~ – 3 j~

+ 5 j~

= –4 i~ + 2 j~

28 ⎯→OB =

⎯→OA +

⎯→OC

= 3 i~ + 4 j~

+ 6 i~ – 2 j~

= 9 i~ + 2 j~

29 a~ – 3b~ =

0 � � 10 – 3

8� �1

= –24� � 7

|a~ – 3b~| = (–24)2 + 72

= 625

= 25 units The unit vector which is parallel to

a~ – 3b~

1= ——

–24� � 7 25

=

24 –—— 25� �

7 —— 25

24 7= –—— i~ + —— j

~

25 25

30 (a) ⎯→PQ =

⎯→PO +

⎯→OQ

=

2 � � –7 +

3 � � –5

= 5� � –12

5� � –12

= m

2 � � –2 + n

1� �6 2m + n = 5 1… –2m + 6n = –12 2… 1 + 2 :7n = –7 n = –1 Substitute n = –1 into 1 : 2m – 1 = 5 2m = 6 m = 3

(b) |⎯→PQ | = 52 + (–12)2

= 169

= 13 units


The unit vector parallel to ⎯→PQ

1 = ——

5� � –12

13

=

5 —— 13� � –12 —— 13

5 12 = —— i~ – —— j

~

13 13

31 (a) ⎯→AC = p

⎯→AB

⎯→AO +

⎯→OC = p(

⎯→AO +

⎯→OB)

–2 i~ – j~

+ k i~ + 4 j~

= p(–2 i~ – j~

+ 4 i~ + 2 j~

)

(k – 2) i~ + 3 j~

= p(2 i~ + j~

)

By comparison, p = 3 and k – 2 = 2p k – 2 = 6 k = 8

(b) |⎯→AC | = 3|

⎯→AB |

(k – 2)2 + 32 = 3 22 + 12

(k – 2)2 + 9 = 9(5) k2 – 4k + 4 + 9 = 45 k2 – 4k – 32 = 0 (k + 4)(k – 8) = 0 k = –4 or k = 8

32 (a) (i) ⎯→AB =

⎯→AO +

⎯→OB

= –2u~ – v~ + 3u~ – 2v~ = u~ – 3v~

(ii) ⎯→AC =

⎯→AO +

⎯→OC

= –2u~ – v~ + mu~ + 5v~ = (m – 2)u~ + 4v~

(b) ⎯→AC = k

⎯→AB

(m – 2)u~ + 4v~ = k(u~ – 3v~) By comparison 4 = –3k –4 k = —— 3 and m – 2 = k –4 m = —— + 2 3 2 = — 3

SPM Appraisal Zone

1 w~ = 2u~ – 3 v~

ha~ + (h + k + 3)b~

= 2(5a~ + 4b~) – 3(3a~ – b~)

= 10a~ + 8b~ – 9a~ + 3b~

= a~ + 11b~ By comparison, h = 1and h + k + 3 = 11 k + 4 = 11

k = 11 – 4 = 7∴ h = 1, k = 7

2 (a) (i) ⎯→AB =

⎯→AO +

⎯→OB

= –2a~ – b~ + 3a~ – 2b~ = a~ – 3b~

(ii) ⎯→AC =

⎯→AO +

⎯→OC

= –2a~ – b~ + ha~ + 5b~

= (h – 2)a~ + 4b~ (b) Since A, B and C are collinear

⎯→AC = k

⎯→AB

(h – 2)a~ + 4b~ = k(a~ – 3b~) = ka~ – 3kb~ so –3k = 4 –4 k = —— 3 –4 and h – 2 = —— 3 4 h = 2 – — 3 2 = — 3 2 ∴ h = — 3

3 4a~ + p(a~ – b~) = a~ + b~ + q(a~ + b~)4a~ + pa~ – pb~ = a~ + b~ + qa~ + qb~(4 + p)a~ – pb~ = (1 + q)a~ + (1 + q)b~So, 4 + p = 1 + q p – q = –3 1… and –p = 1 + q p + q = –1 2…1 + 2 : 2p = –4 p = –2Substitute p = –2 into 1 : –2 – q = –3 q = 1∴ p = –2, q = 1

4 (a) a~ + 2ub~ = (4 + v)a~ + b~

So, 4 + v = 1 v = –3

and 2u = 1 1 u = — 2 1 ∴ u = —, v = –3 2(b) p

~ = kq

~

a~ + 2ub~

= k[(4 + v)a~ + b~]

= (4 + v)ka~ + kb~ By comparison, k = 2u and (4 + v)k = 1 (4 + v)(2u) = 1 1 4 + v = —— 2u 1 v = —— – 4 2u 1 – 8u = ———— 2u

1 5 (a)

⎯→AP = —

⎯→AC

2 1

= — (–a~ + b~) 2

⎯→OP =

⎯→OA +

⎯→AP

1 = a~ + — (–a~ + b~) 2 1 1 = — a~ + —b~ 2 2 1 = — (a~ + b~) 2 2(b)

⎯→AQ = —

⎯→AB

3 2 = —(–a~ + 2b~) 3

⎯→OQ =

⎯→OA +

⎯→AQ

2 = a~ + — (–a~ + 2b~) 3 1 4 = —a~ + —b~ 3 3 1 = — (a~ + 4b~) 3

6 (a) ⎯→MD =

⎯→MC +

⎯→CD

= 2b~ – a~

3(b)

⎯→DN = —

⎯→DA

4 3 = — (–4b~) 4 = –3b~

(c) ⎯→MN =

⎯→MD +

⎯→DN

= 2b~ – a~ – 3b~ = –a~ – b~

7 (a) ⎯→OP =

⎯→OQ +

⎯→QP

= 2a~ + b~ – a~ + b~ = a~ + 2b~

(b) ⎯→SR = b~ – a~

(c) ⎯→OS =

⎯→OQ +

⎯→QS

= 2a~ + b~ – 2⎯→SR

= 2a~ + b~ – 2b~ + 2a~ = 4a~ – b~

8 (a) ⎯→PR =

⎯→PQ +

⎯→QR

= a~ + b~ 1(b)

⎯→PT = —

⎯→PS

3 1 = —b~ 3

(c) ⎯→QT =

⎯→QP +

⎯→PT

1 = –a~ + —b~ 3

9 (a) ⎯→QP =

⎯→QO +

⎯→OP

= –q~

+ 2p~

= 2p~

– q~

1(b)

⎯→MP = —

⎯→QP

2


1 = — (2p

~ – q

~)

2 1 = p

~ – — q

~ 2

1

⎯→NP = —

⎯→MP

2 1 1 = —(p

~ – —q

~)

2 2 1 1 = — p

~ – — q

~ 2 4

⎯→ON =

⎯→OP +

⎯→PN

1 1 = 2p

~ – — p

~ + — q

~

2 4 3 1 = — p

~ + — q

~

2 4

10 (a) ⎯→OP = 3a~ + b~

(b)

11 (a) ⎯→OR = 2p

~ + q

~

(b) ⎯→RS = p

~ – 2q

~

12 |a~| = 5 units

k2 + (–3)2 = 5 k2 + 9 = 25 k2 = 16 k = ±4 k = 4 or –4

13 |a~| = |b~|

(–5)2 = 42 + m2

m2 + 16 = 25 m2 = 9 m = ±3 m = 3 or –3

14 (a) ⎯→OA = 6 i~ + 2 j

~

(b) ⎯→OB =

2� �8

15 (a) ⎯→AB =

⎯→AO +

⎯→OB

=

2� � –2 +

2� �5

= 4� �3

1(b)

⎯→AB = —(4 i~ + 3 j

~)

5

16 (a) ⎯→AB =

⎯→AO +

⎯→OB

=

4� � –3 +

8� �6

=

12� � 3

(b) ⎯→AC = k

⎯→OB

4� �m

= k8� �6

So, 8k = 4 1 k = — 2 and m = 6k = 6

1�— � 2 = 3 1 ∴ k = —, m = 3 2

17 (a) ⎯→OQ =

⎯→OP +

⎯→OR

=

–3� � 3 +

6� �3

= 3� �6

(b) ⎯→PR =

⎯→PO +

⎯→OR

=

3� � –3 +

6� �3

= 9� �0

18 (a) ⎯→PQ = k

⎯→AB

2� �p

= k4� �2

so 4k = 2 1 k = — 2 and p = 2k 1 = 2(—) 2 = 1

(b) |⎯→AB | = |

⎯→PQ |

42 + 22 = 22 + p2

20 = p2 + 4 p2 = 16 p = ±4 p = 4 or –4

19 (a) |p~

| = 92 + (–12)2

= 225 = 15 units(b) p

~ = kq

~

9� � –12

= k6� �m

so 6k = 9 3 k = — 2 and km = –12 3 — m = –12 4 2 m = –12 (—) 3 ∴ m = –8

20 (a) ⎯→AB =

⎯→AO +

⎯→OB

8� � –6 =

–4� � –5

+ ⎯→OB

⎯→OB =

8� � –6

–

–4� � –5

= 12� � –1

∴ B(12, –1)

(b) ⎯→CD = k

⎯→AB

–4� � p = k

8� � –6

so 8k = –4 1 k = –— 2 and p = –6k 1 = –6(–—) 2 = 3 ∴ p = 3

21 ⎯→AB

= k⎯→AC

⎯→AO +

⎯→OB

= k(⎯→AO +

⎯→OC)

– i~ – j~

+ 3 i~ – 2 j~

= k(– i~ – j~

+ 4 i~ + m j~

)

2 i~ – 3 j~

= k[3 i~ + (m – 1) j~

]

So, 3k = 2 2 k = — 3and (m – 1)k = –3 2 (m – 1)(—) = –3 3 9 m – 1 = –— 2 9 m = 1 –— 2 7 = –— 2 7 ∴ m = –— 2

22 m2� �1

+ n1� �3

= 8� �9

2m + n = 8 1… m + 3n = 9 2…1 × 3 : 6m + 3n = 24 3…3 – 2 : 5m = 15 m = 3Substitute m = 3 into 1 : 2(3) + n = 8 n = 8 – 6 = 2 ∴ m = 3, n = 2

23 p1� �3

+ q2� �1

=

1 � � 13 p + 2q = 1 1… 3p + q = 13 2…1 × 3 : 3p + 6q = 3 3…3 – 2 : 5q = –18 q = –2

Q

O

3a

1– —b 2


Substitute q = –2 into 1 : p + 2(–2) = 1

p = 5∴ p = 5, q = –2

24 2⎯→AB +

⎯→BC = 8 i~ – 2 j

~

2(⎯→AO +

⎯→OB) +

⎯→BO +

⎯→OC = 8 i~ – 2 j

~ 2(3 i~ – 5 j

~ + 4 i~ + 6 j

~) – 4 i~ – 6 j

~ +

p i~ + q j~

= 8 i~ – 2 j~

14 i~ + 2 j~

+ (p – 4) i~ + (q – 6) j~

= 8 i~ – 2 j~

(p + 10) i~ + (q – 4) j~

= 8 i~ – 2 j~

So, p + 10 = 8 p = –2and q – 4 = –2 q = 2∴ p = –2, q = 2

25 (a) ⎯→PQ =

⎯→PO +

⎯→OQ

=

–2� � 5 +

8� �3

= 6� �8

(b) |⎯→PQ | = 62 + 82

= 100 = 10 units The unit vector in the direction of

⎯→PQ

1 = ——

6� �8 10

=

3 — 5� � 4 — 5

3 4 = — i~ + — j

~

5 5

26 (a) (i) ⎯→DB =

⎯→DO +

⎯→OB

= –y~

+ 4 x~ = 4 x~ – y

~

(ii) ⎯→CA =

⎯→CO +

⎯→OA

= –2 x~ + 3⎯→OD

= –2 x~ + 3y~

(b) (i) ⎯→OE =

⎯→OD +

⎯→DE

= y~

+ h(4 x~ – y~

)

= 4h x~ + (1 – h)y~

(ii) ⎯→OE =

⎯→OC +

⎯→CE

= 2 x~ + k(–2 x~ + 3y~

)

= (2 – 2k) x~ + 3ky~

From (i) and (ii), 4h x~ + (1 – h)y

~

= (2 – 2k) x~ + 3ky~

So, 4h = 2 – 2k 4h + 2k = 2 2h + k = 1 1… and 1 – h = 3k h + 3k = 1 2… 1 × 3: 6h + 3k = 3 3… 3 – 2 : 5h = 2 2 h = — 5 2

Substitute h = — into 1 : 5 2 2(—) + k = 1 5 4 k = 1–— 5 1 = — 5

2 1∴ h = —, k = — 5 5

27 (a) ⎯→CD =

⎯→CO +

⎯→OD

1 = –3a~ + —b~ 3

⎯→AB =

⎯→AO +

⎯→OB

= –a~ + b~

(b) (i) ⎯→OP =

⎯→OC +

⎯→CP

1 = 3a~ + p(–3a~ + —b~) 3 1 = (3 – 3p)a~ + —pb~ 3

(ii) ⎯→OP =

⎯→OA +

⎯→AP

= a~ + q(–a~ + b~)

= (1 – q)a~ + qb~(c) From (i) and (ii), 1 (3 – 3p)a~ + — pb~ = (1 – q)a~ + qb~ 3 So, 3 – 3p = 1 – q

3p – q = 2 1… 1 and — p = q 3 p – 3q = 0 2… 1 × 3: 9p – 3q = 6 3… 3 – 2 : 8p = 6 3 p = — 4 3 Substitute p = — into 1 : 4 3 3(—) – q = 2 4 9 q = — – 2 4 1 = — 4 3 1 ∴ p = —, q = — 4 4

28 (a) (i) ⎯→AC =

⎯→AO +

⎯→OC

= – x~ + y~

(ii) ⎯→OE =

⎯→OA +

⎯→AE

1 = x~ + — (– x~ + 2y

~)

3 2 2 = — x~ + — y

~ 3 3

(b) (i) ⎯→OD =

⎯→OA +

⎯→AD

= x~ + m(– x~ + y~

)

= (1 – m) x~ + my~

(ii) ⎯→OD = n

⎯→OE

2 2 = n(— x~ + —y

~)

3 3 2 2 = —n x~ + —ny

~ 3 3(c) From (i) and (ii), 2 (1 – m) x~ + my

~ = — n x~ +

3 2 —ny

~

3 2 So, 1 – m = — n 3 3 – 3m = 2n 3m + 2n = 3 1…

2 and m = —n 3 3m – 2n = 0 2…1 + 2 : 6m = 3 1 m = — 2 1Substitute m = — into 1 : 2 1 3(—) + 2n = 3 2 3 2n = 3 –— 2 3 = — 2 3 n = — 4 1 3∴ m = —, n = — 2 4

29 (a) ⎯→AB =

⎯→AO +

⎯→OB

= –a~ + 2b~

(b) (i) ⎯→OD =

⎯→OA +

⎯→AD

= a~ + h(–a~ + 2b~)

= (1 – h)a~ + 2hb~

(ii) ⎯→OD = k

⎯→OC

5 = —k(a~ + b~) 3 5 5 = — ka~ + — kb~ 3 3(c) From (i) and (ii), 5 (1 – h) a~ + 2h b~ = — k a~ + 3 5 — kb~ 3 5 So, 1 – h = — k 3 3 – 3h = 5k 3h + 5k = 3 1… 5 2h = — k 3


6h – 5k = 0 2… 1 + 2 : 9h = 3 1 h = — 3 1 Substitute h = — into 1 : 3 1 3(—) + 5k = 3 3 5k = 3 – 1 = 2 2 k = — 5 1 2 ∴ h = —, k = — 3 5

30 (a) ⎯→PR =

⎯→PO +

⎯→OR

= –3u~ + v~

(b) (i) ⎯→OS =

⎯→OR +

⎯→RS

= v~ + m(3u~ – v~)

= 3mu~ + (1 – m) v~

(ii) ⎯→OS = n

⎯→OQ

4n = —— (2u~ + v~) 5 8 4 = — nu~ + — n v~

5 5(c) From (i) and (ii), 8 4 3mu~

+ (1 – m)v~ = — nu~ + — nv~ 5 5 8 So, 3m = — n 5 15m – 8n = 0 1… 4 1 – m = — n 5 and 5 – 5m = 4n

5m + 4n = 5 2…

2 × 2: 10m + 8n = 10 3… 1 + 3 : 25m = 10 2 m = — 5 2 Substitute m = — into 1 : 5 2 15(—) – 8n = 0 5 8n = 6 3 n = — 4 2 3 ∴ m = —, n = — 5 4

31 (a) (i) ⎯→PR =

⎯→PQ +

⎯→QR

= 4a~ + 2b~

(ii) ⎯→SM =

⎯→SR +

⎯→RM

= 4a~ – b~

(b) (i) ⎯→PN = λ

⎯→PR

= λ(4a~ + 2b~)

= 4λ a~ + 2λ b~

(ii) ⎯→PN =

⎯→PS +

⎯→SN

= 2b~ + �⎯→SM

= 2b~ + �(4a~ – b~)

= 4ua~ + (2 – u)b~(c) From (i) and (ii), 4λ a~ + 2λ b~ = 4�a~ + (2 – �)b~ So, 4λ = 4� λ = �

32 (a) ⎯→XY = – x~ + y

~

⎯→OP =

⎯→OX +

⎯→XP

1 = x~ + —(– x~ + y

~)

5 4 1 = — x~ + —y

~ 5 5

⎯→YQ =

⎯→YO +

⎯→OQ

4 1 = –y

~ + 3(— x~ + —y

~)

5 5 12 2 = —— x~ – —y

~ 5 5

(b) (i) ⎯→YZ =

⎯→YO +

⎯→OZ

= –y~

+ h x~ = h x~ – y

~

(ii) ⎯→YZ = k

⎯→YQ

12 2 = k(——x~ – —y

~)

5 5 12 2 = ——k x~ – —ky

~

5 5(c) From (i) and (ii), 12 2 h x~ – y

~ = ——k x~ – —ky

~

5 5 2 So, – — k = –1 5 5 k = — 2 12 and h = ——k 5 12 5 = —— (—) 5 2 = 6 5 ∴ h = 6, k = — 2

33 (a) ⎯→ON =

⎯→OA +

⎯→AN

= a~ + h(–a~ + b~) = (1 – h)a~ + hb~

(b) ⎯→OL = k

⎯→OM

1 = k(a~ + —b~) 2 1 = ka~ + —kb~ 2 1(c) (1 – h)a~ + hb~ = ka~ + —kb~ 2 So, 1 – h = k h + k = 1 1… 1 and h = — k 2 2h – k = 0 2…1 + 2 : 3h = 1 1 h = — 3

1Substitute h = — into 1 : 3 1 — + k = 1 3 1 k = 1–— 3 2 = — 3 1 2∴ h = —, k = — 3 3

34 (a) (i) ⎯→OB =

⎯→OA +

⎯→OC

= –2 i~ + 2 j~

+ 5 i~ + 2 j~

= 3 i~ + 4 j

~

(ii) |⎯→OB | = 32 + 42

= 25

= 5 units The vector unit in the direction

of ⎯→OB

1 = — (3 i~ + 4 j

~)

5

(iii)

From the diagram,

52 = ( 8 )2 + ( 29 )2 – 2( 8 )( 29 )

cos ∠O⎯→AB

12 cos ∠O

⎯→AB = ——————

2( 8 )( 29 )

= 0.3939

∠O⎯→AB = 66° 48'

(b) (i) ⎯→AD =

⎯→AC +

⎯→CD

= 2 i~ – 2 j~

+ 5 i~ + 2 j~

– 11 i~ + 4 j~

= –4 i~ + 4 j~

(ii) ⎯→OA = 2(– i~ + j

~)

⎯→AD = 4(– i~ + j

~)

So, ⎯→AD = 2

⎯→OA

∴The points O, A andD lie on the same straight line.

35 (a) ⎯→RQ =

⎯→RP +

⎯→PQ

= –4b~ + 4a~

= 4(a~ – b~)

⎯→ST = 2a~ – 2b~

1

⎯→SV = —

⎯→ST

2 = a~ – b~

B

A

O

29

8

5


⎯→PV =

⎯→PS +

⎯→SV

= 2b~ + a~ – b~ = a~ + b~

(b) (i) ⎯→RU = m

⎯→RQ

= m(4a~ – 4b~) = 4ma~ – 4mb~

(ii) ⎯→PU = n

⎯→PV

= n(a~ + b~) = na~ + nb~

(c) ⎯→PU =

⎯→PR +

⎯→RU

na~ + nb~ = 4b~ + 4ma~ – 4mb~ na~ + nb~ = 4ma~ + (4 – 4m)b~ So, 4m = n 4m – n = 0 1… and 4 – 4m = n 4m + n = 4 2…

1 + 2 : 8m = 4 1 m = — 2 1Substitute m = — into 1 : 2 1 4(—) – n = 0 2 n = 2 1 ∴ m = —, n = 2 2

16 Trigonometry

Booster Zone

1 (a) (b)

(c) (d)

(e) (f)

(g) (h)

(i)

2 (a) 480° (b) 380° (c) –510° (d) –600° (e) 710° (f) –650 3 (a) sin 260° = –sin (260° – 180°) = –sin 80° (b) cos 130° = –cos (180° – 130°) = –cos 50° (c) tan 330° = –tan (360° – 330°) = –tan 30° (d) tan (–250°) = –tan(250° – 180°) = –tan 70° (e) cos (–580°) = cos 580° = cos (580° – 360°) = cos 220° = – cos (220° – 180°) = –cos 40° (f) sin (–675°) = –sin 675° = –sin (675° – 360°) = –sin 315° = –(–sin 45) = sin 45°

4

tan θ � 0 ⇒ θ is in the 2nd quadrant

∴ sin θ = 5

—–13

and tan θ = –5

—–12

5

180° � θ � 360° ⇒ θ is in the 4th quadrant

∴ sin θ = –3—5

and cos θ = 4—5

6

180° � θ � 270° ⇒ θ in the 3rd quadrant

∴ tan θ = 24—–7

and sin θ = –24—–25

7 Since A and B are angles in the same quadrant where sin A � 0 and tan B � 0 ⇒ A and B are in the 3rd quadrant.

(a) cos A = –3—5

(b) tan A = 4—3

(c) cos B = –15—–17

(d) sin B = –8

—–17

8 Since tan θ and sin θ have opposite signs, tan θ � 0 ⇒ sin θ � 0, θ is in the 2nd quadrant.

∴ cos θ = –3

—–10

and sin θ = 1

—–10

9

(a) cos 155° = –cos (180° – 155°) = –cos 25° = –k

(b) sin 25° = 1 – k2

(c) tan (–155°) = –tan 155° = – [–tan (180° – 155°)] = – (–tan 25°) = tan 25°

= 1 – k2

——–k

(d) cos 65° = sin 25° = 1 – k2

10 (a) cos (–600°) = cos 600° = – cos 60°

= –1—2

(b) cos 315° = cos 45°

= 1—2

(c) sin (–210°) = – sin 210° = – (– sin 30°)

= 1—2

(d) tan 480° = – tan 60° = – 3(e) tan (–135°) = – tan 135° = – (– tan 45°) = 1(f) sin (–120°) = – sin 120° = – sin 60

= – 3—–2

(g) tan 420° = tan 60° = 3(h) cos (–45°) = cos 45°

= 1—2

(i) sin 495° = sin 45°

= 1—2

11

Since θ is an acute angle, so θ is in the 1st quadrant.

x

yP

α = 70° 470°x

y

Pα = 20°

520°

x

y

P

α = 30°

390°x

y

Pα = 20°

560°

x

y

P

α = 80°

620°x

y

Pα = 20°700°

x

y

P

α = 40°–680°

x

y

P

α = 40°

–400°x

P

α = 10°

–530°

y

x

y

O

P

Qα

θ

12

13

x

y

O

P

Q

αθ

4

3

x

y

O

P

Qθ

α7

25

x

y

O

P

Q

α

A

45

x

y

Oα

B

8

15

x

y

O

P

Qα

θ1

3

x

y

O

1 1 – k2

k25°

x

y

O

P

Qθ

35



(a) sec θ = 1

—–—cos θ

= 5—4

(b) cot θ = 1

—–—tan θ

= 4—3

12 cot θ = –3

1

—–—tan θ

= –3

tan θ = –1—3

Since θ is an obtuse angle, so θ is in the 2nd quadrant.

