MODUL Math Tam

MODUL ADD MATH TOK JIRING 1 by CYY

KEM TEKNIK MENJAWAB

MATEMATIK TAMBAHAN

SMKA TOK JIRING

KUMPULAN HAK

(50 MARKAH)

2013

TOPICS

1. Functions

2. Quadratics Equation/Functions

3. Simultaneous Equation

4. Indices & Logaritme

5. Geometry Coordinate

6. Index Number

7. Solutions Of Triangles

8. Trigonometric Functions

DISEDIAKAN OLEH : CIKGU YUSRI BIN YAAHMAT

yusriyaahmat.blogspot.com


ALGEBRA

1. x = 2 4

2

b b ac

a

− ± −

2. am × a

n = a

m + n ----pan-pen

3. am ÷ a

n = a

m − n ----pan-pen

4. (am)n = a

m n

5. loga mn = loga m + loga n-----pan-pen

6. loga n

m = loga m − loga n -----pan-pen

7. loga mn = n loga m-----kuda pan

8. log

loglog

ca

c

bb

a= ---- kuda

9. ( 1)nT a n d= + − ------- kaki atok gbai

10. [2 ( 1) ]2

n

nS a n d= + −

11. 1n

nT ar

−=

12. nS = 1

)1(

−

−

r

ran

= r

ra n

−

−

1

)1(, 1≠r

13. ∞S = r

a

−1, | r | < 1

KALKULUS

1. y = uv, dx

dy = u

dx

dv + v

dx

du-----sida

2. y = v

u,

dx

dy =

2v

dx

dvu

dx

duv −

3. dx

dy =

du

dy ×

dx

du----3 p/u

4. Area under a curve

=

b

a

y dx∫ or = ∫b

a

dyx

5. Volume generated

= ∫b

a

dxy 2π or = ∫b

a

dyx2π

STATISTIK

1. x = N

x∑----tiada f

2. x = ∑

∑

f

fx

3. σ = N

xx∑ − 2)( =

22)(

xx

N

∑− tiada f

4. σ = ∑

∑ −

f

xxf 2)( =

22)(

fxx

f

∑−

∑

5. m = CLmf

FN

+

−21

6. I = 0

1

Q

Q × 100-2 thn

7. I = ∑∑

i

ii

W

IW----fungsi gubahan

8. rn P =

!)(

!

rn

n

−

9. rnC =

!!)(

!

rrn

n

−

10. P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

11. )( rXp = = rnrr

n qpC − , p + q = 1

12. Mean / Min = np

13. σ = npq

14. Z = σ

µ−X

=0 > 0

< 0 ≥ 0

Paksi-y

Paksi-x

1

1

4Q N=

3

3

4Q N=

DaTo TaBah


GEOMETRI

1. Distance = 221

221 )()( yyxx −+−

2. Midpoint

(x, y) =

++

2,

2

2121 yyxx

3. A point dividing a segment of a line

(x, y) = 1 2 1 2,nx mx ny my

m n m n

+ +

+ +

4. Area of triangle / Luas segi tiga

)()( 3123121332212

1yxyxyxyxyxyx ++−++

5. r = 22yx +

6. r̂ = 22

yx

yx

+

+ ji

TRIGONOMETRI

1. Arc length, s = rθ

2. Area of sector = 2

1 2r θ

3. AA 22 cossin + = 1

4. A2sec = A2tan1 +

5. A2cosec = A2cot1 +

6. sin 2A = 2 sinA cosA

7. cos 2A = cos2 A − sin

2 A

= 2 cos2 A − 1

= 1 − 2 sin2 A

8. )(sin BA ± = sinA cosB ± cosA sinB

9. )(cos BA ± = cosA cosB m sinA sinB

10. )(tan BA ± = BA

BA

tantan1

tantan

m

±

11. tan 2A = A

A2tan1

tan2

−

12. A

a

sin =

B

b

sin =

C

c

sin

13. a2 = b

2 + c

2 − 2bc cosA

14. Area of triangle = 2

1 ab sin C

FORMULA TAMBAHAN

1. x2 – (SOR)x + POR = 0 2 b c

xa a

− + = 0

2. f (x) = 2 2

2 4

b ba x c

a a

+ + −

3. 100

X YZ

×=

Z

Year C

X

Year A

Y

Year B

Sila t2

Vektor unit

SiLa

Anak

Panah


FUNCTIONS

1 :

Given that 43: −→ xxf and xxg 2: → ,

find fg(3).

