Mathematics T for new STPM syallabus

Mathematics (T)

Worked Examples

Loo Soo Yong

1/27/2013

This document contains various questions with detailed workings and explanations from the new STPM syllabus. A PDF version of this document may be obtained from Dropbox.

Mathematics (T)

Binomial Expansions

1. Expand (1+x )−1 up to and include the term x3. Hence, by subsisting x=−0.01, find an estimate for 10.99

, and give your answer to 6 decimal places.

Solution:

(1+x )−1=1+(−1 ) ( x )+ (−1 ) (−2 )2!

( x )2+ (−1 ) (−2 ) (−3 )3 !

( x )3+…

¿1−x+x2−x3+…

¿ (1+x )−1=1−x+x2−x3

Substituting x=−0.01... (1−0.01 )−1=1−(−0.01 )+ (−0.01 )2−(−0.01 )3

∴ 10.99

≈1.010101

2. f ( x ) is given by the expression f ( x )=√4+x

(a) Show that f ( x )can be written as f ( x )=2(1+ x4 )12

(b) When xis so small that x4 and higher powers of x is ignored, show that2+ x4− x2

256+ x3

512.

(c) Hence, state the range of values of x for which the expansion is valid.(d) By substituting x=1 into the equation, find an estimate for √5, giving your answer in the

form pq

, where p and q are integers to be determined.

Solution:

(a) f ( x )= (4+x )12

f ( x )=[4 (1+ x4 )]

12

f ( x )=412(1+ x

4 )12

(c) 2012-2013 All rights reserved

Mathematics (T)

f ( x )=2(1+ x4 )12 (shown)

(b)f ( x )=2[1+ 12 ( x4 )+ ( 12 )( 12−1)

2! ( x4 )2

+( 12 )( 12−1)( 12−2)

3! ( x4 )3]

f ( x )=2[1+ x8− x2

128+ x3

1024 ] f ( x )=2+ x

4− x2

64+ x3

512 (shown)

(c) The expansion is valid for |x4|<1 , i.e. −4<x<4.

(d) √5≈2+ 14−

(1 )2

64+

(1 )3

512

≈1145512

3. For n>0, the expansion of (1+kx )−n in ascending powers of x is given by 1+6 x+27 x2+… where k

and n are integers. Show that k=−3and n=2

Solution:

(a) (1+kx )−n=1−knx+(−n ) (−n−1 )

2 !(kx )2+…

By comparing coefficient of x and x2 , −kn=6

k=−6n

k2=36

n2

(n ) (n+1 ) (k2 )

2=27

n(n+1)2 (36n2 )=27


Mathematics (T)

18 (n+1 )

n=27

9n=18

n=2 (shown)

Subst n=2...

k=−3 (shown)

4. By taking the natural logarithm to both sides or otherwise, show that ddx

(2 x−5 )4 ( x+4 )6 can be

written in the form of A (2 x−5 )p ( x+4 )q+B (2 x−5 )r ( x+4 )s, where A, B, p, q, r and s are constants

to be determined. [No credit will be given if the derivative is obtained using product and chain rule]

Solution:

Let y= (2x−5 )4 (x+4 )6

Taking natural log to both sides, ln y=ln [ (2x−5 )4 ( x+4 )6 ]

Using properties of logarithms, ln y=4 ln (2 x−5 )+6 ln (x+4 )

Differentiating y with x, 1ydydx

=4 [ 12 x−5

(2 )]+6[ 1x+4

(1 )] 1ydydx

= 82 x−5

+ 6x+4

dydx

= y [ 82 x−5

+ 6x+4 ]

dydx

=(2x−5 )4 ( x+4 )6[ 82 x−5

+ 6x+4 ]


Mathematics (T)

dydx

=8 (2x−5 )3 ( x+4 )6+6 (2 x−5 )4 ( x+4 )5

Therefore, A=8, B=6, p=3, q=6, r=4, s=5.

5. Express f ( x )= 3x+5( x+1 ) (x+2 ) ( x+3 ) in partial fractions. If x is so small that x3 and higher powers of x

are ignored, show that f ( x )≈ 56−3736

x+ 227216

x2. Hence, use the quadratic approximation to

estimate the value of ∫0

0.1

f ( x )dx, giving your answer to 4 decimal places.

Solution:

3 x+5

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

+ Bx+2

+ Cx+3

3 x+5=A ( x+2 ) (x+3 )+B ( x+1 ) ( x+3 )+C (x+1)( x+2)

Letting x=−2 ,−1=−B ,B=1

Letting x=−3 ,−4=2C ,C=−2

Letting x=−1 ,2=2 A , A=1

∴ 1x+1

+ 1x+2

− 2x+3

If x is so small that x3 and greater powers of x are ignored,

(1+x )−1+(2+x )−1−2 (3+x )−1

(1+x )−1+[2−1(1+ x2 )

−1]−2[3−1(1+ x3 )

−1] [1+ (−1 ) ( x )+

(−1 ) (−2 )2 !

( x )2]+ 12 [1+ (−1 )( x2 )+ (−1 ) (−2 )2! ( x2 )

2]−2¿ [1−x+x2 ]+1

2 [1− x2+ x2

4 ]−2[ 13 (1− x3+ x2

9 )](c) 2012-2013 All rights reserved

Mathematics (T)

(1−x+x2 )+ 12− x4+ x

2

8−2( 13− x

9+ x2

27 ) 1−x+x2+1

2− x4+ x2

8−23+ 2x9

−2x2

27

f (x)≈56−3736

x+227216

x2 (shown)

∫0

0.1

f ( x )dx ≈∫0

0.156−3736

x+ 227216

x2dx

≈∫0

0.156dx−∫

0

0.13736

x dx+∫0

0.1227216

x2dx

≈ [ 56 x ]00.1

−3736

∫0

0.1

x+ 227216

∫0

0.1

x2

≈ [ 56 x ]00.1

−3736 [ 12 x2]0

0.1

+ 227216 [ 13 x3]0

0.1

≈ [ 56 (0.1 )−56

(0 )]−3736 [ 12 (0.1 )2−12

(0 )2]+ 227216 [ 13 (0.1 )3−13

(0 )3] ≈56

(0.1 )− 377200

+0.0003503

≈0.0785

6. Show that (2−x )−1=12+ 14x+ 18x2+ 1

16x3+…Hence by using the information given,

(a) Expand 4

(2−x ) √1+3 x in ascending powers of x, up to and including the term x2.

