Revision Package BT1 2013 Solutions

2013 Revision Package BT1 Solutions

Chapter 2: Binomial Expansion Solution

1 [AJC/2005/Prelim/P1/Q3]

Solution

( )( )

( )( ) ( )

+++++=

−−

−+=−

The series is valid for 2

112 <⇒< xx

When 11

1=×==

3209411

+= �

2 [TJC/2005/Prelim/P1/Q7]

Solution

( )( )

( ) ( )( ) ( )22

12111232

2112112

xCxBAxxx

++−+=−+

When 2

When 0=x , ( ) ( )

When 1=x , ( ) ( )( ) ( )

12111123

++−+=−+

( )( ) ( )xx

112322

( )( ) ( )

��

+++++−+=

−+++=

−−

The expansion is valid for 2

112x and 12 <⇒<< xx

3 [RJC/2005/Prelim/P1/Q5]

Solution

++−+=

The expansion is valid for 414

<⇒< yy

Let 28 kxxy +=

( ) ( )

( ) ( ) ( )

( ) ( )

��

++−=

+++++−=

++++−++=

++−+=

512512

xxkkxx

kxxkxxkxx

4 [NYJC/2005/Prelim/P1/Q1]

Solution

( ) �++++=−− 322

43211 xxxx

The expansion is valid for 1<x .

expansion into 2

1subst ,

=− ∑∑

5 [HCI/2005/Prelim/P1/Q2]

Solution

( ) ( )( )

( )( )( )

−−−−−

−−−+

−−+

−=−

−−

The series is valid for 212

<⇒< xx

The term of rx is the ( )th

r 1+ term in the series.

( ) ( )( )( ) ( )

( ) ( )( )( ) ( )( ) ( )

1222122

1 term1

+−−−−−−−=+

6(i) [CJC/2005/Prelim/P1/Q2]

Solution

( )( )( )

( ) ( )( )( )11

−−+

−−=

( )( ) ( )( ) ( )22

1111111 ++

( ) ( )( ) ( )111122 −++−++=− xDxxCxBx

When 1=x , ( )4

2−=⇒=− BB

When 1−=x , ( )2

1111 =⇒−−=− BD

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

2−=⇒−+−+= CC

( )( )( ) ( ) ( ) ( )22

−−=

6(ii) ( )

( ) ( ) ( )

( ) ( )

2 3 2 3 2 3

1 3 11

4 1 4 1 2 1

1 3 11 1 1 1

1 3 11 1 1 1 2 3 4

1 1 1 1 3 3 3 3 1 31 ... ... 2 ...

4 4 4 4 4 4 4 4 2 2

x x x x x x x x x

− − −

= − − +− + +

= + − − + + +

= + + + + + − − + − + + + − + + − +

= + + + + + − + − + + + − + − +

= + − +

� � �

The series is valid for 1<x .

2005x of ( )xf is the 2006th term of ( ) 11

1 −− x + 2006th term of ( ) 1

3 −+− x + 2006th term of

( ) 21

1 −+ x

( ) ( ) ( ) 200520052005120062005120062005 100210034

120061

1xxxxx −=

−+=−+−−

Therefore coefficient of 2005x is -1002.

7 [ACJC/2005/Prelim/P1/Q2]

Solution

Using sine rule

( )( )

+−≈

++−≈

−−≈

−−−+

+−−=

+−+=

coscos6

��

π x+6

Chapter 3: Arithmetic & Geometric Progressions Solution

[RJC/2007/Prelim/P1/Q13]

Solution

Let Sn and Hn denote the amount of savings that Selina and Hebe have at the end of the nth year

respectively.

1000 200n

S n= +

( ) ( ) ( )

1000 100 100 1.5 100 1.5 ... 100 1.5

100 1.5 11000

1000 200 1.5 1

= + + + + +

−= +

= + −

where 5n >

When n n

H S> ,

( )51000 200 1.5 1 1000 200

−+ − > +

i.e. ( )5200 1.5 1 0

−− − >

From the GC, the value of k is 12.

u denote the interest that Hebe gets in the kth

Then ( )12 5 1

12100 1.5 1139.0625u

− −= = ,

so Hebe receives $1140 in interest at the end of the 12th

Then for 12n > ,

The total investment yield from the 13th year onwards therefore

cannot exceed 12

Hence, the maximum value of Hebe’s investment is

1$4790

2H u+ = (to 3 s.f.).

2 [AJC/2008/Prelim/P2/Q2]

Solution

Let T7, T3, T1 be the seventh, third and first term of an arithmetic series with first term a and

common difference d.

7 3 16 , 2 ,T a d T a d T a= + = + =

( ) ( )2

a d a d

a a d a d

a ad a ad d

+ = + +

Since 02

1Common ratio

T a ar

T a d a a

≠ ⇒ =

= = = =+ +

Since common ratio < 1, the geometric progression is convergent.

23 32 1 100

12 4 8 12

9 3 16 1 100 0

2 8 8 2

− + − − ≥

+ − − − ≥

Using GC, 22.3n ≥ ∴ Least n = 23.

[DHS/2008/Prelim/P2/Q4]

Solution

Amount saved in 1st

year = ( )1

36000 $180002

2904.1

20104.1

1000000104.1

)104.1(18000

≥−

3 3Since 3 3 and

ad a a a d

= ⇒ + = ⇒ = =

8.2904.1ln

9 years = 108 months

108 = 21(5) + 3 ∴n = 21

total amount = ( ) ( )21

10000 2 25 20 32

= $11155

4 [JJC/2008/Prelim/P2/Q1]

Solution

Let the first term of the AP be a and common difference be d.

( )1 2 9 10 11 18

2 T T ... T T T ... T+ + + = + + +

( ) ( )9 9

2 8 9 172 2

a a d a a a d+ + = + + +

[ ]2 2 8 2 26a d a d+ = +

2d = (reject 2d = − , since a is positive)

Hence, 10a =

[NYJC/2008/Prelim/P1/Q4]

Solution

11 1ln( ) ln( ) ln n

n n n n

ab b a a

− = − =

Since na is a geometric progression, thus 1n

+ = for all n+∈� . Thus 1 ln( )n n rb b+ − = for all

n+∈� . Thus

nb is an arithmetic progression with common difference ln( )r .

