ACJC 2013 JC2 H2 Mathematics REVISION SET G

COMPLETE SOLUTIONS

Pure Maths Part II

VECTORS

1.

PJC07/1/6 1l :

1

3 1

7 0

p

r ,

(i) Given 1q and 4p , since C lies on 1l ,

4 1

3 1

7 0

OC

and

4 1

0 1

4 0

AC OC OA

2AC l

4

4

1

0 0

2

12 8 9 7OC i j k

(ii) 2AB q i k

Given acute angle between 1l and

2l is 60 ,

2

1

1 0cos60

0 2

2 4

q

q

22 4 2q q 2q

SAJC/97/1/8

1

TJC07/1/4 By Ratio theorem,

1

2OM (a + c) and

1

2OP (5b – 3a)

1OS OP OM 2

(5b – 3a) +

1

2

(a + c)

= 1

22

a + 5

2

b +

1

2

c

C, S and B are collinear OS c + 1 b

Since a, b and c are non-parallel vectors, 1 5 1

2 0; 1 ;2 2 2

1 3,

4 8

1

8OS (5b + 3c) (shown)

N1996/2/15

3.

1l :

1 1

2 1 ,

0 1

r

2l :

2 3

1 1 ,

7 5

r

3

1

If 1 2 3 ----(1)

and 2 1 ----(2)

and 7 5 ----(3),

we have from (1) & (2), 3 and 2 which satisfies (3) since

R.H.S of (3) = 7 + 5( 2) = 3 = L.H.S of (3) .

andl

2 intersect and the coordinates of is (4, 1, 3). l E

2

1

3 1

1 1

5 1 9 3 3cos

35 3 35 3 35

27 2 2sin 1 cos 1

35 35

6

sin , where is the shortest dist from to and 2

10

2 2140 4 2

35

.

pp A l AE

AE

p

ACJC 2004

5 (i) 0 1 1

1 2 2 (2 3) (1 4 3) 5

1 1 3

Therefore the line l lies on 1

5 (ii) 2 2 0

1 3 4

4 3 1

AB OB OA

1

2 2 2 2 2 21

0 1

4 2

1 3 11cos

2380 4 1 1 2 3

AB

AB

d n

d n

1 11cos 44.5185

238

Hence angle between line AB and plane 1 90 44.5185 45.4815 45.5o

Alternatively, Let be the acute angle between the line and the plane

1

2 2 2 2 2 21

0 1

4 2

1 3 11sin

2380 4 1 1 2 3

AB

AB

d n

d n

1 11sin 45.5

238

5 (iii)

(iv)

Normal of plane 2 , 2

1 1 8 4

2 2 2 2 1

3 1 4 2

n

Equation of plane 2 ,

8 2 8 8 4

2 1 2 2 2 or 1 1

4 4 4 4 2

r r r

(Accept parametric form)

The 3 planes do not have a common point (or line) of intersection.

6

1l :

1 3

5 2 ,

12 2

r = and 2l :

1 8

5 11 ,

12 6

r =

Consider

3 8 34 2

2 11 34 17 2

2 6 17 1

A normal to 1Π is

2

2

1

. 1

2 1 2 2

2 5 2 2 4

1 12 1 1

Π

: r r

6 Method 1

Line passing through Q and perpendicular to 1Π ,

4 2

: 0 2 ,

8 1

l

r

Since R lies on l,

4 2

2

8

OR

for some

At point of intersection of l and 1Π ,

4 2 2

2 2 4

8 1

43

8 4 4 8 4

43

83

203

OR

(shown)

Method 2

RQ = Projection of PQ

onto the normal = 1 1

1 1

PQ

n n

n n

3 2 2

85 2 23

4 1 18

3

43

24

234 4 1 4 4 1

1

8 43 3

8 83 3

2043 3

4

0

8

OR OQ RQ

(shown)

2 possible scenarios

1 : 2 2 4Π x y z

2 : 2 5Π x ay bz

3 : 3 7Π x y z

Direction vector of line of intersection of 1Π & 3Π =

1 2 5

3 2 3

1 1 4

5 2

3 0 10 3 4 0 3 4 10

4

a a b a b

b

1

Q (4,0,8)

R (1,5,12)P

2Π

1Π

3Π

7

Given 1

2 4 2

: . 3 2 , 1 and 5

0

OA OB

p q

r

(i) Equation of line AX : r =

4 2

1 3

0p

where .

X is the point of intersection between line AX and plane 1 , therefore

4 2 2

1 3 3 2

0p

8 4 3 9 2 1 . Therefore

2

2OX

p

If OX makes an angle of 45o with z-axis

2

2 0

2 0

1cos 45

8

p

p

2

2

18

28

pp

p

2 2p

Since p > 0, we have 2 2p

(ii) 2

1

: 1 1

1

r .

2

6AB OB OA

q p

Plane 2 parallel to vector 2normal of plane AB AB .

2 1

6 1 0

1q p

2 6 0q p 2 2 4q

(iii) 3

2

: 3 0

0

r

Perpendicular distance

= length of projection of OX onto the normal of both planes

=

2 2

2 3

02 2 2

13 13

X

O

Alternative method:

1

2

: . 3 2

0

r

2 22 2

2

. 3

0 2 2

132 3 2 3

r

Therefore the perpendicular distance is 2

13.

(iv) 1

2

: . 3 2

0

r and 4

1 2

: 3 3

2 0

r .

Since plane 4 is parallel to -2i +3j which is the normal of 1 4 1

normal of 4 =

1 2 6

3 3 4

2 0 3

4

6

Equation of : 4 0

3

r

To find common point of intersection, solve the following equations:

1

2

4

: 2 3 2

: 1

: 6 4 3 0

x y

x y z

x y z

the position vector of the point of intersection is 1

7 12 3011

i j k .

