1.Function

8/22/2019 1.Function

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1

Additional Mathematics SPM Chapter 1

Penerbitan Pelangi Sdn. Bhd.

1Functions

1. Diagram (i): Arrow diagram

Diagram (ii): Ordered pairs

Diagram (iii): Graph

2. (a) 3

(b) 2(c) (i) {2, 1, 1}

(ii) {3, 3, 6}

(iii) {3, 3, 6}

(d) one-to-one

3. (a) many-to-one

(b) (i) {8, 9, 16}

(ii) {9, 16}

4. Sincex-coordinate is one more than y, then

(a) x = 12,

(b) y = 14.

5. (a) one-to-many

(b) Since image in setB is the square root of object

in setA,

theny = 9= 3

6. (a) many-to-many

(b) Range = {d, e,f}

7. Diagram (i): one-to-one

Diagram (ii): one-to-one

Diagram (iii): is not a function because object c has

no image.

Diagram (iv): is not a funct ion because it is

one-to-many relation.

Diagram (v): many-to-one

8. (a) 5

(b) 4

(c) f:xx + 3 orf(x) = x + 3

(d) (i) {2, 4, 5}

(ii) {5, 7, 8}

9. f(x) = 3x2

Image for object 1 =f(1)

= 3(1)2

= 3

10. Givenf(x) = 2x 5 andf(x) = 10.

2x 5 = 10

x =15

2

Therefore, the object is

15

2 .

11. h(x) = sinx

h(90) = sin 90

= 1

12. g(x) = x 7

5

(a) g(2) = 2 7

5

= |1|

= 1

(b) g(x) = 4

x 75 = 4

x 7

5= 4 or

x 7

5= 4

x 7 = 20 x 7 = 20

x = 27 x = 13

13. f(x) = |x 5|

Whenx = 2,

f(x) = |2 5|

= |7|

= 7

Whenx = 7,

f(x) = |7 5|= 2

Whenx 5 = 0

x = 5

Whenx = 0,

f(x) = 0 5= 5= 5


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x 2 0 5 7

f(x) 7 5 0 2

f(x)

x2 5 7

2

0

5

7

Range = 0 f(x) 7

14. (a)f(x)

x

1

090 180 270 360

f(x) = cos x

(b) Whenx = 120

f(120) = cos 120

= 12 =

12

Hence, the range off(x) is 0 f(x) 1.

15. (a) (i) f(x) = x2 4

Image for object 3 =f(3)

= 32 4= 5

(ii) Image for object 4 =f( 4)

= ( 4)2 4

= 12

(b) (i)f(x)

x4 2 0 2 3

5

12

4

Hence, the range off(x) is 0

f(x)

12.(ii) f(x) = 5

x2 4 = 5

x2 = 9

x = 9= 3

16. f(x) = 2x 1Whenx = 0,

f(0) = 1= 1

Whenx = 3,

f(3) = 6 1= 5

When 2x 1 = 0, x =

12

x 01

2

3

f(x) 1 0 5

f(x)

x1

5

1

2

3

0

f(x) = 2x 1

Therefore, the range off(x) is 5 f(x) 0.

17. (a) f(7) = 3(7) 1

= 20

(b) f(3) = 5 +3

3

= 6

18. Since the graph is a straight line,

the gradient =3 0

0 (2)

=3

2

,f(x)-intercept = 3

Therefore, the equation isf(x) =3

2

x + 3.

19. (a) fg(x) =f(1 6x)

= 3(1 6x)

= 3 18x

(b) gf(x) =g(3x)

= 1 6(3x)

= 1 18x

(c) f2(x) = ff(x)

=f(3x)

= 3(3x)

= 9x

(d) g2(x) =gg(x)

=g(1 6x)

= 1 6(1 6x)

= 1 6 + 36x= 36x 5

20. (a) hp(x) = h(x2 2x)

= 2(x2 2x) + 3

= 2x2 4x + 3

(b) ph(x) =p(2x + 3)

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

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

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


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21. (a) fg(2) =f[1 4(2)2]

=f[1 16]

=f(15)

= 152

+ 3

= 9

2

(b) gf(2) =g 22 + 3=g(2)

= 1 4(2)2

= 15

22. (a) fg(x) = 4

f(1 3x) = 4

5(1 3x) = 4

5 15x = 4

15x = 1

x =1

15

(b) gf(x) = 3 +x

g(5x) = 3 +x

1 3(5x) = 3 +x

1 15x = 3 +x

16x = 4

x =4

16

=1

4

(c) f2(x) = 8x + 1

ff(x) = 8x + 1 f(5x) = 8x + 1

5(5x) = 8x + 1

25x = 8x + 1

17x = 1

x =1

17

23. hg(x) = 4x2 2x + 5

3g(x) 1 = 4x2 2x + 5

3g(x) = 4x2 2x + 6

g(x) =2

3

(2x2 x + 3)

24. gh(x) =x

2 + 1

g(2 + 5x) =x

2

+ 1 ........................