Since nb is an arithmetic progression

1[ln( ) ln( )]

1ln( )

Nb b b

Since ( )1

1 2 11

b a a a+

= × × ×∑ � , thus

1exp ln( )

a a a b

× × ×

∑�

[TJC/2008/Prelim/P1/Q11]

Solution

(i) S8 = T29

4(2a + 7d) = a + 28d + 98

7a = 98

a = 14

(ii) Given that u14 ≤ 196 and u15 > 196,

14 + 13d ≤ 196 and 14 + 14d > 196

Hence d ≤ 14 and d > 13

i.e. 13 < d ≤ 14.

(i) wn

wn − 1 =

v2n − 1 + v2n

v2n − 3 + v2n − 2 =

ar2n − 2

+ ar2n − 1

ar2n − 4

+ ar2n − 3

Hence the sequence {wn} is a g.p. with common ratio r2.

(ii) w1 + w3 + w5 + … = 4 + 4r

1 − r4 =

From GC, r = − 1

2 (since r ≠ −1)

Hence ∑r = 1

∞ vn =

[TPJC/2008/Prelim/P2/Q2]

Solution

( 1) 70

7 ( 1) 70

( 1) 63.......(1)

+ − =

[2 ( 1) ] 3852

[2(7) ( 1) ] 385.......(2)2

na n d

+ − =

Subt (1) into (2)

(14 63) 770

number of terms = 10

Subt into (1)

common difference = 7

Given GP. Common ratio = 1

x +− .

If the sum exists, 1

x +− < 1

3 1 x− < ≤

S∞ = 4

− −

8 [MJC/2006/Prelim/P1/Q10(a)]

Solution

nS n n= −= −= −= −

117( 1) 3( 1) 3 23 20

nS n n n n

−−−−= − − − = − + −= − − − = − + −= − − − = − + −= − − − = − + −

1(17 3 ) ( 3 23 20)

n n nT S S n n n n

−−−−= − = − − − + −= − = − − − + −= − = − − − + −= − = − − − + −

20 6n= −= −= −= −

(20 6( 1)) (20 6 ) 6n n

T T n n++++

− = − + − − = −− = − + − − = −− = − + − − = −− = − + − − = −

∴∴∴∴ the series is an A.P. with common difference 6d = −= −= −= −

9 [SRJC/2006/Prelim/P1/Q8]

Solution

Let the first term of the AP be a = 21 h, and the common difference d =

605 = h.

(i) ( ) 166121

7 =+=+= daT h

(ii) [ ]=+= 12

7S 415 hours

(iii) Let ( ) ( )[ ] 60)1(22 12

121 =−+ n

( )[ ] 60)1(12 12

1 =−+ nn

( )[ ] 120)1(1121 =−+ nn

1440)1(12 =−+ nnn

01440112 =−+ nn

=±−

=+±−

588111

576012111n 32.8 or – 43.8

Hence during the 33rd lesson, Betty will have completed a total of 60 hours.

10 [SAJC/2006/Prelim/P1/Q2]

Solution

GP 1: 2, , ,a ar ar …

( ) ( )

= − �

GP 2: 2 2 2 2 4, , ,a a r a r …

Substitute (1) into (2)

( )( )( )

29 1 1

− = +

11 [TJC/2006/Promo/Q6]

Solution

Given that 2T5 = S2, we have 2ar4 = a + ar, i.e., 2r

4 – r – 1 = 0.

From G.C., r = - 0 .648 or r = 1. (rejected as |r| < 1 for S to exist).

For Sn is within 5% of S, i.e., |Sn – S| < 0.05S ,

we have (1 ) 0.05

na r a a

−− <

− − − --------------(1)

Since 1

r− > 0 (as a > 0 and r = −0 .648), (1) simplifies to

(1 ( 0.648) ) 1 0.05n− − − <

(0.648) 0.05n <

( )lg(0.648) lg 0.05n <

lg(0.05)6.90

lg(0.648)n > = (correct to 3 significant figures)

Hence, minimum value of n is 7.

Solution

The first element in the thn row

= 1 2 ... ( 1)2 n+ + + −

n n−

The first element in the 10th

row =452 and the first element in the 21

st row =

The sum of all the elements from the 10th to 20th row = 45 46 47 2092 2 2 ... 2+ + + + = ( )45 1652 2 1−

[SAJC/2005/Prelim/P1/Q9]

Solution

Row 1 2 3 … n n+1 n+2 … n+m

No. of tiles 1 3 5 … 2n - 1 2n + 1 2n + 3 …

Total number of tiles

( )( )

terms,2,1:

−++++=

��

Extra tiles needed

( ) ( )( )[ ]

[ ]( )mnm

−++=

211222

square 1 2 3 … n

length

Area of nth square

−−

2 2 2 2

4 4100 17 0.17

1 ln 0.172 2 ln 0.8 ln 0.17 2

2 ln 0.8

− −

< ⇒ <

− < ⇒ > +

Chapter 4: Series, Sequences, Mathematical Induction

1 [DHS/Prelim/P1/2]

Prove by induction that1

(2 1)(2 3) 3(2 3)

r r n=

=+ + +

∑ . [5]

Hence state the value of 1

(2 1)(2 3)r r r

= + +∑ . [1]

[Ans: 1/6]

2 [HCI/Prelim/P1/8]

(i) Express 4

( 1) ( 2)

− + in the form

r r r+ +

− +. [2]

(ii) Hence find 2

( 1) ( 2)

r r r=

− +∑ . [3]

Give a reason why the series is convergent, and state its limit. [2]

Use your answer to part (ii) to find 2

( 1)( 3)

r r r=

+ +∑ . [2]

[Ans: (i) 1, 2, 1A B C= = − = ; (ii)1 1 1 1

6 1 2n n n− + +

1 1 1 1

12 1 2 3n n n− − + +

+ + +]

3 [NJC/2010/Prelim/P2/2(b)]

(i) By expressing ( )

2 7 11

+ in the form

( ) ( )2 ! 4 !