8 Sub , ,0 into 1 and 2.

1

2

: 1( ) 3( ) (0) 8

: 3( ) 1( ) (0) 0

a

b

Using GC, 1, 3

1 3 3

3 1 3

8

b a

a b

a b

1

1 3

: r 3 3 ,

0 8

b a

l a b

1 3

3 3

0 8

b a

a b

5

2

2

+

4

1

0

1

4

Sub a b and 1

4 , we have

11 4 5 4

4

13 4 2

4

a

a

2, 2a b

Since the 3 planes have no point in common, l1 cannot intersect 3.

1

1 1

: r 3 1 ,

0 1

l

Condition 1: 1,3,0 is not on 3. Hence ( 1) (3) (0) 11

4

p q

p

Condition 2:

1

p

q

and

1

1

1

are perpendicular.

1 1

1 0 1 0 1

1

.p p q p q

q

COMPLEX NUMBERS

1

2(a)

3 3

2 2

3 i 3 i

1 i 1 iz

p p

3

22

3 1

1 p

2

8

1 p

2z 2

82

1 p

24 1 3 or 3 (rejected)p p

3

2

3 iarg arg

1 iz

p

3 2

arg 3 i arg 1 3i

3arg 3 i 2arg 1 3i 7

3 26 3 6

For arg( )z ,

7 5arg( ) 2

6 6z

(Shown)

2(b) 5 2z 2i i

2e 2e , 0, 1, 2k

k

By De Moivre’s Theorem,

21

510i

2 e , 0, 1, 2

k

z k

3 31 1 1 1 1

5 5 5 510 10 10 10 10i i i i

i2 e , 2 e ,2 e ,2 e ,2 ez

1 1 3 1 1 1 3

10 5 10 5 10 10 5 10 5i i i i

2 e , 2 e , 2 , 2 e , 2 ez

3(a) 2z

π 5πarg π

6 6z

5πi

62ez

4 ii ea bz 4 4 ii e ea bz

5πi 4

4 6 i2 e e ea b

4e 2

4ln 2

a

a

5πi 4

6ie e

20π 2π

6 3

b

b

3(b)

2 +i

i 2 +3 3 8 8e 2e

k

kw w

i3

i

i3

π π0, 2e 2 cos isin 1 i 3

3 3

1, 2e 2 cos π i sin π 2

π π 1, 2e 2 cos isin 1 i 3

3 3

k w

k w

k w

2,1 i 3,1 i 3w

4 3 8 8 0w w w 31 8 0w w

1, 2, 1 i 3, 1 i 3w

For 4 3iz 8i 8 0z z , let iw z . i, 2i, 3 i, 3 iz

4(a)

2

2

42i 2i 4 0

2i 2i 4( 4)(1) 2i 12i 3

2 2

z z zz

z

Since 1 2 1Re( ) Re( ), 3 iz z z

In trigonometric form, 1

π π2 cos isin

6 6z

Using de Moivre’s theorem, 1

π π2 cos isin

6 6

n n n nz

For 1

nz to real,

πsin 0

6

n

.

Thus smallest possible positive integer is 6.

4(b) (i) i1 e

i i i2 2 2e [e e ]

i2

i2

e cos isin cos isin2 2 2 2

2cos e2

(ii) Let the fourth roots be z. Then

π π

i i 2 π4 4 48 2 8 2 i 16e 16e

k

z

π πi

16 22e

k

z

1, 0 , 1, 2k

(iii) Let 2z w . Then π π

i16 22 2e

k

w

π π π π

i i16 2 32 4π π

2 1 e 4cos e32 4

k kk

w

, 1, 0 , 1, 2k

Thus when 1k ,

π π 9πi i

32 4 32π π 9π4cos e 4cos e

32 4 32w

.

The other roots are

π 7πi i

32 32π 7π4cos e ,4cos e

32 32

and

15πi

3215π4cos e

32

.

5(i) Given 2 8iz , we can rewrite it as

2 i2 28

k

z e

.

12 i i

2 2 4 8e 2 2ek k

z

, where 0,1k .

i.e. i

4 2 2ez

or 3

i42 2e

.

Converting to Cartesian form,1 i

2 22

z

or 1 i

2 22

Hence 2 2iz or 2 2i .

5(ii) Given 2 4 4 2i 0w w , using formula for solving using quadratic formula,

12 8i

2w .

Applying the result in (i), 1

2 2 2i2

w or 1

2 2 2i2

(note that there should only be two solutions for w)

Hence 1 iw or 3 i .

5(iii)

5(iv) The two loci are parallel to each other but are not the same line, so they do not

intersect. They therefore have no points in common.

6

Minimum value of 2

22 2 3 2 2 2z i = 2.

1 2cos 3 2sin i = 2 3 2 i.4 4

z

7

2cos8

a

Hence 2 cos sin i8 8

z a a

= 2 2cos cos 2cos sin i8 8 8 8

= 1 cos sin i4 4

(by double angle formula)

= 1 1

1 i2 2

.

From right angle triangle in diagram,

1sin

128tan18 2 1cos 1

8 2

a

a

.

Hence 3

tan tan cot 2 18 2 8 8

.

8(i) 2 cos 1 1 2 sinz i

1 2 cos sinz i i

1 2z i since cos sin 1i

The locus of P is a circle with radius 2 and centre (1, 1). (shown)

Method 2

Let z x iy , 2 cos 1 1 2 sinx iy i

Compare real and imaginary parts, we get

2 cos 1x and 1 2 siny

11cos and sin

2 2

yx

Since 2 2cos sin 1 ,

2 2

2 21 11 1 1 2

2 2

x yx y

The locus of P is a circle with radius 2 and centre (1, 1). (shown)

x (2,0) O

(i)

(ii)

(1,0)

7

8

a

8(ii)

8(iii) Min. value of 6 4z i

226 1 4 1 2 74 2 units

8(iv) 2sin 0.16515 rad

74

5

tan 0.62025 rad7

Max. arg 6 4 0.455 radz i (3sf)

9(a) Given w* = z − 2i, z = w* + 2i, we obtain |w|2 = w* + 2i + 6.

Let w = ia b .

2 2 i 2i + 6 6 i 2a b a b a b

Comparing real and imaginary parts,

2 − b = 0 b = 2

2

2

4 6

2 0

2 1 0

2 or 1(N.A.)

a a

a a

a a

a

Therefore w = 2 + 2i.