Let 2 + 5x =y

x =y 2

5

From ,

g(y) =y 25

2

+ 1

=y 2

10+ 1

=y + 8

10

Therefore, g(x) =x + 8

10

25. (a) f(x) = 2x + 1

f(2) = t

2(2) + 1 = t

t= 5

(b) f1(t) = 2

26. (a) g1(5) = 0

(b) g(r) = 8

5 + 2r= 8

2r= 3

r= 32

(c) Let g1(7) =x,

then 7 =g(x)

= 5 + 2x

2x = 2

x = 1

Hence,g1(7) = 1

27. (a) Letf1(x) =y,

then x =f(y)

= 2y

y =x

2Hence,f1(x) =

x2

(b) Letg1(x) = y,

then x =g(y)

=y

5

y = 5x

Hence,g1(x) = 5x

(c) Let h1(x) = y,

then x = h(y)

= 3y + 1

3y =x 1

y =x 13

Hence, h1(x) =x 1

3

(d) Letp1(x) = y,

then x =p(y)

=y

2

+ 1

y

2

=x 1

y = 2(x 1)

Hence,p1(x) = 2(x 1)

= 2x 2


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28. (a) t 3 = 0

t= 3

(b) Letf1(x) = y,

then x =f(y)

=2

y 3

y 3 =2

x

y =2

x + 3

Therefore, f1(x) =2

x + 3,x 0

(c) Let g1(x) = y,

then x =g(y)

=y

y + 1

x(y + 1) =y

xy +x =y y xy =x

y(1 x) = x

y =x

1 x

Therefore, g1(x) =x

1 x

,x 1

29. (a) The inverse function off, f1, does not exist

for the domain 5 x 5 because f1 is a

many-to-one type of relation.

(b) The inverse function off, f1, exists because f1

is a one-to-one type of relation.

1. (a) a

(b) r

(c) one-to-one

2. (a) f(2) = 1

(b) g(1) = 3

(c) gf(2) =g(1)

= 3

3. (a) The objects of 5 are 0 and 1.

(b) The images of 2 are 9 and 13.

4. (a) one-to-one

(b) f:xx2 orf(x) = x2

5. f(x) =x n

x

f(4) = 2

3

4 n

4

= 2

3

4 n = 8

3

n = 4 +8

3

=20

3

6. (a) fg(x) = 5x

4g(x) 1 = 5x

4g(x) = 5x + 1

g(x) =1

4

(5x + 1)

(b) gf(x) = 9

g(4x 1) = 9

1

4

[5(4x 1) + 1] = 9

5(4x 1) + 1 = 36

5(4x 1) = 35

4x 1 = 7

x = 84

= 2

7. (a) hf(3) = 2

(b) h1(2) = 6

8. Letf1(x) = y,

then x =f(y)

= 3 4y

4y = 3 x

y =3 x

4

Therefore, f1(x) =3 x

4Hence,f1g(x) =f1(1 + 2x)

=3 (1 + 2x)

4

=3 1 2x

4

=2 2x

4

=2(1 x)

4

=1 x

2

9. f2

(x) = ff(x)= f(p qx)

=p q(p qx)

=p pq + q2x

f2(x) = q2x +p pq ................Compare to thef2(x) = x + 2.

Therefore, q2 = 1

q = 1

and ppq = 2

When q = 1, p p(1) = 2

0 2

Therefore, q = 1 is rejected.


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When q = 1, p p(1) = 2

p +p = 2

2p = 2

p = 1Hence,p = 1 and q = 1.

10. (a) Letp1(x) = y,

then x =p(y)

=4

y + 1

x(y + 1) = 4

y + 1 =4

x

y =4

x 1

Hence,p1(x) =4

x

1,x 0

(b) p1(5) = 45

1

= 1

5

11. Letf1(x) = y,

then x =f(y)

=p 3y

3y = x +p

y = x

3

+p3

f1(x) = x

3

+p3

.....................