+ + where A and B are

real constants, show that ( ) ( )

7 11 5 5

4 ! 4! 4 !

+ + += −

+ +∑ . [3]

(ii) Use the method of mathematical induction to prove your result in (i). [5]

(iii) Hence, find ( )

∞ + +

+∑ . [3]

[Ans: (i) ( ) ( )

2 ! 4 !r r−

+ +; (iii)

4 The sequence of numbers x1, x2, x3, ... is such that x1 = −2 and xn + 1 =

xn − 1.

(i) Find the value of x5 x10 correct to 2 decimal places. [2]

(ii) As n → ∞, xn → θ . Find the exact value of θ . [3]

(ans : (i) 7.40 (ii) 1 1

− − )

5 (VJC Promo 2008) A sequence of real numbers x1 , x2 , x3 , ... satisfies the relation

xn + 1 = 4 + 15xn − 10

4xn − 4

for n ≥ 1. As n → ∞, xn → l.

(i) By solving x = 4 + 15x − 10

4x − 4 , find the value(s) of l. [3]

(ii) Prove that (xn + 1 − 4)2 − (l − 4)2 = 6 − xn

4xn − 4 . [2]

(iii) Hence show that if xn > l, then xn + 1 < l. [4]

(ans : (i) 6 )

6 (JJC/CT/09)

The rth term of a sequence is given by ur = ( )

1r r + for r = 1, 2, 3, … .

(i) Evaluate 1

∑ for n = 1, 2 and 3. By observing the pattern of values, make a

conjecture for a formula for 1

∑ in terms of n. [3]

(ii) Prove, using mathematical induction, your conjecture for 1

∑ in (i), in terms of n.

(iii) Using the formula, evaluate 2

−∑ . [2]

[Ans: (iii) 1

( 1)n +]

7 [RVHS/2010/Prelim/P1/5]

The sequence of numbers u u u1 2 3, , ,… is given by 1 2u = and 1

+= for all positive

integers.

(i) By writing down the terms 2u and 3u , make a conjecture for nu in terms of n . [2]

(ii) Prove your conjecture by mathematical induction. [4]

(iii) Write down the limit of 1n nu u − as n tends to infinity. [1]

[Ans: (iii) 1]

8 [HCI/Prelim/2010]

The sequence of numbers nu , where n = 0, 1, 2, 3, …, is such that 0u = 2− and 1

Prove by induction that, for 0,n ≥

−. [5]

9 [NYJC/2010/Prelim/P2/2]

The sequence of numbers {ur} where r = 1, 2, 3, …, is such that it satisfies the recurrence relation

+ − =+

and 1u = 1.

(i) By dividing the above recurrence relation by r and using the method of difference, show that

( )2 22

nu n n= − + for n = 1, 2, 3, …. [5]

(ii) Prove the result in (i) using mathematical induction. [4]

(iii) Find the exact value of 3

n as n → ∞ . [1]

[Ans: (iii) ½]

10 [CJC/Prelim/P1/4]

The diagram below shows the graph of xxy −+= 2ln2 . The two roots of the equation

2ln2 =− xx are denoted by α and β , where βα < .

α βx

(i) Find the values of α and β , correct to 3 decimal places. [2]

A sequence of real numbers x1, x2, x3, … satisfies the recurrence relation

2)ln( 2

1 ++++====++++ nn xx for 1≥n .

(ii) Prove that if the sequence converges, it must converge to either α or β . [2]

(iii) By considering nn xx −+1 and the graph above, prove that

βα <<>+ nnn xxx if1 ,

βα ><<+ nnnn xxxx or if1 . [2]

(iv) Hence deduce the value that xn converges to for 21 =x , giving your answer correct to 3 decimal

places. [2]

[Ans: (i) 357.5,464.0 == βα ; (iv) 357.5=→ βnx ]

11 [DHS/2010/Prelim/P1/3]

(i)Show that 2 24 5 ( )n n n a b− + = − + , where a and b are constants to be determined. [1]

(ii) Show that ( )2 2 2 2

1 4 5 1 ( 1) 1 5 2.N

n n n N N=

+ − − + = + + − + − −∑ [3]

(iii) Without the use of a graphic calculator, deduce that the sum in (ii) is strictly less than 2 1.N +

[Ans: (i) 2( 2) 1n − + ]

Chapter 5, 7: Graphing Techniques & Transformations Solutions

bx= + +

x bx= −

= − ±

Since a and b are positive constants, ab > 0 and thus there are 2 distinct real solutions for x.

Hence, C has exactly two stationary points. (shown)

(ii) Given that C passes through the point (0,3) ,

b= −

'f ( )y x=

−O ( )

__2 2b−

[MJC/2010/Prelim/P1/Q8]

Let the curve be 3 2y ax bx cx d= + + + .

Since the points ( )2,75− , ( )0,3 and ( )1,12 lie on the curve.

Using ( )0,3 , 3d =

Using ( )2,75− ,

( ) ( ) ( )3 2

2 2 2 75a b c d− + − + − + =

8 4 2 72 (1)a b c− + − = ��

Using ( )1,12 ,

12a b c d+ + + =

9 (2)a b c+ + = ��

Since ( )1,12 is a maximum point, d

( )f 1 2y x= −

1' , 3

3' , 75

( )' 0,12C

( )" 1, 3C

( )" 0, 0B

( )" 2, 72A −

( )"' 1, 3C −

( )"' 2, 72A − −

23 2 0ax bx c+ + =

3 2 0 (3)a b c+ + = ��

Using GC to solve (1),(2) and (3),

8, 7, 10a b c= − = =

Thus the equation of the curve is 3 28 7 10 3y x x x= − + + + .