(b) State the set values of x for which the expansion is valid.

(c) By taking x=−127

, find an estimate for √2, giving your answer to 4 decimal places.


Separate the constants first for easier integration.

Mathematics (T)

Solution:

(2−x )−1=2−1(1− x2 )

−1

(2−x )−1=12(1+(−1 )(−x

2 )+ (−1 ) (−2 )2 ! (−x

2 )2

+(−1 ) (−2 ) (−3 )

3 ! (−x2 )

3

+…)

(2−x )−1=12 (1+ x

2+ x2

4+ 18x3)

(2−x )−1=12+ 14x+ 18x2+ 1

16x3 (shown)

(a)4

(2−x ) √1+3 x=(4 )( 1

2−x)( 1

√1+3 x)

¿4 (2−x )−1(1+3 x)−12

¿4 ( 12+ 14 x+ 18 x2)(1+(−12 ) (3 x )+

(−12 )(−12 −1)2!

(3 x )2) ¿(2+ x+

12x2)(1−3

2x+ 278x2)

¿2−3 x+274x2+x−3

2x2+ 1

2x2

¿2−2 x+234x2

(b) The expansion is valid for |−x2 |<1∩|3x|<1.

∴|x|<2∩|x|< 13

From the graph below,

-2−13

13

2

Therefore the set values of x are {x :|x|< 13, x∈ R }

(c) Let x=−127

,


You only need to expand until x2 as the question specifies you to do so.

Mathematics (T)

4

(2−(−127 ))√1+3(−127 )≈2−2(−127 )+ 234 (−127 )

2

81√255

≈60712916

∴√2≈1.4137

7. Consider the curve y=x3+4 x2+x−6 .(a) Find the equation of this tangent at the point where ¿ –1 .(b) Find the coordinates of the point where this tangent meets the curve again.

Solution:

(a) y=x3+4 x2+x−6

dydx

=3 x2+8 x+1

At point x=−1, dydx

=3 (−1 )2+8 (−1 )+1

dydx

=−4

When x=−1 , y= (−1 )3+4 (−1 )2+(−1 )−6 , y=−4Therefore the coordinates is (−1 ,−4 )The equation of the tangent, y− (−4 )=(−4)(x+1) ∴ y=−4 x−8

(b) y=x3+4 x2+x−6(1) y=−4 x−8 (2 ) Subst. (2) to (1), −4 x−8=x3+4 x2+ x−6 x3+4 x2+5 x+2=0Factoring, ( x+1 )2 ( x+2 )=0 x=−1(rejected) or x=−2When x=−2 , y=(−2 )3+4 (−2 )2+(−2 )−6 y=0Therefore the point is (−2,0).


Mathematics (T)

8. The polynomial p ( x )=(ax+b )3 leaves a remainder of −1 when divided by (x+1) and a remainder

of −27 when divided by (x−2). Find the real numbers a and b.

Solution:

p (−1 )=−1

(−a+b )3=−1

−a+b=−1

b=a−1 (1 )

p (2 )=27

(2a+b )3=27

2a+b=3(2)

Subst (1 ) into (2 ), 2a+a−1=3

3a=4 , a=43

Subst a=43

into (1)

b=43−1 , b=1

3.

∴a=43, b=1

3.

9. Two complex numbers are defined as z1=a1+ iand z2=

b1−2 i . Find the real numbers a and b such

that z1+ z2=3.

Solution:

z1=a1+ i


Mathematics (T)

z1=a1+ i ( 1−i

1−i ) z1=

a−ai2

z1=a2−a2i

z2=b

1−2 i

z2=b

1−2 i ( 1+2i1+2i ) z2=

b+2bi5

z2=b5+2b5i

Given z1+ z2=3,

a2−a2i+ b5+ 2b5i=3

( a2 + b5 )+(−a

2+ 2b5 )i=3

Comparing real and imaginary parts,

a2+b5=3(1)

−a2

+2b5

=0

a2=2b5

(2)

Subst. (2) to (1),

2b5

+ b5=3


If there is a complex number as the denominator, always multiply by its

conjugate.

Mathematics (T)

3b5

=3 , b=5

Subst b=5into (2)

a2=105, a=4

∴a=4 , b=5.

10. A curve is defined parametrically as x=t 2+sin 2t , y=t+sin t , t∈R .(a) Find the gradient of the curve where t=0.(b) Hence, find the equation of the tangent to the curve at the point where t=0.

Solution:

(a) x=t 2+sin 2t

dxdt

=2t+2cos2 t

y=t+sin t

dydt

=1+cos t

dydx

=( dydt )( dtdx ) ¿ (1+cos t ) ¿ ¿¿¿

When t=0 , dydx

=¿¿

dydx

=1

(b) When t=0,

x=(0 )2+sin2 (0 ) , x=0When t=0 y=0+sin 0 , y=0The point is (0,0)The equation of the tangent is y−0=1 ( x−0 ) ∴ y=x .


Mathematics (T)

11. Find the set values of x for which 1

x−√x≥415

Solution:

1

x−√x≥415

x−√ x≤ 154

(√ x )2−√ x≤ 154

Let √ x=u , u2−u≤

154

4 u2−4u−15≤0

(2u+3 ) (2u−5 )≤0

−32

52

Therefore, −32≤u≤

52

−32≤√x ≤ 5

2

0≤ x≤254

However, we need to enforce the condition that x>1.

Therefore, the solution set is {x :1<x≤254, x∈ R }


If a>b , then 1a< 1b

Since 1

x−√x is undefined for

x≤1, hence this condition must be enforced.

x is always positive, because of squaring a number always produces a positive number. The minimum value of x2 is zero.

Mathematics (T)

12. Let f to be a cubic polynomial. Given that f (0 )=2, f ' (0 )=−3, f (1 )=f '(1) and f ' ' (−1 )=6 , find the

polynomial f (x).

Solution:

Let f ( x )=a x3+b x2+cx+d

f (0 )=2 , a (0 )3+b (0 )2+c (0 )+d=2

d=2

f ' ( x )=3a x2+2bx+c

f ' (0 )=−3 ,3a (0 )2+2b (0 )+c=−3

c=−3

f ( x )=a x3+b x2−3 x+2

f ' ( x )=3a x2+2bx−3

f (1 )=f ' (1 ) ,a+b−3+2=3a+2b−3

2a+b=2(1)

f ' ' ( x )=6ax+2b

f ' ' (−1 )=6 ,−6a+2b=6

−3a+b=3 (2)

(2 )− (1 ) :−5a=1, a=−15

Subst. a=−15

into (1),

−25

+b=2 , b=125

∴ f ( x )=−15

x3+ 125x2−3 x+2

13. Show that p=2 is a root to the equation p3+ p2−5 p−2=0. Hence find the other two solutions in exact form.


Mathematics (T)

An arithmetic sequence has p as its common difference. A geometric sequence has p as its common ratio. Both sequences has 1 as the first term.