(a) cosec θ = 1

—–—sin θ

= 10

(b) sec θ = 1

—–—cos θ

= –10—–3

13 (a) sec θ = 1

—–—cos θ

= 5—3

cosec θ = 1

—–—sin θ

= 5—4

cot θ = 1

—–—tan θ

= 3—4

(b) sec θ = 1

—–—cos θ

= 5—3

cosec θ = 1

—–—sin θ

= –5—4

cot θ = 1

—–—tan θ

= –3—4

(c) sec θ = 1

—–—cos θ

= –25—–7

cosec θ = 1

—–—sin θ

= 25—–24

cot θ = 1

—–—tan θ

= –7

—–24

(d) sec θ = 1

—–—cos θ

= –17—–15

cosec θ = 1

—–—sin θ

= –17—–8

cot θ = 1

—–—tan θ

= 15—–8

14 (a) cos θ � 0 ⇒ θ is in the 2nd or 3rd quadrant.

cos α = 0.73 α = 43° 7' ∴ θ = 180° – 43° 7', 180° + 43° 7' = 136° 53', 223° 7' (b) tan θ � 0 ⇒ θ is in the 2nd or 4th

quadrant. tan α = 2 α = 63° 26' ∴ θ = 180° – 63° 26', 360° – 63° 26' = 116° 34', 296° 34' (c) sin θ � 0 ⇒ θ is in the 1st or 2nd

quadrant. sin α = 0.6712 α = 42° 10' θ = 42° 10', 180° – 42° 10' = 42° 10', 137° 50' (d) tan θ � 0 ⇒ θ is in the 1st or 3rd

quadrant. tan α = 0.3346 α = 18° 30' ∴ θ = 18° 30', 180° + 18° 30' = 18° 30', 198° 30' (e) cos θ � 0 ⇒ θ is in the 1st or 4th

quadrant. cos α = 0.46 α = 62° 37' ∴ θ = 62° 37', 360° – 62° 37' = 62° 37', 297° 23' (f) sin θ � 0 ⇒ θ is in the 3rd or 4th

quadrant. sin α = 0.866 α = 60° ∴ θ = 180° + 60°, 360° – 60° = 240°, 300° (g) 2 sin2 θ = 1

sin2 θ = 1—2

sin θ = ± 1—2

So, θ is in all the quadrants.

sin α = 1—2

α = 45° ∴ θ = 45°, 135°, 225°, 315° (h) 4 tan2 θ = 1

tan2 θ = 1—4

tan θ = ± 1—2

So, θ is in all the quadrants.

tan α = 1—2

α = 26° 34' ∴ θ = 26° 34', 153° 26', 206° 34',

333° 26'15 (a) sin 2θ � 0 ⇒ 2θ is in the 1st or

2nd quadrant and 0° � θ � 360° ⇒ 0° � 2θ � 720°

sin α = 0.866 α = 60° For angle 2θ in this interval 2θ = 60°, 120°, 60° + 360°, 120°

+ 360° = 60°, 120°, 420°, 480° ∴ θ = 30°, 60°, 210°, 240° (b) tan 2θ � 0 ⇒ 2θ is in 2nd or 4th

quadrant and 0° � θ �360° ⇒ 0° � 2θ � 720° tan α = 1.264 α = 51° 39' 2θ = 128° 21', 308° 21', 488° 21',

688° 21' ∴ θ = 64° 11', 154° 11', 244° 11',

344° 11' (c) cos 2θ � 0 ⇒ 2θ is in 2nd or 3rd

quadrant and 0° � θ � 360° ⇒ 0° � 2θ � 720°. cos α = 0.74 α = 42° 16' 2θ = 137° 44', 222° 16', 497° 44',

582° 16' ∴ θ = 68° 52', 111° 8', 248° 52',

291° 8'

(d) sin 1—2

θ � 0 ⇒ 1—2

θ is in the 1st or

2nd quadrant and 0° � θ � 360°

⇒ 0° � 1—2

θ � 180°

sin α = 4—7

α = 34° 51'

1—2

θ = 34° 51', 145° 9'

∴ θ = 69° 42', 290° 18'

(e) tan 1—2

θ � 0 ⇒ 1—2

θ is in the 1st or

3rd quadrant and 0° � θ � 360°

⇒ 0° � 1—2

θ � 180°

tan α = 0.3 α = 16° 42'

1—2

θ = 16° 42', 196° 42'

θ = 33° 24', 393° 24' ∴ θ = 33° 24' (f) cos 3θ � 0 ⇒ 3θ is in the 2nd or

3rd quadrant and 0° � θ � 360 ⇒ 0° � 3θ � 1080°

cos α = 1—2

α = 45°

x

y

O

P

Q

θ1

3

10

x

y

O

P

Qθ3

54

x

y

O

P

θ

Q3

45

x

y

O

P

Qθ

8

15

17

x

y

O

P

Q

θ24

25

7

α

α


3θ = 135°, 225°, 495°, 585°, 855°, 945°

∴ θ = 45°, 75°, 165°, 195°, 285°, 315°

(g) sin (2θ + 30°) � 0 ⇒ 2θ + 30° is in the 1st or 2nd quadrant and0° � θ � 360°

⇒ 30° � 2θ + 30° � 750° sin α = 0.75 α = 48° 35' 2θ + 30° = 48° 35', 131° 25',

408° 35', 491° 25' 2θ = 18° 35', 101° 25', 378° 35',

461° 25' ∴ θ = 9°18', 50°43', 189°18',

230°43' (h) tan (2θ – 40°) � 0 ⇒ 2θ – 40° is

in 2nd or 4th quadrant and 0° � θ � 360°

⇒ –40° � 2θ – 40° � 680° tan α = 0.7 α = 35° 2θ – 40° = –35°, 145°, 325°, 505° 2θ = 5°, 185°, 365°, 545° ∴ θ = 2° 30', 92° 30', 182° 30',

272° 30'

16 (a) 2 sin θ cos θ = sin θ 2 sin θ cos θ – sin θ = 0 sin θ (2 cos θ – 1) = 0

sin θ = 0 or cos θ = 1—2

sin θ = 0 θ = 0°, 180°, 360°

cos θ = 1—2

θ = α, 360° – α = 60°, 300° ∴ θ = 0°, 60°, 180°, 300°, 360° (b) 2 cosec2 θ + cot θ = 8 2(1 + cot2 θ) + cot θ = 8 2 cot2 θ + cot θ – 6 = 0 (2 cot θ – 3)(cot θ + 2) = 0

cot θ = 3—2

or cot θ = –2

cot θ = 3—2

1

——–tan θ

= 3—2

tan θ = 2—3

θ = 33° 41', 213° 41' cot θ = –2

1

——–tan θ

= –2

tan θ = –1—2

θ = 180° – α, 360° – α = 153° 26', 333° 26' ∴ θ = 33° 41', 153° 26', 213° 41',

333° 26' (c) 4 tan θ = sin θ

4� sin θ——–cos θ � = sin θ

4 sin θ = sin θ cos θ

4 sin θ – sin θ cos θ = 0 sin θ (4 – cos θ) = 0 sin θ = 0 or cos θ = 4 sin θ = 0 θ = 0°, 180°, 360° cos θ = 4 (no solution) ∴ θ = 0°, 180°, 360° (d) 3 cos θ = cot θ

3 cos θ = cos θ——–sin θ

3 sin θ cos θ = cos θ 3 sin θ cos θ – cos θ = 0 cos θ (3 sin θ – 1) = 0

cos θ = 0 or sin θ = 1—3

cos θ = 0 θ = 90°, 270° or 3 sin θ – 1 = 0

sin θ = 1—3

θ = 19° 28', 160° 32' ∴ θ = 19° 28', 90°, 160° 32', 270° (e) 2 cos2 θ = 3 cos θ – 1 2 cos2 θ – 3 cos θ + 1 = 0 (2 cos θ – 1)(cos θ – 1) = 0

cot θ = 1—2

or cot θ = 1

cos θ = 1—2

θ = 60°, 300° or cos θ = 1 θ = 0°, 360°

∴ θ = 0°, 60°, 300°, 360° (f) 2 tan2 θ + 1 = 3 tan θ 2 tan2 θ – 3 tan θ + 1 = 0 (2 tan θ – 1)(tan θ – 1) = 0

tan θ = 1—2

or tan θ = 1

tan θ = 1—2

θ = 26° 34', 206° 34' or tan θ = 1 θ = 45°, 225° ∴ θ = 26° 34', 45°, 206° 34', 225° (g) sec θ = 3 cos θ

1

——–cos θ

= 3 cos θ

3 cos2 θ = 1

cos2 θ = 1—3

cos θ = ± 1—3

∴ θ = 54° 44', 125° 16', 234° 44', 305° 16'

(h) 4 cot2 θ + 6 = 11 cot θ 4 cot2 θ – 11 cot θ + 6 = 0 (4 cot θ – 3)(cot θ – 2) = 0

cot θ = 3—4

or cot θ = 2

cot θ = 3—4

tan θ = 4—3

θ = 53° 8', 233° 8'

cot θ = 2

tan θ = 1—2

θ = 26° 34', 206° 34' ∴ θ = 26° 34', 53° 8', 206° 34',

233° 8'17 (a)

(b)

(c)

(d)

(e)

(f)

x

y

y = sin x + 23

2

1

0 90° 180° 270° 360°

x

y

y = 4 cos x4

0

–4

90° 180° 270° 360°

x

y

y = 3 cos x + 5

8

7

6

5

4

3

2

1

O 90° 180° 270° 360°

x

y

y = 3 tan x 135° 315°

3

O–3

45° 90° 180° 225°

x

y

y = tan x – 1

O

–1

–2

45° 135° 225° 315° 90° 180° 270° 360°

x

y

y = 2 sin x2

O

–2

90° 180° 270° 360°


(g)

18 (a)

(b)

(c)

(d)

(e)

(f)

(g)

19 (a) y = 3 cos 2x

(b) y = –2 cos 1—2

x

(c) y = |tan x| (d) y = 2 sin 3x + 220 (a)

∴ The number of solutions = 4

(b)


(c)

∴ The number of solutions = 4 (d)


(e)

∴ The number of solutions = 3 (f)

∴ The number of solutions = 5 21 (a) sec x – cos x =

1——–cos x

– cos x

= 1 – cos2 x————

cos x

= sin2 x

———cos x

= sin x � sin x———cos x �

= sin x tan x(b) (cot A + tan A) cos A

= � 1———tan A

+ tan A� cos A

= � 1 + tan2 A

————–tan A � cos A

= �(sec2 A) cos A

———sin A � cos A

= cos2 A

————–—cos2 A sin A

= 1

——–sin A

= cosec A

x

y

y = 4 tan 2 x

4

0–4

67.5° 157.5°

22.5° 45° 90° 112.5° 135° 180°

x

y

y = |2 sin x – 1|

3

2

1

0

–1

–2

–3

90° 180° 270° 360°

x

y

y = |3 cos 2x + 1|

4

3

2

1

0

–1

–2

45° 90° 135° 180°

x

yy = |5 cos x – 2|

7

6

5

4

3

2

1

01

2

3

4

5

6

7

90° 180° 270° 360°

1y = –3 cos —x 2

x

y

3

0

–3

180° 360° 540° 720°

x

y

y = 4 – 2 sin 2x6

4

2

0 45° 90° 135° 180°

x

yy = 1 – cos 2x

2

1

0 45° 90° 135° 180°

x

y

xy = 1 – —– 2π

y = |sin x|1

O

–1

π—2

π 2π 3—π 2

3xy = —– 2π

x

y

y = |2 cos x|

2

O

–2

π 2π 3—π 2

π—2

xy = —– 4π

x

y

y = cos 2x

1

O

–1

π 2π 3—π 2

π—2

3xy = 4 – —– π

x

y

y = 3 sin x + 14

3

2

1

O–1

–2

π 2π 3—π 2

π—2

x

y

3y = 3 cos —x 2

3xy = — – 3 π

3

O

–3

π—3

π 2π 2—π 3

4—π 3

5—π 3

xy = — π

x

y

y = 1 – sin 2x2

1

O π 2π 3—π 2

π—2

x

y

y = 3 cos x – 21

O

–1

–2

–3

–4

–5

90° 180° 270° 360°


(c) cosec A – sin A = 1

——–sin A

– sin A

= 1 – sin2 A————

sin A

= cos2 A———sin A

= cos A � cos A———sin A �

= cos A cot A

(d) sin2 θ(1 + cot2 θ)——————–—

cos2 θ

= sin2 θ(cosec2 θ)————–——

cos2 θ

= sin2 θ

———cos2 θ

� 1———sin2 θ �

= 1

———cos2 A

= sec2 θ

(e) sin A

————1 + cos A

+ 1 + cos A

————–sin A

= sin2 A + (1 + cos A)2

————–——–——–sin A(1 + cos A)

= sin2 A + 1 + 2 cos A + cos2 A————–——–—————–

sin A(1 + cos A)

= 2 + 2 cos A

————–——–sin A(1 + cos A)

= 2(1 + cos A)

————–——–sin A(1 + cos A)

= 2

——–sin A

= 2 cosec A

(f) tan A

————–1 + tan2 θ

= � sin θ——–cos θ �

——––sec2 θ

= sin θ——–cos θ

÷ 1

——–cos2 θ

= sin θ——–cos θ

(cos2 θ)

= cos θ sin θ22 (a) 2 cos2 θ + sin θ = 1 2(1 – sin2 θ) + sin θ = 1 2 sin2 θ – sin θ – 1 = 0 (2 sin θ + 1)(sin θ – 1) = 0

sin θ = –1—2

or sin θ = 1

sin θ = –1—2

θ = 210°, 330° sin θ = 1 θ = 90° ∴ θ = 90°, 210°, 330° (b) 2 cosec2 θ – 1 = 3 cot θ 2(1 + cot2 θ) – 1 = 3 cot θ 2 + 2 cot2 θ – 1 = 3 cot θ 2 cot2 θ – 3 cot θ + 1 = 0 (2 cot θ – 1)(cot θ – 1) = 0

cot θ = 1—2

or cot θ = 1

cot θ = 1—2

tan θ = 2 θ = 63° 26', 243° 26'

cot θ = 1 tan θ = 1 θ = 45°, 225° ∴ θ = 45°, 63° 26', 225°, 243° 26' (c) 2 cot2 θ + cosec θ = 4 2(cosec2 θ – 1) + cosec θ = 4 2 cosec2 θ – 2 + cosec θ = 4 2 cosec2 θ + cosec θ – 6 = 0 (2 cosec θ – 3)(cosec θ + 2) = 0

cosec θ = 3—2

or cosec θ = –2

cosec θ = 3—2

sin θ = 2—3

θ = 41° 49', 138° 11' cosec θ = –2

sin θ = –1—2

θ = 210°, 330° ∴ θ = 41° 49', 138° 11' 210°, 330°

(d) 3 sec2 θ = 5�1 + 1

——–cot θ �

3(1 + tan2 θ) = 5(1 + tan θ) 3 tan2 θ + 3 = 5 + 5 tan θ 3 tan2 θ – 5 tan θ – 2 = 0 (3 tan θ + 1)(tan θ – 2) = 0

tan θ = –1—3

or tan θ = 2

tan θ = –1—3

θ = 161° 34', 341° 34' tan θ = 2 θ = 63° 26', 243° 26' ∴ θ = 63° 26', 161° 34', 243° 26',

341° 34' (e) 5 tan2 θ = 11 sec θ – 7 5(sec2 θ – 1) = 11 sec θ – 7 5 sec2 θ – 11 sec θ + 2 = 0 (5 sec θ – 1)(sec θ – 2) = 0

sec θ = 1—5

or sec θ = 2

sec θ = 1—5

cos θ = 5 (no solution) sec θ = 2

cos θ = 1—2

θ = 60°, 300° ∴ θ = 60°, 300° (f) 3 cosec2 θ = 2 sec θ

3

——–sin2 θ

= 2

——–cos θ

2 sin2 θ = 3 cos θ 2(1 – cos2 θ) = 3 cos θ 2 cos2 θ + 3 cos θ – 2 = 0 (2 cos θ – 1)(cos θ + 2) = 0

cos θ = 1—2

or cos θ = –2

cos θ = 1—2

θ = 60°, 300° cos θ = –2 (no solution) ∴ θ = 60°, 300°

23

(a) sin (A – B) = sin A cos B – cos A sin B

= � 4—5 �� 5

—–13 � – � 3

—5 �� 12

—–13 �

=

20—–65

– 36—–65

= –16—–65

(b) cos (A + B) = cos A cos B – sin A sin B

= � 3—5 �� 5

—–13 � – � 4

—5 �� 12

—–13 �

=

15—–65

– 48—–65

= –33—–65

(c) tan (A – B)

= tan A – tan B

————–——–1 + tan A tan B

=

4—3

– 12—–5————–——–

1 + � 4—3 �� 12

—–5 �

= �– 16

—–15 �

———–

�– 11—–5 �

=

16—–33

24

(a) sin (α + β)—————cos (α – β)

= sin α cos β + cos α sin β———————————cos α cos β + sin α sin β

h�——––�1 + h2

1�——––�1 + k2 +

1�——––�1 + h2

k�——––�1 + k2

= ——————————

1�——––�1 + h2

1�——––�1 + k2 +

h�——––�1 + h2

k�——––�1 + k2

= h + k

————————( 1 + h2 )( 1 + k2 )

÷

1 + hk

————————( 1 + h2 )( 1 + k )

=

h + k——––1 + hk

(b) tan (α + β) = tan α + tan β

————–——–1 – tan α tan β

= h + k

——––1 – hk

x x

y

B0 5

1213

y

A0 3

45

x

1 + h2

y

O

h

1α

x

1 + k2

y

O

k

1β


25

(a) sin 2A = 2 sin A cos A

= 2� 4—5 �� 3

—5 �

= 24—–25

(b) cos 2A = 1 – 2 sin2 A

= 1 – 2� 4—5 �2

= 1 – 32—–25

= –7

—–25

(c) tan 2A = 2 tan A

————–1 – tan2 A

=

2� 4—3 �

————––1 – � 4

—3 �

2

= � 8—3 �

——–

� –7—–9 �

= –

24—–7

26

(a) sin 2θ = 2 sin θ cos θ = 2m 1 – m2

(b) cos 4θ = 1 – 2 sin2 2θ = 1 – 2 (2m 1 – m2 )2

= 1 – 2 (4m2 [1 – m2)] = 1 – 8m2 + 8m4

(c) cos θ = 1 – 2 sin2 1—2

θ

m = 1 – 2 sin2 1—2

θ

2 sin2 1—2

θ = 1 – m

sin2 1—2

θ = 1 – m

—–—–2

Since θ is an acute angle,

sin 1—2

θ = 1 – m——–2

27

(a) sin 2B = 2 sin B cos B

= 2� 4—5 ��– 3

—5 �

= –24—–25

sin 4B = 1 – 2 sin2 2B

= 1 – 2�– 24—–25 �

2

= –527—––625

(b) cos B = 2 cos2 1—2

B – 1

–3—5

= 2 cos2 1—2

B – 1

2 cos2 1—2

B = 2—5

cos2 1—2

B = 1—5

(c) tan B = 2 tan B

—2————––

1 – tan2 B—2

–4—3

= 2 tan B

—2————––

1 – tan2 B—2

4 tan2 B—2

– 4 = 6 tan B—2

2 tan2 B—2

– 3 tan B—2

– 2 = 0

(2 tan B—2

+ 1)(tan B—2

– 2) = 0

tan B—2

= –1—2

or tan B—2

= 2

Since B is an obtuse angle,

90° � B � 180° ⇒ 45° � B—2

� 90

∴ tan B—2

= 2

28 (a) tan 2A = 2 tan A

————–1 – tan2 A

4—3

= 2 tan A

————–1 – tan2 A

4 – 4 tan2 A = 6 tan A 2 tan2 A + 3 tan A – 2 = 0 (2 tan A – 1)(tan A + 2) = 0

tan A = 1—2

or tan A = –2

Since A is an acute angle,

tan A = 1—2

(b) tan 3A = tan (2A + A)

= tan 2A + tan 2A

————–———1 – tan 2A tan A

=

4—3

+ 1—2————––—

1 – � 4—3 �� 1

—2 �

= � 11—–6 �

——–

� 1—3 �

=

11—–2

29 (a) sin 3A = sin (2A + A) = sin 2A cos A + cos 2A sin A = (2 sin A cos A)(cos A) + (1 – 2 sin2 A)(sin A) = 2 sin A cos2 A + sin A – 2 sin3 A = 2 sin A (1 – sin2 A) + sin A – 2 sin3 A = 2 sin A – 2 sin3 A + sin A – 2 sin3 A = 3 sin A – 4 sin3 A