Im(z)

Re(z)

(1, -1)

(-6, 4)

9(b)

arg(a) = 4

a = 1 + i

10

10 (i) Minimum value will be the perpendicular distance from origin to the line segment

CB. Let the point of intersection of this perpendicular and CB be D. By observing

the right angle triangle formed by O, C, and D, OD = | a | sin 4

=

| |

2

a.

10 (ii) The locus of z satisfying both relations is the line segment CB, excluding C. The

argument of the point represented by B is arg( )2

a

, while the argument of the

point represented by C is arg( )a .

Hence the range of values is arg( ) arg( ) arg( )2

a z a

.

11(i) Locus of P is a half-line from and excluding the point representing the complex

number1 2i that makes an angle of with the positive real axis.

11(ii)

11(iii)

5EF and radius 2EG

2sin

5 0.287

Locus of P meets the locus of Q more than once when 0.287 0.287 .

DIFFERENTIATION

1 2 26 2 21 0x xy y

Differentiate with respect to x

d d 2 6 4 0

d d

d4 6 6 2

d

d 6 2

d 4 6

3

2 3

y yx y x y

x x

yy x y x

x

y y x

x y x

y x

y x

2(a) 4e

1tan

xyx

0d

de

1

1 1tan

2

yx

yx

x

x

x

yx

x

yx

1tan

2e

1

1

d

d

)1(

)1(e2

2tan 1

xx

xyx

2(b) xxy )12( )12ln(ln xxy

)12ln(12

21

d

d

x

x

x

yx

y

)12ln(

12

2

d

dx

x

xy

x

y

)12ln(

12

2)12( x

x

xx x

3(a) 2

2

2

2

2

ln 2

cosd(ln( ))

2 (ln co )

2 sin 1

cos

12 ta

d

n

s x

x

dx xx d

x x

x x

x x

x

x

3(b) 2

1

2

2 2

2 2

(1 2 ) (2 )( 2)

2 (1 2tan (

(

))

21 21 )

1 2

1 2 4 2

(1 2 ) (2 )

5

5 15

1

x x

x x

xx

x

x x

x x

x x

d

dx

4.

2

2 2 2

3

2 3 2 2

f ( ) e

f '( ) 3 e (2 e ) 1 e (3 2 ) 1

x

x x x

x x x

x x x x x x

For all values of x , we have

22 2 0, e 0, (3 2 ) 0xx x , and so22 2 e (3 2 ) 0xx x .

Hence f '( ) 1 0x , and so 23f ( ) exx x x is strictly increasing for all values of

x .

DIFFERENTIATION and its APPLICATIONS

1. ln lny xy x

Differentiate implicitly w.r.t x,

1 d dln

d d

1y yxy x x y

y x xx

1 d dln ln

d d

y yy x x y x

y x x

1 dln ln

d

yx x y y x

y x

2 2d (1 ln ) (1 ln )

d 1 ln 1 ln

y y x y x

x xy x y

For the tangent to be parallel to the y–axis, d

0d

x

y

2

1 ln0

(1 ln )

y

y x

1 ln 0y ey

Hence from xyy x , we have ee xx e e 0xx

Hence x = 1.32, and since the tangent parallel to the y-axis takes the form x c ,

hence equation of tangent is 1.32x

2. 2 1

2dy

x at y atdx t

,

Equation of tangent is 212 ( )y ap x ap

p

2py x ap

Equation of normal is 22 ( )y ap p x ap 3 2y px ap ap

(ii) A = (9a, 6a) 29 3a ap p

Equation of normal is 23 (3) 2 (3) 3 33y x a a x a

Since Q lies on the normal, sub 2( ,2 )aq aq into eqn of normal to get:

22 3 33aq aq a 23 2 33 0q q

113 or

3q q

Coordinates of Q are 121 22

,9 3

a a

(iii) Since 3

( , )2

T a a lies on the tangent, it satisfies the equation

2py x ap

23

2ap a ap

22 3 2 0p p

1(2 1)( 2) 0 or 2

2p p p

Points B and C are 1

( , ) and (4 , 4 )4

a a a a

(Gradient of TB) × (gradient of TC) =1

2 ( ) 12

Thus TB is perpendicular to TC

3 By Pythagoras’ Theorem,

22 2

2

15 15

30 (shown)

r h

r h h

1515 h

3(i) 2 345

3V h h

2

2

d d30

d d

d20 30 5 5

d

d0.050929 0.0509 cm/min (3.s.f)

d

V hh h

t t

h

t

h

t

Alternative

2 2

2

d90 3 30

d 3

d d d

d d d

d 120

d 30 5 5

d0.050929 0.0509 cm/min (3.s.f)

d

Vh h h h

h

h h V

t V t

h

t

h

t

3(ii)

2

12 2

30

d 130 30 2

d 2

r h h

rh h h

t

2

d d d

d d d

15 5dWhen 5, 0.050929

d 30 5 5

d0.045552 0.0456 cm/min (3.s.f)

d

r r h

t h t

rh

t

r

t

4(i) Applying the cosine rule to ASB, we have

2 2 2 2 2 2 2 2 2 2 2(3 2) ( 2 ) ( 3 ) 2 2 3 cos .s s s s

Rearranging,

2 2 2

2 2 2 2

4 9 1 6cos ,

2 4 9 4 9

s s s

s s s s

which upon squaring both sides gives the required result.

4(ii) 2 2

2 2 2 2

( 6)1

( 4)( 9) 4 9

s P Q

s s s s

So

2 2 2 2 2 2( 6) ( 4)( 9) ( 9) ( 4).s s s P s Q s

Comparing coefficients gives

9 4 0

1,

P Q

P Q

4 9

and .5 5

P Q

Thus

2 2

1 4 9f ( ) 1

5 4 9s

s s

and

2 2 2 2

2 9 4f '( ) .