(a) Write down the first four terms of each sequence.(b) If the sum of the third and fourth terms of the arithmetic sequence is equal to the sum of the

third and fourth terms of the geometric sequence, find the possible values of p.(c) Hence, state the value of p for which the geometric sequence has a sum to infinity, and find the

value, expressing your answer in the form of a+b√5, where a, b are constants.(d) For the same value stated in (c), find the sum of the first 20 terms of the arithmetic sequence,

giving your answer in the form p+q √5 , where p and q are constants.

Solution:

p=2 is a root, therefore,

23+22−5 (2 )−2=0 (shown)

( p−2) is a factor.

( p−2 ) ( p2+kp+1 )=0

Comparing p2,

k−2=1 , k=3.

( p−2 ) ( p2+3 p+1 )=0

p=2∨p2+3 p+1=0

p=−3±√32−4 (1)(1)

2

∴ p=2 , p=−3+√52

, p=−3−√52

(a) Arithmetic sequence, 1 ,1+ p ,1+2 p ,1+3 pGeometric sequence, 1 , p , p2 , p3

(b) (1+2 p )+(1+3 p )=p2+ p3

p3+ p2−5 p−2=0

( p−2 ) ( p2+3 p+1 )=0

p=2∨p2+3 p+1=0


Mathematics (T)

∴ p=2 , p=−3+√52

, p=−3−√52

(c) For the geometric sequence to have a sum to infinity, |p|<1. Therefore, p=−3+√52

.

S∞=

1

1−−3+√52

¿ 12−(−3+√5)

2

¿2

5−√5

¿ 25−√5 (5+√5

5+√5 ) ¿2(5+√5)20

,12+ 110

√5

(d) Sum of first 20 terms of the arithmetic sequence ¿ 202 [2 (1 )+(20−1 ) −3+√5

2 ] ¿10[2+19(−3+√5

2 )] ¿10[2+ 192 (−3+√5 )] ¿20+95 (−3+√5 ) ¿−265+95√5

14. A curve, C is defined implicitly as 2 xy+6 x2−3 y2=6.

(a) Show that the tangent at point A(1 , 23) has gradient

203.

(b) The line x=1cuts the curve at point A(1 , 23) and at point B. Determine the coordinates of point

B.

Find, in the form of r=(ab)+s (cd )(i) The equation of the tangent at A.(ii) The equation of the normal at B.

(c) Hence, find the acute angle between tangent at A and the normal at B.

Solution:

(a) 2 xy+6 x2−3 y2=6


Mathematics (T)

Differentiating y with x, 2 x ( dydx )+ y (2 )+12 x−6 y ( dydx )=0At (1 , 23 ), 2( dydx )+ 43 +12−4 ( dydx )=0 2dydx

= 43+12

dydx

=203

(shown)

(b) (i) The equation of the tangent is r=( 123 )+s( 320)(ii) Subst. x=1 into 2 xy+6 x2−3 y2=6 2 y+6−3 y2=6 3 y2−2 y=0 y (3 y−2 )=0

y=0∨ y=23

(1,0 )∨(1 , 23 )(rejected )

∴B(1,0)

At B, 2 (1 )( dydx )+2 (0 )+12 (1 )−6 (0 )( dydx )=0 dydx

=−6

Gradient of the normal ¿− 1

dydx

Gradient of the normal at point B ¿16

The equation of the normal at B (10)+s(61)

(c) Acute angle between two lines cos−1(61) ∙( 320)

√62+(−1 )2√32+202

¿cos−10.3089


This point is rejected due to

(1 , 23 ) is point A.

Mathematics (T)

¿72 °

∴ The acute angle between two lines is 72 °

15. By using de Moivre’s theorem, prove that cos 4θ=8sin4θ−8sin2θ+1 . Hence,

(a) Show that one of the roots of the equation 8 x4−8x2+1=0 is sin18π and express the other

roots in trigonometric form.

(b) Deduce that sin18π=12

√2−√2 and find an exact expression for sin118π .

Solution:

¿¿

(c+is )4=cos 4θ+i sin 4θ

1 (c )4+4 (c )3 ( is )+6 ( c )2 ( is )2+4 ( c ) ( is )3+1 (is )4=cos 4θ+ isin 4 θ

c4+4c3 si−6c2 s2−4c s3i+1 s4=cos 4θ+i sin 4θ

Comparing real parts, cos4θ−6cos2θ sin2θ+4sin4θ=cos4θ

cos4θ−6cos2θ ¿¿¿

cos4θ−6cos2θ+6cos4θ+¿¿¿¿

cos4θ−6cos2θ+6cos4θ+1−2cos2θ+cos4θ=cos 4θ

∴cos 4θ=8cos4θ−8cos2θ+1 (shown)

(a) Let x=sin θ and consider cos 4θ=0 , 8 x4−8x2+1=0

Solving cos 4θ=0, 4 θ=π2,3π2

,−π2,−3π

2

θ=π8,3π8,− π8,−3 π

8

∴ The solutions are x=sinπ8, sin

3 π8, sin

−π8

, sin−3 π8

(b) Let θ=π8

, cos 4( π8 )=8sin4 ( π8 )−8sin2( π8 )+1


¿¿ may be written as (c+is )4 for

simplicity.

Use the binomial theorem.

Value of θ must be within −π<θ≤π

Mathematics (T)

0=8sin4 ( π8 )−8sin2( π8 )+1 sin2( π8 )=8±√ (−8 )2−4(8)(1)

2 (8 )

¿ 8±√3216

¿ 8±4√216

¿ 8−4√216

sin( π8 )=√ 8−4 √216

¿√ 4(2−√2)16

¿12√2−√2 (shown)

sin( 11 π8 )=sin( 12π8 −18π )

¿ sin12π8cos

π8−cos 12π

8sin

π8

¿ sin3π2cos

π8−cos 3 π

2sin

π8

¿−cosπ8

Since sin2( π8 )+cos2( π8 )=1

cos2( π8 )=1−sin2( π8 )

cos2( π8 )=1−( 12 √2−√2)2

cos2( π8 )=1−14 (2−√2)

cos2( π8 )=1−12+ √2

4

cos2( π8 )=12+ √2

4

cos2( π8 )=2+√2

4


The positive root is rejected because of the requirement of the question.