[DHS/2010/Prelim/P1/10b]

[TJC/2010/Prelim/P2/Q4]

1x = −

( )fy x′=

f '( )y x=

A’(-2, 0.5)

(b) Note:

2 2 5 23

x xy x

− − += = − − +

− −

x= − − A→

2 24 4y x x

x x= − − − = − − +

B→ ( )

( )2 2

1 4 31 1

y x xx x

= − − − + = − − +− −

A: a reflection about y-axis

B: a translation of 1 unit in the positive direction of x-axis.

2 2 5 23

x xy x

− − += = − − +

− − ,

Asymptotes are: x = 1 and y = −x−3

Adding graph of ( )

−+ + =

to part (ii), there are 4 roots for the

equation ( )

1 2 54 1

− − − ++ + =

5(i) [RVHS/2010/Prelim/P1/Q10]

1x = −

1 2−

( )2 fy x=

− − +=

− ( )

−+ + =

y x= − −

Method 1:

Using long division:

( )( )2 2 2

c b aax bx cy ax b a

− −+ += = + − +

1a∴ = , 2b = −

(ii) 2 2

x x cy

− +=

At ( )0, 4.5− : 4.52

c− =

9c∴ = −

(0,-4.5)

( )2 2 24x y k− + = is a circle centre (4, 0), radius k

2 24 5 4

145 145 or

⇒ < − >

(4, k)

[HCI/2010/Prelim/P1/Q12]

2 2 2( 2) (1 )x a y− = −

−⇒ + = reps an ellipse.

Method 1:

Sequence of transformations:

Scale // to x-axis by factor a.

Translate in the positive x-direction by 2 units.

Method 2:

Sequence of transformations:

Translate in the positive x-direction by 2

a units.

Scale // to x-axis by factor a.

(bi) 2

− −

− − −

(bii) Sub

x into

2 2 2( 2) (1 )x a y− = − :

2 2 4( 2) 1

− − = − +

( ) ( ) ( )22 22 2 2 21 ( 2) 1 4⇒ + − = + − −x x a x a x --- (*)

(shown)

Hence the x-coordinate of the points of intersection of 1C and 2C satisfy equation (*).

(iii) From (ii), number of intersection points between 1C and 2C gives the number of real roots of the

1= −x

1= −y x

4− 2

−+ =

2 − a 2 a+ 2−

equation (*). From the graphs, there are 2 points of intersection between 1C and 2C . Hence 2

real roots.

7(i) [NYJC/2010/Prelim/P1/Q6]

y = f(x)

8a [AJC/2010/Prelim/P1/Q10]

1ln ln 3 2ln(3 )

6 9y x x

− = = − = − −

The graph can be obtained from lny x= by

1. translate of 3 unit in the negative x direction ln( 3)y x= +

2. reflect in the y-axis ln( 3)y x= − +

3. scale by factor 2, parallel to the y-axis

4. reflect in the x-axis

Chapter 6: Functions Solution

(1 ) 2 ( 1)f ( )

x x xx

+ − −′ =

− + +=

For stationary point, f ( ) 0 1 2,1 2x x′ = ⇒ = − +

Coordinates of stationary points 1 2

1 2,2 2

− − −

1 21 2,

+ − +

Range of f is 1 2 1 2

,2 2 2 2

− − − +

(iii) 1

fg( )1

Domain of fg is { }: 0x x∈ ≥�

Range of g is [0, )∞

Thus, range of fg is 1 2

− − +

2 1e , 0xy x

+= < .

−−=

Range of f = ( )e,∞

( )1f ln 1, ex x x−∴ = − − >

1g f fg( )

ln e 1 1x

= − − = − −

Alternative:

fg( ) e e , 1

since domain of f , 0

x xx x

+ = = >

= − −

= −∞

fR is ( )∞− ,1

(ii) gD is ( ) }2{\,1 ∞

since gf DR ⊄

as ( )∞− ,1 ⊄ ( ) }2{\,1 ∞

gf∴ does not exist

(iii) gf DR ⊂

f(x) > 1

122 >+ xx

12 −=∴k

222 ≠+ xx

0222 ≠−+ xx

Solving 31 ±−≠x i.e. 1 3a = − +

4 (i) Range of f is (0, 1]

(ii) gf(x) = g(cos x) =1

cos x, 0

2x≤ <

(iii) Area = 3 3

cos dx x dx

=∫ ∫

ln sec tanx xπ

1ln 3 ln(1 0)

= + − +

ln( 3 2)= +

5 (i) An inverse function exists if and only if the function is a one-one function.

)24( xy +=

124 22 −=⇒+= yxxy

y = cos x

y y = sec x

62),4(

1: 21 ≤≤−→∴ −

...2!2

41 2 +−+ x

Expansion is valid for

2<⇒ x

x1（　2　x3)x(f −=+⇒−=

2x =⇒

( )2 2

2 32 2 2

x x kx

k k kx x

= − +

= − + − +

= − + −

The minimum point is at

k k −

f-1(x)

f(x) y = x y

= − ∞

From the graph of f , it is not one-one and hence 1f −

doesn’t exist. For example ( ) ( )f 0 f 3k= =

[ 1, )

= − ∞

When domain of f is restricted to the set of negative real numbers, 1f −

exist. The minimum point

at 0 0x k≥ ⇒ ≥ . Therefore 4k = .

2f : 4 3, x x x x−− + ∈ �

( )3,fR = ∞

Let ( ) 2f 4 3x x x y= − + =

2 1 since 0

− + =

− − =

− = +

− = ± +

= ± +

= − + <

( )-1 1 32 ,f x xx += − >

( )fy x=

k k −

The vertical asymptote is 1x = − .

The horizontal asymptote is 2y = .

When ( )g 0x = ,

+ = −

1fg ln 2 1

1 4fg ln 3 ,

= + < −

( )gy x=

( ), ln 3fgR = −∞

( ) ( )

h 3 14

3 9h 14

= − + +

= − − +

= − − − +

= − −

Maximum point is 3 65

Greatest 3

2b = .