5 ( 9) ( 4)

ss

s s

4(iii) When f '( ) 0s , we have

2 2 2 2

2 2 2 2

2 2 2 2

2 2

2 9 40

5 ( 9) ( 4)

9 40 (as 0)

( 9) ( 4)

9 4

( 9) ( 4)

3 2(as all quantities are positive)

9 4

s

s s

ss s

s s

s s

Thus 2 6,s giving 6s as 0.s 2

2 2 2 2 2 2 2 2

2 9 4 2 5( 30)f '( ) .

5 ( 9) ( 4) 5 ( 9) ( 4)

s s ss

s s s s

This implies that

s 6

6 6

f '( )s 0 0 0

f ( )s __

Since 2cos

is decreasing for 0,2

, when f is minimum, 2cos

is

minimum and this implies is maximum. Hence the soccer player should shoot

when 6.s

Alternative solution : use graph

5

Let x be the distance between A and P. Then 2 2 28 64PW x x

Cost of laying the pipe, 260000 64 45000 12C x x

d0

d

C

x

2

6000045000 0

64

x

x

2

2

2 2

2

60000 45000 64

4 3 64

16 9 64

576

7

576 576 (reject as 0)

7 7

x x

x x

x x

x

x x

To show that C is minimum when 576

7x

Method 1: First Derivative Test

x 576

7x

576

7x 576

7x

2

d 6000045000

d 64

C x

x x

< 0 0 > 0

C is a minimum when 576

7x .

Method 2: Second Derivative Test

22

2 2

2 2

2 2

3 32 22 2

6000060000 64

d 64

d 64

60000 64 60000 38400000 for all

64 64

xx

C x

x x

x xx

x x

C is a minimum when 576

7x .

Land

W

P B A

8 Ocean

x 12 – x

BINOMIAL EXPANSION and MACLAURIN’S SERIES

1

111 222

2

3

3cos 2 4 3 cos 2 4 1

4

1 3

1 1 3 32 2cos 2 1

2 2 4 2! 4

1 3 5

32 2 2

3! 4

1

2

xx x x

x xx

x

2 2

3

2 3 2 3

2 3 27 1351 1

2 8 128 1024

1 3 229 633 1 3 229 6331

2 8 128 1024 2 16 256 2048

x x xx

x x x x x x

Replace x with x ,

21 1

2 2

1

2

1 3

2 2

2cos 2 4 3 1 4 3

2

1 2 4 3

1 2 1 3 229 633Thus

2 16 256 20484 3

xx x x

x x

xx x x

x

2(i)

1144

314 4

16

214 16 2 16

2 23 31 164 8192 32 4096

16 2 1

2 1 ...

2 1 ... 2 ...

x

x x

x

x x x x

This expansion is valid for 16

1 16 16.x x

2(ii)

14

22 2 231

32 4096

2 4 3 23 3132 32 4096

23 27132 32 4096

2 23 10132 4096

16 3 2 3 3 ...

2 6 9 ...

2 ...

2 (up to term in )

t t t t t t

t t t t t

t t

t t t

2(iii) Replacing t with –t in the above expansion, 142 23 101

32 409616 3 2t t t t

3 sin 2 sin cos 2 cos sin 2

4 4 4

cos cos

x x x

x x

1 1cos 2 sin 2

2 2

cos

x x

x

2

2

1 4 11 (2 )

22 2

12

xx

x

2

2

1 4 11 (2 )

22 2

12

xx

x

12

211 2 2 1

22

xx x

221

1 2 2 1 ...22

xx x

2 21 11 2 2

22x x x

21 31 2

22x x

4(a)

2

2

2 3 2

3 2

3 2

πtan cos2 10

4

tan 1cos2 10

1 tan

Given that is sufficiently small,

1 41 10

1 2

1 1 2 1 10 1

1 1 2 2 10 10

2 12 10 2 0

6 5 1 0 (Shown)

x x x

xx x

x

x

x xx

x

x x x x x

x x x x x x

x x x

x x x

Solving the cubic equation with GC,

4.95(N.A.) or 1.22(N.A.) or 0.166 (3 s.f.)x x x

4(b)

1

2 1

2

222

2

22 2

22

22 2

1 tan

1 tan

d 1Differentiate w.r.t. , 2

d 1

d d 2Differentiate w.r.t. , 2 2 1 2

d d 1

d d (Shown)

d d 1

y x

y x

yx y

x x

y y xx y x x

x x x

y y xy

x x x

22 2 23 2 2

43 2 2 2

2 2

32

Differentiate w.r.t. ,

1 4 1d d d d d2

d d d d d 1

1 4

1

x

x x xy y y y yy

x x x x x x

x x

x

2 3

2 3

2 3

2 3

d 1 d 1 d 5When 0, 1, , , .

d 2 d 4 d 8

511 8412 2 6

1 1 51

2 8 48

y y yx y

x x x

y x x x

x x x

1

11

2

2 2

2 2 2

2

1 tan

1

1 tan 1

1 3

1 1 1 2 21 1

2 8 2 2

1 3 1 1 11

2 8 2 4 8

11

2

x

x

x x

x x x x

x x x x x

x x

5 2 22 2 2

2 4

( 2)( 3)1 1 ( 2)

2 2 2! 2

3 1

4

x x x

x x

5(i) Since x is sufficiently small for 3x and higher powers of x to be neglected,

2

2

22

22

2

2

sin 2sec

2sin

cos

2

12

2 12

2 1

2 2

x x

xx

xx

xx

x x

x x

OR

2

2

2

2

sin 2sec

sin 2(1 tan )

2(1 )

2 2

x x

x x

x x

x x

5(ii) 1 0 2 3

1

2

1 2 3 4

2 2 2 2 2

3 4 1 1

4 8

1 1 4

2

rr

r

6(i) 1

2

4 2 2 12

xx

2 331 11 11 2 2 22 2

2 1 ...2 2 2! 2 3! 2

x x x

2 31 1 12 ...