Use the identity

sin (a−b )=sinacosb−cos a sinb

sin( π8 ) is on the first quadrant, therefore sin( π8 )is

positive.

Use the identity

sin2θ+cos2θ≡1

Mathematics (T)

cos ( π8 )=√2+√22

−cos ( π8 )=−√2+√22

∴sin (11 π8 )=−√2+√22

16. A curve is given as 3 x2+4 y2=7. Find the gradient of tangent of the curve at the point where x=1 and y>0.

Solution:

3 x2+4 y2=7

When x=1

3 x2+4=7

x2=1

x=1

∴ The point is (1,1)

6 x+8 y ( dydx )=0

At (1,1), 6+8( dydx )=0

dydx

=−68

dydx

=−34

17. Let f ( x )=√ 1x2−2. Find the set values of x for which f is real and finite.

Solution:

Let f ( x )>0


f (x) is only defined for x>0

Mathematics (T)

√ 1x2−2>0 1

x2−2>0

1−2 x2

x2>0

1−2 x2>0

2 x2−1<0

x2−12<0

(x+√ 12 )(x−√ 12 )<0

x2>0 , x>0

- + - +

−√ 12 0 √ 12 ∴ The set values of x is {x :−√2

2< x< √2

2, x≠0 , x∈ R }

18. Find, in the form of y=mx+c ,the equation of the tangent to the curve y=x2 ln x at the point with x-coordinate e.

Solution:

y=x2 ln x

When x=e,

y=e2 ln e

y=e2


Use the product rule

Mathematics (T)

The coordinates is (e , e2 )

dydx

=x2( 1x )+¿

dydx

=x+2x ln x

At the x-coordinate e , dydx

=e+2e ln e

dydx

=3e

Equation of tangent, y−e2=3e(x−e)

y=3 ex−2e2

19. The equation of a curve is x2 y−x y2=2.

(a) Show that dydx

= y2−2xyx2−2 xy

(b) Show also, if dydx

=0 ,then y=2x

(c) Hence, find the coordinates of the point on the curve where the tangent is parallel to the x-axis.

Solution:

(a) x2 y−x y2=2

x2( d ydx )+ y (2 x )−x (2 y dydx )− y2(1)=0

x2( dydx )−2 xy ( dydx )= y2−2 xy

( dydx ) (x2−2 xy )= y2−2xy

dydx

= y2−2xyx2−2 xy

(shown)

(b) When dydx

=0 , y2−2 xy

x2−2 xy=0

y2−2 xy=0 y ( y−2 x )=0


ddxln x=1

x

Use the formula y− y1=m (x−x1 )

Using a combination of implicit differentiation and product rule

Since dydx

is undefined for y=0 ,

therefore the solution y=0 is rejected.

Mathematics (T)

y=0∨ y=2x Since y ≠0 ,∴ y=2 x (shown)

(c) Subst. y=2xinto x2 y−x y2=2 x2 (2 x )−x (2x )2=2 2 x3−4 x3=2 −2 x3=2 x3=−1 x=−1Subst. x=−1 into y=2x, y=−2 ∴ The coordinates is (−1 ,−2 )

20. The expansion (2−px )6 in ascending powers of x, as far as the term in x2 is 64+Ax+135 x2. Given p>0, find the value of p and the value of A.

Solution:

(2−px )6=26(1− p2x)6

¿64¿

¿64 (1−3 px+ 154 p2

x2) ¿64−192 px+240 p2 x2

Comparing the coefficient of xand x2,

135=240 p2

p2= 916

p=34

−192 p=A

A=−192( 34 ) A=−144


Do not write it as 2(1− p2x )6

!

(1+x )n=1+ nx1 !

+n (n−1 ) x2

2!+…

Mathematics (T)

∴ A=−144 , p= 34

21. Write down an identity for tan2θ and use this result to show that tan3θ=3 tan θ−tan3θ

1−3 tan2θ.

(a) It is given that 0<θ< π2

and θ=sin−1 1

√10. Without using a calculator, show that tan3θ=

139

.

(b) Hence, show that the solutions to the equation tan¿¿ for 0<x<2 π are

x=√1010

or ¿ √10 (1+3√3)20

.

Solution:

tan (3θ )=tan(2θ+θ)

¿tan 2θ+ tanθ1−tan2θ tan θ

¿

2 tan θ

1−tan2θ+ tanθ

1−( 2 tanθ1−tan2θ )¿¿

¿2 tan θ

1−tan2θ+ tanθ ¿¿¿¿

¿

2 tanθ1−tan2θ

+ tan θ−tan3θ

1−tan2θ1−tan2θ−2 tan2θ

1−tan2θ

¿

3 tanθ−tan3θ1−tan 2θ1−3 tan 2θ1−tan 2θ

¿ 3 tan θ−tan3θ

1−3 tan2θ (shown)

(a) θ=sin−11

√10


You may draw a triangle and use Pythagoras’ Theorem to find the value

of tanθ

Mathematics (T)

sin θ=¿ 1

√10¿

csc θ=√10 1+cot2θ=csc2θ 1+cot2θ=10 cot2θ=9 cot θ=3

tanθ=13

From tan3θ=3 tan θ−tan3θ

1−3 tan2θ,

tan3θ=3 ( 13 )−( 13 )

3

1−3( 13 )2

tan3θ=139

(shown)

(b) Let x=1

√10 ∴θ=sin−1 x 3sin−1 x=3θ

Subst. 3sin−1 x=3θ into tan3θ=3 tan θ−tan3θ

1−3 tan2θ tan¿¿

It is given that tan3θ=139

∴ 3 tan θ−tan3θ

1−3 tan2θ=139

27 tanθ−9 tan3θ=13−39 tan2θ 9 tan3θ−39 tan 2θ−27 tan θ+13=0 Let t=tanθ 9 t 3−39 t2−27 t+13=0From part (b), it is known that tanθ=1/3 ∴t=1 /3 3 t−1=0 (3 t−1) is a factor of the cubic equation.