Compare to thef1(x) =q

2

x +2

3

Therefore,q

2

= 1

3

q = 2

3

and p = 2

12. (a) Let f1(x) = y,

then x =f(y)

= 4 + 5y

y =x 4

5

Hence,f1(x) =x 4

5

(b) gf1(x) = gx 45

=x 45

2

1

=x 4

10 1

=x 4 10

10

=x 14

10

(c) hg(x) = 6 5x

hx2 1 = 6 5x ................

Let x2

1 =y,

thenx

2

= 1 + y

x = 2 + 2y...............

Therefore, becomes h(y) = 6 5(2 + 2y)= 6 10 10y

= 4 10y

Hence, h(x) = 4 10x

1. (a)

1

2

(b) x = 12 since element in setB is half of setA.

2. (a) f1(15) = 3

(b) codomain = {10, 15, 20, 30}

3. Let f(x) = y,

then x =f1(y)

x =y + 1

4

y + 1 = 4x

y = 4x 1

Hence,f(x) = 4x 1

4. fg(x) =fx2 + 3= 1 + 5x2 + 3= 1 +

52

x + 15

=5

2

x + 16

5. gf(3) =g(5)

= 6

6. (a) many-to-one

(b) f:xx2 orf(x) = x2

7. f(x) = 2x 2f(2) = 2(2) 2

= 6= 6

Therefore, the range is 2 f(x) 6.


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8. h(x) =x + 5

2

h(q) = 8

q + 52

= 8

q + 5 = 16

q = 11

9. fg(x) =x

4

+ 7

3g(x) 1 =x

4

+ 7

3g(x) =x

4

+ 8

g(x) =x

12

+8

3

10. h2(x) = hh(x)

= h(ax + b)

= a(ax + b) + b

= a2x + ab + b..................................

Compare with h2(x) = 4x + 9Therefore, a2 = 4

a = 2

and ab + b = 9 ......................................Since a 0,

hence a = 2

Substitute a = 2 into ,2b + b = 9

3b = 9

b = 3Hence, a = 2 and b = 3.

11. hg(x) = hx2 = 3x2 1=

32

x 1

Hence,p(x) = hg(x)

12. (a) (i) Let h1(x) =y,

then x = h(y)

= 4y 3

y =x + 3

4

h1(x) =x + 3

4

h1g(2) = h11 22 = h1(0)

=0 + 3

4

=3

4

(ii) Let g1(x) =y,

then x =g(y)

= 1 y

2

y2

= 1 x

y = 2 2x

g1(x) = 2 2x

hg1(x) = h(2 2x)

= 4(2 2x) 3

= 8 8x 3

= 5 8x

(b) gh(x) =g(4x 3)

= 1 4x 32

= 1 2x 32 = 1 2x + 3

2

=5

2

2x

=5 4x

2

...........................

Comparegh(x) =m nx

2

with ,

Hence, m = 5 and n = 4.

13.x 1 0

23

4

f(x) 5 2 0 10

f(x) = 2 3x

Whenx = 1,

f(1) = 2 3(1)= 2 + 3= 5

Whenx = 4,

f(4) = 2 12= 10= 10

When f(x) = 0,

2 3x = 0

x =

2

3

Whenx = 0,

f(0) = 2 0= 2

f(x)

x1 42

3

0

5

10

2


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(a) The range for 0 x 4 is 0 f(x) 10.

(b) Whenf(x) = 5

2 3x = 5 or 2 3x = 5

x = 73

3x = 3 x = 1

Therefore, the domain is 1 x7

3

.

14. (a) f(2) = 7

2

a

+ b = 7 ...................................

f(2) = 2

2

a + b = 2..................................

+ , 2b = 9

b =9

2

Substitute b =9

2

into ,

2

a

+9

2

= 7

2

a = 7

92

=14 9

2

=5

2

5a = 4

a =4

5

(b) f(x) = x4

5

+ 92

=5x

4

+9

2

Letf1(4) =p,

then 4 =f(p)

=5

4

p +9

2

5

4

p = 4 9

2

= 1

2

p = 12 45 =

25

Hence,f1(4) = 2

5

15. (a) fg(x) = 4x 7

f(x 3) = 4x 7 ..........................