2 16 64

Expansion is valid if 1 2 2 2.2

x x x

xx x

6(ii) d:

dx

de 3cos3

d

y yx

x .

d:

dx

22

2

d de e 9sin3 9 e 1

dd

y y yy yx

xx

22

2

d d9e 9 0 Shown .

d d

yy y

x x

d:

dx

3 2

3 2

d d d d2 9e 0

d dd d

yy y y y

x xx x

.

When 0x , 0y , d

3d

y

x ,

2

2

d9

d

y

x ,

3

3

d27

d

y

x .

2 33 9 27...

1! 2! 3!y x x x

2 39 9

3 ...2 2

x x x .

6(iii)

2 1eln 2 1 ln 1 sin 3

1 sin 3

x

x xx

2 3 2 3

2 1 4 4 4 2 2

1 1 1 1 12 2 2 2 ... 2

2 16 64 4 8

x x x

x x x x x x

2 3 2 3

2 3

4 4 ln 1 sin 3

1 1 9 92 ... 3 ...

4 8 2 2

17 352 2 ...

4 8

x x

x x x x x x

x x x

2 10.1 0.1

2 3

0 0

e 17 35ln d 2 2 d

1 sin 3 4 8

0.19131 to 5 dp

x

x x x x xx

INTEGRATION and ITS APPLICATION

1 1 2cos dx x x

2 21 2

4

2 31 2

4

21 2 4

2cos d

2 2 1

1 4cos d

2 4 1

1cos 1

2 2

x x xx x

x

x xx x

x

xx x C

2 (i) tant x 2 2sec 1

dtx t

dx

= 2 2

2

1 1

11

1

dtt t

t

= 2

1

1 2dt

t

= 11tan 2

2t c

= 11tan 2 tan

2x c

(ii)

volume is same as

Exact volume =

2

420

1d

1 sinx

x

41

0

1

1tan 2 tan

2

tan 22

x

3 Method 1:

sin 2 cos dx x x

1sin 3 sin d

2

1 cos3cos

2 3

1 cos3cos

2 3

x x x

xx C

xx C

Method 2:

sin 2 cos dx x x

12

3

[f ( )]2sin cos d [use f ( )[f ( )] d ]

1

2cos

3

nn x

x x x x x x cn

x C

(i)

d dcos 1, 2cos 2

d d

x yt t

t t

d d d 2cos 2

d d d cos 1

y y t t

x t x t

When d

0,d

y

x

3 3 1 1 3cos 2 0 2 , , ,

2 2 4 4 4 42 2t t t x

At point A, 1

, 142

x y

1y is the equation of the tangent to the curve at point A.

Or

2

12

1 siny

x

2

1

1 siny

x

y =2

Since 0 t , the maximum and minimum values of y (i.e. sin 2y t ) is 1 and -1. The

y-coordinate of point A is 1 and since the tangent to this max pt is a horizontal line

(d

0,d

y

x ), therefore the equation of the tangent to the curve at point A is y = 1.

(ii)

1

42

4

4

44

Area 1 d

3 1 sin 2 cos 1 d

4 2

3 1 sin 2 cos sin 2 d

4 2

3 1 1 cos3 cos 2 cos

4 2 3 22

3 1 2 1 1

4 3 22 3 2

3 1 2 2

4 6 3

y x

t t t

t t t t

x tx

4 22 3131)31)(31(

102

x

CBx

x

A

xx

x

By cover-up rule, when 1 3

1 Ax

)31)(()31( 1102 2 xCBxxx

When x = 0, C = 3

When x = 1, B = 1

xx

x

xx

xx

xd

31

3

31

1d

)31)(31(

1021

0

2

1

0

2

y

x

A

34ln

6

1

3tan34ln6

14ln

3

1

3tan331ln6

131ln

3

1

1

1

0

12

xxx

xx exex

coscos )(sind

d

xxxe

xxe

x

x

d )cossin2(

d 2sin

cos

cos

xxxe x d ))(cossin(2 cos

cxe

ceex

xexex

x

xx

xx

)cos1(2

2)(cos2

d )(sin)(cos2

cos

coscos

coscos

5 Equation of Tangent: y = 2x – 2

When x = 3, y = 4

Area )4)(2(

2

1d )1(

3

1

2 xx

43

3

1

3

x

x

3

8 or 2.67 (3 s.f.)

Volume

8

0

24

0

2

d 1)4()3(d 2

2yyy

y

8

0

24

0

23

23642

34

y

yyy

y

3

40

6 2 2

2

1 2 (1 )1 1

1 (1 )

x x

x x x

x x

2 2

1 1 d d1 2 1 (1 )

1 ln 1

(1 )

xx x

x x x x

x Cx

2 2 2 2

2

1e sin(e ) d (2e )sin(e ) d

2

cos(e )

2

x x x x

x

x x

C

1 1

2

1

1 2 2

1 2

1sin d sin . d

11

sin ( 2 ).(1 ) d2

sin 1

x x x x x xx

x x x x x

x x x C

2

2

2

2 2

2 1

d2 3

2 31 d

2 32 2 1

1 d2 3 ( 1) 2

1 1ln 2 3 tan

2 2

xx

x xx

xx x

xx

x x xx

x x x C

7(a)

(i)

Total area of all n rectangles, A

3 3

3 3

33 3 3 3

3 13

1

2

4

1 1 1 1 20 2 2 2

1 3 1 1 2 ... 2

1 12 2 ... 2 1 2 3 ... ( 1)

1 12

1 ( 1)2 (

4

n

r

n n n n n

n

n n n n

nn n

n rn n

nn

n

2

2

2

2 2

1 1)

12 2 1

4

1 1 1 9 1 12 (shown)

4 2 4 24 4

n nn

n nn n

(ii) Area under the curve 2

2

9 1 1 9lim unit

4 2 44n n n

(iii) For

2

9 1 1 90.99

4 2 44n n

2

1 10.0225

2 4n n

Using GC,

0.512 or 21.7n n

Since n , least n = 22

(b)(i) By GC, ( 0.340,1.162) and (0.340,1.162)

(ii) Since both curves are symmetrical about the y-axis,

(I)