(b) sin 2A

————–1 – cos 2A

= 2 sin A cos A

————–———1 – (1 – 2 sin2 A)

= 2 sin A cos A

————–——2 sin2 A

= cos A

———sin A

= cot A

(c) cos (A + B)—————cos A cos B

=

cos A cos B – sin A sin B———————————–

cos A cos B

= cos A cos B—————cos A cos B

– sin A sin B—————cos A cos B

= 1 – tan A tan B

(d) cot (A + B) = 1

————–tan (A + B)

= 1

———————

� tan A + tan B——————–1 – tan A tan B �

=

1 – tan A tan B———————

tan A + tan B

= � 11 – ————— cot A cot B �

——–————— 1 1——– + ——– cot A cot B

= � cot A cot B – 1———————

cot A cot B �——–—————

� cot A + cot B——————

cot A cot B � =

cot A cot B – 1———————

cot A + cot B30 (a) 4 sin 2θ = sin θ 4(2 sin θ cos θ) – sin θ = 0 sin θ (8 cos θ – 1) = 0

sin θ = 0 or cos θ = 1—8

sin θ = 0 θ = 0°, 180°, 360°

cos θ = 1—8

θ = 82° 49', 277° 11' ∴ θ = 0°, 82° 49', 180°, 277° 11', 360° (b) tan 2θ = 2 cot θ

2 tan θ

————–1 – tan2 θ

= 2

—–—–tan θ

2 tan2 θ = 2 – 2 tan2 θ 4 tan2 θ = 2

tan2 θ = 1—2

tan θ = ± 1—2

x

y

Oθ

m

11 – m2

x

y

O

B

3

45

x

y

AO 3

45


∴ θ = 35° 16', 144° 44', 215° 16', 324° 44'

(c) cos 2θ = cos θ + 2 3(2 cos2 θ – 1) = cos θ + 2 6 cos2 θ – cos θ – 5 = 0 (6 cos θ + 5)(cos θ – 1) = 0

cos θ = –5—6

or cos θ = 1

cos θ = –5—6

θ = 146° 27', 213° 33' cos θ = 1 θ = 0°, 360° ∴ θ = 0°, 146° 27', 213° 33°, 360° (d) 4 sin θ cos θ = 1 2(2 sin θ cos θ) = 1

sin 2θ = 1—2

sin 2θ � 0 ⇒ 2θ is in the 1st or 2nd quadrant and 0° � θ � 360° ⇒ 0° � 2θ � 720°

For angle 2θ in this interval 2θ = 30°, 150°, 390°, 510° ∴ θ = 15°, 75°, 195°, 255° (e) 3 tan θ = 2 tan (45° – θ)

= 2� tan 45° – tan θ———————–1 + tan 45° tan θ �

= 2� 1 – tan θ————–1 + tan θ �

3 tan θ + 3 tan2 θ = 2 – 2 tan θ 3 tan2 θ + 5 tan θ – 2 = 0 (3 tan θ – 1)(tan θ + 2) = 0

tan θ = 1—3

or tan θ = –2

tan θ = 1—3

θ = 18° 26', 198° 26' tan θ = –2 θ = 116° 34', 296° 34' ∴ θ = 18° 26', 116° 34', 198° 26',

296° 34' (f) sin 2θ + 3 cos2 θ = 0 2 sin θ cos θ + 3 cos2 θ = 0 cos θ (2 sin θ + 3 cos θ) = 0 cos θ = 0 or 2 sin θ = –3 cos θ cos θ = 0 θ = 90°, 270° 2 sin θ = –3 cos θ

sin θ

—–—–cos θ

= –3—2

tan θ = –3—2

θ = 123° 41', 303° 41' ∴ θ = 90°, 123° 41', 270°, 303° 41'

SPM Appraisal Zone

1 (a) 0.87(b) –0.5

(c) tan 120° = 0.87—–––0.5

= –1.74

2 270° � θ � 360° ⇒ θ is in the 4th quadrant.

(a) cos (–θ) = cos θ

= 5

—–41

(b) sin 2θ = 2 sin θ cos θ

= 2�– 4—–41 �� 5

—–41 �

= –40—–41

3

(a) cot A = 1

———tan A

= 1—p

(b) cos (90° – A) = sin A

= p

—–—–1 + p2

4

(a) tan2 θ = � 1– m2

—–—–m �

2

= 1– m2

—–—–m2

(b) cos θ = 2 cos2 1—2

θ – 1

m = 2 cos2 1—2

θ – 1

cos2 1—2

θ = m + 1

—–—–2

5 (a) cos 2A = 2 cos2 A – 1

= 2� 3—5 �

2

– 1

= –7

—–25

(b) cos 4 A = 2 cos2 2A – 1

= 2�– 7—–25 �

2

– 1

= –527—––625

6

(a) cos 65° = sin (90° – 65°) = sin 25° = 1 – h2

(b) tan (–155°) = –tan 155° = –(–tan 25°) = tan 25°

= 1– h2

—–—–h

7

(a) sec A = 1

———cos A

= 1

———

�– 3—5 �

= –

5—3

(b) cosec A = 1

———sin A

= 1

——

� 4—5 �

=

5—4

(c) cot A = 1

———tan A

= 1

——

�– 4—3 �

= –

3—4

8 (a) cos x = 1—2

x = 45°, 315° (b) 2 sin x cos x = cos x 2 sin x cos x – cos x = 0 cos x (2 sin x – 1) = 0

cos x = 0 or sin x = 1—2

cos x = 0 x = 90°, 270°

sin x = 1—2

x = 30°, 150° ∴ x = 30°, 90°, 150°, 270°

9 2 cos2 x = 3 cos x – 2 cos 60°

2 cos2 x – 3 cos x + 2� 1—2 � = 0

2 cos2 x – 3 cos x + 1 = 0 (2 cos x – 1)(cos x – 1) = 0

cos x = 1—2

or cos x = 1

cos x = 1—2

x = 60°, 300° cos x = 1 x = 0°, 360°

∴ x = 0°, 60°, 300°, 360°

x

y

Oθ

5

441

x

y

OA

p

1

1 + p2

x

y

Oθ

1

m

1 – m2

x

y

O

1

h

1 – h2

25°

x

y

O3

45

A


10

From the sketch, a = 5 and b = 3.11 a = 4, b = 4 and c = 1 ∴ y = 4 cos 4x + 1

12 (a) a = 3 and period = 360—––

2 = 180°

(b) (90°, 1)13

∴ Range = 2 � y � 4

14 a = 2, b = 315 a = 3, b = 416 y = –2 cos 3x

17 (a) a = 2, b = 3—2

and c = 1

(b) y = 1 – 2 sin 3—2

x

18 y = |4 cos 2x|

19 cot θ – tan θ

——————cot θ + tan θ

=

1——–tan θ

– tan θ————––—

1——–tan θ

+ tan θ

= � 1 – tan2 θ

—————tan θ �

——–————

� 1 + tan2 θ—————

tan θ �=

1 – tan2 θ—————1 + tan2 θ

= 1 – tan2 θ

—————sec2 θ

= 1

———sec2 θ

– � sin2 θ———cos2 θ �

——–——

� 1———cos2 θ �

= cos2 θ – sin2 θ

20 1 – cot2 θ

—————1 + cot2 θ

= 1 – cot2 θ

—————cosec2 θ

= 1

———–cosec2 θ

– � cos2 θ———sin2 θ �

——–——1

———sin2 θ

= sin2 θ – cos2 θ = 1 – cos2 θ – cos2 θ = 1 – 2 cos2 θ ∴ a = 1 and b = –221 5 + cot y = 2 cosec2 y 5 + cot y = 2(1 + cot2 y) 2 cot2 y – cot y – 3 = 0 (2 cot y – 3)(cot y + 1) = 0

cot y = 3—2

or cot y = –1

cot y = 3—2

1

——–tan y

= 3—2

tan y = 2—3

y = 33° 41', 213° 41' cot y = 1

1

——–tan y

= 1

tan y = 1 y = 45°, 225°∴ y = 33° 41', 45°, 213° 41', 225°

22 3 cos2 θ + 3 cos θ = sin2 θ 3 cos2 θ + 3 cos θ = 1 – cos2 θ 4 cos2 θ + 3 cos θ – 1 = 0 (4 cos θ – 1)(cos θ + 1) = 0

cos θ = 1—4

or cos θ = –1

cos θ = 1—4

θ = 75° 31', 284° 29' cos θ = –1 θ = 180°

∴ θ = 75° 31', 180°, 284° 29'23 sin (A + B)

= sin A cos B + cos A sin B

= � 15—–25 �� 24

—–25 � + � 20

—–25 �� 7

—–25 �

= 360——625

+ 140——625

= 500——625

= 4—5

24 tan (α – β) = tan α – tan β

——————–1 + tan α tan β

=

3—4

– 12—–5————––—

1 + � 3—4 �� 12

—–5 �

= �– 33

—–20 �

——––

� 14—–5 �

= –

33—–56

25 cos 2A = 2 cos2 A – 1

= 2� 1—–——1 + m2 �2

– 1

= 2

———1 + m2

– 1

= 2 – (1 + m2)—————–

1 + m2

= 1 – m2

———1 + m2

26

sin A = 2 sin A—2

cos A—2

= 2� t—–——

1 + t2 �� 1—–——

1 + t2 � =

2t———1 + t2

27

(a) sin (A + B) = sin A cos B + cos A sin B

= �– 15—–17 ��– 3

—5 � + �– 8

—–17 ��– 4

—5 �

=

45—–85

+ 32—–85

= 77—–85

(b) cos B = 2 cos2 B—2

– 1

–3—5

= 2 cos2 B—2

– 1

2 cos2 B—2

= 1 – 3—5

cos2 B—2

= 1—5

cos 1—2

B = ± 1—5

180° � B � 270°

⇒ 90° � B—2

� 135°

∴ cos 1—2

B = –1—5

28 (a) cos 2A = 2 cos2 A – 1

= 2� 3—5 �

2

– 1

= –7

—–25

(b) tan A = 2 tan

A—2————––

1 – tan2 A—2

4—3

= 2 tan

A—2————––

1 – tan2 A—2

x

y

O 1

t

1 + t2

A—2

x x

y

O

A8

1517

y

O

B3

45

x

y

y = a sin x + b

8

3

O

–2π 2π 3—π

2π—2

x

yy = 3 – cos 2x

4

3

2

1

0 π 3—π 2

π—4

π—2


4�1 – tan2 A—2 � = 6 tan

A—2

4 tan2 A—2

+ 6 tan A—2

– 4 = 0

2 tan2 A—2

+ 3 tan A—2

– 2 = 0

�2 tan A—2

– 1��tan A—2

+ 2� = 0

tan A—2

= 1—2

or tan A—2

= –2

0° � A � 90° ⇒ 0° � A—2

� 45°

∴ tan 1—2

A = 1—2

29

(a) cos (A + B) = cos A cos B – sin A sin B

= � 3—5 ��– 24

—–25 � – � 4

—5 �� 7

—–25 �

= –72

—––125

– 28

——125

= –100—––125

= –4—5

(b) tan (B – A) = tan B – tan A

——————–1 + tan B tan A

= –

7—–24

– 4—3————––—–

1 – �– 7—–24 �� 4

—3 �

=

�– 13—–8 �

——––

� 25—–18 �

= –117—––100

30 1 + cos 2θ

—————sin 2θ

= 1 + 2 cos2 θ – 1————–——–

2 sin θ cos θ

= 2 cos2 θ

————–—2 sin θ cos θ

= cos θ——–sin θ

= cot θ31 2 cos 2x = 2 – 3 sin x

2(1 – 2 sin2 x) = 2 – 3 sin x 4 sin2x – 3 sin x = 0 sin x (4 sin x – 3) = 0

sin x = 0 or sin x = 3—4

sin x = 0 x = 0°, 180°, 360°

sin x = 3—4

x = 48° 35', 131° 25'∴ x = 0°, 48° 35', 131° 25', 180°, 360°

32 sin 2θ + 3 cos2θ = 0 2 sin θ cos θ + 3 cos2 θ = 0 cos θ (2 sin θ + 3 cos θ) = 0

cos θ = 0 or 2 sin θ = –3 cos θ cos θ = 0 θ = 90° 270°

2 sin θ = –3 cos θ

tan θ = –3—2

θ = 123° 41', 303° 41' ∴ θ = 90°, 123° 41', 270°, 303° 41'

33 (a), (b)

x = π |sin 2x|

|sin 2x| = x—π

y = x—π

x 0 πy 0 1


34 (a), (b)

x = π (1 + cos 2x)

x—π = 1 + cos 2x

2 + 2 cos 2x = 2x—π

y = 2x—π

x 0 πy 0 2


35


36 (a) 1 – tan2 x

———–—1 + tan2 x

= � sin2 x1 – ——– cos2 x�——–——–

� sin2 x1 + ——– cos2 x �

= cos2 x – sin2 x

————–——cos2 x + sin2 x

= cos2 x – (1 – cos2 x) ————–———–cos2 x + 1 – cos2 x

= 2 cos2 x – 1 = cos 2x (b) (i), (ii)

1 – tan2 x

———–—1 + tan2 x

+ 2x—π = 1

cos 2x = 1 – 2x—π

–y = 1 – 2x—π

y = 2x—π – 1

x 0 πy –1 1


∴ The number of solutions = 638 (a), (b)

3x—π + tan 2x = 2

y = 2 – 3x—π

x 0 πy 2 –1


x

y

O

A 3

45

x

y

O

B

24

725

y

xO

1

π—4

π—2

3—π 4

π

y = |sin 2x| xy = — π

y

xO

3

π—2

3—π 2

π

y = –3 cos x

xy = 1– — π

–3

2π

y

xO

1

π—4

π—2

3—π 4

π

2xy = — – 1 π

–1

y

xO

2

π—3

2—π 3

π

xy = 2 – — π

4—π 3

5—π 3

2π

3y = |2 sin —x| 2

y

π—4

3—π 4

π

y = tan 2x

3xy = 2 – — π

Oxπ—

2

y

xO

4

π—4

π—2

3—π 4

π

y = 2 + 2 cos 2x

2xy = — π2


39


(a) 1

——–sin A

+ 1

——–cot B

= 1

——–sin A

+ tan B

= 1

——–

� 12—–13 �

+ �– 3—4 �

=

1—3

(b) sin A cos B

————————–sin A sin B + cos B

= � 12—–13 ��– 4

—5 �

————––—–—–

� 12—–13 �� 3

—5 � + �– 4

—5 �

= �– 48

—–65 �

——––

�– 16—–65 �

= 3

(c) sin (A – B)

—————–cos (A + B)

= sin A cos B – cos A sin B————————–——–cos A cos B – sin A sin B

=

� 12—–13 ��– 4

—5 � – �– 5

—–13 �� 3

—5 �

————––—–—–———

�– 5—–13 ��– 4

—5 � – � 12

—–13 �� 3

—5 �

=

�– 33—–65 �

——––

�– 16—–65 �

=

33—–16

41

(a) cos 2A = 1 – 2 sin2 A

= 1 – 2� 4—5 �

2

= –7

—–25

(b) sin 2A = 2 sin A cos A

= 2� 4—5 ��– 3

—5 �

= –24—–25

(c) cos A = 1 – 2 sin2 A—2

–3—5

= 1 – 2 sin2 A—2

2 sin2 A—2

= 1 + 3—5

= 8—5

sin2 A—2

= 4—5

sin A—2

= ± 2—5

90 � A � 180 ⇒ 45 � A—2

� 90

∴ sin 1—2

A = 2—5

42

(a) cos 2θ = 4—5

(b) cos 2θ = 1 – 2 sin2 θ

4—5

= 1 – 2 sin2 θ

2 sin2 θ = 1—5

sin2 θ = 1

—–10

sin θ = ± 1—10

Since θ is an acute angle,

sin θ = 1—10

(c)

tan θ = 1—3

(d) tan 3θ = tan (θ + 2θ)

= tan θ + tan 2θ

———————–1 – tan θ tan 2θ

=

1—3

+ 3—4————––—

1 – � 1—3 �� 3

—4 �

= � 13—–12 �

——––

� 3—4 �

=

13—–9

x

y

O

A

3

45

x

y

0θ

101

3

y

x0

53

4

2θ

y

xO

3

π

xy = — – 3 π

2π

1y = 3 cos —x 2

–3

3π 4π

x x

y

O

A

5

12

13

y

O

B3

4

5

17 Permutations and Combinations

1

Booster Zone

1 (a) 9! = 362 880(b) 8! = 40 320(c) 7! = 5040(d) 6! = 720

2 (a) 5P4 = 120

(b) 6P4 = 360

(c) 8P4 = 1680

(d) 7P4 = 840

3 (a) 5! = 120(b) (i) 24 × 2! = 48

(ii) 120 – 48 = 72

4 (a) 6 × 4! = 144(b)

The number of arrangements = 4 × 4 × 3 × 2 × 1 × 2 = 192

5 (a) 8! = 40 320(b) (i)

The number of arrangements = 5 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 25 200

(ii)

The number of arrangements = 7 × 6 × 5 × 4 × 3 × 2 × 1 × 3 = 15 120

(iii) 720 × 3! = 4320

6 (a) 7! = 5040(b) (i)

The number of arrangements = 1 × 6 × 5 × 4 × 3 × 2 × 1 = 720(ii)

or

The number of arrangements = 5! × 2 = 240

7 (a)

The number of arrangements = 60 480

(b)

The number of arrangements = 100 800(c) 120 × 5! = 14 400(d)

= 5! × 4! = 2880

8 6P4 = 360

9

The number of arrangements = 2 × 6 × 5 × 4 × 3= 720

10 (a) 7P5 = 2520

(b) (i)

The number of arrangements = 4 × 6 × 5 × 4 × 3 = 1440(ii)

The number of arrangements = 6 × 5 × 4 × 3 × 4 = 1440

11 30C5 = 142 506

12 20C4 = 4845

13 5C3 × 10C

8 = 450

14 6C3 × 4C

2 = 120

15 (a) 6C4 × 2C

2 = 15

(b) 8C5 × 3C

3 = 56

16 (a) 16C7 = 11 440

(b) 10C4 × 6C

3 = 4200

(c) 11 440 – (10C7 × 6C

0) = 11 320

17 (a) (8C4 × 4C

0) + (8C

0 × 4C

4)

= 70 + 1 = 71(b) 8C

2 × 4C

2 = 168

18 (a) 16C4 = 1820

(b) 1 × 14C2 = 91

(c) (8C3 × 8C

1) + (8C

4 × 8C

0)

= 448 + 70 = 518

19 (a) 6C2 × 3C

2 = 45

(b) 6C4 × 3C

0 = 15

20 (a) 10C3 = 120

(b) (1 × 10C4) + (1 × 10C

4) = 420

SPM Appraisal Zone

1 (a) 8! = 40 320(b) 5040 × 2! = 10 080

2

The number of arrangements = 3 × 4 × 3 × 3= 108

3

The number of arrangements = 7 × 7 × 6 × 5 × 4 × 3 × 2 × 1= 35 280

4 (a) 8P5 = 6720

(b) = 240

or = 240

or = 240

or = 240

∴ The number of ways that 5-digit numbers can be formed

= 240 + 240 + 240 + 240 = 960

5 (a) 6! = 720(b)

The number of permutations = 1 × 4 × 3 × 2 × 1 × 1 = 24

6 = 12

or = 18

The number of arrangements = 12 + 18= 30∴ There are 30 odd 4-digit numbers

less than 3000 which can be formed using the digits 1, 2, 3, 4 and 5 without repetition.

7

The number of arrangements = 6 × 5 × 3= 90

C V

4 4 3 2 1 2

V

5 7 6 5 4 3 2 1

C

7 6 5 4 3 2 1 3

1 6 5 4 3 2 1

RA

1 5 4 3 2 1 1

R A

1 5 4 3 2 1 1

BB

4 3 7 6 5 4 3 2 1

5 4 4 3 3 2BG

2 1 1BG BG BG G

V

2 6 5 4 3

6 5 4 3 4O

4 6 5 4 3

7 7 6 5 4 3 2 1

6

6

6 5 4 2 1

5 2 1 4

2 1 5 4

2 1 6 5 4

1 3 2 2

1 3 2 3

01

02

6 5 3E

G B

5 7 6 5 4 3 2 1 4

3 4 3 3CV

T

1 4 3 2 1 1C



8 (a)

The number of permutations = 3 × 2 × 2 × 1 × 1 = 12(b) 24 × 2! = 48

9 6P4 × 5P

3 = 21 600

10 (a) 9C3 = 84

(b) (5C2 × 4C

1) + (5C

3 × 4C

0)

The number of arrangements = 40 + 10 = 50

11 (a) 8C4 = 70

(b) 1 × 7C3 = 35

12 (a) 7C3 × 8C

2 = 980

(b) 7C5 + 8C

5 = 21 + 56

= 77

13 (a) 12C5 = 792

(b) The number of arrangements = 792 – [(7C

0 × 5C

5) + (7C

1 × 5C

4)

+ (7C2 × 5C

3)]

= 792 – (1 + 35 + 210) = 546

14 (6C3 × 4C

2) + (6C

4 × 4C

1)

= 120 + 60= 180

15 (a) 1 × 12C6 = 924

(b) 4C2 × 9C

5 = 756

(c) (8C2 × 5C

5) + (8C

3 × 5C

4) +

(8C4 × 5C

3) + (8C

5 × 5C

2)

= 28 + 280 + 700 + 560 = 1568

M

3 2 2 1 1M SMS

18 Probability

1

Booster Zone

2 9 1 (a) —– (b) —– 13 26

1 2 2 (a) — (b) — 2 3 1 1 (c) — (d) — 3 2

1 3 3 (a) — (b) — 4 8 7 1 (c) —– (d) —– 24 12

1 5 4 (a) — (b) — 2 6

24 1 5 (a) —– (b) — 35 7 3 (c) — 8

k 2 6 ——–– = — k + 4 3 3k = 2k + 8 k = 8

x 3 7 (a) —– = — 30 5 3 x = — (30) 5 = 18 ∴ The number of yellow beads = 30 – 18 = 12

18 + y 5 (b) ——––– = — 30 + y 8 144 + 8y = 150 + 5 3y = 6 y = 2 ∴ 2 blue beads should be added

to the bag.