Answer : f(x) = 3x - 4 , g(x) = 2x

g(3) = 2( )

= ( )

fg(3) = f [ g(3) ]

= f ( )

= 3 ( ) - 4

= ( )

2 :

Given that xxf 23: −→ and 2: xxg → , find

gf(4).

Answer : f(x) = 3 – 2x , g(x) = x2.

f( ) = 3 – 2( )

= ( )

gf(4) = g ( )

= ( )2

= ( )

3:

Given that 23: −→ xxf , find f 2(2).

Answer : f(x) = 3x - 2

f(2) = 3( ) – 2 = ( )

f2(2) = f [ f(2) ]

= f ( )

= 3 ( ) – 2

= ( )

4:

Given that xxg 43: −→ , evaluate gg(1).

Answer : g(x) = 3 – 4x

g(1) = 3 – 4( ) =

gg(1) = g [g(1)]

= g ( -1)

= 3 – 4 ( )

= ( )

5 :

Given that 43: −→ xxf and xxg 2: → ,

find fg(x).

Answer : f(x) = 3x - 4 , g(x) = 2x

fg(x) = f [ g(x) ]

= f ( )

= 3 ( ) - 4

= 6x – 4

6 :

Given that xxf 23: −→ and 2: xxg → , find the

composite fuction gf.

Answer: f(x) = 3 – 2x , g(x) = x2.

gf(x) = g[f(x)]

= g ( )

= (3- 2x)2

=

EXAMPLE 1 :

Given that f(x) = 4x – 6 , find

÷ x + ( )

∴ f –1

(x) = 4

6+x

EXAMPLE 2 :

Given that f(x) = 2x + 3 , find f –1

(x).


QUADRATIC EQUATIONS/FUNCTIONS

1

Given that the roots of the quadratic equation

2x2 + (p+1)x + q - 2 = 0 are -3 and ½ . Find the value

of p and q.

x = -3 , x = ½

x + 3 = 0 or 2x – 1 = 0

(x + 3) ( 2x – 1) = 0

2x2 + 5x – 3 = 0

Comparing with the original equation :

p + 1 = , q - 2 =

p = , q =

2

Given that the roots of the quadratic equation

3x2 + kx + p – 2 = 0 are 4 and

- ⅔. Find k and p.

(Ans : k = -10 , p = -6)

3

The roots of the quadratic equation

2x2 + px + q = 0 are - 6 and 3.

Find

(a) p and q,

(b) range of values of k such that

2x2 + px + q = k does not have real roots.

Answer :

(a) x = -6 , x = 3

(x + 6) (x – 3) = 0

x2 + 3x - 18 = 0

2x2 + 6x – 36 = 0

Comparing : p = , q =

(b) 2x2 + 6x – 36 – k = 0

a = 2, b = 6, c = -36 - k

62 – 4(2)(-36 – k) < 0

k < – 40.5

4

The roots of the quadratic equation

2x2 + px + q = 0 are 2 and -3.

Find

(a) p and q,

(b) the range of values of k such that

2x2 + px + q = k does not have real roots.


Solve the inequality x2 + x - 6 ≥ 0

x2 + x - 6 ≥ 0

(x + 3) ( x – 2) ≥ 0

Consider f(x) = 0. Then x = -3 , x = 2

Range of x is : x ≤ -3 atau x ≥ 2

L4. Solve the inequality x2 + 3x - 10 ≥ 0.

x ≤ -5 , x ≥ 2

SIMULTANEOUS EQUATIONS

1. Solve x + y = 3, xy = – 10 .

x + y = 3 ........ (1)

xy = – 10 ........ (2)

From (1), y = ( )......... (3)