Volume required 2

20.34001 0.52

20 0.34001

12π 2 cos(5 ) d 1 d

4x x x

x

32.61 unit or 0.832π (3 s.f.) by GC

(II) 2

2 1

2 1

2 cos(5 )

5 cos (2 )

1cos (2 )

5

y x

x y

x y

2

2

2

11

4

11

4

1

4( 1)

yx

yx

xy

Volume required 1.162 1.162 1

0 1

1 1π d cos (2 ) d

4( 1) 5y y y

y

30.567 unit or 0.180π (3 s.f.) by GC 8(a)

2 2 2

2

2

2 1

1 1 6 1d d d

1 3 6 1 3 1 3

1 1 1 ln 1 3 d

6 3 1/ 3

1 3 ln 1 3 tan 3

6 3

x xx x x

x x x

x xx

x x C

(b) 2 2 22de e 2 e

d

x x xx xx

Using by parts with 22' 1 2 e ,xv x u x ,

2 2

2 2

2 2

2 2

2

2

2 1 d 1 2 d

d

1

2

x x

x x

x x

x x e x x x e x

x e xe x

x e e C

9(i) 6/5 1/2 6/5

0 0 1/2

1/2 6/52 2

0 1/2

2 1 d 1 2 d 2 1 d

37

50

x x x x x x

x x x x

(ii) d2sin , 2cos

d

xx

1

0 0

6 3sin

5 5

x

x

26/5

0

21

0

2

0

2

0

2

0 0

00

2

1

1 d4

4sin 31 2cos d where sin

4 5

4 4sin2cos d

4

1 sin 2cos d

2cos d cos 2 1 d

1sin 2 sin cos

2

sin cos sin 1 sin

3 4 3sin

5 5 5

a

a

a

a a

aa

xx

a

a a a a a a

1

3 (Since sin )

5

12 3sin (shown)

25 5

a

(iii)

(a)

26/5

0

26/5 6/5

0 0

6/5

1

0

1

1 2

3Area of 1 2 1 d

4 5

32 1 d 1 d

5 4

37 3 12 3sin

50 5 25 5

37 18 12 3sin from (i) and (ii)

50 25 25 5

23 3sin units

50 5

xR x x

xx x x

x

(iii)

(b)

Volume generated

y

x

Idea: + –

2

1 4/52

4/5 3/5

22/5

3/5

3 3

44 1 d d

2 5

1 d

5 2

8513.56 units (to 3 s.f) by GC or units

750

yy y y

yy

11 (i) Total Area of rectangles

11 2

01e e e e

n

n n n

n

1

1

e 11

e 1

n

n

n

n

1

e 1

e 1

nn(shown).

10(i) 2 2

2 21

24

3

x y

2 29 4 16x y

(ii)

(iii) Required volume

=

1.08729 1.08729 22

2

0 0

16 9 3d d

4 4

x xx x

x

= 9.487 (3 d.p.)

y

y = 0

x

-2 2

x = -2 x = 2

(1.09, 1.16)

(-1.09, -1.16)

(ii) Actual Area =

11

00

e d e e 1

x xx

Actual Area > Total Area of rectangles

1

e 1e 1

e 1

nn

1

e 1 1 nn

1 1

e 1 n

n (shown)

e (1) xy

e(2)y

x

Point of intersection is 1, e

Volume = 2e 4

2

1 e

eln d d

y y yy

= 4.99 (to 2 d.p)

12 (i) Let 1 y x , then

d 1

d 2

y

x x .

4 3

1 2

3

2

3

2

1 1 d 2 1 d1

12 1 d

2 ln

2 3 ln 3 2 2 ln 2

22 2ln

3

x y yyx

yy

y y

(ii)

Let

d 1ln(1 ),

d

vu x

x x

4

1

4 4

1 1

4

1

1ln(1 ) d

1 12 ln(1 ) 2 d

1 2

14ln 3 2ln 2 d

1

24ln 3 2ln 2 2 2ln

3

6ln 3 4ln 2 2

x xx

x x x xx x

xx

13

14(a)

2 d

2 4d

xx t t

t . when 0, 0x t . when

2

, 08 4

x t t

.

2

8

0

4

0

44

0 0

Area d

sin 4 d

4 cos 4cos d

y x

t t t

t t t t

4

0

24sin

2

2 24 0

2 2

22 2

2

t

15 (i)

4 4

2 2

2

2

3 1

1 1

1( 1

1

1tan

3

x xx x

x x

x xx

x x x C

d d

) d

(ii) Let tanu x

2 2 2

2

dsec tan 1 1

d

1d d

1

ux x u

x

x uu

When

, tan 1

4 4

0, tan 0 0

x u

x u

4144

20 0

3 1 10

1

tan d du1

1=[ tan ]

3

1 21 tan 1

3 4 3

ux x

u

u u u

(iii) 4 44 4

04

2 4tan d 2 tan d 2( )

4 3 2 3x x x x

A parametric 2tan , secy x , where 0 2 .

(iv) When 1, tan 1 , 24

y x

Area of region R =

2

1

2 dy x

24

0

2 24

0

2 24

0

4 24

0

2 tan 2sec tan d

4 tan sec d

4 tan (tan 1)d

4 (tan tan )d (shown)

4 24 4

0 0

24

0

40

4 (tan )d 4 (tan )d

24( ) 4 (sec 1)d

4 3

84[tan ]

3

8 44[1 ]

3 4 3

(v) 1

2 2

0

(2) 2 2 dyV x y

2 2 24

0

8 2 (sec ) sec d 13.4

16 (b)

(i) 2d( sin 2 )

dx x

x =

22 sin 2 2 cos2x x x x

(ii) 2 2(2 cos 2 2 sin 2 )d sin 2x x x x x x x

2 2

2

2

2

(2 cos 2 ) d 2 sin 2 d sin 2

[ cos 2 cos 2 d ] sin 2

1[ cos 2 sin 2 ] sin 2

2

1cos 2 sin 2 sin 2

2

x x x x x x x x

x x x x x x

x x x x x C

x x x x x C

2 21 1 1(2 cos 2 ) d cos 2 sin 2 sin 2

2 4 2x x x x x x x x C

17(i)