1 8 8 (a) — (b) — 3 9

1 5 9 (a) — (b) — 3 6

1 710 (a) — (b) — 2 8

2 3 11 (a) — (b) — 3 5 11 (c) 0 (d) —– 15

12 P(A � B) = P(A) + P(B) 0.8 = 0.55 + x x = 0.25

2 113 (a) — (b) — 5 3 11 (c) —– (d) 0 15

2 113 (a) — (b) — 5 3 11 (c) —– (d) 0 15

2 114 (a) — (b) — 9 9 1 (c) — (d) 1 3

15 (a) P(Zhao Ming or Mukhriz wins) = P(Zhao Ming wins) + P(Mukhriz

wins) 2 1 = — + —– 5 10 1 = — 2 1 (b) P(some one else wins) = 1 – — 2 1 = — 2

16 (a) P(P or Q wins)

= P(P wins) + P(Q wins) 1 1 = — + — 4 6 5 = —– 12 (b) P(Q or R wins) = P(Q wins) + P(Q wins) 1 1 = — + —– 6 12 1 = — 4 P(neither Q nor R wins) = 1 – P(Q or R wins) 1 = 1 – — 4 3 = — 4

17 (a) P(motorcycle or car) = P(motorcycle) + P(car) = 0.15 + 0.55 = 0.7

(b) P(none of these vehicles) = 1 – 0.15 – 0.55 – 0.2 = 0.1

18 (a) 0 (b) P(Zaki or Hong Ping or Raju

wins) = P(Zaki wins) + P(Hong Ping

wins) + P(Raju wins) = 0.4 + 0.3 + 0.2 = 0.9

319 (a) P(white) = —– 12 1 = — 4 (b) P(neither white nor black) = 1 – P(white or black) 3 4 = 1 – (—– + —–) 12 12 5 = —– 12 (c) P(either black or red) = P(black) + P(red) 4 5 = —– + —– 12 12 3 = — 4

220 (a) —– 5 (b) P(either odd or divisible by 4) = P(odd) + P(divisible by 4) 5 2 = —– + —– 10 10 7 = —– 10

21 P(A � B) = P(A) × P(B) 5 2 —– = — × P(B) 12 3 5 3 P(B) = —– × — 12 2 5 = — 8

22 P(C � D) = P(C) × P(D) 0.1 = k(k + 0.3) k2 + 0.3k – 0.1 = 0 10k2 + 3k – 1 = 0 (5k – 1)(2k + 1) = 0 1 1 k = — or k = – — 5 2 1 ∴ k = — 5

2 323 (a) P(RB) = — × — 9 9 2 = — 3



(b) P(different colours) = 1 – P(same colour) 4 9 16 = 1 – (—– + —– + —– 81 81 81 52 = —– 81

24 (a) P(BG) = P(B) × P(G) 1 2 = — × — 5 5 2 = —– 25 (b) P(the same colour) = P(RR or GG) 3 3 = —– + —– 10 25 21 = —– 50

25 (a) P(at least one wins) = P(WW or WL or LW) 6 9 2 = —– + —– + —– 20 20 20 17 = —– 20 (b) P(both win) = P(WW) 3 2 = — × — 4 5 3 = —– 10

26 (a) P(all qualify) = P(QQQ) 1 2 5 = — × — × — 2 3 6 5 = —– 18 (b) P(only one qualifi es) = P(QQ'Q' or Q'QQ' or Q'Q'Q) 1 1 5 = —– + —– + —– 36 18 36 2 = — 9

27 (a) P(R � G) = P(RRR or RRG or RGR or

GRR) 1 1 1 1 = —– + —– + —– + — 15 21 10 6 8 = —– 21 (b) P(at most two red apples) = 1 – P(RRR or GGG) 1 5 = 1 – (—– + —–) 15 28 317 = —–– 420

28 (a) P(Hamdan passes both subjects) 3 2 = — × — 4 3 1 = — 2 (b) P(Xue Ming passes only one

subject)

= P(PF or FP) 4 2 1 1 = (— × —) + (— × —) 5 3 5 3 3 = — 5

29 (a) P(a Mathematics book and a history book)

= P(MH or HM) 6 9 9 6 = (—– × —–) + (—– × —–) 15 15 15 15 12 = —– 25 (b) P(at least one Mathematics book) = 1 – P(HH) 9 9 = 1 – (—– × —–) 15 15 16 = —– 25

30 (a) P(the same colour) = P(WW or BB) 5 1 = — + —– 9 18 11 = —– 18 (b) P(one white) = P(WB or BW) 1 5 = — + —– 9 18 7 = —– 18

SPM Appraisal Zone

4 3 1 (a) P(RR) = —– × —– 12 11 12 = —–– 132 1 = —– 11 (b) P(at least one rotten) = 1 – P(GG) 8 7 = 1 – (—– × —–) 12 11 19 = —– 33

2 (a) P(at least one brown) = P(BrBr or BrB1 or BrR or B1Br

or RBr) 3 2 3 8

= (—– × —–) + (—– × —–) + 12 11 12 11 3 1 8 3 (—– × —–) + (—– × —–) 12 11 12 11 1 3 + (—– × —–) 12 11 5 = —– 11 (b) P(of the same colour) = P(BrBr or B1B1) 3 2 8 7 = (—– × —–) + (—– × —–) 12 11 12 11

31 = —– 66

3 (a) P(selecting a point in the circle) area of the circle = —–—–—–—–—–—– area of the square 22

49π 49(—–)

77 7 = —–– = —–––––––– = —––

256 256 128 (b) P(selecting a point outside the

circle) area of the shaded region = —––––––––—––––––––––– area of the square 256 – 49π = —–––––––– 256 22 256 – 49 (—–) 7 = —–—––—––—––– 256 102 = —–– 256 51 = —–– 128

4 P(of the same colour) = P(RR or BB or GG) 5 4 3 2 2 1 = (—– × —) + (—– × —) + (—– × —) 10 9 10 9 10 9 14 = —– 45

6 2 5 (a) — = — k 3 2k = 18 k = 9 3 2 (b) P(GG) = — × — 9 8 6 = —– 72 1 = —– 12

6 (a) P(begins with a vowel) 240 = —–– 720 1 = — 3 (b) P(ends with a consonant) 480 = —–– 720 2 = — 3

7 (a)

O

DO

0.3

0.7

C

DC

0.6

0.4

C

DC

0.2

0.8


O : Oversleeps DO : Does not oversleep C : Cycles to school DC : Does not cycle to school

(b) P(does not cycle to school) = (0.3 × 0.4) + (0.7 × 0.8) = 0.12 + 0.56 = 0.68

8 (a) P(choosing a letter I) 2 = —– 14 1 = — 7 (b) P(choosing a consonant) 9 = —– 14

2 2 9 (a) P(EE) = — × — 4 4 4 = —– 16 1 = — 4 (b) 2nd spin

1 2 3 41st spin

4

3

2

1

The ∆ indicates the event where the sum is odd

∴ Probability that the sum is odd 8 = —– 16 1 = — 2

2 110 (a) P(B1B1) = — × — 3 4 1 = — 6

(b) P(choosing a pair of brown shoes)

= P(BrB or BrB1 or BrBr) 1 1 1 1 1 2 = (— × —) + (— × —) + (— × —) 3 4 3 4 3 4 1 = — 3

11 (a) S A B

16 3 9

2

2

6

4

C8

(b) (i) P(reads at least one) = 1 – P(reads none) 8 = 1 – —– 50 21 = —– 25

(ii) P(reads only one) = P(reads only A) + P(reads

only B) + P(reads only C) 16 9 6 = —– + —– + —– 50 50 50 31 = —– 50

12 L: person had bought laptops D: person had bought digital cameras

(a) P(L or D) = 1 – P(neither L nor D) = 1 – 0.15 = 0.85(b) P(L or D) = P(L) + P(D) – P(L and D) 0.85 = 0.7 + 0.5 – P(L and D) P(L and D) = 0.7 + 0.5 – 0.85 = 0.35

13 (a) P(passes the point P)

1 1 = — × — 4 3

1 = —– 12 (b) P(passes the point Q)

3 1 1 1 = (— × —) + (— × —) 4 3 4 3 1 = — 3

14 (a) (i) P(one is dotted and the other is stripped)

= P(DS or SD) 3 4 4 3 = (—– × —–) + (—– × —–) 12 12 12 12 1 = — 6

(ii) P(taking of the same pattern)

P(DD or SS or SiSi) 3 3 4 4

= (—– × —–) + (—– × —–) 12 12 12 12 5 5 + (—– × —–) 12 12 25 = —– 72 (b) P(all three are silk ties)

= P(SiSiSi) 5 5 5 = —– × —– × —– 12 12 12 125 = —–– 144

15 (a) P(all three complete) = P(A) × P(B) × P(C) = 0.8 × 0.7 × 0.5 = 0.28 (b) P(at least two complete) = P(ABC or ABC ' or AB 'C or A'BC) = 0.28 + (0.8 × 0.7 × 0.5) + (0.8 ×

0.3 × 0.5) + (0.2 × 0.7 × 0.5) = 0.28 + 0.28 + 0.12 + 0.07 = 0.75

19 Probability Distributions

1

Booster Zone

1 p = 0.2, q = 1 – 0.2 = 0.8 and n = 6.(a) P(X = 3) = 6C

3 (0.2)3(0.8)3

= 0.0819

(b) P(2 � X � 5) = P(X = 3) + P(X = 4) + P(X = 5) = 0.0819 + 6C

4(0.2)4(0.8)2 +

6C5(0.2)5(0.8)

= 0.0819 + 0.0154 + 0.0015 = 0.0988

(c) P(X � 1) = 1 – P(X � 1) = 1 – P(X = 0) – P(X = 1) = 1 – 6C

0(0.2)0(0.8)6 –

6C1(0.2)(0.8)5

= 1 – 0.2621 – 0.3932 = 0.3447

2 X ~ B(8, 0.4)(a) P(X = 4) = 8C

4(0.4)4(0.6)4

= 0.2322

(b) =(X � 3) = P(X = 0) + P(X = 1) + P(X = 2) = 8C

0(0.4)0(0.6)8 + 8C

1(0.4)(0.6)7

+ 8C2(0.4)2(0.6)6

= 0.0168 + 0.0896 + 0.2090 = 0.3154

3 X ~ B(n, p) with n = 7 and p = 0.25.(a) P(X � 2) = 1 – P(X � 1) = 1 – P(X = 0) – P(X = 1) = 1 – 7C

0(0.25)0(0.75)7

– 7C1(0.25)(0.75)6

= 1 – 0.13348 – 0.31146 = 0.5551

(b) P(X � 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.13348 + 0.31146 +

7C2(0.25)2(0.75)5

= 0.44494 + 0.31146 = 0.7564

4 X ~ B(n, 0.5) with p = 0.5 and q = 0.5. P(X � 1) � 0.95 1 – P(X = 0) � 0.95 1 – nC

0(0.5)0(0.5)n � 0.95

1 – 0.5n � 0.95 0.5n � 0.05 n log

10 0.5 � log

10 0.05

n � log

10 0.05

–––––––––log

10 0.5

n � 4.32

∴ The least number of shots is 5.

5 (a) P(X � 1) = 0.0081 P(X = 0) = 0.0081 nC

0(0.7)0(0.3)n = 0.0081

0.3n = 0.34

∴ n = 4(b) P(X = 2) = 4C

2(0.7)2(0.3)2

= 0.2646

6 X ~ B(4, 0.5)P(X � 3)= P(X = 3) + P(X = 4)= 4C

3(0.5)3(0.5) + 4C

4(0.5)4(0.5)0

= 0.25 + 0.0625= 0.3125

7 X ~ B(n, p) with n = 20 and p = 0.1The mean, E(X ) = np = 20(0.1) = 2Standard deviation, σ = npq

= 2(0.9) = 1.8 = 1.342

8 (a) E(X ) = 3.2 0.4n = 3.2 n = 8

(b) Standard deviation, σ = npq = 3.2(0.6) = 1.92 = 1.386

9 np = 5 … 1 npq = 4 … 2

2 ÷ 1 :

q = 4––5

So, p = 1 – 4––5

= 1––5

Substitute p = 1––5

into 1 :

n� 1—5 � = 5

n = 25

10 (a) np = 2 … 1 npq = 1.6 … 2 2 – 1 : q = 0.8 So p = 1 – q = 1 – 0.8 = 0.2 Substitute p = 0.2 into 1 : n(0.2) = 2 n = 10 ∴ p = 0.2

(b) P(X = 4) = 10C4(0.2)4(0.8)6

= 0.0881

11 (a) P(Z � 1.9) = 0.0287

(b) P(Z � – 0.75) = P(Z � 0.75) = 0.2266

(c) P(Z � –2.45) = 1 – P(Z � –2.45) = 1 – P(Z � 2.45) = 1 – 0.00714 = 0.99286

f(z)

zO–2.45

P(Z > –2.45)

(d) P(Z � 1.34) = 1 – P(Z � 1.34) = 1 – 0.0901 = 0.9099

f(z)

zO 1.34

P(Z < 1.34)

(e) P(0.83 � Z � 1.85) = P(Z � 0.83) – P(Z � 1.85) = 0.2033 – 0.0322 = 0.1711

f(z)

zO 0.83 1.85

P(0.83 < Z < 1.85)

(f) P(–1.764 � Z � – 0.246) = P(Z � 0.246) – P(Z � 1.764) = 0.4029 – 0.0388 = 0.3641

f(z)

zO

P(–1.764 < Z < –0.246)

–0.246–1.764



(g) P(–2.57 � Z � 0.132) = 1 – P(Z � – 2.57) – P(Z � 0.132) = 1 – 0.00508 – 0.4475 = 0.5474

f(z)

zO

P(–2.57 < Z < 0.132)

0.132–2.57

(h) P(–1.68 � Z � 1.725) = 1 – P(Z � – 1.68) – P(Z � 1.725) = 1 – 0.0465 – 0.0423 = 0.9112

f(z)

zO

P(–1.68 < Z < 1.725)

1.725–1.68

(i) P(–2.05 � Z � 0) = 0.5 – P(Z � 2.05) = 0.5 – 0.0202 = 0.4798

f(z)

zO

P(–2.05 < Z < 0)

–2.05

(j) P(0 � Z � 1.76) = 0.5 – P(Z � 1.76) = 0.5 – 0.0392 = 0.4608

f(z)

zO

P(0 < Z < 1.76)

1.76

(k) P(|Z | � 1.64) = P(–1.64 � Z � 1.64) = 1 – 2P(Z � 1.64) = 1 – 2(0.0505) = 0.8990

f (z)

zO 1.64

P(|Z| < 1.64)

–1.64

(l) P(|Z| � 2.326) = 2P(Z � 2.326) = 2(0.01) = 0.02

f(z)

zO 2.326–2.326

12 (a) 0.0668(b) 0.00714(c) 1 – 0.0344 = 0.9656(d) 1 – 0.0104 = 0.9896(e) 0.2119 – 0.0082 = 0.2037(f) 1 – 0.1056 – 0.0139 = 0.8805(g) 0.5 – 0.3372 = 0.1628(h) 2(0.1587) = 0.3174

13 (a) f(z)

zO a

0.0778

∴ a = 1.42

(b) f(z)

zOa

0.1515

∴ a = –1.03

(c) f(z)

zOa

0.229 0.771

∴ a = –0.742

(d) f(z)

zO a

0.47720.5228

∴ a = 0.057

(e) f(z)

zO a

0.05 0.05

–a

0.9

∴ a = 1.645

(f) f(z)

zO a–a

0.0485

0.903

0.0485

∴ a = 1.66

14 (a) z = 0.44(b) z = – 1.85(c) z = – 1.2(d) z = 2(e) z = 1.1(f) z = 0.7(g) z = 1.2(h) z = – 1

15 (a) x – 12–––––––3

= 1.2

x – 12 = 3.6 x = 15.6

(b) x – 6––––––2

= – 0.7

x – 6 = – 1.4 x = 4.6

(c) x – 50–––––––5

= – 0.742

x – 50 = – 3.71 x = 46.29

(d) x – 200––––––––6

= 1.4

x – 200 = 8.4 x = 208.4

16 X is the height, in centimeters, of a boy.X ~ N(150, 25)(a) P(X � 158)

= P�Z � 158 – 150–––––––––

5 � = P(Z � 1.6) = 0.0548

f(z)

zO 1.6


(b) P(X � 146) = P�Z � 146 – 150–––––––––

5 � = P(Z � – 0.8) = P(Z � 0.8) = 0.2119

f(z)

zO–0.8

17 X is the examination markX ~ N(45, 202)

(a) P(X � 40) = P�Z � 40 – 45–––––––

20 � = P(Z � – 0.25) = 1 – P(Z � – 0.25) = 1 – 0.4013 = 0.5987

f(z)

zO–0.25

Since there are 200 candidates, the number of candidates who pass the examination is 200 × 0.5987 = 119.74

∴ 120 candidates passed

(b) Given P(X � x) = 0.05 Standardising

P�Z � x – 45––––––

20 � = 0.05

So, x – 45––––––20

= 1.645

x – 45 = 32.9 x = 77.9 = 78 (1 d.p.)

f(z)

zO z

0.05

∴ A distinction is awarded for a mark of 78 or more.

18 P(X � 106) = 0.8849

P�Z � 106 – 100–––––––––

σ � = 0.8849

So, 106 – 100–––––––––σ

= 1.2

1.2σ = 6 σ = 5

f (z)

zO z

0.8849

0.1151

19 X is the volume, in m�, of a can of drinks.X ~ N(350, σ 2)(a) P(X � 357) = 0.2 Standardising

P�Z � 357 – 350–––––––––

σ � = 0.2

So 357 – 350–––––––––

σ = 0.842

7 = 0.842σ σ = 8.31(2 d.p.)∴ The standard deviation, σ = 8.31 m�

O z

f(z)

z

0.2

(b) P(X � 340) = P�Z � 340 – 350–––––––––

8.31 � = P(Z � – 1.203) = 0.1145

∴ The percentage of cans that contain less than 340 m� is 0.1145 × 100 = 11.45%.

O

f(z)

z–1.203

20 X is life, in hours, of a batteryX ~ N(160, 302)P(150 � X � 175)

= P� 150 – 160–––––––––30

� Z � 175 – 160–––––––––

30 �= P(–0.3 � Z � 0.5)= 1 – P(Z � 0.3) – P(Z � 0.5)= 1 – 0.3821 – 0.3085= 0.3094∴ The percentage of batteries which

have a life between 151 hours and 175 hours is 0.3094 × 100 = 30.94%.

O

f(z)

z–0.3 0.5

SPM Appraisal Zone

1 (a) P(X = 0) = 1 – [P(X = 1) + P(X = 2) +

P(X = 3) + P(X = 4) + P(X = 5)] = 1 – (0.05 + 0.1 + 0.25 + 0.35

+ 0.2) = 1 – 0.95 = 0.05

(b) P(X � 2) = 1 – P(X = 0) – P(X = 1) = 1 – 0.05 – 0.05 = 0.9

2 X ~ B(4, p) P(X = 4) = 0.0256 4C

4(p4)(1 – p)0 = 0.0256

p4 = 0.44

∴ p = 0.4

3 X ~ B(8, 0.4)(a) P(X = 3) = 8C

3(0.4)3(0.6)5

= 0.2787

(b) P(X � 6) = P(X = 7) + P(X = 8) = 8C

7(0.4)7(0.6) + 8C

8(0.4)8(0.6)0

= 0.00852

4 np = 20 … 1 npq = 4 npq = 16 … 2

Substitute 1 into 2 : 20q = 16

q = 4––5

So, p = 1 – q

= 1 – 4––5

p = 1––5

Substitute p = 1––5

into 1 : 1––5

n = 20

n = 100

∴ p = 1––5

, n = 100

5 P(X � 1) � 0.95 1 – P(X = 0) � 0.95 1 – nC

0(0.1)0(0.9)n � 0.95

1 – 0.9n � 0.95 0.9n � 0.05 n log

10 0.9 � log

10 0.05

n � log

10 0.05

–––––––––log

10 0.9

n � 28.43∴ The least value of n is 29.

6 (a) P(Z � b) = 0.3 + 0.6 = 0.9

(b) P(Z � a) = 1 – 0.3 = 0.7

7 (a) P(Z � a) = 0.45 From the table, a = 0.125

(b) P(a � Z � b) = 0.45 – 0.3 = 0.15

8 P(X � a) = 0.242

a – 12–––––––

2 = 0.7

a – 12 = 1.4 a = 13.4

9 (a) P(X � 56) = P�Z � 56 – 50–––––––

5 � = P(Z � 1.2) = 0.1151


(b) P(X � 42) = P�Z � 42 – 50–––––––

5 � = P(Z � – 1.6) = P(Z � 1.6 = 0.0548

10 X is the mass, in grams, of a cabbageX ~ N(1000, 1502)P(715 � X � 1264)

= P� 715 – 1000––––––––––150

� Z � 1264 – 1000––––––––––

150 �= P(–1.9 � Z � 1.76)= 1 – 0.0287 – 0.0392= 0.9321∴ There are 500 × 0.9321 = 466

cabbages have a mass between 715 g and 1264 g.