( )fy x=

( )1,2gR = −

= −∞

Let ( )h x y=

3, since

− − =

− = −

− = ±

( )-1 3 65h

4x y x− − <=

( )2 1 1

f 21 1

+= = −

The horizontal asymptote is 2y = .

( )f 0 1=

( )hy x=

( ) ( )1 ln 2g x x= − + +

( )g 0 1 ln 2= − +

When ( )g 0x = ,

1 ln 2 0

ln 2 1

− + + =

ii) ( )1, 2fR =

Let ( )2 1

( )fy x=

1 ln 2− +

2e −

( )fy x=

x xy y

− = −

( )-1 1f , 1 2

−= < <

( ) ( )

( )( ) ( )

h 1 ln 2

− + +

− + + =

Alternative

Let ( ) ( )g 1 ln 2x x y= − + + =

1 ln 2

ln 2 1

− + + =

( )-1 1g 2, 1 ln 2xx e x

+= − > − +

( )( ) ( )

( ) ( )( )

iv) ( ) ( )1 ln 2, 0,g fR D= − + ∞ ⊄ = ∞ therefore fg doesn’t exist.

Chapter 8: Inequalities Answers

1 22 3x x+ <

23 2 3x x− < + <

23 2x x− < + and 22 3x x+ <

22 3 0x x+ + > and 22 3 0x x+ − <

+ + > and ( )( )2 3 1 0x x+ − <

x∈� and 3

x− < <

2x∴− < <

2 2 22 3 2 3x x x

xe e e e

+ < ⇒ + <

∴replace x with

1 0 12

e e− < < ⇒ < <

0x∴ <

From graph, to solve for values of x at the points of intersection:

5 -1 0

542 −−= xxy

( )542 −−−= xxy

5542 =−− xx or ( ) 5542 =−−− xx

01042 =−− xx or 042 =− xx

142 ±=x or

0=x or 4=x

142 −≤∴ x or 142 +≥x or 40 ≤≤ x

Otherwise method :

.|1||5|

( )( )515 −+≤ xx

( ) ( ) 222551 ≥−+ xx

( )( )[ ] ( )( )[ ] 0551551 ≥−−++−+ xxxx

Consider

( )( ) 0551 =+−+ xx or ( )( ) 0551 =−−+ xx

142 ±=x or

0=x or 4=x

142 −≤∴ x or 142 +≥∴ x or 40 ≤≤ x

+− ≤

+ + + - -

142 − 142 + 0 4

7 4 30

+ − + −≤

7 4 30

+ − − +≤

− − −

− +≥

− −

Since 2 22 3 ( 1) 2 0 for ,x x x x− + = − + > ∈�

2 3 4 0x x− − >

( ) ( )4 1 0x x− + >

1 or 4x x< − >

22 5 2exx

x+ − ≤

25 2e 2x

⇒ − ≤ −

x⇒ − ≤ −

From the diagram,

or 0 or 0x x xα β≤ ≤ < >

Using G.C.,

1− 4

− + +

α β x

exy x= −

x= −

2.18 or 0.920 0 or 0x x x≤ − − ≤ < > (3s.f.)

Therefore solution set is ( , 2.18] [ 0.920,0) (0, )−∞ − ∪ − ∪ ∞ .

5 2 21 34 2 1 4( )

4 4x x x+ + = + +

Since x is real, 21

( ) 04

x + ≥

Thus 24 2 1x x+ + is always positive.

4 2 10

Since 4 2 1 0,

2 2 11 1 1

++ > ⇒ + + >

i.e. 1 1

Replace x by 1

x, we have

x> −

2 or 0x x∴ < − >

12 294

−⇒

(12 29) 4(5 )0

+ − −≤

⇒ 2( 5 )( 5 )(2 3) 0x x x− + + ≤

⇒ x= -3/2, 5x ≤ − or 5x ≥ x= -3/2, 5x < − or 5x > ( 5x ≠ ± )

(12 29 )4

Replace x by 1/x, ⇒ x= -2/3, 1

< − or 1

⇒ x= -2/3, 1

x− < < or 1

⇒ x= -2/3, 1 1

5 5x− < < as x = 0 is a solution to the inequality

7 ln 2 0

− ≥

where 1a > .

1 1 8 ) 1 1 8 )2

1 1 8 1 1 80

a ax x

a ax or x

− ≥

− −≥

+ + − +− −

− + + +≤ < ≥

Chapter 9: System of Linear Equations

1. [CJC/2010/Prelim/P1/Q1]

Let x be the price of high heels, y be the price of facial mask and z be the price of

handbag in dollars.

We have,

70.18375153

30.1158872

20.12983105

Solving, x = 29.9, y = 99.9, z = 49.9

Total cost of gift 20.629$90.49290.99590.29 =×+×+=

2. [NJC/2010/Prelim/P1/Q1]

Let the unit digit be z.

Let the tenth digit be y.

Let the hundredth digit be x.

15 (1)x y z+ + = −

(100 10 ) (100 10 ) 594

99 99 594 (2)

x y z z y x

+ + − + + =

⇒ − = −

4 5 (3)

⇒ − + + = −

Using GC to solve the equations simultaneously,

8, 5, 2x y z= = = .

Thus the number is 852.

3. [IJC/2010/Prelim/P2/Q1]

Let V 2ms− , M 2ms− , S 2ms− be the gravitational pull on each planet Venus, Mars

and Saturn respectively.