Let x 3 =y,

then x =y + 3

Therefore, becomesf(y) = 4(y + 3) 7= 4y + 12 7

= 4y + 5

Hence,f(x) = 4x + 5

(b) Let f1(5) =p,

then 5 =f(p)

= 4p + 5

p = 0

Therefore, gf1(5) =g(0)

= 0 3

= 3

(c) Let g1(x) =y, and Let f1(x) =y,

then x =f(y)

= 4y + 5

y =x 5

4f1(x) =

x 5

4

then x =g(y)

=y 3

y =x + 3

g1

(x) =x + 3

f1g1(x) = 2x 9

f1(x + 3) = 2x 9

(x + 3) 5

4

= 2x 9

x 2 = 8x 36

7x = 34

x =34

7

16. (a) f2(x) =ff(x)

=f xx + 2

= xx + 2

xx + 2 + 2

=

xx + 2

x + 2(x + 2)

x + 2

=

xx + 2

3x + 4

x + 2

=x

3x + 4

,x 4

3

(b) Let f1(x) = y,

then x =f(y)

=y

y + 2

x(y + 2) =y

xy + 2x =y

y xy = 2x

y(1 x) = 2x

y =2x

1 x

Hence,f1(x) =2x

1 x

,x 1


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17. (a) f(x) =x

x2 3x =x

x2 4x = 0

x(x 4) = 0 x = 0, 4

(b) f(x) = 2g(x)

x2 3x = 2x2 + 2

x2 + 3x + 2 = 0

(x + 1)(x + 2) = 0

x = 1, 2

18. (a) f(x) =x + 1

2 f(r) = 4

r+ 1

2

= 4

r+ 1 = 8

r= 7(b) g(4) =p

4 =g1(p)

4 =p4

+ 5

p4

= 1

p = 4

(c) Let g(x) =y,

then x =g1(y)

=y

4

+ 5

y

4

=x 5

y = 4x 20

g(x) = 4x 20

gf(x) =gx + 12 = 4x + 12 20= 2(x + 1) 20

= 2x + 2 20

= 2x 18

19. (a) f(x) 8

x 5 + 1 8 x 5 7

7 x 5 77 + 5 x 7 + 5

2 x 12

(b) (i) f2(x) =ff(x)

=f(x 2)

= (x 2) 2

=x 4

(ii) f3(x) =f2f(x)

=f2(x 2)

= (x 2) 4

=x 6

Hence,f30(x) =x 2 30

=x 60

1. f(x) = |1 x|, 0 x 2

x 0 1 2

f(x) 1 0 1

f(x)

x0 1

1

2

f(x) = 7 3x, 2 x 3

x7

3

3

f(x) 0 2

f(x)

x0 7

33

2

When f(x) = 0,

7 3x = 0

x =7

3

The graph of the function is

f(x)

x0

7

3

321

2

1

1

Therefore, the range is 2 f(x) 1.

2. f(x) = 2x + 1

f2(x) =ff(x)

=f(2x + 1)

= 2(2x + 1) + 1

= 4x + 3

f3(x) =ff2(x)

=f(4x + 3)

= 2(4x + 3) + 1

= 8x + 7


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f4(x) =f2f2(x)

=f2(4x + 3)

= 4(4x + 3) + 3

= 16x + 15Therefore, fn(x) = 2nx + 2n 1

3. Sincef2(x) = 4x + 9 is a linear function.

Letf(x) = ax + b, where a and b are constants.

f2(x) = ff(x)

=f(ax + b)

= a(ax + b) + b

= a2x + ab + b

Compare tof2(x) = 4x + 9

a2 = 4 and ab + b = 9

a = 2

When a = 2, 2b + b = 93b = 9

b = 3

When a = 2, 2b + b = 9

b = 9

b = 9

Therefore, the possible expressions are f(x) = 2x + 3

orf(x) = 2x 9.

4. f(x) = px q

Givenf(4) = 5

4p q = 5 .....................................

Given f1(9) = 6

9 =f(6)

9 = 6p q

6p q = 9 ................................

, 2p = 4 p = 2

Substitutep = 2 into ,6(2) q = 9

q = 12 9

= 3

5. Let f1(x) = y,

then x =f(y)

= ay by + 4

xy + 4x = ay b

ay xy = 4x + b

y(a x) = 4x + b

y =4x + b

a x

Therefore, f1(x) =4x + b

a x

Compare tof1(x) = 4x 3

x 2

= (4x + 3)

(2 x)

=4x + 3

2 x

Hence, a = 2, b = 3

f(x) = 3x

2x 3

x + 4

= 3x

2x 3 = 3x2 + 12x

3x2 + 10x + 3 = 0

(3x + 1)(x + 3) = 0

3x + 1 = 0 or x + 3 = 0

x = 1

3

or x = 3

6. g(1) = 55

1 b(1)