11 X ~ N(4.5, 1.21)

(a) z = 6.7 – 4.5––––––––

1.1

= 2

(b) P(3.4 � X � 6.7)

= P� 3.4 – 4.5––––––––1.1

� Z � 6.7 – 4.5––––––––

1.1 � = P(–1 � Z � 2) = 1 – 0.1587 – 0.0228 = 0.8185

12 P(X � 20) = 0.0228

P�Z � 20 – �

–––––––15 � = 0.0228

So, 20 – �

–––––––15

= – 2

20 – � = – 30 � = 50

13 P[|Z | � a] = 0.6 P(0 � Z � a) = 0.3So, P(Z � a) = 0.5 – 0.3 = 0.2From the table, a = 0.842.

14 X ~ N(�, 52) P(X � 42) = 0.0808

P�Z � 42 – �

–––––––5 � = 0.0808

So, 42 – �

–––––––5

= 1.4

42 – � = 7 � = 35

15 X is the consultation time, in minutes, of a patient.X ~ N(10, 52)

(a) P(X � 15) = P�Z � 15 – 10–––––––

5 � = P(Z � 1) = 0.1587

(b) P(8 � X � 14)

= P� 8 – 10––––––5

� Z � 14 – 10–––––––

5 � = P(–0.4 � Z � 0.8) = 1 – 0.3446 – 0.2119 = 0.4435

16 (a) X ~ N(�, σ2) P(X � 6.35) = 0.02

P�Z � 6.35 – �––––––––

σ � = 0.02

6.35 – �––––––––

σ = – 2.054

� – 2.054σ = 6.35 … 1

f(x)

x6.35

0.02

P(X � 7.55) = 0.05

P�Z � 7.55 – �––––––––

σ � = 0.05

7.55 – �––––––––

σ = 1.645

� + 1.645 σ = 7.55 … 2

f(x)

x7.55

0.05

2 – 1 : 3.699σ = 1.2 σ = 0.324Substitute σ = 0.324 into 1 :

� – 2.054(0.324) = 6.35 � = 7.015

∴ � = 7.015, σ = 0.324

(b) (i) P(X = 3) = 10C

3(0.02)3(0.98)7

= 0.00083

(ii) P(X � 2) = 1 – P(X = 0) + P(X = 1) = 1 – 10C

0(0.05)0(0.95)10

– 10C1(0.05)(0.95)9

= 1 – 0.59874 – 0.31512 = 0.0861

17 X ~ N(50, 102)

(a) (i) P(X � 45) = P�Z � 45 – 50–––––––

10 � = P(Z � – 0.5) = P(Z � 0.5) = 0.3085

(ii) P(42 � X � 56)

= P�42 – 50–––––––

10 � Z �

56 – 50–––––––

10 � = P(–0.8 � Z � 0.6) = 1 – P(Z � 0.8) – P(Z � 0.6) = 1 – 0.2119 – 0.2743 = 0.5138

∴ 0.5138� 365––––7 � = 26.79

= 27 weeks

(b) P(X � n) = 0.6

P�Z � n – 50

–––––––10 � = 0.6

So, n – 50

–––––––10

= – 0.253

n – 50 = – 2.53 n = 47.47 = 47

f(z)

0.60.4

nx

18 (a) (i) P(X = 4) = 10C4(0.15)4(0.85)6

= 0.0401

(ii) P(2 � X � 5) = P(X = 3) + P(X = 4) = 10C

3(0.15)3(0.85)7 + 0.0401

= 0.1298 + 0.0401 = 0.1699

(b) Mean, � = np = 10(0.15) = 1.5

Variance, σ 2 = npq = 1.275

(c) P(X � 1) � 0.95 1 – P(X = 0) � 0.95

1 – nC0(0.15)0(0.85)n � 0.95

1 – 0.85n � 0.95 0.85n � 0.05 n log

10 0.85 � log

10 0.05

n � log

10 0.05

–––––––––log

10 0.85

n � 18.43 ∴ The least value of n is 19.

19 X is the Mathematics mark.X ~ N(55, 82)(a) P(45 � X � 60)

= P� 45 – 55–––––––

8 � Z �

60 – 55–––––––

8 � = P(–1.25 � Z � 0.625) = 1 – P(Z � 1.25) – P(Z � 0.625) = 1 – 0.1056 – 0.2660 = 0.6284


(b) P(X � 70) = P�Z � 70 – 55–––––––

8 � = P(Z � 1.875) = 0.0303

∴ The number of candidates with distinction

= 0.0303 × 200 = 6.06 = 6 students

(c) P(X � x) = 0.97

P�Z � x – 55

–––––––8 � = 0.97

So, x – 55

–––––––8

= – 1.881

x – 55 = – 15.048 x = 39.95 = 40 (2 s.f.) ∴ Lowest passing mark = 40

f (x)

0.03 0.97

55x

x

P(X = 4) = 5C4(0.97)4(0.03)

= 0.132820 X is the length, in centimeters, of a

bottleX ~ N(25, 22)

(a) (i) P(X � 28) = P�Z � 28 – 25–––––––

2 � = P(Z � 1.5) = 0.0668

(ii) P(23 � X � 28)

= P� 23 – 25–––––––2

� Z � 1.5� = P(–1 � Z � 1.5) = 1 – P(Z � 1) – P(Z � 1.5) = 1 – 0.1587 – 0.0668 = 0.7745

(b) P(X � 23) = P(Z � – 1) = 0.1587 ∴ The number of bottles = 0.1587 × 500 = 79.35 = 79(2 s.f.)

(c) P(X � L) = 0.1

P�Z � L – 25–––––––

2 � = 0.1

So, L – 25–––––––2

= – 1.281

L – 25 = – 2.562 L = 22.44 = 22 cm (2 s.f.)

f(z)

25x

L

0.1

21 (a) (i) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 5C0� 1––

4 �0� 3––4 �5

–

5C1� 1––

4 �� 3––4 �4

= 1 – 0.2373 – 0.3955 = 0.3672

(ii) P(X � 1) � 0.9 1 – P(X = 0) � 0.9

1 – nC

0� 1––4 �0� 3––

4 �n � 0.9

1 – � 3––4 �n

� 0.9

� 3––4 �n

� 0.1

n log10� 3––

4 � � log10

0.1

n � log

10 0.1

–––––––––––log

10 � 3––

4 � n � 8 ∴ n = 9

(b) X is the Chemistry mark. X ~ N(65, 102)

(i) P(X � 40) = P�Z � 40 – 65–––––––

10 � = P(Z � – 2.5) = 0.00621

(ii) P(X � 80) = P�Z � 80 – 65–––––––

10 � = P(Z � 1.5) = 0.0668 ∴ The percentage of students

who obtained Al grade = 0.0668 × 100 = 6.68%

22 (a) (i) P(X = 2) = 3C2� 3––

7 �2� 4––7 �

= 0.3149

(ii) P(X � 2) = 1 – P(X = 3)

= 1 – 3C3� 4––

7 �3� 3––7 �0

= 0.8134

(b) X is the mass, in kilograms, of a student.

X ~ N(55, 42) (i) P(X � 62)

= P�Z � 62 – 55–––––––

4 � = P(Z � 1.75) = 0.0401

(ii) P(X � m) = 0.22

P�Z � m – 55–––––––

4 � = 0.22

So, m – 55–––––––4

= 0.772

m – 55 = 3.088 m = 58.09 (2 d.p.)

f(x)

55x

m

0.22

23 (a) np = 2 … 1 npq = 1.5 … 2

2 ÷ 1 : q = 1.5––––2

= 0.75 So, p = 1 – q = 1 – 0.75 = 0.25 Substitute p = 0.25 into 1 : n(0.25) = 2 n = 8 ∴ p = 0.25, n = 8

(b) (i) P(30 � X � 42) = 1 – 0.1587 – 0.0808 = 0.7605

(ii) P(X � 42) = 0.0808

P�Z � 42 – �

––––––––σ � = 0.0808

So, 42 – �

––––––––σ

= 1.4

42 – � = 1.4σ � + 1.4σ = 42 … 1

and P(X � 30) = 0.1587

P�Z � 30 – �

––––––––σ � = 0.1587

So, 30 – �

––––––––σ

= –1

30 – � = –σ

� – σ = 30 … 2

1 – 2 : 2.4σ = 12 σ = 5

Substitute σ = 5 into 1 : � + 1.4(5) = 42 � = 35 ∴ � = 35, σ = 5

24 (a) X ~ B(6, 0.9) (i) P(X = 3) = 6C

3(0.9)3(0.1)3

= 0.0146

(ii) P(X � 3) = P(X = 3) + P(X = 4) +

P(X = 5) + P(X = 6) = 0.0146 + 6C

4(0.9)4(0.1)2

+ 6C5(0.9)5(0.1) +

6C6(0.9)6(0.1)0

= 0.0146 + 0.09842 + 0.35429 + 0.53144

= 0.9988


(b) X is the length, in centimeters, of a pencil.

X ~ N(�, 0.52) P(X � 15) = 0.05

P�Z � 15 – �

––––––––0.5 � = 0.05

So, 15 – �

––––––––0.5

= 1.645

15 – � = 0.8225 � = 14.18 (2 d.p.)

f(x)

15xµ

0.05

25 (a) (i) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 10C0� 3––

5 �0� 2––5 �10

–

10C1� 3––

5 �1� 2––5 �9

= 0.9983

(ii) Mean, � = np = 500 × 3––

5 = 300 Standard deviation, σ = npq

= 300� 2––5 �

= 120 = 10.95 ∴ � = 300, σ = 10.95

(b) (i) P(X � a) = 1 – 0.9772 = 0.0228

(ii) P(X � a) = 0.9772

P�Z � a – 50

–––––––5 � = 0.9772

So, a – 50

–––––––5

= – 2

a – 50 = – 10 a = 50 – 10 = 40

Motion along a Straight Line

1

Booster Zone

1 (a) s2 = 2(2 – 4)2

= 2(–2)2

= 8 m

(b) s4 = 4(4 – 4)2

= 0

(c) s6 = 6(6 – 4)2

= 6(4) = 24 m

(d) s9 = 9(9 – 4)2

= 9(25) = 225 m

2 (a) s1 = 6(1)2 – 13

= 5 m

(b) s4 = 6(4)2 – 43

= 32 m

(c) s5 = 6(5)2 – 53

= 25 m

(d) s8 = 6(8)2 – 83

= –128 m

3 (a) (i) s2.5

= 5(2.5) – (2.5)2

= 12.5 – 6.25 = 6.25 m

(ii) s5 = 5(5) – 52

= 0 m

(b) s = –24 5t – t 2 = –24 t 2 – 5t – 24 = 0 (t – 8)(t + 3) = 0 t = 8 or t = –3 ∴ t = 8 s

(c) s = 0 m 5t – t 2 = 0 t 2 – 5t = 0 t(t – 5) = 0 t = 0 or t = 5 ∴ The particle passes through O

again when t = 5 s.

4 (a) s2 = 2(2)2 – 8(2)

= 8 – 16 = –8 m

(b) (i) s = 0 2t 2 – 8t = 0 2t(t – 4) = 0 t = 0 or t = 4 ∴ t = 4 s

(ii) s = 10 2t 2 – 8t = 10 t 2 – 4t – 5 = 0 (t – 5)(t + 1) = 0 ∴ t = 5 s

(c) s � 0 2t 2 – 8t � 0 2t(t – 4) � 0

40t

∴ t � 4

5 (a)

0

t = 0s = 0

s

t = 6s = 12

t = 2s = –4

(b) (i) s = (2)2 – 4(2) = –4 = 4 m

(ii) s3 = 32 – 4(3)

= 9 – 12 = –3 m ∴ The distance in 4th second = s

4 – s

3

= 0 – (–3) = 3 m

(iii) The distance in the interval 0 � t � 6

= 2(4) + 12 = 20 m

Average velocity during the fi rst 6 seconds

= distance travelled

––––––––––––––––time taken

= 20–––6

= 31––3

m s–1

6 (a)

6520

7

5

9

t (s)

s (m)s = 5 + 4t – t2

(b) (i) s1 = 5 + 4(1) – (1)2

= 8 s

0 = 5 + 4(0) – (0)2

= 5 ∴ s

1 – s

0 = 8 – 5

= 3 m

(ii) s4 = 5 + 4(4) – 42

= 5 m∴ The distance in the 5th

second = |s

5 – s

4|

= |0 – 5| = 5 m

Average speed during the fi rst 6 seconds

= distance travelled

––––––––––––––––time taken

= 4 + 9 + 7

–––––––––6

= 20–––6

= 31––3

m s–1

7 (a) (i) 8 m

(ii) 20 m

(iii) –8 m

(b) Average speed

= (20 – 8) + 20 + 8

––––––––––––––––10

= 40–––10

= 4 m s–1

8 (a) (i) s0 = (0 – 2)2 + 5

= 9 m

s2 = (2 – 2)2 + 5

= 5 m∴ The distance during the

next 2 seconds = �5 – 9� = 4 m

(ii) s4 = (4 – 2)2 + 5

= 9 m∴ The distance during the

next 2 seconds = 9 – 5 = 4 m

(b) Average velocity

= (9 – 5) + (30 – 5)

––––––––––––––––7

= 41––7

m s–1

9 (a) v = ds

–––dt

= 3t 2 – 6t ∴ v

2 = 3(2)2 – 6(2)

= 12 – 12 = 0 m s–1

20



(b) s = 0 t 3 – 3t 2 = 0 t 2(t – 3) = 0 t = 0 or t = 3 ∴ v

3 = 3(3)2 – 6(3)

= 27 – 18 = 9 m s–1

(c) v = 9 3t 2 – 6t = 9 t 2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3 or t = –1 ∴ t = 3 s

10 (a) v = ds

–––dt

= 4t – 5 ∴ v

0 = 4(0) – 5

= –5 m s–1

(b) s = 8 2t 2 – 5t + 5 = 8 2t 2 – 5t – 3 = 0 (2t + 1)(t – 3) = 0

t = – 1––2

or t = 3

∴ v3 = 4(3) – 5

= 12 – 5 = 7 m s–1

(c) v = 3 4t – 5 = 3 4t = 8 t = 2 ∴ s

3 = 2(3)2 – 5(3) + 5

= 18 – 15 + 5 = 8 m

11 (a) s = t 3 – 3t 2 – 9t + 5

v = ds

–––dt

= 3t 2 – 6t – 9 When the particle reverses its

direction, v = 0 3t 2 – 6t – 9 = 0 t 2 – 2t – 3 = 0 (t + 1)(t – 3) = 0 t = –1 or t = 3 ∴ t = 3 s

(b) v � 0 3t 2 – 6t – 9 � 0 t 2 – 2t – 3 � 0 (t + 1)(t – 3) � 0

t3–1

∴ t � 3

12 (a) v = ds

–––dt

= 3t 2 – 18t + 24 ∴ v

3 = 3(3)2 – 18(3) + 24

= –3 m s–1

(b) When the particle is instantaneously at rest,

v = 0 3t 2 – 18t + 24 = 0 t 2 – 6t + 8 = 0 (t – 2)(t – 4) = 0 t = 2 or t = 4

(c) v � 0 t 2 – 6t + 8 � 0 (t – 2)(t – 4) � 0

42t

∴ 2 � t � 4

13 (a) v = 8 – 4t s = �v dt = 8t – 2t 2 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = 8t – 2t 2

When t = 1, s = 8(1) – 2(1)2

= 6 ∴ The displacement of Q from O

is 6 m.

(b) For maximum displacement, v = 0 8 – 4t = 0 t = 2 ∴ Maximum displacement = 8(2) – 2(2)2

= 16 – 8 = 8 m

(c) s4 = 8(4) – 2(4)2

= 32 – 32 = 0

0s

2

t = 0s = 0

t = 4s = 0

t = 2s = 8

∴ The total distance travelled during the fi rst 4 seconds

= 2(8) = 16 m

14 (a) v = 3t 2 – 6t

s = �v dt = t 3 – 3t 2 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = t 3 – 3t 2

When t = 5, s = (5)3 – 3(5)2

= 50 ∴ The displacement of particle

when t = 5 is 50 m.

(b) When the particle is momentarily at rest,

v = 0 3t(t – 2) = 0 t = 0 or t = 2 ∴ s

2 = 23 – 3(2)2

= 8 – 12 = –4 m

15 (a) v = 12t – 3t 2

s = �v dt = 6t 2 – t 3 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = 6t 2 – t 3

s4 = 6(4)2 – 43

= 32 m s

3 = 6(3)2 – 33

= 54 – 27 = 27 m ∴ The distance travelled in the 4th

second = 32 – 27 = 5 m

(b) When t = 2, s = 6(2)2 – 23

= 24 – 8 = 16 m When t = 6, s = 6(6)2 – 63

= 216 – 216 = 0 m ∴ The motion of the particle is as

shown below.

0

t = 0s = 0

s

t = 6s = 0

t = 2s = 16

t = 4s = 32

∴ The distance travelled from t = 2 to t = 6 = (32 – 16) + 32 = 48 m

16 (a) v = 10 – 2t s = �v dt = 10t – t 2 + c When t = 0, s = 0 and so c = 0 Hence at time t, s = 10t – t 2

(i) When the velocity is 8 m s–1, 10 – 2t = 8 2t = 2 t = 1 ∴ s

1 = 10(1) – 12

= 9 m

(ii) When the particle in instantaneously at rest,

v = 0 10 – 2t = 0 t = 5 ∴ s

5 = 10(5) – 52

= 25 m

(b) When the particle passes through O again,

s = 0 10t – t 2 = 0 t (10 – t) = 0 t = 0 or t = 10 ∴ v

10 = 10 – 2(10)

= 10 – 20 = –10 m s–1


17 (a) When the particle is instantaneously at rest,

v = 0 t 2 – 4t + 3 = 0 (t – 1)(t – 3) = 0 t = 1 or t = 3

(b) s = �(t 2 – 4t + 3)dt

= t 3

–––3

– 2t 2 + 3t + c

When t = 0, s = 0 and so c = 0 Hence at time t,

s = t 3

–––3

– 2t 2 + 3t

When t = 1, s = 1––3

– 2 + 3

= 1 1––3

m

When t = 3, s = 9 – 18 + 9 = 0 m

∴ The distance between the positions when the particle is

at rest = 1 1––3

m

18 s = �v dt

= 3t + 5t 2

–––2

– 2t 3

–––3

+ c

When t = 0, s = 0 and so c = 0

Hence, s = 3t + 5––2

t 2 – 2––3

t 3

(a) s3 = 3(3) + 5––

2(3)2 – 2––

3(3)3

= 131––2

m

s2 = 3(2) + 5––

2(2)2 – 2––

3(2)3

= 102––3

m

∴ The distance travelled in the 3rd second

= 131––2

– 102––3

= 25––6

m

(b) s6 = 3(6) + 5––

2(6)2 – 2––

3(6)3

= –36 m The motion of the particle is as

shown below.

0

t = 0s = 0

s

t = 6s = –36

t = 3s = 13

1––2

∴ The distance travelled during the fi rst 6 seconds

= 2�131––2 � + 36

= 63 m

19 s = t 3 – 4t 2 – 12t

v = ds

–––dt

= 3t 2 – 8t – 12

a = dv

–––dt

= 6t – 8

(a) When t = 0, a = 6(0) – 8 = –8 m s–2

(b) When t = 2, a = 6(2) – 8 = 4 m s–2

(c) When the particle passes O again, s = 0 t 3 – 4t 2 – 12t = 0 t (t 2 – 4t – 12) = 0 t (t + 2)(t – 6) = 0 t = 0 or t = –2 or t = 6 ∴ a

6 = 6(6) – 8

= 28 m s–2

(d) a � 0 6t – 8 � 0 6t � 8

t � 4––3

20 v = t 2 + 2t – 8

a = dv

–––dt

= 2t + 2

(a) When the particle is instantaneously at rest,

v = 0 t 2 + 2t – 8 = 0 (t + 4)(t – 2) = 0 t = –4 or t = 2 ∴ a

2 = 2(2) + 2

= 6 m s–2

(b) t 2 + 2t – 8 = 16 t 2 + 2t – 24 = 0 (t + 6)(t – 4) = 0 t = –6 or t = 4 ∴ a

4 = 2(4) + 2

= 10 m s–2

21 a = 8 – 4tv = �a dt = 8t – 2t 2 + cWhen t = 0, v = 3 and so c = 3∴ At time t, v = 8t – 2t 2 + 3

(a) v3 = 8(3) – 2(3)2 + 3

= 9 m s–1

(b) For maximum velocity, a = 0 8 – 4t = 0 t = 2 ∴ Maximum velocity = 8(2) – 2(2)2 + 3 = 11 m s–1

22 (a) a = 2t – 6 ∴ a

0 = 2(0) – 6

= –6 m s–2

(b) v = �a dt = t 2 – 6t + c When t = 0, v = 2 and so c = 2 Hence at time t, v = t 2 – 6t + 2 For minimum velocity, a = 0 2t – 6 = 0 t = 3 ∴ Minimum velocity = 32 – 6(3) + 2 = –7 m s–1

23 a = 6 – 2tv = �a dt = 6t – t 2 + cWhen t = 0, v = 16 and so c = 16Hence, v = 6t – t2 + 16s = �v dt

= 3t 2 – t 3

––3

+ 16t + c


Hence, s = 16t + 3t 2 – 1––3

t 3

(a) For maximum displacement, v = 0 6t – t 2 + 16 = 0 t 2 – 6t – 16 = 0 (t – 8)(t + 2) = 0 t = 8 or t = –2 ∴ Maximum displacement

= 16(8) + 3(8)2 – 1––3

(8)3

= 128 + 192 – 512––––3

= 149 1––3

m

(b) s3 = 16(3) + 3(3)2 – 1––

3(3)3

= 48 + 27 – 9 = 66 m

24 a = 3 – tv = �a dt

= 3t – 1––2

t 2 + c

When t = 0, v = 8 and so c = 8

Hence, v = 8 + 3t – 1––2

t 2

s = �v dt

= 8t + 3––2

t 2 – 1––6

t 3 + c


Hence, s = 8t + 3––2

t 2 – 1––6

t 3

(a) v4 = 8 + 3(4) – 1––

2(4)2

= 12 m s–1

(b) For maximum velocity, a = 0 3 – t = 0 t = 3 ∴ Maximum velocity

= 8 + 3(3) – 1––2

(3)2

= 12 1––2

m s–1


(c) When the particle is at rest, v = 0

8 + 3t – 1––2

t 2 = 0

t 2 – 6t – 16 = 0 (t + 2)(t – 8) = 0 t = –2 or t = 8

∴ s8 = 8(8) + 3––

2(8)2 – 1––

6(8)3

= 64 + 96 – 256––––3

= 74 2––3

m

(d) s10

= 8(10) + 3––2

(10)2 – 1––6

(10)3

= 80 + 150 – 500––––3

= 63 1––3

m

∴ Average speed during the fi rst 10 seconds

=

74 2––3

+ �74 2––3

– 63 1––3 �

–––––––––––––––––––––10

= 8.6 m s–1

25 a = 12 – 6tv = �a dt = 12t – 3t 2 + cWhen t = 0, v = 0 and so c = 0Hence, v = 12t – 3t 2

s = �v dt = 6t 2 – t 3 + cWhen t = 0, s = 0 and so c = 0Hence, s = 6t 2 – t 3

(a) When the particles is at rest, v = 0 12t – 3t 2 = 0 3t(4 – t) = 0 t = 0 or t = 4 ∴ The time taken to reach A is

4 s.