(3 t−1 ) (3 t2+kt−13 )=0 Comparing x2, 3k−3=−39 k=−12


Mathematics (T)

(3 t−1 ) (3 t2−12 t−13 )=0

t=13∨3 t2−12 t−13=0

t=12±√ (−12 )2−4 (3 )(−13)

2(3)

t=12±10√36

t=2+53√3 or t=2−

53

√3

22. Expand (z+ 1z )(z−1z ) . Hence or otherwise, expand (z+ 1z )4

(z−1z )2

.


Mathematics (T)

(a) By using de Moivre’s theorem, if z=cos θ+isin θ , show that zn+ 1

zn=2cosnθ and find a similar

expression for zn− 1

zn.

(b) Hence, express cos4θ sin2θ in the form of A cos6θ+B cos 4θ+C cos2θ+D, where A ,B ,C ,Dare constants to be determined.

(c) By using the result in (b), find ∫cos4θ sin2θdθ

Solution:

(z+ 1z )(z−1z )=z2− 1

z2

(z+ 1z )4

(z−1z )2

=(z+ 1z )2

(z−1z )2

(z+1z )2

¿ [(z+ 1z )( z−1z )]2

(z+1z )2

¿( z2− 1z2 )

2

(z+1z )2

¿( z4−2+ 1z4 )(z2+2+ 1z2 ) ¿( z6+2 z4+z2−2 z2−4− 2z2+ 1z2+ 2z4+ 1z6 ) ¿( z6+ 1z6+2 z4+ 2z4−z

2−1

z2−4)

(a) z=cos θ+isin θ zn=cos nθ+i sinnθ (1)

z−n=cos (−nθ )+i sin (−nθ )

1

zn=cosnθ−isinnθ (2)

(1)+(2): zn+ 1

zn=2cosnθ (shown)

(1)-(2): zn− 1

zn=2i sinnθ (shown)

(b) cos4θ sin2θ


Using de Moivre’s Theorem

cos (−θ)=cos θ

Mathematics (T)

¿ ¿

−64 cos4θ sin2θ=[(z6+ 1z6 )+2(z4+ 1z4 )−(z2+ 1z2 )−4 ] −64 cos4θ sin2θ=¿¿¿

cos4θ sin2θ=−1

64¿¿¿

cos4θ sin2θ=−1

32cos6θ− 1

16cos 4θ+ 1

32cos 2θ+ 1

16

∴ A=−132

,B=−116

,C= 132

,D= 116

(c) ∫cos4θ sin2θdθ=∫− 132cos6θ− 1

16cos 4θ+ 1

32cos2θ+ 1

16

¿−132∫ cos6θ−

116∫ cos4 θ+

132∫ cos2θ+∫

116

¿−132

¿

¿−1192

sin 6θ− 164sin 4θ+ 1

64sin 2θ+ 1

16θ+c

23. f ( x ) is defined as f ( x )=12sin 2 x+cos x

(a) Find f ' (x)(b) Hence, find the possible values of sin xfor which f ' ( x )=0

Solution:

(a) f ' ( x )=12¿

f ' ( x )=cos 2x−sin x(b) f ' ( x )=0

cos2 x−sin x=0


ddxsin 2 x=2cos2x

ddxcos x=−sin x

∫cos aθ=1a sinaθ+c

Mathematics (T)

1−2sin2 x−sin x=0 2sin2 x+sin x−1=0 ¿

sin x=−1 or sin x=12

24. A curve is defined parametrically as x=t+ln tand y=t+e t, t>0.

(a) Find dxdt

and dydt

in terms of t. Show that dydx

=t (1+e t )t+1

and hence deduce that the curve has no

turning points.(b) Find, in exact form, the equation of the normal of the curve at the point where t=1.

Solution:

(a) x=t+ln t

dxdt

=1+ 1t

dxdt

= t+1t

dydt

=1+e t

dydx

=dydt ( dtdx )

¿ (1+e t ) ( tt+1 )

¿t (1+e t )t+1

(shown)

To find the turning point dydx

=0

t(1+e t )t+1

=0

t (1+et )=0 t=0 or e t=−1 ∵t>0 , ∴ The curve has no turning points.

(b) When t=1,

dydx

=1 (1+e )1+1

dydx

=1+e2


You do not need to solve for x as the question requires the values of sin x

only.

Mathematics (T)

Gradient of normal¿− 1

dydx

¿− 11+e2

¿−21+e

When t=1 , x=1+ ln1 y=1+e1

x=1 y=1+e

Equation of the normal y− (1+e )= −21+e

( x−1 )

y=−21+e

x+ 21+e

+1+e

25. A curve C is defined as y=xsin x , x>0.

(a) Find dydx

in terms of x

(b) Hence, find the equation of the tangent to the curve C at the point where x=π2

Solution:

(a) y=xsin x

ln y=sin x ln x

1ydydx

=sin x( 1x )+ ln x ¿¿¿

dydx

= y ¿

dydx

=xsin x¿

(b) When x=π2

, y=( π2 )sinπ2

y=( π2 )When x=

π2, y=π

2


Mathematics (T)

dydx

=( π2 )¿ dydx

=( π2 )( 2π ) dydx

=1

Equation of tangent, y−( π2 )=1(x− π2 )

∴ y=x

Newton-Raphson method

The Newton-Raphson method is used when an equation f ( x )=0 cannot be solved using simple algebraic methods. The formula for Newton-Raphson method is given by

xn+1=xn−f (xn )f ' (xn )

Consider the graph of the function y=f ( x ) for the interval [a ,b ]

y=f ( x )

a b

From the graph, it is known that f ( x )<0 for x=a and f ( x )>0 for x=b. There is a change in sign of

f (x) for the equation. Hence, a root exists in the interval [a ,b] for the equation f ( x )=0. Refer to question 26 for an example.

26. Show that the equation x ln x−1=0 has a root between the interval [1,2 ]. Hence, by using

Newton-Raphson method with x0=1.5 as the first approximation, find the root of the equation, giving your answer to 5 decimal places.


f ( x )>0

f ( x )<0

Mathematics (T)

Solution:

Let f ( x )=x ln x−1

f (1 )=(1)¿

f (1 )=−1

f (2 )=2 ln 2−1

f (2 )=0.38629

Since there is a change in sign, therefore a root exists between 1 and 2.