2

3 10

1

3 1 0 , 1

x x

x

x x x

Hence, 1 3x , 1x . (ii)

2

2 2

2

2

2

2

2

3 1 2 3 d d

1 1

2 1 4 d

1

1 4 d

1

41 d

1

4

1

x x x xx x

x x

x xx

x

xx

x

xx

x Cx

(iii)

2

2

3

2 2

2 3

3

2 3

2

3 1 7 d

61

3 1 3 1 7 d d

61 1

4 4 7

1 1 6

4 4 4 4 73 2 3

3 1 2 1 1 3 1 6

4 75 6 5

1 6

4 31

1 6

6 4 31

p

p

p

x xx

x

x x x xx x

x x

x xx x

pp

pp

pp

p p p

2

31

6 37 55 0

3 11 2 5 0

11 5 or (N.A.)

3 2

p p

p p

p

−1 1 3

+ + −

18(i)

4(ii)

2 2

2

2

2

2

2 31

2 1

2( 3) 1

2

23 1

2

y x

yx

yx

y

x

y

x

y = 2

222 2

2

0

2 22

0

2 2

22

0

2 23 1 3 1 d

2 2

2 29 6 1 1

2 2

2 29 6 1 1 d

2 2

212 1 d (shown)

2

y yV y

y y

y yy

yy

d2 2sec , 2sec tan

d

yy

22

0

23

0

3

0

3

0

212 1 d

2

12 sec 1 2sec tan d

=24 tan sec tan d

24 2 3 sec d from (*)

124 2 3 3 ln 2 3

2

12 2 3 ln 2 3

yV y

19 (i)

2

2 2

2 2

2 2

2

1 2 1 1 4 4

2 241

2

4 4

4 4

2 4

xx x

x x

x x

x x

x

(ii)

2 2 1

0 0

2 1

1 1 14 d 4 4sin

2 2 2 2

1 4 4sin

4 2 4

kkx

x x x x

k kk k a

(iii) 2 24 4y x

22

21

2

xy

2

1

k

R

2

0

14 d

2

k

R x x

(iv)

1

Required area 4 with 1

1 3 4sin

2

3 46

2 3

3

R k

20

ln dx x x

= 2 2 1

ln d2 2

x xx x

x

= 2 2

ln where is an arbitrary constant.2 4

x xx c c

22 xu d

2 1d

uu

x

When x = 2 , u = 2. When x = 7, u = 3.

7

2

2d

11

xx

x

=

3

22

(2 ) d9

uu u

u

=

3 2

22

2 d9

uu

u

=

3

22

92 1 d

9u

u

=

3

1

2

2 3tan3

uu

= 1 22 3 3 2 3tan

4 3

= 13 22 6 tan

2 3

21 (a) Then

d

d

x xuu e e u

x .

d 1

d

x

u u

1 1 1 d d

22x xx u

e e uu

u

2

1 d

2u

u

-1

22

1 1 d tan

2 22

uu c

u

12

tan2 2

xec

(b)

2 2 2

1

0 0 1

4 3 d 4 3 d 4 3 da a

x x x x x x x x x

13 3

2 2

0 1

2 3 2 33 3

a

x xx x x x

321 1

2 3 2 3 2 33 3 3

aa a

322

2 2 33 3

aa a

22

2

1 d

1 ( 2)y

y = 1 1 ( 2)

ln2 1 ( 2)

yc

y

=

1 1ln

2 3

yc

y

222

x

xy

22

2

2 2 2 2

22

2

2

( 2)2

2( 2) ( 2)

2( 2)

1 ( 2)

2 2 where 2 and 2

1 ( 2)

xy

x

y x y x

yx

y

a by

When x = 2

1 ,

3

7y

Volume required

=

73

2

22

1 7 22 d

2 3 1 ( 2)y

y

=

73

2

2

7 22 d

12 1 ( 2)y

y

= 7

12

72 2

3

73

22

12 d

1 ( 2)y

y

= 5

4

73

2

1ln

3

y

y

= 5

4

3

4ln

3

2ln

= 5

ln 24

23

The intersection points are (2,1/3) and (3,1/2).

1/22 2 2

1/3

21/2

1/3

3

1 1Volume required= 3 2

2 3

9 1 4

2 1 3

191.0159

6

6.76 units

x dy

ydy

y

y

x

y=1

1x x=2 x=3

1/3 1/2

When , ,x t y 1

3 13

When , ,x t y 1 1

22 2

dx

dt t

2

1

dy x1

2

1

3

d dt

d

x

tt

2

23

1

4

dttt

2

223

1 1

4

a 3 b= 2

Using t = 2sin

When ,t

33

When ,t

24

1 1

2

1

2

1

3

x

y

cosdt

d

2

dttt

2

223

1 1

4

cos dsinsin

4

3

2 22

1 12

24 4

cos dec

4

3

21

4

cot

4

3

1

4

1 11

4 3

1 11

4 3

unit2

(=0.382)

24

25

At point P, t =1, x = 3, y = 4.

At point Q, t = 1 3 13

, ,2 4 8

x y

Area of the region

= 4

13

8

1 3 133 4 d

2 4 8x y

= 1

2 21

2

675(3 )(3 3) d

64t t t

= 5.32 units2

y = ex – 2,

When the graph crosses the x-axis, x = ln2.