(b) s4 = 6(4)2 – 43

= 32 m ∴ The distance OA is 32 m.

(c) For maximum speed, a = 0 12 – 6t = 0 t = 2 ∴ Maximum speed = 12(2) – 3(2)2

= 24 – 12 = 12 m s–1

26 a = 6 – 2tv = �a dt = 6t – t 2 + cWhen t = 1, v = 10 and so c = 5Hence, v = 5 + 6t – t 2

(a) v0 = 5 + 6(0) – 02

= 5 m s–1

(b) a = 0 6 – 2t = 0 t = 3 ∴ v

3 = 5 + 6(3) – 32

= 14 m s–1

27 a = 1 – 2tv = �a dt = t – t 2 + cWhen t = 0, v = 6 and so c = 6Hence, v = t – t 2 + 6s = �v dt

= 6t + 1––2

t 2 – 1––3

t 3 + c


Hence, s = 6t + 1––2

t 2 – 1––3

t 3

(a) When the particle reverses its direction,

v = 0 t – t 2 + 6 = 0 t 2 – t – 6 = 0 (t + 2)(t – 3) = 0 t = –2 or t = 3 ∴ The particle reverses its direction

when t = 3 s.

(b) s3 = 6(3) + 1––

2(3)2 – 1––

3(3)3

= 18 + 9––2

– 9

= 13 1––2

m

28 a = 6t – 12v = 3t 2 – 12t + cWhen t = 0, v = 9 and so c = 9Hence, v = 3t 2 – 12t + 9s = t 3 – 6t 2 + 9t + cWhen t = 0, s = 0 and so c = 0Hence s = t 3 – 6t 2 + 9t

(a) When the particle is at rest, v = 0 3t 2 – 12t + 9 = 0 t 2 – 4t + 3 = 0 (t – 1)(t – 3) = 0 t = 1 or t = 3 ∴ The particle is at rest when

t = 1 s and t = 3 s.

(b) s1 = (1)3 – 6(1)2 + 9(1)

= 4 m s

2 = (2)3 – 6(2)2 + 9(2)

= 2 m ∴ The distance of the particle

from O is = 4 + (4 – 2) = 6 m

SPM Appraisal Zone

1 v = 12t – 3t 2

s = �v dt = 6t 2 – t 3 + cWhen t = 0, s = 0 and so c = 0Hence at time t, s = 6t 2 – t 3

a = dv

–––dt

= 12 – 6t

(a) For maximum velocity, a = 0 12 – 6t = 0 t = 2

Furthermore, da–––dt

= –6 � 0

⇒ v is maximum when t = 2 ∴ Maximum velocity = 12(2) – 3(2)2

= 24 – 12 = 12 m s–1

(b) When the particle reverses its direction,

v = 0 12t – 3t 2 = 0 3t (4 – t) = 0 t = 0 or t = 4 ∴ The particle reverses its direction

when t = 4 s.

(c) s4 = 6(4)2 – 43

= 96 – 64 = 32 m s

3 = 6(3)2 – 33

= 54 – 27 = 27 m ∴ The distance travelled in the

4th second = 32 – 27 = 5 m

(d) When the particle passes O again, s = 0 6t 2 – t 3 = 0 t 2(6 – t) = 0 t = 0 or t = 6 ∴ The particle passes O again

when t = 6 s.

2 a = 8 – 2tv = 8t – t 2 + cWhen t = 0, v = 9 and so c = 9Hence, v = 8t – t 2 + 9

(a) (i) For maximum velocity, a = 0 8 – 2t = 0 t = 4 ∴ Maximum velocity = 8(4) – 42 + 9 = 32 – 16 + 9 = 25 m s–1


(ii) When the particle stops, v = 0 8t – t 2 + 9 = 0 t 2 – 8t – 9 = 0 (t + 1)(t – 9) = 0 t = –1 or t = 9 ∴ The particle stops after

9 s, so m = 9.

(b)

Ot (s)

v = 9 + 8t – t 2

v (m s–1)

9

25

4 9

From the graph,

s = �9

0 (9 + 8t – t 2)dt

= �9t + 4t 2 – 1––3

t 3�9

0

= 81 + 324 – 243 = 162 m ∴ The total distance travelled is

162 m.

3 a = 12 – 6tv = 12t – 3t 2 + cWhen t = 0, v = 15 and so c = 15Hence, v = 12t – 3t 2 + 15s = �v dt = 15t + 6t 2 – t 3 + cWhen t = 0, s = 0 and so c = 0Hence, s = 15t + 6t 2 – t 3

(a) (i) s1 = 15(1) + 6(1)2 – (1)3

= 20 m

(ii) v = 0 12t – 3t 2 + 15 = 0 t 2 – 4t – 5 = 0 (t + 1)(t – 5) = 0 t = –1 or t = 5 ∴ s

5 = 15(5) + 6(5)2 – (5)3

= 75 + 150 – 125 = 100 m

(iii) a = 0 12 – 6t = 0 t = 2 ∴ s

2 = 15(2) + 6(2)2 – (2)3

= 30 + 24 – 8 = 46 m

(b) The motion of the particle is shown below.

O

t = 2s = 46

s

t = 5s = 100

t = 0s = 0

t = 6s = 90

∴ The total distance from t = 2 to t = 6 is

= (100 – 46) + (100 – 90) = 64 m

4 (a) v = t 2 + pt + 8

a = dv

–––dt

= 2t + p When t = 4, a = 2 2t + p = 2 8 + p = 2 p = –6

(b) (i) When the particle is at rest, v = 0 t 2 – 6t + 8 = 0 (t – 2)(t – 4) = 0 t = 2 or t = 4 ∴ The par t ic le i s

instantaneously at rest when t = 2 s and t = 4 s.

(ii) s = �v dt

= 1––3

t 3 – 3t 2 + 8t + c


Hence s = 1––3

t 3 – 3t 2 + 8t

s2 = 1––

3(8) – 3(4) + 8(2)

= 6 2––3

m

s4 = 1––

3(4)3 – 3(16) + 8(4)

= 5 1––3

m

∴ The distance of AB

= 6 2––3

– 5 1––3

= 1 1––3

m

(c)

Ot (s)

v = t2 – 6t + 8

v (m s–1)

7432

8

15

From the graph, the acceleration is negative when 0 � t � 3.

5 (a) v � 0 2t – 6 � 0 t � 3

(b) s = �v dt = t 2 – 6t + c When t = 0, s = 10 and so c = 10 Hence, s = t 2 – 6t + 10

(i) When the particle reverses its direction,

v = 0 2t – 6 = 0 t = 3 s

3 = 32 – 6(3) + 10

= 9 – 18 + 10 = 1 m

The motion of the particle is as shown below.

Y Xs

t = 3s = 1

t = 0s = 10

∴ The particle will not reach Y.

(ii) s8 = 82 – 6(8) + 10

= 26 m ∴ The total distance travelled

during the fi rst 8 seconds = (10 – 1) + (26 – 1) = 9 + 25 = 34 m

(iii) s = t 2 – 6t + 10

Ot (s)

s = t2 – 6t + 10

s (m)

3 8

10

1

26

6 (a) s = 5 + 3t 2 – t 3

When t = 0, s = 5 + 3(0)2 – 03

= 5 ∴ The distance of OA = 5 m

(b) v = ds

–––dt

= 6t – 3t 2

When the velocity is negative, v � 0 6t – 3t 2 � 0 3t 2 – 6t � 0 3t (t – 2) � 0

O

t2

∴ t � 2

(c) a = dv

–––dt

= 6 – 6t


For a maximum velocity, a = 0 6 – 6t = 0 t = 1 ∴ Maximum velocity = 6(1) – 3(1)2

= 3 m s–1

(d) s2 = 5 + 3(2)2 – 23

= 9 m s

4 = 5 + 3(4)2 – 43

= –11 m


O

t = 0s = 5

t = 2s = 9

t = 4s = –11

∴ The total distance travelled by the particle in the fi rst 4 seconds

= (9 – 5) + 9 + 11 = 24 m

7 (a) s = 2t 2 – 8 When t = 0, s = 2(0)2 – 8 = –8 ∴ The distance OA = 8 m

(b) When the particle passes O, s = 0 2t 2 – 8 = 0 t 2 = 4 t = ±2 t = 2 s

v = ds

–––dt

= 4t ∴ v = 4(2) = 8 m s–1

(c) s3 = 2(3)2 – 8

= 18 – 8 = 10 m


OA

t = 3s = 10

t = 0s = –8

∴ Average speed = 8 + 10––––––3

= 6 m s–1

8 (a) (i) a = dv

–––dt

= 8 – 6t

(ii) s = �v dt = 5t + 4t 2 – t 3 + c When t = 0, s = 0 and so c = 0 Hence, at time t, s = 5t + 4t2 – t3

(b) (i) When the particle passes O again,

s = 0 5t + 4t 2 – t 3 = 0 t 3 – 4t 2 – 5t = 0 t(t 2 – 4t – 5) = 0 t(t + 1)(t – 5) = 0 t = 0 or t = –1 or t = 5 ∴ t = 5 s

(ii) When the velocity is 2 m s–1, v = 2 5 + 8t – 3t 2 = 2 3t 2 – 8t – 3 = 0 (3t + 1)(t – 3) = 0

t = – 1––3

or t = 3

∴ t = 3 s

(iii) When the acceleration is 6 m s–2,

a = 6 8 – 6t = 6 6t = 2

t = 1––3

∴ t = 1––3

s

9 (a) (i) a = dv

–––dt

= 12t – 4 When t = 0, a = 12(0) – 4 = –4 m s–2

(ii) For minimum velocity, a = 0 12t – 4 = 0

t = 1––3

∴ Minimum velocity

= 6� 1––3 �

2

– 4� 1––3 � – 2

= 2––3

– 4––3

– 2

= –2 2––3

m s–1

(iii) When P moves towards the left,

v � 0 6t 2 – 4t – 2 � 0 3t 2 – 2t – 1 � 0 (3t + 1)(t – 1) � 0

t

11––3

–

∴ 0 � t � 1

(b)

0–2

t (s)1 3

40

v (m s–1)v = 6t2 – 4t – 2

∴ The distance travelled during the fi rst 3 seconds

= ��1

0 (6t 2 – 9t – 2) dt � +

�3

1 (6t 2 – 4t – 2) dt

= [2t 3 – 2t 2 – 2t]1

0 +

[2t 3 – 2t 2 – 2t]3

1

= 2 + [30 – (–2)] = 34 m

10 (a) a = 4t – 16 When t = 0, a = 4(0) – 16 = –16 m s–2

(b) v = �a dt

= � (4t – 16)dt = 2t 2 – 16t + c When t = 0, v = 30 and so c = 30 Hence, at time t, v = 2t 2 – 16t + 30

(c) When the particle reverses its direction of motion,

v = 0 2t 2 – 16t + 30 = 0 t 2 – 8t + 15 = 0 (t – 3)(t – 5) = 0 t = 3 or t = 5 ∴ t = 3 s and t = 5 s

(d) s = �v dt

= 2––3

t 3 – 8t 2 + 30t + c

When t = 0, s = 0 and so c = 0 Hence, at time t,

s = 2––3

t 3 – 8t 2 + 30t

When t = 3,

s = 2––3

(3)3 – 8(3)2 + 30(3)

= 36 m

When t = 5,

s = 2––3

(5)3 – 8(5)2 + 30(5)

= 250––––3

– 200 + 150

= 33 1––3

m


Os

t = 3s = 36

t = 0s = 0

t = 5

s = 331––3

∴ The distance travelled by the particle in the fi rst 5 seconds

= 36 + �36 – 33 1––3 �

= 38 2––3

m

21 Linear Programming

1

Booster Zone 1 (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

2 (a)

(b)

(c)

(d)

(e)

(f)

3 (a) y � – 1 (b) x � 5(c) y � 2x (d) x + y � 4(e) y – 2x � 8 (f) 2x – y � 10

4 (a)

(b)

(c)

(d)

5 (a) x � 5, y � x + 5, x + y � 5(b) y � 2, y � 2x, y � x(c) y � 6, y � x, x + y � 6(d) 2y � x, y � 3x, 3y + 2x � 24

6 x = curry puffs, y = doughnutsI : x + y � 120II : x � 2yIII : x – y � 40∴ x + y � 120, x � 2y, x – y � 40

7 I : x � yII : x � 2yIII : 8x + 5y � 5000∴ x � y, x � 2y, 8x + 5y � 5000

8 (a) 2 � x � 4(b) At point (3, 6), 3 + 2(6) = 15

9 (a) Maximum value of x = 10 Maximum value of y = 12(b) 2x + 3y = 6 At point (5, 2), 2(5) + 3(2) = 16 ∴ The minimum value of

2x + 3y is 16.10 (a) I : 60x + 30y � 50 × 60

2x + y � 100 II : 30x + 60y � 30 × 60 x + 2y � 60

III : y

—x � 2—1

y � 2x ∴ 2x + y � 100, x + 2y � 60, y � 2x

O 2

y

x

x = 2

O–1

y

x

x = –1

O3

y

xy = 3

O–3

y

xy = –3

O

5

5

y

xx + y = 5

O

(1, 2)

y

x

y = 2x

O

–6

6

y

xx – y = 6

O–2

1

y

x

2y = x + 2

O

–4

3

y

x

4x – 3y = 12

O

2

y

x

y = 2

O–4

y

x

x = –4

O

6

3

y

xy + 2x = 6

O

–4

8

y

xx – 2y = 8

O

(1, 3)

y

x

y = 3x

O

6

6

y

x

O

2

y

x

y = xy = 2

x + y = 2

O

3

y

x

y = 3

4y = 3x +

12

y = 3x – 4

O

y

xx = 1 y + 2x = 6

2y = x

O

y

xx + 2y = 4

y = 2x

2y + 3x = 12



(b)

(c) (i) 47 chairs (ii) 20 chairs

(d) 20x + 25y = k At point (25, 50), 20(25) + 25(50) = 1750 ∴ The maximum profit can

be obtained by the factory is RM1750.

SPM Appraisal Zone

1 (a) I : x + y � 150

II : y � 1—2

x

III : 400x + 200y � 80 000 2x + y � 400 ∴ x + y � 150, y �

1—2 x,

2x + y � 400(b)

(c) (i) When x = 100, y = 200 (ii) When y = x, the minimum

number printer A = 75 and printer B = 75.

(d) 120x + 60y = k At point (160, 80), 120(160) + 60(80) = 24 000 ∴ The maximum profi t = RM24 000

2 (a) I : x + y � 30 II : x – y � 20 III : 20x + 40y � 1600 x + 2y � 80 ∴ x + y � 30, x – y � 20, x + 2y � 80(b)

(c) (i) 3x + 2y = 6 (ii) At point (40, 20), 3(40) + 2(20) = 160 ∴ The maximum amount = RM160

3 (a) I : x + y � 140 II : y � 3x III : x – y � 40 ∴ x + y � 140, y � 3x, x – y � 40(b)

(c) (i) When y = 60, 20 � x � 80 (ii) 60x + 40y = 2400, x = 50. At point (50, 90), 60(50) + 40(90) = 6600 ∴ The maximum fees

collected = RM6600

4 (a) I : x � 2y

II : x—y

� 2—7

x � 2—7

y

III : 5x + 10y � 800 x + 2y � 160

∴ x � 2y, x � 2—7

y, x + 2y � 160(b)

(c) (i) Maximum point is (20, 70). Number of slippers = 20 Number of sandals = 70

(ii) 3x + 8y = k At point (20, 70), 3(20) + 8(70) = 620 ∴ The maximum amount

= RM620 5 (a) I : x + y � 80

II : y – 2x � 20 III : 5x + 10y � 400 x + 2y � 80 ∴ x + y � 80, y – 2x � 20, x + 2y � 80(b)

(c) (i) When x = 30, 25 � y � 50 (ii) 4x + 8y = k At point (20, 60), 4(20) + 8(60) = 560 ∴ The maximum profit

obtained = RM560

10

100

20

20

30

30

40

40

50

50

60

60

70

70

80

80

90

100

y

x

2x + y = 100

(25, 50)

x + 2y = 60

y = 2x

R

50

500

100

100

150

150

200

200

250

250

300

300

350

350

400

400

450

y

x

(160, 80)

x + y = 150

1y = — x 2

y = x

R

2x + y = 400

x = 100

10

100

20

20

30

30

40

40

50

50

60

60 70 803x + 2y = 6

x + y = 30

y

x

(40, 20)

x + 2y = 80

x – y = 20

R

200

20

40 60

40

80 100

60

120 140

80

160

100

120

140

160

60x + 40y = 2400

y

x

y = 3x

x = 50

R

x + y = 140

y = 60

x – y = 40

100

10

20 30

20

40 50

30

60 70

40

80

50

60

70

80

y

x

y – 2x = 20

(20, 60)

x = 30

R

x + 2y = 80

x + y = 80

200

20

40 60

40

80 100

60

120 140

80

160

100

120

140

y

x

(20, 70)

2x = — y 7

Rx + 2y = 160

x = 2y

1

SPM-Cloned Questions

Chapter 12: Progressions 1 (a) 2m + 1 – m = 5m – 1 – (2m + 1)

m + 1 = 3m – 2 2m = 3 m = 3—

2 (b) The fi rst three terms are

3—2

, 4, 13—–2

, ...

So, a = 3—2

and d = 4 – 3—2

= 5—2

∴ S

11 =

11—–2

�2� 3—2 � + 10 � 5—

2 �� =

11—–2

(3 + 25) = 154

2 For –9, –5, –1, ... d = –5 – (–9) = 4 S

4 = 100

4—2

[2a + 3(4)] = 100 2(2a + 12) = 100 2a + 12 = 50 2a = 38 a = 19∴ The four consecutive terms which sum up to 100 are 19, 23, 27 and 31.

3 (a) d = 9 – 4 = 5(b) The next four terms are 19, 24, 29

and 34. So, the sum of these terms are

19 + 24 + 29 + 34 = 106.

4 For 18, 6, 2, ... a = 18 and r =

6—–18

= 1—3

∴ S∞ = 18———1 – 1—

3 = 18——

� 2—3 �

= 27

5 x—–16

= 16—––8

x = –2(16) = –32

6 (a) 8, 16, 24(b) d = 16 – 8 = 8

7 (a) T4 = 155

k + 3m = 155 … 1 S

8 = 1340

8—2

(2k +7m) = 1340

2k + 7m = 335 … 2 1 × 2 2k + 6m = 310 … 3 2 – 3 m = 25 Substitute m = 25 into 1 : k + 3(25) = 155 k = 155 – 75 = 80(b) 80 + (n – 1)25 = 120 + (n – 1)15 80 + 25n – 25 = 120 + 15n – 15 25n + 55 = 105 + 15n 10n = 50 n = 5

8 (a) T1 = 6� 2—

3 � = 4m T

2 = 4� 2—

3 � = 8—3

m This is in the form of a geometric

progression with a = 4 and r = 8—

3 ÷ 4

= 2—

3

∴ T10

= 4� 2—3 �

9

= 0.104 m(b) The total distance travelled = 6 + 2(4) + 2� 8—

3 � + ... = 6 + 8———

1 – 2—3

= 6 + 24 = 30 m

Chapter 13: Linear Law

1 y = p x – q

—–x

y

—–x

= –q� 1—x � + p

From the graph, –q = 4 – 1———

1—2

– 2 =

3——–

� 3– — 2 �

–q = –2 q = 2

y

—–x

= –2� 1—x � + p At (2, 1), 1 = –2(2) + p

p = 1 + 4 = 5∴ p = 5, q = 2

2 3y = 8x3 + x 3y

—–x

= 8x2 + 1 y—x

= 8—3

x2 + 1—3

At (h, 3), 3 = 8—3

h + 1—3

8—3

h = 8—3

h = 1 At (4, k), k = 8—

3(4) + 1—

3

= 33—–3

= 11∴ h = 1, k = 11

3 y = 2x2 – 4 y

—x2

= 2 – 4—x2

y

—x2

= 2 – 4� 1—x2 �

4 x + a—

x = by

bxy = x2 + a xy = 1—

bx2 + a—

b From the graph, 1—b

= 2——–12 – 6

= 1—3

b = 3 At (6, 0), 0 = 1—

3(6) + a—

3 a—

3 = –2

a = –6∴ a = –6, b = 3

5 (a) y = k4x

log10

y = log10

k + log10

4x

log10

y = log10

4(x) + log10

k(b) log

10 k = 2

k = 102

= 100

6 (a) x2 xy

1 15.5

4 20.0

9 27.6

16 38.0

25 51.5

36 67.8



(b) y = hx + k—x xy = hx2 + k

From the graph,

(i) h = 38 – 14———–16 – 0

= 1.5 (ii) k = 14

7 (a) 1—x1—y

1.00 0.75

0.50 1.00

0.33 1.09

0.25 1.12

0.20 1.15

(b)

(c) a—y = – b—x + 4 1—y = – b—a � 1—x � + 4—a From the graph,

(i) 4—a = 1.26 a =

4——1.26

= 3.175

(ii) – b—a = 0.75 – 1.26—————

1.0 – 0 –

b——–3.175

= –0.51

b = 1.619

Chapter 14: Integration 1 Area = 9 unit2

�k

0x2 dx = 9

�x3

—3 �

k

0

= 9

k3

—3

= 9

k3 = 27 k = 3

2 �

3

1 f(x) dx + �

5

3[f(x) + 2] dx

= �3

1f(x) dx + �

5

3f(x) dx + �

5

32 dx

= �5

1f(x) dx + �2x�

5

3

= 8 + (10 – 6)= 12

3 (a) �

2

3 –2h(x) dx = –2 �

2

3h(x) dx

= –2(–4) = 8

(b) �3

2[6 – h(x)] dx = �

3

26 dx – �

3

2h(x) dx

= �6x�3

2 – 4

= (18 – 12) – 4 = 2

4 (a) �(3x2 – 5) dx

= x3 – 5x + c Compare with x3 – px + c So, p = 5

(b) �(3x2 – 5) dx = 4 x3 – 5x + c = 4 When x = 2, 23 – 5(2) + c = 4 8 – 10 + c = 4 c = 6

5 (a) At turning point (2, 5), k(2) – 4 = 0 2k = 4 k = 2 (b)

dy—–dx

= 2x – 4 y = �(2x – 4) dx

= x2 – 4x + c Since (2, 5) lies on the curve, 5 = 22 – 4(2) + c 5 = 4 – 8 + c c = 9

∴ The equation of the curve is y = x2 – 4x + 9.