Formula for Newton-Raphson method: xn+1=xn−f (xn )f ' (xn )

, x0=1.5

f' ( x )=x ( 1x )+ln x

f ' ( x )=1+ ln x

x0=1.5

x1=1.77877

x2=1.76327

x3=1.76322

x4=1.76322

Therefore, the root of the equation is x=1.76322

27. A function is defined parametrically as x=5 t 2 , y=t 5+ 20t3

3.

(a) Find d ydx

in terms of t .

(b) It is given that dydx

=1. Show that t 3+4 t−2=0

(c) Show that the equation t 3+4 t−2=0 has a root between 0 and 1.

(d) Hence, by using Newton-Raphson method with x0=0 as the first approximation, find the root of

the equation t 3+4 t−2=0, giving your answer correct to 5 decimal places.


The full working is not required when finding the root

Stop when the value starts to converge

Mathematics (T)

Solution:

(a) x=5 t 2

dxdt

=10 t

y=t5+ 20 t3

3

dydt

=5 t4+20 t2

dydx

=( dydt )( dtdx ) ¿5 t 4+20 t2( 110 t ) ¿5 t2 (t 2+4 )( 110t ) ¿12t (t 2+4 )

(b)dydx

=1

12t 3+2 t=1

t 3+4 t−2=0 (shown)

(c) Let f (t )=t 3+4 t−2 f (0 )=−2 f (1 )=3Since there is a change in sign, therefore there is a root between 0 and 1.

(d) xn+1=xn−f (xn )f ' (xn )

, x0=0

f ' ( x )=3 t 2+4 x0=0 x1= 919 x2=21684579 x3=0.47347 x4=0.47347 ∴ The root of the equation is x=0.47347

28. The parametric equations of curve C are x=t 2 , y=2t


Mathematics (T)

(a) Show that the normal to C at the point with parameter p has equation

x+ py=p3+2 p.(b) The normal to C at the point P intersects the x-axis at A and the y-axis at B. Given that O is the

origin and OA=2OB , find the value of p.

Solution:

(a) x=2 t , y=t 2

x=2 p , y=p2

dxdp

=2 , dydp

=2 p

dydx

=dydp ( dpdx )

¿2 p ( 12 ) ¿ p

Gradient of the normal ¿− 1

dydx

=−1p

¿−1p

At the point x=2 p , y=p2 ,

Equation of normal is y−p2=−1p

( x−2 p )

y−p2=−1p

x+2

y+1px=p2+2

py+ x=p3+2 p (shown)

(b) The x-intercept is given by point A.x-intercept, y=0. x=p3+2 p A ( p3+2 p ,0 )The y-intercept is given by point B.y-intercept, x=0 py=p3+2 p y=p2+2 B (0 , ( p2+2 ))


Mathematics (T)

Distance of OA ¿√ [ ( p3+2 p )−(0 ) ]2+ [0−0 ]2

¿ ( p3+2 p )Distance of OB ¿√ [0−0 ]2+¿¿¿ ¿ p2+2Since OA=2OB , ∴ p3+2 p=2 [ p2+2 ] p3+2 p=2 p2+4 p3−2 p2+2 p−4=0Using a calculator, we know that p=2is a root of the equation. Therefore p−2 is a factor.

( p−2 ) ( p2+kp+2 )=0Comparing p2 , k−2=−2 k=0 ( p−2 ) ( p2+2 )=0 ∴ p=2.

29. The function f defined on the domain (0 , π2 ) is given by

f ( x )=xcos x

(a) Find f ' ( x ) in terms of x.The x-coordinate of the maximum point is denoted by α .

(b) Show that α ln α tanα=1(c) Verify the root lies between 1.27 and 1.28.

Solution:

(a) f ( x )=xcos x

ln [ f (x ) ]=cos x ln x

1

f ( x )[ f ' ( x ) ]=cos x ( 1x )+ ln x ¿¿¿

f ' ( x )=[ f (x ) ][ cos xx −ln x sin x ] f ' ( x )=xcosx [ cos xx −ln x sin x ]

(b) To find the maximum point, f ' ( x )=0 when x=α .When x=α ,

α cosα [ cos αα −ln α sinα ]=0 α cosα=0 (undefined)


Mathematics (T)

cosαα

−ln α sin α=0

cosαα

=ln α sin α

a ln α sinα=cos α a ln α tanα=1 (shown)

(c) Let f (α )=α ln α tanα−1 f (1.27 )=(1.27 ) ln (1.27 ) tan (1.27¿−1¿ ¿−0.0215 f (1.28 )=(1.28 ) ln (1.28 ) tan (1.28 )−1 ¿0.0558Since there is a change of sign of f (α ) , therefore a root exists between 1.27 and 1.28.

30. Show that 3

1+ xp+ 3

1+x−p simplifies to a constant, and find the constant.

Solution:

31+ xp

+ 31+x−p=

3 (1+x−p )+3 (1+x p )(1+x p ) (1+x−p )

¿ 3+3 x−p+3+3 xp

1+ x−p+ xp+1

¿ 6+3 xp+3x−p

2+ xp+x−p

¿3 (2+ xp+x−p )2+x p+x−p

¿3 (shown)

31. Show that p3+q3−( p+q )3=−3 pq (p+q ). Hence or otherwise, find, in terms of a and b, the three

values of x for which (a−x )3+ (b−x )3−(a+b−2 x )3=0

Solution:

LHS: p3+q3−[ p3+3 p2q+3 pq2+q3]

¿−3 p2q−3 pq2


α must be in radians.

Mathematics (T)

¿−3 pq ( p+q )(RHS ) (shown)

Given that p3+q3−( p+q )3=−3 pq (p+q ) ,

Replacing p with (a−x ) and q with (b−x )...,

(a−x )3+ (b−x )3−(a−x+b−x )3=−3 (a−x ) (b−x ) (a−x+b−x )

(a−x )3+ (b−x )3−(a+b−2 x )3=−3 (a−x ) (b−x )(a+b−2 x)

∴ (a−x )3+(b−x )3=−3 (a−x ) (b−x ) (a+b−2x )+(a+b−2x )3

Hence, −3 (a−x ) (b−x ) (a+b−2 x )+ (a+b−2x )3− (a+b−2 x )3=0

−3 (a−x ) (b−x ) (a+b−2 x )=0

a−x=0 , x=a

b−x=0 , x=b

a+b−2 x=0 , x=a+b2

∴ x=a , x=b , x=a+b2

32. Solve the following system of linear equations without using a calculator.3 x+ y+ z=0x− y+z=22 x−3 y−z=9

Solution:

(3 1 11 −1 12 −3 −1|

029)