Substitute (3) into (2),

x ( ) = – 10

3x – x2

= – 10

x2 – 3x – 10 = 0

(x ) (x ) = 0

x = ( ) atau x = ( )

From ( ), when x = ( ) , y = 3 – ( ) =

x = ( ) , y = 3 – ( ) =

2. Solve x + y = 5, xy = 4 .

(Ans : x = 1, y = 4 ; x = 4, y = 1)

3. Solve x + y = – 2 , xy = – 8 .

(Ans : x = – 4 , y = 2 ; x = 2, y = – 4 )

4. Solve 2x + y = 6, xy = – 20 .

(Ans : x = – 2 , y = 10 ; x = 5, y = – 4 )

x

-3 2

y=f(x)


INDICES N LOGARITME

Solve the equation log2 (x+1) = 3.

Answers: log2 (x+1) = 3

x + 1 = 23

x =

L1. Solve the equation log2 (x – 3 ) = 2.

Jawapan:

Ans : x = 7

Solve the equation log10 (3x – 2) = – 1 .

Jawapan: 3x – 2 = 10-1

3x – 2 = 0.1

3x = 2.1

x = 0.7

L2. Solve the equation log5 (4x – 1 ) = – 1 .

Ans : x = 0.3

Solve the equation log3 (2x – 1) + log2 4 = 5 .

Ans : x = 14

L6. Solve the equation

log4 (x – 2) + 3log2 8 = 10.

Ans : x = 6

Solve the equation

log2 (x + 5) = log2 (x – 2) + 3.

Ans : x = 3

L8. Solve the equation

log5 (4x – 7) = log5 (x – 2) + 1.

Ans : x = 3

Solve log3 3(2x + 3) = 4

Ans : x = 12

L10 . Solve log2 8(7 – 3x) = 5

Ans : x = 1


Given log 2 T - log4 V = 3, express T in terms of

V. [4]

(Ans: T = 8V ½

)

L8. Given log 4 T + log 2 V = 2, express

T in terms of V. [4]

(Ans: 16V-2

)

Solve 42x – 1

= 7x. [4]

( Ans: x = 1.677 )

L10. Solve 42x – 1

= 9x. [4]

( Ans: x = 2.409 )

Solve the equation 3 3log 9 log (2 1) 1x x− + = .

[3]

(Ans: x = 1 )

6. Given that log 2m p= and log 3m r= ,

express 227

log16

m

m

in terms of p and r. [4]

(Ans: 3r – 4p +2 )

Solve the equation 5 3 68 32x x− += .

[3]

(Ans : x = 3.9 )

8. Given that 5log 2 m= and 5log 3 p= ,

express 5log 2.7 in terms of m and p. [4]

(Ans: 3p – m – 1 )


GEOMETRY COORDINATE

1

Given two points A(2,3) and B(4,7)

Distance of AB = 2 2( ) ( )+

= 20 unit.

2.

P(4,5) and Q(3,2)

PQ =

[ 10 ]

3.

The point P internally divides the line segment

joining the point M(3,7) and N(6,2) in the ratio 2

: 1. Find the coordinates of point P.

P =

+

+

+

+

12

)2(2)7(1,

12

)6(2)3(1

4.

The point P internally divides the line segment

joining the point M (4,5) and N(-8,-5) in the ratio

1 : 3. Find the coordinates of point P.

51,

2

5

P(0, 1), Q(1, 3) and R(2,7)

Area of ∆ PQR = 2

1 0 1 2 0

1 3 7 1

= 2

1[ ( ) – ( ) ]

= 1 unit 2

6

P(2,3), Q(5,6) and R(-4,4)

Area of ∆ PQR =

17

2

unit2

7

. Find the equation of a straight line that passes

through the point (5,2) and has a gradient of -2.

y = -2x + 12

8

. Find the equation of a straight line that passes

through the point (-8,3) and has a gradient of 4

3.

4y = 3x + 36

●

1

2

N(6, 2)

M(3, 7)

P(x, y)


INDEX NUMBER

Notes :

a. Price Index, I = 1

0

100P

P× , P1 = price at a specific time, P0 = price at the base year.

b. Composite Index , IW

IW

=∑∑

, I = price index, W = weightage

1. Table below shows the price indices and percentage of usage of four main

ingredients ,P,Q,R and S, in the production of a type of cake.