6 �5

hf(y) dy = 3(5) – 11

= 15 – 11 = 4 unit2

7 (a) y = k(x + 1)3

dy—–dx

= 3k(x + 1)2

At x = –2,

dy—–dx

= 6

3k(–2 + 1)2 = 6 3k = 6 k = 2(b) (i) Area of the shaded region, P

= �–1

–2 2(x + 1)3 dx

= (x + 1)4�———�

–1

–2 2

= 0 – 1—2

= – 1—2

= 1—2

unit2

(ii) Volume generated = π�

0

–14(x + 1)6 dx

= π 4(x + 1)7�————�

0

–1 7 = 4—

7 π unit3

8 (a) y = x2 + 4 dy

—–dx

= 2x At A(–1, 5), dy

—–dx

= 2(–1) = –2 ∴ Equation of the tangent at A is y – 5 = –2 (x + 1) y – 5 = –2x – 2 y = –2x + 3(b) Area under the curve = �

0

–1(x2 + 4) dx

= �x3

—3

+ 4x�0

–1

= 0 – �– 1—3

– 4� = 13—–

3 unit2

Area of trapezium

= 1—2

(1)(3 + 5)

= 4 unit2


= 13—–3

– 4

= 1—3

unit2


= π�6

4(y – 4) dy

= π�y2

—2

– 4y�6

4 = π[18 – 24 – (8 – 16)] = 2π unit3

Chapter 15: Vector

1 (a) →OP = �2

3 �(b)

→PQ =

→PO +

→OQ

= –2 i~ – 3 j~

+ 4 i~ + 4 j~

= 2 i~ + j~

2 →PQ =

→PO +

→OQ

= 8 i~ – 4 j~

+ 5 i~ + 3 j~

= 13 i~ – j~

3 →OD = 3—

4

→OB

= 3—4

(8x~ + 4y~

)

= 6x~ + 3y~

xy

x2

70

60

50

40

30

20

10

0 5 10 15 20 25 30 35 40

(0, 14)

(16, 38)

1—y(0, 1.26)

1.2

1.0

0.8

0.6

0.4

0.2

00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1—x

(1.0, 0.75)


4 (a) a~ – b~ = 12 i~ + j~

– (9 i~ – k j~

) = 3 i~ + (1 + k) j

~(b) �a~ – b~� = 5

32 + (1 + k)2 = 5 9 + 1 + 2k + k2 = 25 k2 + 2k – 15 = 0 (k + 5)(k – 3) = 0 k = –5 or k = 3

5 (a) →QR =

→QP +

→PR

= –2a~ + 5b~ (b)

→QS =

1—4

→QR

=

1—4

(–2a~ + 5b~)

= – 1—2

a~ + 5—4

b~

→PS =

→PQ +

→QS

= 2a + �– 1—2

a~ + 5—4

b~� =

3—2

a~ + 5—4

b~

6 (a) (i)

→OP =

1—3

→OA

= 1—3

(6x~)

= 2x~

→BP =

→BO +

→OP

= 2x~ – 4y~

(ii) →AQ =

1—2

→AB

= 1—2

(–6x~ + 4y~

)

= –3x~ + 2y~

→OQ =

→OQ +

→AQ

= 6x~ + (–3x~ + 2y~

) = 3x~ + 2y

~

(b) →OR =

→OB +

→BR

h(3x~ + 2y~

) = 4y~

+ k(2x~ – 4y~

) 3hx~ + 2hy

~ = 2kx~ + (4 – 4k)y

~ Equating the coeffi cients, we have: 3h – 2k = 0 … 1 2h = 4 – 4k 2h + 4k = 4 h + 2k = 2 … 2 1 + 2 : 4h = 2 h = 1—

2 Substitute h = 1—

2 into 1 :

3� 1—2 � – 2k = 0

2k = 3—2

k = 3—4

∴ h = 1—2

, k = 3—4

(c) � →AB � = 122 + 42

= 160

= 12.649

7 (a) (i) →AP =

1—3

→AD

→AD = 3

→AP

= 3y~

→DB =

→DA +

→AB

= –3y~

+ x~ = x~ – 3y

~

(ii) →BR =

1—3

→BD

= 1—3

(–x~ + 3y~

)

= – 1—3

x~ + y~

→AR =

→AB +

→BR

= x~ + �– 1—3

x~ + y~�

= 2—3

x~ + y~

(b) →AR = h

→AC

= h(→AD +

→DC)

= h�3y~

+ kx~ – 3—2

y~�

= hkx~ + 3—2

hy~

2—3

x~ + y~

= hkx~ + 3—2

hy~

Equating the coeffi cients,

3—2

h = 1 and hk = 2—3

h = 2—3

2—3

k = 2—3

k = 1 ∴ h =

2—3

, k = 1

8 (a) (i) →AP =

→AO +

→OP

= 2p – 2q

(ii) →PC =

1—3

→PQ

= 1—3

(–2p + 4q)

= – 2—3

p + 4—3

q

→OC =

→OP +

→PC

= 2p – 2—3

p + 4—3

q

= 4—3

p + 4—3

q

(b) →AB =

→AO +

→OB

h(2p~

– 2q~

) = –2q~

+ k� 4—3

p~

+ 4—3

q~�

2hp~

– 2hq~

= 4—3

kp~

+ � 4—3

k – 2�q~

Equating the coeffi cients,

2h = 4—3

k

6h – 4k = 0 3h – 2k = 0 … 1

–2h = 4—3

k – 2

–6h = 4k – 6 6h + 4k = 6 3h + 2k = 3 … 2

1 + 2 : 6h = 3

h = 1—2

Substitute h = 1—2

into 1 :

3� 1—2 � – 2k = 0

2k = 3—2

= 3—4

∴ h = 1—2

, k = 3—4

Chapter 16: Trigonometry 1

(a) sec θ = 1——–

cos θ = 1————–

1 + m2

m�———–� =

1 + m2

———–m

(b) cos (90° – θ) = sin θ

= 1 + m2

1———–

2 2 cos 2x + 4 sin x + 1 = 0 2(1 – 2 sin2 x) + 4 sin x + 1 = 0 2 – 4 sin2 x + 4 sin x + 1 = 0 4 sin2 x – 4 sin x – 3 = 0 (2 sin x + 1)(2 sin x – 3) = 0 2 sin x + 1 = 0 sin x = – 1—

2 x = 210°, 330° or sin x = 3—

2 (no solution)

∴ x = 210°, 330°

3 3 sin2 x – 5 cos x = 5 3(1 – cos2 x) – 5 cos x = 5 3 – 3 cos3 x – 5 cos x = 5 3 cos2 x + 5 cos x + 2 = 0 (3 cos x + 2)(cos x + 1) = 0 3 cos x + 2 = 0

cos x = – 2—3

x = (180° – 48° 11'), (180° + 48° 11') = 131° 49', 228° 11'or cos x + 1 = 0 cos x = –1 x = 180°∴ x = 131° 49', 180°, 228° 11'

4

m

1

θ

1 + m2

pA

1 – p21


(a) sin 2A = 2 sin A cos A = 2p 1 – p2

(b) cos A = 2 cos2 A—2

– 1

p = 2 cos2 A—2

– 1

p + 1 = 2 cos2 A—2

cos2 A—2

= p + 1——–

2

cos A—2

= p + 1——–

2

5

(a) tan A = – 3—4

(b) cos (A + B) = cos A cos B – sin A sin B

= �– 4—5 �� 5—–

13 � – � 3—5 �� 12—–

13 � = – 20—–

65 – 36—–

65

= – 56—–65

6 (a),(b)

2 sin2 x = x——180°

1 – 2 sin2 x = 1 – x——180°

cos 2x = 1 – x——180°

y = 1 – x——180°

x 0 180°

y 1 0

Number of solutions = 2

7 (a) cos4 x – sin4 x = (cos2 x + sin2 x)(cos2 x – sin2 x) = cos2 x – sin2 x = 1 – sin2 x – sin2 x = 1 – 2 sin2 x = cos 2x

(b) (i),(ii)

2(cos4 x – sin4 x) =

x—π – 1 2 cos 2x = x—

π – 1

cos 2x = x—–

2π – 1—

2

y = x—–2π

– 1—2

x 0 π

y – 1—2

0


8 (a),(b)

2x—–3π

– sin 2x = 1

sin 2x = 2x—–3π

– 1

3—2

sin 2x = x—π – 3—2

y = x—π – 3—2

x 0 π

y – 3—2

– 1—2


Chapter 17: Permutations and Combinations

1 4P3 × 5P

4 = 24 × 120

= 2880

2 (a) 6C4 × 5C

2 = 15 × 10

= 150(b) (6C

2 × 5C

4) + (6C

1 × 5C

5)

= 75 + 6 = 81

3 (a) 9! = 362 880

(b) 5 7 6 5 4 3 2 1 4 V C = 5 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 4 = 100 800

4 (a) 18C11

= 31 824(b) 3C

1 × 5C

3 × 4C

4 × 6C

3

= 600

5 (a) 5 4 3 = 5 × 4 × 3 = 60(b) 4 3 4 = 4 × 3 × 4 C = 48

6 (a) 5C4 × 4C

3 = 5 × 4

= 20(b) 4! × 4! = 576

7 (a) 6 5 4 3 = 6 × 5 × 4 × 3 = 360(b) 5 4 3 4 = 5 × 4 × 3 × 4 O = 240

8 (a) 9! = 362 880(b) 6! × 4! = 17 280

Chapter 18: Probability 1 Area of big circle = π(6)2

= 36π cm2

Area of small circle = π(3)2

= 9π cm2

(a) Probability of hitting the unshaded area

area of small circle = ———————— area of big circle

= 9π——36π

= 1—4

(b) Area of shaded area = (36π – 9π) cm2

= 27π cm2

Probability of hitting the shaded area

Area of the shaded area = —————————— Area of big circle

= 27π——36π

= 3—4

2 The tree diagram is drawn and shown as follows:

1st set 2nd set 3rd set

y

5

x

3

O

A

y

x

1312

OB

x

y

y = cos 2x

45° 90° 135° 180°

1

0

–1

y = 1 – x——180°

yy = cos 2x

π—4

3—4

π

1

0

–1

π—2

xπy = x—–

2π – 1—

2

2—5

3—5

Win

Not win

2—5

3—5

2—5

3—5

Win

Not win

Win

Not win

Win

2—5

Win

2—5

0xπ—

4π—2

3—4

π 5—4

ππ

y = x—π – 3—2

3—2

y

3—2

–

y = 3—2

sin 2x

3—2

π


From the tree diagram,(a) P(the match ends in two sets

only)

= � 2—5

× 2—5 � + � 3—

5 × 3—

5 � = 4—–

25 + 9—–

25

= 13—–25

(b) P(Pawaz wins after competing three sets)

= � 2—5

× 3—5

× 2—5 � + � 3—

5 × 2—

5 × 2—

5 � = 12—–

125 + 12—–

125

= 24—–125

3 We defi ne the following events as follows:A : getting the same number on the two

spinners.B : the fi rst spinner shows the larger

numbers.

The possibility diagram is shown below:

From the possibility diagram,(a) The dots enclosed in the loop above

represent the possible outcomes in event A.

n(A) = 4.

∴ P(A) = n(A)——n(S)

= 4—–16

= 1—4

(b) The triangle in the possibility diagram above contains the dots representing the outcomes in event B.

n(B) = 6 ∴ P(B) = n(B)——

n(S)

= 6—–16

= 3—8

4 P(G) = 5—

8 x—–

32 = 5—

8

x = 5—8

× 32

= 20

The number of marbles which are not green = 32 – 20 = 12

5 P(two chips of the same colour)= P(RR) + P(BB)

= � 4—7 �� 3—

5 � + � 3—7 �� 2—

5 �= 12—–

35 + 6—–

35

= 18—–25

6 (a) P(Mukhriz or Haziq wins) = P(Mukhriz wins) + P(Haziq wins)

= 1—3

+ 1—4

= 7—–12

(b) P(neither Mukhriz nor Haziq wins) = 1 – P(Mukhriz or Haziq wins)

= 1 – 7—–12

= 5—–12

7 (a) P(neither of them is chosen) = 2—

5 × 7—–

15

= 7—–25

(b) P(one of them is chosen)

= � 2—5

× 7—–15 � + � 3—

5 × 8—–

15 � = 14—–

75 + 24—–

75

= 38—–75

8 (a) P(all three are prefects) = 0.4 × 0.4 × 0.4 = 0.064(b) P(only one of them is prefect) = (0.4 × 0.6 × 0.6) + (0.6 × 0.4

× 0.6) + (0.6 × 0.6 × 0.4) = 0.432

Chapter 19: Probability Distributions

1 (a) X ~ B(10, 0.6) P(x � 8) = P(X = 8) + P(X = 9) + P(X = 10) = 10C

8(0.6)8(0.4)2 + 10C

9(0.6)9(0.4)

+ 10C10

(0.6)10(0.4)0

= 0.12093 + 0.04031 + 0.00605 = 0.1673(b) X ~ N(55, 52)

(i) z = 62 – 55———–

5 = 1.4 (ii) z = –1.4

x – 55———5

= –1.4

x = 55 – 7 = 48 kg

(iii) P(46 � X � 58)

= P� 46 – 55———–5

� Z � 58 – 55———–5 �

= P(–1.8 � Z � 0.6) = 1 – 0.0359 – 0.2743 = 0.6898

2 (a) X ~ B(8, 0.1) (i) P(X = 3) = 8C

3(0.1)3(0.9)5

= 0.0331 (ii) P(X � 3) = P(X = 0) + P(X = 1) + P(X = 2) = 8C

0(0.1)0(0.9)8 + 8C

1(0.1)

(0.9)7 + 8C2(0.1)2(0.9)6

= 0.43047 + 0.38264 + 0.1488 = 0.9619

(b) X ~ N(600, 202) (i) P(X � 575)

= P�Z � 575 – 600—–———20 �

= P(Z � –1.25) = 0.1056 (ii) P(570 � X � 610)

= P�570 – 600————20

� Z � 610 – 600————20 �

= P(–1.5 � Z � 0.5) = 1 – 0.0668 – 0.3085 = 0.6247

(c) The number of cabbages in the market

= 50———0.6247

= 80

3 P(0 � Z � k) = 0.5 – 0.2842 = 0.2158

4 X ~ B(8, 0.3)P(X = 3) = 8C

3 (0.3)3(0.7)5

= 0.2541

5 (a) z = 1.5

x – 220———–20

= 1.5

x – 220 = 30 x = 250

(b) P(X � 210) = P�Z � 210 – 220————–

20 � = P(Z � –0.5) = 1 – P(Z � –0.5) = 1 – 0.3085 = 0.6915

2nd spinner

1st spinner

4

3

2

1

1 2 3 4

f(z)

z–1.8 0.6O

f(z)

z–1.5 0.5O


6 (a) (i) P(X � 2) = 1 – P(X � 2) = 1 – P(X = 0) – P(X = 1) = 1 – 10C

0 (0.05)0(0.95)10

– 10C1 (0.05)(0.95)9

= 1 – 0.59874 – 0.3151 = 0.0862 (ii) npq = 5.7 n(0.05)(0.95) = 5.7

n = 5.7———0.0475

= 120(b) X ~ N(62, 82) P(55 � x � 72)

= P� 55 – 62———–8

� Z � 72 – 62———–8 �

= P(–0.875 � Z � 1.25) = 1 – P(Z � –0.875) – P(Z � 1.25) = 1 – 0.1908 – 0.1056 = 0.7036 ∴ The total number of workers

= 76———0.7036

= 108

7 (a) X ~ B�6, 3—5 �

(i) P(X = 4) = 6C4 � 3—

5 �4

� 2—5 �

2

= 0.3110 (ii) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 6C0 � 3—

5 �0

� 2—5 �

6

– 6C1 � 3—

5 �� 2—5 �

5

= 1 – 0.004096 – 0.036864 = 0.9590

(b) X ~ N(70, 252) (i) P(65 � X � 85)

= P�65 – 70———25

� Z � 85 – 70———25 �

= P(–0.2 � Z � 0.6) = 1 – P(Z � –0.2) – P(Z � 0.6) = 1 – 0.4207 – 0.2743 = 0.3050 (ii) P(X � 100)

= P�Z � 100 – 70———–20 �

= P(Z � 1.5) = 0.0668 The total number of workers

= 25———0.0668

= 374

8 (a) (i) E(X) = np

= 20� 1—4 �

= 5 (ii) σ = npq

= 5� 3—4 �

= 3.75 = 1.9365

(b) (i) P(X = 3) = 8C3 � 1—

4 �3

� 3—4 �

5

= 0.2076 (ii) P(X � 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 8C0�1—

4 �0

�3—4 �

8

– 8C1�1—

4 ��3—4 �

7

= 1 – 0.1001129 – 0.2669678 = 0.6329

Chapter 20: Motion along a Straight Line

1 (a) a = dv—–dt

= 6t – 8 When the particle is at rest, v = 0 3t2 – 8t – 3 = 0 (3t + 1)(t – 3) = 0 Since t � 0, t = 3 ⇒ a = 6(3) – 8 = 18 – 8 = 10 m s–2

(b) s = ∫∫v dt = t3 – 4t2 – 3t + c When t = 0, s = 0 and so c = 0 Hence, s = t3 – 4t2 – 3t When t = 2, s = (2)3 – 4(2)2 – 3(2) = 8 – 16 – 6 = –14 When t = 3, s = (3)3 – 4(3)2 – 3(3) = 27 – 36 – 9 = –18 Thus, the distance of AB is = 18 – 14 = 4 m

2 v = ds—–dt

= 3pt2 + 2qt + 4

a = dv—–dt

= 6pt + 2qWhen t = 2, a = 0 6p(2) + 2q = 0 12p + 2q = 0 6p + q = 0 … 1and v = –8 3p(2)2 + 2q(2) + 4 = –8 12p + 4q = –12 3p + q = –3 … 21 – 2 : 3p = 3 p = 1

Substitute p = 1 into 1 : 6(1) + q = 0 q = –6∴ p = 1, q = –6

3 (a) v = ∫a dt = 12t – 3t2 + c When t = 0, v = 15 and so c = 15 Hence, v = 12t – 3t2 + 15(b) s = ∫v dt = 6t2 – t3 + 15t + c When t = 0, s = 0 and so c = 0 Hence, s = 6t2 – t3 + 15t When t = 5, s = 6(5)2 – 53 + 15(5) = 150 – 125 + 75 = 100 Thus, the distance travelled in the

fi rst 5 seconds is 100 m.

4 (a) (i) a = 6 – 2t v = ∫a dt = ∫(6 – 2t) dt = 6t – t2 + c When t = 0, v = 16 and so c = 16 Hence, v = 6t – t2 + 16 For maximum velocity, a = 0 6 – 2t = 0 t = 3

Moreover, da—–dt

= –2 � 0, v is

maximum when t = 3. Thus, maximum velocity = 6(3) – 32 + 16 = 18 – 9 + 16 = 25 m s–1

(ii) When the particle stops, v = 0 6t – t2 + 16 = 0 t2 – 6t – 16 = 0 (t – 8)(t + 2) = 0 Since t � 0, t = 8 ∴ k = 8

(b)

Total distance travelled

= �8

0(6t – t2 + 16) dt

= �3t2 – 1—3

t3 + 16t�8

0

= 3(64) – 1—3

(512) + 16(8)

= 149 1—3

m

f(z)

z–0.5 O

f(z)

z–0.875 1.25O

v

25

16

0 3 8t

v = 6t – t2 + 16


5 (a) v = 2t (4 – t) = 8t – 2t2

a = dv—–dt

= 8 – 4t For maximum velocity, a = 0 8 – 4t = 0 t = 2


= –4 � 0 ⇒ v is

maximum when t = 2. Thus, maximum velocity = 8(2) – 2(2)2

= 16 – 8 = 8 m s–1

(b) s = ∫v dt

= 4t2 – 2—3

t3 + c


Hence, s = 4t2 – 2—3

t3

s4 = 4(4)2 – 2—

3(4)3

= 64 – 128—–3

= 21 1—3

s3 = 4(3)2 – 2—

3 (3)3

= 36 – 18 = 18 m ∴ The distance travelled in the 4th

second

= 21 1—3

– 18

= 3 1—3

(c) When the particle passes O again, s = 0

4t2 – 2—3

t3 = 0

2t2�2 – 1—3

t� = 0

t = 0 or = 6 ∴ t = 6 s(d) When the particle reverses its

direction of motion v = 0 8t – 2t2 = 0 2t (4 – t) = 0 t = 0 or t = 4 ∴ t = 4 s

6 (a) v = t2 – 8t + 12 When t = 0, v = 02 – 8(0) + 12 = 12 Therefore, the initial velocity of

the particle is 12 m s–1.

(b) a = dv—–dt

= 2t – 8 For minimum velocity, a = 0 2t – 8 = 0 t = 4


= 2 � 0, v is

minimum when t = 4. Thus, minimum velocity = 42 – 8(4) + 12 = 16 – 32 + 12 = –4 m s–1

(c) When the particle moves to the left,

v � 0 t2 – 8t + 12 � 0 (t – 2)(t – 6) � 0

∴ 2 � t � 6(d) s = ∫v dt

= ∫(t2 – 8t + 12) dt

= 1—3

t3 – 4t2 + 12t + c When t = 0, s = 0 and so c = 0

Hence, s = 1—3

t3 – 4t2 + 12t

s2 = 1—

3(23) – 4(2)2 + 12(2)

= 8—3

– 16 + 24

= 10 2—3

m

s6 = 1—

3(6)3 – 4(6)2 + 12(6)

= 72 – 144 + 72 = 0 m The motion of the particle is shown

as follows:

∴ The total distance travelled in the fi rst 6 seconds

= 2�10 2—3 �

= 21 1—3

m

7 (a) v = 8 + 2t – t2

a = dv—–dt

= 2 – 2t When the particle is at rest, v = 0 8 + 2t – t2 = 0 t2 – 2t – 8 = 0 (t – 4)(t + 2) = 0 Since t � 0, t = 4 ∴ The acceleration of the particle

at R = 2 – 2(4) = 2 – 8 = –6 m s–2

(b) For maximum velocity, a = 0 2 – 2t = 0 t = 1


= –2 � 0 ⇒ v is

maximum when t = 1. Thus, maximum velocity = 8 + 2(1) – (1)2

= 9 m s–1

(c) s = ∫v dt = ∫(8 + 2t – t2) dt

= 8t + t – 1—3

t3 + c When t = 0, s = 0 and so c = 0

Hence, s = 8t + t – 1—3

t3

s4 = 8(4) + 4 – 1—

3 (4)3

= 36 – 64—–3

= 14 2—3

m

s6 = 8(6) + 6 – 1—

3(6)3

= 54 – 72 = –18 m The motion of the particle is shown

as follows:

∴ The total distance travelled in the fi rst 6 seconds

= 2�14 2—3 � + 18

= 47 1—3

m

8 (a) v = 3t – 1—2

t2 + 8

When t = 0, v = 3(0) – 1—2

(0)2 + 8 = 8 Therefore, the initial velocity of

the particle is 8 m s–1.