R1−3 R2→R2=(3 1 10 4 −22 −3 −1|

0−69 )


Express the system of linear equations in the form of an augmented matrix

Using elementary row operations

Mathematics (T)

2 R1−3 R3→R3 ¿(3 1 10 −4 20 11 5|

0−6−27)

11R2+4 R3→R3=(3 1 10 −4 20 0 −42|

0642)

−42 z=42 , z=−1 −4 y+2 z=6 −4 y−2=6 −4 y=8 , y=−2 3 x+ y+ z=0 3 x−2−1=0 x=1 ∴ x=1 , y=−2 , z=−1

33. f ( x )=2x+1 , x∈R. Show that a real number, k exists such that for all values of x,

f (x+ f ( x ) )=kf ( x )

Solution:

f ( x )=2x+1

x+f ( x )=3 x+1

f (x+ f ( x ) )=f (3 x+1)

¿2 (3x+1 )+1

¿6 x+3

¿3 (2x+1 )

¿3 f ( x )

∴ k=3

34. Solve the equation sin 5 x−cos5 x=cos x−sin x for 0≤ x≤π , giving your answer in terms of π .

Solution:


Mathematics (T)

sin 5 x−cos5 x=cos x−sin x

sin 5 x+sin x=cos5 x+cos x

2sin( 5 x+x2 )cos ( 5x−x2 )=2cos( 5 x+x2 )cos( 5 x−x

2 ) 2sin 3x cos2 x=2cos3 xcos 2x

sin 3 xcos 2x−cos3 xcos2 x=0

cos2 x ¿¿

cos2 x=0 sin 3 x−cos3 x=¿0¿

2 x=π2,3 π2

sin 3 x=cos3 x

x=π4,3 π4

tan3 x=1

3 x=π4,5 π4

x=π12

,5 π12

35. In the binomial expansion of (1−4 x )p ,|x|< 14, the coefficient of x2 is equal to the coefficient of x4,

and the coefficient of x3 is positive. Find the value of p.

Solution:

(1−4 x )p=[1+( p ) (−4 x )+ ( p ) ( p−1 )2 !

(−4 x )2+ p ( p−1 ) ( p−2 )3 !

(−4 x )3+ p (p−1 ) (p−2 ) (p−3 )4 !

(−4 x )4+…]

Since the coefficient of x2 is equal to the coefficient of x4 ,

p ( p−1 )2

(16 )= p ( p−1 ) ( p−2 ) ( p−3 )24

(256 )

8 p (p−1 )=323p ( p−1 ) ( p−2 ) ( p−3 )

8 p (p−1 )−323p ( p−1 ) ( p−2 ) ( p−3 )=0


Mathematics (T)

p (p−1 )[8−323

(p−2 ) (p−3 )]=0

p (p−1 )[8−323 (p2−5 p+6 )]=0 p (p−1 )[8−323 p2+ 160

3p−64]=0

p (p−1 )=0

p=0∨p=1 (rejected)

8−323p2+ 160

3p−64=0

−323

p2+1603

p−56=0

−32 p2+160 p−168=0

4 p2−20 p+21=0

(2 p−7 ) (2 p−3 )=0

p=72∨p=3

2

Since the coefficient of x3is given by p ( p−1 ) ( p−2 )

6(−64 )

When p=72,

Coefficient of x3 is ( 72 )( 72−1)( 72−2)6

(−64 )=−140 (rejected)

When p=32,

Coefficient of x3=

( 32 )( 32−1)( 32−2)6

(−64)

¿4


Mathematics (T)

∴ p=32

36. The curve C has parametric equations

x=15 t−t 3 , y=3−2 t2

Find the values of t at the points where the normal at C at t=1 cuts C again.

Solution:

x=15 t−t 3

dxdt

=15−3 t 2

y=3−2 t2

dydt

=−4 t

dydx

=dydt ( dtdx )

¿−4 t ( 1

15−3 t 2 ) dydx

= −4 t15−3 t2

Normal ¿− 1

dydx

¿ 15−3 t2

4 t

When t=1

Gradient of normal¿15−34

¿3


Mathematics (T)

When t=1 ,

x=15 (1 )−(1 )3

¿14

y=3−2 (1 )2

y=1

The point is (14,1).

The equation of normal y−1=3 ( x−14 )

y−1=3 x−42

y=3 x−41

Since the equation of the normal cuts the curve again,

3−2t 2=3 (15 t−t 3 )−41

3−2t 2=45t−3 t 3−41

3 t3−2 t2−45 t+44=0

Since it is known that t=1 lies on the curve, t−1 is a factor.

Factoring gives (t−1 ) ( t+4 ) (3 t−11)=0

t=−4∨t=113

37. Find the coordinates of the turning points of the curve x3+ y3−3 xy=48and determine their nature.

Solution:

x3+ y3−3 xy=48

Differentiating y with x,

3 x2+3 y2( dydx )−3 x( dydx )+ y (−3 )=0


Substitute the parametric equations into y=3 x−41

Mathematics (T)

(3 y2−3 x )( dydx )=3 y−3 x2

dydx

=3 y−3x2

3 y2−3 x

To find the turning point, dydx

=0

3 y−3 x2

3 y2−3x=0

3 y−3 x2=0

y=x2

When y=x2

x3+(x2 )3−3 x (x2 )=48

x6−2 x3−48=0

(x3−8 ) ( x3+6 )=0

x3=8 , x=2

x3=−6 , x=− 3√6

When x=2 , y=4

When x=− 3√6 , y=623

3 x2+3 y2( dydx )−3 x( dydx )+ y (−3 )=0

Differentiating y with x,

6 x+3 y2( d2 yd x2 )+( dydx )(6 y dydx )−3 x ( d2 yd x2 )+( dydx ) (−3 )−3 dydx

=0

Since dydx

=0


Mathematics (T)

6 x+3 y2( d2 yd x2 )−3x ( d2 yd x2 )=0

6 x=−3 y2( d2 yd x2 )+3 x( d2 yd x2 )

d2 yd x2

= 6 x−3 y2+3x

d2 yd x2

=6 ( x )

3 (− y2+x )

d2 yd x2

= 2 xx− y2

When x=2 , y=4

d2 yd x2

=2 (2 )2−16

¿−13<0 (maximum)

When x=− 3√6 , y=623

d2y

d x2=2 (− 3√6 )

−3√6−643

¿0.2857>0 (min)

∴ (2,4 ) is a maximum point, (−3√6 ,623) is a minimum point.