630S + 630V + 630M = 13860 ---------- (1)

900S − 600V = 2880 ---------- (2)

600S + 630V + 900M = 3(4870) = 14610 ---------- (3)

630 630 630 13860

900 600 0 2880

600 630 900 14610

From GC,

M = 3.8, V = 9, S = 9.2

Hence the weight of Probe D on Saturn is ( )500 9.2 N× = 4600 N

4. [PJC/2010Prelim/P1/Q1]

Let the number of diagonals be 2d An Bn C= + +

No of diagonals in a triangle = 0

No of diagonals in a quadrilateral = 2

No of diagonals in a pentagon = 5

Therefore,

16 4 2

25 5 5

Solving using GC, 1 3

, , 02 2

A B C= = − =

Triangle

(3 sides)

Quadrilateral

(4 sides)

Pentagon

(5 sides)

Thus, 21 3

2 2d n n= −

For a polygon of 200 sides,

The number of diagonals = 21 3(200) (200) 19700

2 2− =

5. [RI/2010/Prelim/P1/Q3]

Let x, y, z be the exchange rate quoted for Sterling Pound, Euro Dollar and Swiss

Franc, respectively (i.e. 1 Sterling Pound = x Singapore Dollars, 1 Euro Dollar = y

Singapore Dollars and 1 Swiss Franc = z Singapore Dollars).

36 77 42 269.9x y z+ + =

55 18 63 233.45x y z+ + =

40 31 26 175.5x y z+ + =

Using the GC, x = 2.15, y = 1.78, z = 1.32

59 24 313kx y z+ + =

313 59(1.78) 24(1.32)82

− −∴ = =

6. [RVHS/2010/Prelim/P1/Q1]

Let the no. of small, medium and large bottles manufactured be denoted by s, m and l

respectively.

So, 150 335 475 280400s m l+ + = ,

200 450 600 370700s m l+ + = and

3.8 2(2.5 4.2 )m s l= +

Using GC, solving the augmented matrix:

150 335 475 280400

200 450 600 370700

5 3.8 8.4 0

Ans: s = 166, m = 550 and l = 150

Assumption:

The plastic bottles are of negligible thickness. OR

The plastic bottles are of the same thickness.

7. [SAJC/2010/Prelim/P1/Q1]

Let 3 2

nu an bn cn d= + + +

1 63u = : ( ) ( )

231 1 (1) 63

63 (1)

a b c d

+ + + =

+ + + = −

2 116u = : ( ) ( )

232 2 (2) 116

8 4 2 116 (2)

a b c d

+ + + =

+ + + = −

3 171u = : ( ) ( )

233 3 (3) 171

27 9 3 171 (3)

a b c d

+ + + =

+ + + = −

4 234u = : ( ) ( )

234 4 (4) 234

64 16 4 234 (4)

a b c d

+ + + =

+ + + = −

Using the GC APPL to solve (1), (2), (3), (4) simultaneously, we get:

1, 5, 61, 6a b c d= = − = =

3 25 61 6n

u n n n= − + +

Hence 50u = ( ) ( ) ( )3 2

50 5 50 61 50 6 115556− + + =

8. [TJC/2010/Prelim/P1/Q6]

(a) ( )( )

( )( )( )

f 5 84 61 1 0 0

g 2 2 1

x xx x

x x x x x

++ +≤ ⇒ − ≤ ⇒ ≤

− − − +

( )( )( ) [ ]5 8 2 1 0 1,2x x x x⇒ + − + ≤ ≠ −

8 or 1 2

5x x⇒ ≤ − − < <

( )( )

f 81 or 1 2

− −

−≤ ⇒ ≤ − − < <

0 2xe−⇒ < <

x⇒ >

(b) ( ) ( )h fy x x= +

( ) ( )3 2 2 3 24 6 1 4 6y ax bx c x x ax b x x c⇒ = + + + + + = + + + + +

Since it passes through the points ( ) ( )1, 18 , 1, 14− − − and ( )3,30

( ) ( )( ) ( ) ( )3 2

1 1 1 4 1 6 18a b c− + + − + − + + = −

21 (1)a b c⇒ − + + = − − − −

( ) ( )( ) ( ) ( )3 2

1 1 1 4 1 6 14a b c+ + + + + = −

25 (2)a b c⇒ + + = − − − −

( ) ( )( ) ( ) ( )3 2

3 1 3 4 3 6 30a b c+ + + + + =

27 9 3 (3)a b c⇒ + + = − − − [M2 – -1 for any one wrong equation]

Solving (1),(2) and (3) gives 2, 10, 33a b c= − = = −

( ) ( )3 2h 2 10 33 h 101 1958625x x x= − + − ⇒ = −

Chapter 12: Applications of Differentiation Solution

1 [TPJC Promo 08]

2 [NJC promo 08]

2V x y=

Surface Area: 24 48xy x+ =

−∴ =

(48 )4

V x x⇒∴ = −

Using GC, max volume occurs when 4x = and hence 2y = .

Dimension of box is 4 cm x 4 cm x 2 cm.

3 [JJC promos 08]

= − ⇒

22( 10 )2

= + −

22 400x x= + −

xx= − [shown]

2 221 1

400 ( 2 ) 4002 2 2

dA xx x x

− = − − + −

22 400

x= − + −

For stationary value of A, 0dA

22 400

x⇒ = −

2 2 400x x⇒ = −

10 2 ( 0)x x⇒ = >∵

2(10 2) 400 200 30 2p∴ = + − = (cm)

4 [RJC promos 08]

By Pythagoras’ Theorem,

2 2 2 2

( ) ( ) ( ) ( )

3 (5 ) 4 (Shown)

x AP PB

OA OP NP NB

= + + +

= + + − +

2 2 2 2 2 2 2 2

df( ) 0.5(2 ) 0.5(2)(5 ) 5

d 3 (5 ) 4 3 (5 ) 4

(5 ) 16 ( 5) 9

9 (5 ) 16

x x x x x

x x x x

− − −= + = +

+ − + + − +

− + + − +=

+ − +

Let df( )

x= , and hence we are solving

2 2 2 2

(5 ) 16 ( 5) 9

((5 ) 16)) (5 ) ( 9)

x x x x

− + = − − +

⇒ − + = − +

On simplification, we have

16 9(5 )

7 90 225 0

(7 15)( 15) 0

+ − =

− + =

When df ( )

x= x = 15/7 (since x > 0) satisfies the quadratic equation given.