Ingredients

Price index for the

year 2012

(2010=100)

Percentage of

usage

P m 20

Q 105 30

R 108 10

S 120 40

(a) Calculate

(i) the price of ingredient Q in the year 2010 if its price in the year 2012 is RM 50.00,

(ii) the price index of R in the year 2012, based on the year 2008, given that its price

index in the year 2010, based on the year 2008 is 110.

(b) The composite index number of the cost of production of this type of cake in the year

2012,based on the year 2010 is 112.8. Calculate.

(i) the value of m,

(ii) the cost of these ingredients for the production of this type of cake in

the year 2012 if the corresponding cost in year 2010 is RM60.00.

Guided Solutions

a (i)

,

(ii) ( ) ( )

( )1 0 0

×=

b (i)

(ii)

5 0( ) 1 0 5 , ( )× = =x

x

( )( ) 1 0 5 (3 0 ) 1 0 8 (1 0 ) 1 2 0 ( 4 0 )( ) , ( )

1 0 0

+ + += =

mm

2012100 ( ) , ( )2012

60

× = =Q

Q


SOLUTION OF TRIANGLES

(1) Diagram 1 shows the triangle ABC.

Calculate the length of BC.

Answer :

8 . 2

( ) ( )

B C=

( )

( )( )

BC = ×

Using the scientific calculator,

BC = ( )

(1) Diagram 1 shows the triangle ABC.

Find ∠ACB.

( ) ( )

( ) ( )=

(3)

Find the value of x.

2 2 2( ) ( ) 2( )( ) cos( )x = + −

x =

(4)

Find the value of x.

2 2 2( ) ( ) 2( )( ) cos( )x = + −

[ 7.475 ]

Diagram 1

A

B C

600

15 cm

Diagram 1

10 cm

12.3

P

16.4 cm

x cm

Q R 67

0

5cm

x cm

7

cm

P Q

R

750

5

cm

12.3

P

16.4 cm

x cm

Q R 67

0


(5)

Find BAC∠ .

2 2 215 ( ) ( ) 2( )( ) cos BAC= + − ∠

)14)(13(2

151413 cos

222 −+=∠BAC

(6)

Find BAC∠ .

2 2 2( ) ( ) ( ) 2( )( ) cos BAC= + − ∠

[ 83.17°]

(7)

Find area of PQR

Area = a b s i n c1

2

( ) ( ) s in ( )=1

2

=

(8)

Find area of PQR

C

A

B

13cm 14 cm

15 cm Diagram 1

C

A

B

11cm 13 cm

16 cm

Diagram 2

12.3

P

16.4 cm

x cm

Q R 67

0

12.3

P

16.4 cm

x cm

Q R 67

0

5cm

7 cm P Q

R

750

5 cm


TRIGONOMETRIC FUNCTIONS

GRAPHS OF TRIGONOMETRIC FUNCTIONS

Paper 2

1. (a) Prove that θθθ 2cos2cottan ec=+ . [ 4 marks ]

(b) (i) Sketch the graph xy2

3cos2= for π2≤x≤0 .

(ii) Find the equation of a suitable straight line for solving the equation 1 - 4

3

2

3cos xx

π= .

Hence, using the same axes, sketch the straight line and state the number of

solutions to the equation 1 - 4

3

2

3cos xx

π= for π2≤x≤0 .

[ 6 marks]

2. (a) Sketch the graph y = cos 2x for 00 180≤≤0 x . [ 3 marks ]

(b) Hence, by drawing a suitable straight line on the same axes, find the number of solutions

satisfying the equation 180

-2sin2 2 xx = for 0180≤≤0 xo .

[ 3 marks ]

( SPM P2 No. 3 )

3. (a) Prove that cosec 2 x – 2 sin

2 x – cot

2 x = cos 2x. [ 2 marks ]

(b) (i) Sketch that graph of y = cos 2x for π2≤x≤0 .