38. Find ∫ 11+√ x

dx by using the substitutionu=√x .

Solution:

Let u=√x

dudx

= 12√x


Mathematics (T)

dx=2√ x du

dx=2udu

∫ 11+u

(2u )du

¿2∫ u1+u

du

¿2∫ u+1−1u+1

du

¿2∫ 1− 1u+1

du

¿2u−ln|u+1|+c

¿2√x−ln|√ x+1|+c

Maclaurin’s Theorem

Certain functions, for example sin x , ln x and ex

can be expressed in the form of a polynomial. The Maclaurin’s Theorem (or sometimes referred as Taylor series) is given by

f ( x )≈ f (0 )+ f ' (0 )1!

x+f ' '(0)2 !

x2+f ' ' ' (0)3!

x3+f IV (0)4 !

x4+…

The approximation becomes more accurate when more terms are included in the expansion.

The Maclaurin’s series is used for:

Finding approximate values for an integral Evaluating limits Approximating the value of a function

Refer to question 39 and 40 for an example.


Mathematics (T)

39. If x is so small that x4 and higher powers of xmay be ignored, expand ex as a polynomial in x.

Solution:

Using Maclaurin’s Theorem,

f ( x )≈ f (0 )+ f ' (0 )1!

x+f ' ' (0 )2 !

x2+f ' ' ' (0 )3 !

x3

Let f ( x )=ex

f (0 )=1

f ' ( x )=e x

f ' (0 )=1

f ' ' ( x )=ex

f ' ' (0 )=1

f ' ' ' ( x )=ex

f ' ' ' (0 )=1

∴ f ( x )≈1+x+12x2+ 1

6x3

40. Given that f ( x )=ln ¿¿show that f IV ( x )=2 f ' ' (x)¿Expand ln ¿¿up to and include the term x4. Hence, find an approximation for ln ¿¿ giving your answer to 4 decimal places.

Solution:

Let f ( x )=ln ¿¿

f (0 )=ln¿¿

f (0 )=0

f' ( x )= 1

cos x¿

f ' (0 )=0

f ' ' ( x )=−sec 2 x

f ' ' (0 )=−1


Mathematics (T)

f ' ' ' ( x )=−2¿

f ' ' ' ( x )=−2 sec2 x tan x

f ' ' ' ( x )=2 f ' ' ( x )¿

f ' ' ' (0 )=0

Differentiating f (x) with respect to x,

f IV ( x )=2 f ' ' ( x )¿

f IV (0 )=2 f ' ' (0 ) ( sec20 )+ tan 0 (2 f ' ' ' (0 ) )

¿2(−1)(1)

¿−2

Using Maclaurin’s Theorem,

f ( x )≈ f (0 )+ f ' (0 )1!

x+f ' ' (0)2 !

x2+f ' ' ' (0)3 !

x3+f IV (0)4 !

x4

f ( x )≈0+ 0( x )1

+−12

x2+0 (x3 )6

+−224

x4

f ( x )≈−12x2− 1

12x4

∴ ln cos x≈−12x2− 1

12x 4

When x=π10

,

ln cos ( π10 )≈−12 ( π10 )2

− 112 ( π10 )

4

≈−0.0502

41. Show that ∫1

24 x2+6x3+3 x

dx=ln k , where k is an integer to be found.

Solution:


Mathematics (T)

Let 4 x2+6x( x2+3)

≡Ax

+Bx+Cx2+3

4 x2+6≡ A (x2+3 )+(x )(Bx+C )

Let x=0 ,6=3 A

A=2

Compare x2 , 4=A+B

4=2+B

B=2

Compare x ,C=0

∫1

24 x2+6x3+3 x

dx ≡∫1

2

( 2x + 2 xx2+3 )dx

¿2∫1

21xdx+∫

1

2

( 2xx2+3 )dx ¿2¿¿

¿2¿

¿2 ln 2+ln 7−2 ln 2

¿ ln 7

∴ k=7

42. By using a suitable substitution or otherwise, find ∫ sin x+cos x¿¿¿ ¿.

Solution:Let u=cos x−sin x

dudx

=−sin x−cos x

dx= du−¿¿


Mathematics (T)

¿∫ sin x+cos x¿¿¿ ¿

¿∫−1u2

du

¿−∫ (u−2 )du

¿−[ u−1

−1 ]+c ¿

1u+c

¿1

cos x−sin x+c

43. Show that ¿.Hence, evaluate

∫0

π12

¿¿

Solution:LHS, ¿ ¿¿ ¿¿¿ ¿¿ ¿¿ ¿ sin 4 θ+sin 2θ (RHS) (shown)

∫0

π12

¿¿

¿∫0

π12

¿¿

¿ [−18 cos 8x−14 cos4 x ]0π12


Mathematics (T)

¿ [−18 cos 2π3 −14cos

π3 ]−[−18 cos0−14 cos0]

¿ [−18 (−12 )−14 ( 12 )]−[−18 −14 ]

¿516

44. By using the substitution x=4 tan 2θ+4, for 0<θ< π2

, show that

∫163

16 √x−4x

dx=k∫α

β

tan2θdθ

where k ,α ,β are constants to be determined. Hence evaluate

∫163

16 √x−4x

dx

Solution:

When x=163,

163

=4 tan2θ+4

4 tan2θ= 4

3

tan2θ=1

3

tanθ=1

√3 θ=

π6

When x=16 , 16=4 tan2θ+4 tan2θ=3 tanθ=√3 θ=

π3

x=4 tan 2θ+4

dxdθ

=4(2)¿


Mathematics (T)

dxdθ

=8 tanθ sec2θ

dx=8 tan θ sec2θdθ

¿∫π6

π3 √4 tan2θ+4−4

4 tan2θ+4(8 tan θ sec2θ )

¿∫π6

π32 tan θ4¿¿

¿

¿∫π6

π32 tan θ4 sec2θ

¿¿

¿∫π6

π3

4 tan2θdθ

¿4∫π6

π3

tan2θdθ

∴ k=4 , α=π6, β=π

6

4∫π6

π3

tan2θdθ=4∫π6

π3

¿¿

¿4 ¿¿

¿4 {[ tan π3− π3 ]−[ tan π6−π

6 ]}¿4 [√3− π

3−√33

+ π6 ]

¿ 8√3−2π3


Mathematics (T)


Mathematics T for new STPM syallabus

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