= 5

5 = 5(1 b)

1 b = 1

b = 2

f(1) = 5

a(1) + b = 5

a + 2 = 5

a = 3

Therefore, g(y) =5

1 2y

andf(y) = 3y + 2

Function which mapszontox is

gf1(z) =gz 23 f1(z) = z 23 =

5

1 2z 23

=5

3 2z+ 4

3

=15

7 2z

,z7

2

7. (a) f(x) = 1 2x

f1(x) =x 12

=1 x

2

g(x) = 4 +x

g1(x) =x 4


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g1f1(x) = g1 1 x2 =

1 x

2

4

=1 x 8

2

=x 7

2

(b) g f(x) =g(1 2x)

= 4 + (1 2x)

= 5 2x

Let (gf)1(x) =y,

then gf(y) =x

5 2y =x

2y =x 5

y =

5 x

2

Therefore, (gf)1(x) =5 x

2

8. hg(x2) = 41

h(3x2 2) = 41

4(3x2 2) + 1 = 41

4(3x2 2) = 40

3x2 2 = 10

3x2 = 12

x2 = 4

x = 2

9. gf(x) =g(x 3)

= a(x 3)2 b= a(x2 6x + 9) b

= ax2 6ax + 9a b

Givengf(x) = 2x2 12x + 13

Therefore, a = 2

9a b = 13

9(2) b = 13

18 b = 13

b = 5

10. gf(x) =x2 4x + 8

g(x 2) =x2 4x + 8

= (x 2)2 4 + 8

g(y) =y2 + 4

Therefore, g(x) = x2 + 4

11. fg(x) =f(2 + bx)

= a(2 + bx) 1

= 2a + abx 1

g f(x) =g(ax 1)

= 2 + b(ax 1)

= 2 + abx b

When fg=gf

2a + abx 1 = 2 + abx b

2a 1 + abx= 2 b + abx

2a 1 = 2 b2a + b = 3

12. f2(x) = ff(x)

=f xx 3

=

xx 3

x

x 3

3

=

xx 3

x 3(x 3)

x 3

=

xx 3

x 3x + 9

x 3

=x

9 2x

Letf1(x) = y,

then x =f(y)

=y

y 3

xy 3x =y

xy y = 3x

y(x 1) = 3x

y = 3xx1

Therefore, f1(x) =3x

x1

Whenf2(x) = f1(x)

x

92x

=3x

x1

x(x 1) = 3x(9 2x)

x2 x = 27x 6x2

7x2 28x = 0

7x(x 4) = 0

x = 0 or x 4 = 0

x = 4

13. f(x) =x

2x 1

x + 3

=x

2x 1 =x2 + 3x

x2 +x + 1 = 0

b2 4ac = (1)2 4(1)(1)

= 1 4

= 3 0

Hence,f(x) = x has no real root.


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14. gf(x) =g(2x 1)

=4

a(2x 1) b

= 42ax a b

=

42

2ax a b

2

=2

ax a + b2

Compare to givengf(x) =2

3x 1

,

a = 3 and a + b2

= 1

3 + b2

= 1

3 + b = 2

b = 1

15. y = h(x)

y =x2 + 3

y =fg(x)

=f(x + 1)

= 3(x + 1)

= 3x + 3

01

3

y= fg(x)

y= h(x)

y

x

The number of solutions forh(x) = fg(x) is two.

16. (a) y =f(x)

=x

2

+7

2

x 3 0 4

f(x) 27

2

11

2

y =g(x)= |1 2x|

x 3 01

2

4

g(x) 7 1 0 7

03 41

2

7

7

2

11

2

12

y= g(x)

y= f(x)

y

x

(b)x

2

+7

2

= 1 2x

x + 7 = 2 4x

5x = 5

x = 1

x

2

+7

2

= (1 2x)

= 1 + 2x

x + 7 = 2 + 4x

3x = 9

x = 3

Forx

2

+7

2

|1 2x|, 1 x 3.

17. (a) y =f(x)

= |x2 4|

x 4 2 0 2 4

f(x) 12 0 4 0 12

04 2 42

4

1

12y= f(x)

f(x)

y= g(x)

x

(b) |x2 4| 1 0

|x2 4| 1

x2 4 = 1

x2 = 5

x = 5

x2 4 = 1

x2 = 3

x = 3

For |x2 4| 1 < 0,

5 x 3, 3 x5

1.Function

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