Using the 1st

derivative test,

x −ve 0 +ve

Or alternatively using the second derivative test,

2 3 32 2 2 22 2 2 22 2

2 2 2 2 2

2 2 2 22 2

1 1( )(2 ) 2( 5)

d f( ) 1 12 2

d 3 (5 ) 4( 3 ) ((5 ) 4 )

9 ((5 ) 4 ) ( 5)

( 3 ) ((5 ) 4 )

x x xx

x x xx x

x x x x

− −−

= + + ++ − +

+ − +

+ − − + − −= +

+ − +

which is greater than zero when x = 15/7.

Hence f(x) is minimum when x = 15/7, and on substituting x gives f(x) = 74 or 8.60 (to 3.s.f)

5 [SAJC 2006 Prelims/P1/Q11]

( ) ( )( )

21 1 2

dy tt t

= − + = −+

( ) ( )2 2

dt tdy dy

d t tt x× = − × = −=

Equation of tangent at 2

( )( )

( ) ( )

22 2 2

1 3 +1 shown

ty x t

t y t xt t t

t y xt t t

− = − − − + +

+ − + = − + −

+ + = −

(ii) When 3t = , equation of tangent

( ) ( )2 223 1 3 3 3 3 1

100 3 25

+ + = − +

When the tangent meets the curve at 2

( )( ) ( )

100 3 25

1100 3 2 1 25

100 3 2 1 1 25 1

100 3 2 2 1 25 25

100 6 3 6 3 25 25

6 28 6 72 0

4(From GC) 3,

q q q q

+ − =

+ − + = +

+ + − − = +

+ − + − = +

− + + =

Therefore the point of intersection Q is

4 1 8 92 1, ,

3 3 2541

− − = −

6 [CJC 2006 Prelims/P1/Q11]

c tandx

x a θ

θ θθ

3 secdy

y a θ

23 sec 3 3

sin5 sec tan 5sin5cos

θθ θ θ θθθ

× = = ==

ii) When the normal’s gradient = 1−

The point is 25 9

When the tangent is parallel to the y -axis,

sin 05 θ

The point is ( ) ( )sec0,3 tan 05 5 ,0aa a=

2 25 43− =

7 [AJC 2007 Prelims P1Q13b]

By similar triangles

and 63

( ) ( ) ( )

1 16 6

6 (shown)27

= −+

( ) ( )2 2

9 18020

dt d t hV hπ π× = × =

When 3h = , ( )3

180 20 cms

dt ππ=

8 [HCI 2007 Prelims P1Q2]

t dV xd π π= × = × =

When 3x = , ( )

12 4m s

dt ππ==

∫ ∫

When 5x = , 36

125t C

When 10x = , 1000

Time taken 1000 125

s36 36 36

875C C

π π π = + − + =

Chapter 13: Maclaurin’s Series Solutions

[ACJC/2008/Promo]

Solution

3 22 0

d y dy d y

dx dx dx+ =

24 2 3

4 2 32 2 0

d y d y dy d y

dx dx dx dx+ + =

When 0x = , 0y = 0dy

dx= −

Using Maclaurin’s series,

( ) ( )2 42 41 2 1 1

... ...2! 4! 2 12

x xy x x

− −= + + = − − +

2 4 2 4

1 1ln cos ...

4 2 4 12 4

1 ln 2 ln 2

2 32 3072 16 1536

π π π

π π π π

= − − +

− ≈ − − ∴ ≈ +

( ) 2 41 1ln cos ... 2

2 12y x x x= = − − +

Differentiate y wrt x,

1tan ...

8 tan 2 2 ...

− = − − +

∴ = + +

[MI/2008/Promo]

Solution

21y x x= + +

dx x= +

x x xdx

+ = + +

+ = (shown).

dy x x

y d x =⋅+ ⋅+ +

−)2(1

2(1 ) 1

d y dy dyx x x

dx dx dx+ + = +

ydx =++

22 )1( (shown).

( )3 2 2

3 2 21 (2 ) .1

d y d y d y dy dyx x x

dx dx dx dx dx+ + + + =

( )3 2

3 21 3 . 0

d y d yx x

dx dx+ ⋅ + =

( )4 3 3 2

4 3 3 21 (2 ) 3 .3

d y d y d y d y dyx x x

dx dx dx dx dx+ + + ⋅ + =

( ) 03.512

42 =++⋅+

When ,0=x ,1=y ,1=dx

dx= − .

Maclaurin’s Series: ....8

11 42 +−++= xxxy

1=x is suitable as the terms 0

for all n ≥ 4.

3 [RJC/2008/Promo]

Solution

Let1e x

y− +=

ln 1y x= − + --- (1)

Differentiate (1) with respect to x:

1 d 1 d

2 1d d2 1

y yx y

y x xx

−= ⇒ + = −

+ --- (2)

d 2 d d2 1

d d d2 1

d d d2 1 1

d d d4 1 2 2 1

d d4 1 2

y y yx

x x xx

y y yx x

y yx y

+ + = −+

+ + = − +

d d4 1 2 0

y yx y

x x+ + − = (shown) --- (3)

( )3 2 2

d d d d4 1 4 2 0

d d d d

y y y yx

x x x x+ + + − =

( )3 2

d d d4 1 6 0

y y yx

x x x+ + − =

At 0x = ,

d 1 d 1e , e , e

d 2 d 2

− − −= = − = and

−= −

By Maclaurin’s series,

1 1 2 3

1 7e e

1 2 8e e

2 2! 3!y x x x

− −

= + − + + +

1 2 31 1 7

e 12 4 48

x x x−

≈ − + −

( ) ( )

2 1 2 1

2 2 1 1 7 = 1 2 ... 1 ...