2

0

12 d d

ln

kex

ee x x

x x

ln 2 2

0 ln 2

1

( 2) d ( 2) d dln

kex x

e

xe x e x xx

P

Q

C

2 0

ln 2 2

0 ln 22 2 ln ln

kex x

ee x e x x

2 1(2 2ln 2 1) ( 4 2 2ln 2) ln ln

2e k k

lnk = 2(e2 + 4ln2 – 7)

26

13

321 12

0 0

0

15

2

0

(2 1) 1(b) 2 1 d (2 1) d

3 3

1 1 (2 1)3 3 0

3 3 5

1 2 3 13 9 3 1

15 5 15

xx x x x x x

x

DIFFERENTIAL EQUATIONS

1. DHS 12/P1/Q11

(a)

2 de

d

x yz

x

2 22 2 2

2 2

d d d d de 2e e 2

d d d d d

x x xz y y y y

x x x x x

Given 2

1 4

2

d d2 e

d d

xy y

x x

2 1 4de e

d

x xz

x

1 4 2 1 2d d

e .e e (shown)d d

x x xz z

x x

Hence, 1 2d e dxz x

1 21

e2

xz C

2 1 2

1 4 2

1 4 2

d 1e e

d 2

d 1e e

d 2

1 1e e where and are arbitrary constants

8 2

x x

x x

x x

yC

x

yC

x

y C D C D

2. HCI 12/P1/Q8

(i) 2 2 36x y (ii) d

d

yky

t

Method 1

Implicit Differentiation w.r.t. time

d d2 2 0

d d

x yx y

t t

2 2

d d

d d

= ( )

(36 ) (shown)

x y y

t x t

yky

x

y k xk

x x

Method 2

Implicit Differentiation w.r.t x then using chain rule (rate of change equation)

d2 2 0

d

yx y

x

2 2

d d d d d d or

d d d d d d

d (36 ) = (shown)

d

y y x x x y

t x t t y t

x ky y k xk

xt x x

y

(iii) 2

2

2

2

2

2 4

2 4 4

d 2(36 ) (given)

d

d 2 d36

1 2 d 2 d

2 36

1ln 36 2

2

ln 36 4 '

36 e

36 e 36 e ( 0)

t

t t

x x

t x

xx t

x

xx t

x

x t C

x t C

x A

x A x A x

4

Using initial conditions, when 0, 4

4 3620

36 20e t

t x

AA

x

4

4

For to be 3,

36 9 27

27 36 20e9

e20

1 9ln 0.2 s

4 20

t

t

OY

OX

t

(iv)

3. NJC 12/P1/Q4

(i)

2

2

22 2 2 2

4 2 4

2

d d2

d d

d2 2 0

d

d

d

d shown

d

x ut

x utu t

t t

ut tu t t ut ut

t

ut u t

t

uu

t

(ii) 2

2

2

1 d 1dt

1'

'

where '

uu

t Cu

tt C

x

tx C C

t C

Since there was 0.2 milligrams of bacteria

after 15 minutes, then

2

0.25 10.2 0.05 0.2 0.0625

0.25 16C C

C

216

16 1

tx

t

Jesse’s model is not appropriate.

Based on Jesse’s model, the rod

would never fall flat on the

ground.

When t = 4,

216 4 256

or 3.9416 4 1 65

x

(iii) As

216, .

16 1

tt

t

The particular solution of the DE suggests that the amount of bacteria in the Petri

dish will grow indefinitely as time passes. Hence the model is not a realistic one.

4. RJC 12/P1/Q9

2d2 ( 1)e

d

xyy x

x

--- (1)

Let 2e xz y 2 2d d

e 2ed d

x xz yy

x x

Sub. into (1), 2d

2 ( 1)ed

xyy x

x

2 2de 2 e 1

d

x xyy x

x

d1

d

zx

x

( 1) dz x x

2

2

xz x c ,

22e

2

x xy x c

(i)

When 0, 1, 1.x y c

22The particular solution is e 1 .

2

x xy x

(ii)

2

22 2 2

2

3 22 2 2

3 2

d2 ( 1)e

d

d d2 e 2( 1)e (2 1)e

d d

d d2 2e 2(2 1)e 4 e

d d

x

x x x

x x x

yy x

x

y yx x

x x

y yx x

x x

2 3

2 3

2 3

2 3

d d dWhen 0, 1, 1, 1, 2

d d d

1 2Using Maclaurin's expansion, 1 ...

2! 3!

1 1 1 ...

2 3

y y yx y

x x x

y x x x

x x x

(iii) Replace x by -2x in the standard series expansion of ex and perform an expansion for

22e 1

2

x xx

up to and including the term in x3. Compare the coefficients of this

series with that of the expansion in (ii) to verify the correctness.

5. RJC 12/P2/Q2

(i)

d

, 0d

xkx k

t

1d dx k t

x

ln , 0x kt c x

e ktx A When 0t , 80x , thus 80A .

When 3t , 20x ,

320 80e k

1 1 1

ln ln 43 4 3

k .

Thus ln 4

380et

x

(ii)

2ln 4 1

When 6, 80e (80)16

t x

Just before the (n+1)th injection, the amount of drug present in the blood stream =

1

16nu

1

Immediately after the ( +1)th injection,

1amount of drug present, 80

16n n

n

u u

In the long run, the amount present approaches the value 85.3 (3 s.f.).

6. RVHS 12/P2/Q1

3 2

3 2

2 2 2

d(1 ) 0 (1)

d

d dLet

d d

Subs. into (1):

d0

d

d( 1) (shown)

d

yx x y x

x

u yu xy x y

x x

uy y x y x

x

uux x u x

x

3

3

3

2

3

3

333

1d d

1

ln | 1|3

1 e where e

e 1 1e 1 (shown) and f

3

xC

xx

u x xu

xu C

u A A

Au A y x x

x

(i)

(ii) 1A 1A

7. TJC 12/P1/Q11(b)

(i) Let (15 )dx kx x

dt .

Given that when t = 0, x = 5, 0.1dxdt

. 0.1 = k(5)(15 5) k = 1500

Hence, (15 )

500

x xdxdt

.

(ii) (15 )

500

x xdxdt

1500 1

(15 )dx dt

x x

500 1 115 15

dx t cx x

100 ln ln 153

x x t c

100 ln3 15

x t cx

as 0 15x

When t = 0, x = 5 100 100ln 0.5 ln 23 3

c

Hence, 100 2ln3 15

xtx

is the solution of the DE.

(iii) When x = 0.95(15) = 14.25 m2, 121.25t hours. (2dp)

It required a minimum of 122 hours for 95% of the soil area to become infected.