(b) a = dv—–dt

= 3 – t When t = 0, a = 3 Therefore, the initial acceleration

of the particle is 3 m s–2.(c) For maximum velocity, a = 0 3 – t = 0 t = 3


= –1 � 0 ⇒ v is

maximum when t = 3. Thus, the maximum velocity

= 3(3) – 1—2

(3)2 + 8

= 12 1—2

m s–1

(d) When the particle stops instantaneously, v = 0.

3t – 1—2

t2 + 8 = 0

2 6t

t = 0s = 0

t = 6s = 0

t = 2s = 10 2—

3

OS

t = 0s = 0

t = 6s = –18

t = 4s = 14 2—

3

O


1—2

t2 – 3t – 8 = 0

t2 – 6t – 16 = 0 (t – 8)(t + 2) Since t � 0, t = 8 s = ∫v dt

= ∫∫�3t – 1—2

t2 + 8� dt

= 3—2

t2 – 1—6

t3 + 8t + c


Hence, s = 3—2

t2 – 1—6

t3 + 8t

s8 = 3—

2(8)2 – 1—

6(8)3 + 8(8)

= 96 – 256—–3

+ 64

= 74 2—3

m

Chapter 21: Linear Programming

1 (a) I: (0, 0) and (2, 6)

m = 6—2

= 3 The equation is y – 6 = 3(x – 2) y – 6 = 3x – 6 y = 3x ∴ y � 3x II: (0, 0) and (5, 3)

m = 3—5

The equation is

y – 3 = 3—5

(x – 5)

y – 3 = 3—5

x – 3

y = 3—5

x

∴ y � 3—5

x III: (2, 6) and (5, 3)

m = 3 – 6——–5 – 2

= –1 The equation is y – 6 = –1 (x – 2) y – 6 = –x + 2 y + x = 8 ∴ y + x � 8(b) The maximum value is at point (5, 3). Thus, the maximum value of 2x + y = 2(5) + 3 = 13

2 (a) I: (4, 0) and (6, 4)

m = 4—2

= 2 The equation is y – 4 = 2(x – 6) y – 4 = 2x – 12

y = 2x – 8 ∴ y � 2x – 8

II: (0, 8) and (6, 4)

m = – 4—6

= – 2—3

The equation is

y – 8 = – 2—3

(x – 0)

3y – 24 = –2x 3y + 2x = 24 ∴ 3y + 2x � 24

III: x � 2(b) The number of solutions is 17.

3 (a) I: A(2, 0) and E(5, 12)

m = 12—–3

= 4 The equation is y = 4(x – 2) y = 4x – 8 ∴ y � 4x – 8 y – 4x � –8

II: B(6, 0) and D(0, 8)

m = – 8—6

= – 4—3

The equation is

y = – 4—3

(x – 6)

y = – 4—3

x + 8

∴ y � – 4—3

x + 8

3y + 4x � 24

III: D(0, 8) and E(5, 12)

m = 4—5

The equation is

y – 8 = 4—5

(x – 0)

5y – 40 = 4x 5y – 4x = 40 ∴ 5y – 4x � 40

(b) y – 4x = –8 … 1 3y + 4x = 24 … 2

1 + 2 : 4y = 16 y = 4 Substitute y = 4 into 1 : 4 – 4x = –8 4x = 12 x = 3 ∴ C(3, 4) Thus, the value of z x + 2y = 3 + 2(4) = 11

4 (a) I: x + y � 80 II: y � 5x

III: y � 1—2

x + 5

(b)

(c) (i) 15 � y � 60 (ii) The maximum total fees = 60x + 40y = 60(50) + 40(30) = 3000 + 1200 = RM4200

5 (a) I: 45x + 30y � 600 3x + 2y � 40

II: 50x + 70y � 350 5x + 7y � 35

III: x—y

� 4—5

y � 5—4

x

(b)

(c) (i) 14 shirts (ii) Maximum profi t = 16x + 8y = 16(7) + 8(9) = 112 + 72 = RM184

6 (a) I: x + y � 400 II: x � 2y

y � 1—2

x

III: 8x + 4y � 1200 2x + y � 300

y

80

70

60

50

40

30

2015

10

y = 5x

0 10 20 30 40 50 60 70 80

R

k = 60x + 40y

(50, 30)

y = 1—2

x + 5

x

y

20

18

16

14

12

10

8

6

4

2

0 1 2 3 4 5 6 7 8x

R

3x + 2y = 40

(7.25, 9.1)

y = 5—4

x

k = 16x + 8y 5x + 7y = 35


(b)

(c) (i) 250 tubes (ii) Maximum profi t = 8x + 4y = 8(267) + 4(134) = RM2672

7 (a) I: x + y � 7 II: x � 3y

y � 1—3

x

III: 400x + 200y � 2000 2x + y � 10

(b)

(c) (i) 1 van (ii) Maximum number of students

that can be accommodated = 40x + 8y = 40(4) + 8(1) = 168

8 (a) I: x + y � 160 II: x � 3y

y � 1—3

x

III: y – x � 80(b)

(c) (i) 120 (ii) Maximum cost = 0.5x + 0.8y = 0.5(40) + 0.8(120) = 20 + 96 = RM116

y

R

x

2x + y = 10

x + y = 7

(4.3, 1.4)

10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7

y = 1—3

x

y = 1—3

x

y

R

x

k = 0.5x + 0.8y

160

140

120

100

80

60

40

20

0 20 40 60 80 100 120 140 160

(40, 120)

y – x = 80x + y = 160

y

400

350

300

250

200

150

100

50

0 50 100 150 200 250 300 350 400

x + y = 400

(267, 134)R

x

2x + y = 300

y = 1—2

x


Paper 1 1 (a) (i) k = 2

(ii) 1(b) One to one function

2 (a) h–1(3) = 6 – 3——–3

= 1(b) h–1(y) = x

6 – y——–3

= x

6 – y = 3x y = 6 – 3x h(x) = 6 – 3x ∴ h� 1—

3 � = 6 – 3� 1—3 �

= 5

3 f(x) = 5 – 6—x

y = 5 – 6—x

6—x = 5 – y x = 6——–

5 – y f –1(x) = 6——–

5 – x f(x) = f –1(x)

5 – 6—x = 6——–

5 – x

5x – 6———x

= 6——–5 – x

(5x – 6)(5 – x) = 6x 31x – 5x2 – 30 = 6x 5x2 – 25x + 30 = 0 x2 – 5x + 6 = 0 (x – 2)(x – 3) = 0 x = 2 or x = 3

4 4x2 – (m + 2)x + m – 1 = 0 For equal roots, b2 – 4ac = 0 (–m – 2)2 – 4(4)(m – 1) = 0 m2 + 4m + 4 – 16 m + 16 = 0 m2 – 12m + 20 = 0 (m – 2)(m – 10) = 0 m = 2 or m = 10

5 (a) (i) p = –2 (ii) q = –9 (iii) f(x) = a(x – 2)2 – 9 At (0, –5), –5 = 4a – 9 4a = 4 a = 1

(b) f(x) = (x + 2)2 – 9

6 2(x2 – 2) � 7x 2x2 – 7x – 4 � 0 (2x + 1)(x – 4) � 0

∴ x � – 1—

2 or x � 4

7

�4k4�3—2

———–�27k –3�–

2—3

= 8k6

——1—9

k2

= (8)(9)(k4) = 2332k4

Compare with 2x3yk2, we havex = 3, y = 2, z = 4.

8 log2x – 2 = log

4(x – 4)

log2x –

log2(x – 4)

—————log

24

= 2

log2x – 1—

2 log

2(x – 4) = 2

2 log2x – log

2(x – 4) = 4

log2 x2�——–�x – 4

= 4 x2

——–x – 4

= 16

x2 = 16(x – 4) x2 – 16x + 64 = 0 (x – 8)2 = 0 x = 8

9 (a) 7 – (m + 1) = 2m + 1 – 7 6 – m = 2m – 6 3m = 12 m = 4(b) The fi rst three terms are 5, 7 and

9 with a = 5 and d = 2. ∴ The sum of the next three terms = 11 + 13 + 15 = 39

10 (a) T3 = S

3 – S

2

= 3[2(3) – 3] – 2 [2(2) – 3] = 9 – 2 = 7(b) T

1 = 1[2(1) – 3]

= –1 T

2 = S

2 – S

1

= 2 – (–1) = 3 ∴ d = T

2 – T

1

= 3 – (–1) = 4

11 (a) T1 = 64(3)

= 192 T

2 = 32(3)

= 96

r = T

2—–T

1 = 96—–

192

= 1—2

(b) S∞ = a——1 – r

= 192———1 – 1—

2

= 384

12 xy = p + qx y = p� 1—x � + q From the diagram, p = 8 – 2———–

2 – (–2) = 3—

2

At the point (–2, 2), 2 = 3—

2 (–2) + q

2 = –3 + qq = 5

13 Area of ∆ABC –2 2 4 –21—2 | | –1 –3 3 –1

= 1—2

[6 + 6 – 4 – (–2 – 12 – 6)]

= 1—2

(28)

= 14 unit2

14 2y = x + 6

y = 1—2

x + 3

mPQ

= –2

The equation of PQ: y – 2 = –2 (x – 2) y – 2 = –2x + 4 y + 2x = 6At x-axis, y = 0 So, 2x = 6 x = 3∴ P(3, 0)At y-axis, x = 0 So, y = 6∴ Q(0, 6)

1– — 2

4x



m(0) + 3n 6m + n(0)�————– , ————–� m + n m + n = (2, 2)

So 3n——–m + n

= 2

3n = 2m + 2n 2m = n

m—n

= 1—2

∴ PR : RQ = 1 : 2

15

16 a = kb ~ ~

�–86 � = k� m

3 � So, we have, 3k = 6 and km = –8 k = 2 2m = –8 m = –4∴ m = –4

17 a = 4, b = 3, c = 2

18 �k

l � 4—

x2 � dx = 2 � 4–— x �k

1 = 2

4–— k

– (–4) = 2

4—k

= 2 k = 2

19 (a) ~x =

35—–7

= 5 σ = 287—–

7 – 52

= 16 = 4(b) New σ = 4k

20 (a) 1—2

(10)2θ = 60 50θ = 60 θ = 60—–

50 = 1.2 radians(b) ∠AOB = 3.142 – 1.2 = 1.942 radians s

AB = 10(1.942)

= 19.42 cm Perimeter of the sector AOB = 2(10) + 19.42 = 39.42 cm

21 y = 2x – 3———x + 5

dy—dx

= 2(x + 5) – (2x – 3)————————(x + 5)2

= 13———(x + 5)2

Compare with h———(x + 5)2

, we have

h = 13

22 v = 4—3

πr3

dv

—–dr

= 4πr2

dv

—–dt

= dv

—–dr

× dr

—–dt

12 = 4πr2 dr

—–dt

When surface area is 25π cm2

4πr2 = 25π

r2 = 25—–4

r = 5—2

(r � 0)

So, 12 = 4π � 5—2 �

2

dr

—–dt

12 = 4π � 25—–4 � dr

—–dt

dr

—–dt

= 12——25π

cm s–1

23 (a) 8 7 6 5 4 = 8 × 7 × 6 × 5 × 4 = 6720(b) 2! 6P

3 × 4 = 960

24 P(blue marble) = 1—

5 x———

2x + 15 = 1—

5 5x = 2x + 15 3x = 15 x = 5Total number of marbles= 8 + x + x + 7= 15 + 2x= 15 + 2(5)= 25

25 (a)

From the table, a = 0.7(b) z = 0.7 x – 55———

10 = 0.7

x – 55 = 7 x = 62

Paper 2Section A 1 y = x – 4 … 1

y2 = 2x2 – 17 … 2Substitute 1 into 2 : (x – 4)2 = 2x2 – 17 x2 – 8x + 16 = 2x2 – 17

x2 + 8x – 33 = 0 (x + 11)(x – 3) = 0x = –11 or x = 3

Substitute x = –11 into 1 : y = –11 – 4 = –15

Substitute x = 3 into 1 : y = 3 – 4 = –1∴ x = –11, y = –15; x = 3, y = –1

2 (a)

dy—–dx

= 3 4x – 5 = 3 4x = 8 x = 2 When x = 2, y = 3(2) – 1 = 5 ∴ R(2, 5) (b) dy

—–dx

= 4x – 5 y = ∫(4x – 5)dx = 2x2 – 5x + c When x = 2 and y = 5, 5 = 2(2)2 – 5(2) + c 5 = 8 – 10 + c c = 7 ∴ y = 2x2 – 5x + 7

3 (a), (b)

4π = 3x———

|cos 2x| |cos 2x| = 3x—–

4π y =

3x—–4π

x 0 π

y 03—4

∴ Number of solutions is 4.

4 (a) 4x – 12 = 2x – 8 2x = 4 x = 2 When x = 2, y = 4(2) – 12 = 8 – 12 = –4 ∴ P(2, –4) Equation of AB: y – (–4) = 1(x – 2) y + 4 = x – 2 y = x – 6 At x-axis, y = 0 0 = x – 6 x = 6 ∴ A(6, 0)

b –2a

f(x)

a zO

0.242

0.258

yy = |cos 2x|

π—4

y = 3x—–4π

3—π 4

–1

1

π—2

x


At y-axis, x = 0 y = 0 – 6 = –6 ∴ B(0, –6)(b)

m(0) + 6n————–

m + n = 2

6n = 2m + 2n 2m = 4n

m—n

= 2—1

m : n = 2 : 1 ∴ AP : PB = 2 : 1

5 (a) Median = 47.5 39.5 +

20 + k�———�2 – 7�—————�

k10 = 47.5

20 + k———2

– 7 = 0.8k 20 + k———

2 = 7 + 0.8k

20 + k = 14 + 1.6k 0.6k = 6 k = 10(b) (i) Range = 74.5 – 24.5

= 50

(ii) x = 1425——30

= 47.5

σ = 72 917.5———–30

– (47.5)2

= 174.333 = 13.204

6 (a) Let BC and AB be x cm and y cm respectively.

A = 1—2

xy

A1 = 1—

2 � x—2 �� y—

2 � = 1—

8xy

A2 = 1—

2 � x—4 �� y—

4 � = 1—–

32xy

A3 = 1—

2 � x—8 �� y—

8 � = 1—–

128xy

Since A

1—–A

= A

2—–A

1

= A

3—–A

2

= 1—4

,

the areas of triangles form a geometric


.

(b) (i) A1 = 1—

2(12)(6)

= 36 cm2

36� 1—4 �

n–1

= 9—–64

� 1—4 �

n–1

= 1——256

= � 1—4 �

4

∴ n – 1 = 4 n = 5

(ii) S∞ = 36———1 – 1—

4 = 36——

3�—�4 = 48 cm2

Section B 7 (a) x 1 2 3 4 5 6

y—x 5 9 13 15 19.5 23

(b) y = ax2 + bx

y—x

= ax + b

(i) a = 19.5 – 1.5————5 – 0

= 18—–5

= 3.6 (ii) b = 1.5 (iii) When x = 3.5,

y—x

= 14

y—–3.5

= 14

y = 49

8 (a) 2x2 + 3x = 27 2x2 + 3x – 27 = 0 (2x + 9)(x – 3) = 0 x = – 9—

2 or x = 3

When x = 3, y = 32

= 9 ∴ A(3, 9) At x-axis, y = 0 2(0) + 3x = 27 x = 9 ∴ B(9, 0) (b) Area of the shaded region = �

3

0 x2 dx + 1—

2 (9 – 3)(9)

= x3

�—�3

0 3 + 27

= (9 – 0) + 27 = 36 unit2

(c)

Volume = π ��

9

0

y dy

= π y2

�—�9

0 2

= 40 1—2

π unit3

→ → → 9 (a) (i) BA = BO + OA

= –b + a ~ ~ = a – b ~ ~ → → → OQ = OB + BQ → = b + 1—

3BA

~

= b + 1—3

(a – b) ~ ~ ~

= 1—3

a + 2—3

b ~ ~ → → → (ii) BP = BO + OP → = –b + 1—

2OA ~

= –b + 1—2

a ~

= 1—2

a – b ~ ~ → → → (iii) PQ = PA + AQ → =

1—2

a + 2—3

AB

~

= 1—2

a + 2—3

(–a + b) ~ ~ ~

= 1– — 6

a + 2—3

b ~ ~ → →(b) (i) OR = m OQ

= m � 1—3

a + 2—3

b� ~ ~

= 1—3

ma + 2—3

mb ~ ~ → → (ii) BR = n BP

= n� 1—2

a – b� ~ ~ = 1—

2na – nb ~ ~

→ → →(c) OR = OB + BR

1—3

ma + 2—3

mb = b + 1—2

na – nb

~ ~ ~ ~ ~

1—3

ma + 2—3

mb = 1—2

na + (1 – n)b ~ ~ ~ ~

So 1—3

m = 1—2

n

2m = 3n 2m – 3n = 0 … 1

and

2—3

m = 1 – n

2m = 3 – 3n

A(6, 0)

P(2, –4)

B(0, –6)

n

m

30

25

20

15

10

5

0

y—x

(0, 1.5)

14

1 2 3 4 5 6x

(5, 19.5)

y

xO

9

y = x2


2m + 3n = 3 … 2

1 + 2 : 4m = 3

m = 3—4

Substitute m = 3—4

into 1 :

2� 3—4 � – 3n = 0

3n = 3—2

n = 1—2

∴ m = 3—4

, n = 1—2

10 (a)

tan θ =

7—7

θ = 45° ∴ ∠TSV = 45—–

180 × π

= 0.786 rad(b) Area of the shaded region

= � 1—2

(72)(1.571) – 1—2

(7)(7)� +

1—2

(3.5)2(0.786)

= 13.99 – 4.814 = 9.176 cm2

(c) SSQ

= 7(1.571) = 10.997 cm S

TV = 3.5(0.786)

= 2.751 cm QS = 72 + 72

= 9.899 cm ∴ Perimeter of the shaded region = 10.997 cm + 3.5 cm + 2.751 cm + (9.899 – 3.5) cm = 23.647 cm

11 (a) (i) P(X � 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2)

= 1 – 10C0 � 3—

5 �0

� 2—5 �

10

– 10C1 � 3—

5 �1

� 2—5 �

9

– 10C2 � 3—

5 �2

� 2—5 �

8

= 0.9877 (ii) The number of students who

do not stay in hostel

= 800 × 2—5

= 320

(b) (i) P(X � 73) = P�Z � 73 – 55———

15 � = P(Z � 1.2) = 0.1151

(ii) Given P(X � m) = 0.2 P�Z �

m – 55———15 � = 0.2

So m – 55———15

= 0.842

m – 55 = 12.63 m = 67.63

Section C 12 (a)

P10—–

3 × 100 = 130

P

10 = 130 × 3———–

100 = RM3.90 (b) 130(2n) + 375 + 110 + 120n

————————————3n + 4

= 124.5 380n + 485 = 373.5n + 498 6.5n = 13 n = 2

(c) 24.90——–P

08

× 100 = 124.5

P08

= 24.90 × 100—————

124.5 = RM20 (d) I

12/08 =

130(4) + 125(3) + 110 + 144(2)——————————–

10 = 1293——–

10 = 129.3

13 (a) (i) a = dv—–dt

= 2t + 2 (ii) s = ∫ v dt = t3

—3

+ t2 – 15t + c

When t = 0, s = 8 and so c = 8. Hence at time t,

s = t3

—3

+ t2 – 15t + 8

(b) (i) When Q is instantaneously at rest,

v = 0 t2 + 2t – 15 = 0 (t + 5)(t – 3) = 0 t = –5 or t = 3 When t = 3,

s = 33

—3

+ 32 – 15(3) + 8

= 9 + 9 – 45 + 8 = –19 ∴ The distance of Q from Y = 19 + 8 = 27 m (ii) When t = 9,

s = 93

—3

+ 92 – 15(9) + 8

= 197

∴ The total distance travelled by Q in the time interval from t = 0 to t = 9.

= 8 + 2(19) + 197 = 243 m

14 (a) I: x + y � 180 II: y � 2x III: 2y � x(b)

(c) (i) The minimum number of students for course A is 40.

(ii) k = 500x + 600y ∴ The maximum total fees

per month obtained = 500(56) + 600(110) = RM94 000

15 (a) (i) 8———

sin 40° =

12————sin ∠ABC

sin ∠ABC = 12 × sin 40°—————8

= 0.9642 ∠ABC = 74°37' (ii) 122 = 52 + 92 – 2(5)(9) cos ∠ADC 90 cos ∠ADC = –38 cos ∠ADC = –0.4222 ∠ADC = 114° 58' (iii) The area of ABCD

= 1—2

(8)(12) sin 65°23' +

1—2

(5)(9) sin 114° 58'

= 43.638 + 20.395 = 64.03 cm2

(b) (i)

(ii) AB'————–

sin 34° 37' =

12————–sin 105° 23'

AB' = 12 × sin 34° 37'——————–

sin 105° 23' = 7.07 cm

Q

P S7

7

θ

f(z)

0.2

Oz

m – 55———15

y

180

160

140

120

100

80

60

40

20

0

y = 2x

(56, 110)

y = 802y = x

x + y = 180

k = 500x + 600y

20 40 60 80 100 120 140 160 180x

R

A

105° 23'

40°B'

12 cm

34° 37'

C

B

8 cm

74° 37'

Analysis Spm Additional Mathematics

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