(ii) Hence, using the same axes, draw a suitable straight line to find the number of

solutions to the equation 3( cosec2

x – 2 sin2 x – cot

2 x) = 1-

π

x for π2≤≤0 x .

State the number of solutions. [ 6 marks ]

(SPM P2 No.5)


INTEGRATION

1. 43x dx∫ 2. 52

3x dx∫

3. 6

2

3dx

x∫ 4.

4

7dx

x∫

5. 3(2 3)x dx−∫ 6. ( )42

5 43

x dx−∫

7.

2

1

8x dx =∫

=

[12]

8.

4

3

2

x dx =∫

=

[60]

9.

2

1

(2 1)(2 1)x x dx− +∫

=

[ 25

3]

10.

3

2

1

(3 2)x dx−∫

=

[38]


11. Given that 3

1( )g x dx∫ = 6, find

a) the value of 3

1

( )

2dx

g x∫ ,

b) the value of 1

35 ( ) dxg x∫ ,

c) the value of k such that 3

1[ ( ) ]g x k dx+∫ = 10.

Guided Solutions

a) 3

1

( )

2dx

g x∫

= 3

1( )

1

2g x dx∫

= 1

( )2

= ( )

c) 3 3

1 1( )g x dx k dx+∫ ∫ = 10

( ) + [ ]3

1kx

= 10

( ) − ( ) = 4

k = ( )

12. Given that 5

2( )f x dx∫ = 9, find

a) the value of 5

2

2 ( )

3dx

f x∫ ,

b) the value of 2

54 ( )f x dx∫ ,

c) the value of k such that 5

2[ ( ) ]f x kx dx+∫ = 30.

b) 1

35 ( )g x dx∫

5( ) = ( )


13. Diagram shows the curve y = 5x4 and the straight line x = p.

If the area of the shaded region is 32 unit2, find the value of p.

Guided Solutions

Area = b

ay dx∫ =

4

0

5p

x dx∫ = ( )

[ ]0

p = ( )

( ) – ( ) = 32

p = ( )

14. Diagram shows the shaded region bounded by y-axis, the curve y2 = 4x and

a straight line y = k.

Given that the area of the shaded region is 9

4 unit

2, find the value of k.

Answer :

y2 = 4x

y = k

0 x

y

y = 5x4

x = p

0 x

y


PAPER 2

15. Diagram shows the straight line PQ is normal to the curve 2

13

xy = + at M (3, 4). The straight line

MN is parallel to y-axis.

Find

a) the value of h,

b) the area of the shaded region,

c) the volume of revolution, in terms of π, when the region bounded by the curve, the

y-axis and straight line y = 4 is rotated through 360° about y-axis.

Guided Solutions

a) m1 = dy

dx =

2

3

x =

2( )

3 = ( )

m2 = 1

1

m

− = ( ) =

4 0

3 h

−

− (m2 is gradient of PQ)

h = ( )

b) Area of region A = 23

0

( 1)3

xdx+∫ = ( )

Area of region B = ( )

Hence, the area of the shaded region = A + B = ( )

c) The volume of revolution = π4

2

1

x dy∫ = π4

1

( ) dy∫ = ( )

.

B A

N Q (h, 0) O x

y

y = 2

13

x+

M (3, 4)

P ••••

•

•

•


A O

B

y

P

C

x k

Q

32 += xy

9=+ xy

16. Diagram shows the curve 32 += xy intersects the straight line AC at point B.

It is given that the equation of straight line AC is y + x = 9 and the gradient of the curve at point B is

4. Find

a) the value of k,

b) the area of the shaded region P,

c) the volume of revolution, in terms of π , when the shaded region Q is rotated through

360º about the x–axis .

9. Diagram shows the curve 22 1y x= + intersects the straight line 9 9y x= − at point (2, k).

Find,

a) the value of k,

b) the area of the shaded region,

c) the volume of revolution, in terms of π, when the region bounded by the curve, the

y-axis and straight line y = 9 is rotated through 360° about y-axis.

END OF MODULE

(2, k)

y = 2x2 + 1

y = 9x−9

x

•

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