2! 3! 2 4 48

1 = (16 24 20 11 ...)

x x x xe e e

x xe x x x x

x x xe

− + − +

+ + + + − + − +

+ + + +

[TJC/2008/Promo]

Solution

4 sin 2y x= +

4 sin 2y x⇒ = +

cos 2dy

22sin 2

d y dyy x

dx dx⇒ + = −

2 3 2 2 4 cos 2

dy d y d y dy d yy x

dx dx dx dx dx⇒ + + = −

3 23 4 cos 2

d y dy d yy x

dx dx dx⇒ + = −

When x = 0, y = 2, 1

dx= − ,

dx= −

Maclaurin’s series of y:

1 1 612 ......

2 8 2! 32 3!

2 16 192

x xy x

= + + − + − +

≈ + − −

When x = 0.5, by part (i) result, y ≈ 2.19466 (to 5 d.p.)

This gives an estimated error = 2.20033 2.19466

100% 0.258%2.20033

−× =

Solution

1siny x

−= ⇒

( ) ( )

3 2 22

diff. w.r.t ,

1 3 0 ( )

d y dy xx

dxdx x

d y dyx x

d y d y d y dyx x x

dxdx dx dx

d y d y dyx x Shown

dxdx dx

−− + =

− − =

− + − − + =

− − − =

4 3 22

5 4 32

diff. w.r.t ,

1 5 4 0

diff. w.r.t ,

1 7 9 0

d y d y d yx x

dx dx dx

d y d y d yx x

dx dx dx

− − − =

(0) 0, '(0) 1, ''(0) 0, '''(0) 1, ''''(0) 0, '''''(0) 9f f f f f f= = = = = =

...6 40

x xy x= + + +

3 1 ...

d x xx

= + + +

Using 1

2x = ,

1 2 21 ...

= + + +

1 2 21 ...

= + + +

≈ 147

⇒ 256

[CJC/2008/Prelim/P2/Q1]

Solution

1 sin 1x x+ ≈ +

Since x is small.

By standard Maclaurin’s Series expansion,

1 2 21 1 1

2 2!x x x x

+ = + ≈ + +

2 8x x≈ + − (Shown)

Given 3x − e

x ln a = 0

⇒ 3x = e

x ln a

⇒ 3x = a

⇒ 3 = a

Given that ex ≈ 1 + x +

3x = e

x ln a

⇒ 3x ≈ 1 + xln3 + (xln3)

(xln3)3

( ) (0) 1y f x f= ⇒ =

1 sin 12 1 sin '(0)

dy xy x f

+= + ⇒ = =

By implicit differentiation, 2 2

1cos 4 1 4

22 4 cos "(0) 0

d y dy dxy y x f

dx dx y

− −

+ = ⇒ = = =

By implicit differentiation, 3 2

33 2 2

sin 4 121

12 4 sin '"(0)2 4

dy dy d yx y

d y dy d y dy dx dx dxy x f

dx dx dx dx yy

− − − ⋅

+ ⋅ + = − ⇒ = = −

By Maclaurin’s Expansion,

"(0) '"(0)(0) '(0) .....

f fy f xf x x

= + + + +

≈ + −

Solution

1 sin x

y e−

2 2d(1 )

x− =

d d d d2 (1 ) 2 2

d d d d

y y y yx x y

x x x x− + − =

d d (1 )

y yx x y

x x∴ − − =

Diff wrt x,

d d d d d2 (1 )

d d d d d

y y y y yx x x

x x x x x− + − − − =

d d d3 (1 ) 2

y y yx x

x x x− + − =

When x = 0,

y =1, d

sin.1 ..

x x xe x

= + + + +

1sin 2 3 21(1 ...)(1 )

cos 2 3 2

xe x x x

−= + + + + −

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

x x xx= + + + + + +

2 351 ...

6x x x= + + + +

Chapter 14: Integration Techniques Solutions

1 [AJC 2007 Common Test]

1 8 1 1

8 2 11 4

1 1.2 1 4 sin (2 )

1 11 4 sin (2 )

xdx dx

−= − −

−−

= − − − +

∫ ∫

2 [AJC 2007 Common Test]

cos 2sin cos

10, , ,

cos cos

1 sin sin

cos2sin cos

x dx d

θ θ θ θ

π πθ θ

θθ θ θ

= ⇒ = −

= = = =

= =−

= −−

∫ ∫

(1 cos 2 )

2 4 2 4 2

π π π

= − + = −

3 (AJ 06Prelim P2/Qn 3)

(i) ( ) ( )1

2 2 221

1 1 2 12

dx x x x x x

− − = − − + −

( ) ( )

11 2 1

x x x x

− = − − + −

−= + −

− + −=

(ii) ( )1 1 2 2

14 sin sin 2 2

1x xdx x x x dx

− −⌠⌡

= − ×−∫

1 2 2 1

1 2 1sin 2

sin 2 1 sin

xx x dx dx

x x x x x C

− −

⌠ ⌠

⌡⌡

−= + −

− −

= + − − +

4 (VJ 06FE Qn 6)

( ) ( )sin 3cos 3sin 2cos 3cos 2sinx x A x x B x x+ = + + −

sin 3cos 3 sin 2 cos 3 cos 2 sinx x A x A x B x B x+ = + + −

Comparing the coefficients for xsin : 123 =− BA (1)

Comparing the coefficients for xcos : 232 =+ BA (2)

Solving 13

( ) ( )

( )⌡⌠

⌡⌠

−++=

⌡⌠

cos2sin3

sin2cos3

cos2sin3

sin2cos3

cos2sin3

sin2cos313

7cos2sin3

cos2sin2

cos3sin

Let 3sin 2cosu x x= +

3cos 2sindu

+−+=

⌡⌠

sin2cos3ln13

sin2cos3

1sin2cos3

cos2sin3

sin2cos3

5 (RJ 06Prelim P1/Qn 14)

++−=

Comparing coefficients

1=−= BA

( )( )

( ) Cxxx

+−+−−−=

−−+−−−=

+−+−−+

−−

−−+

−−

+−−=

−−

++−−=

−−

3sin386

6 (NJ 06FE Qn 8)

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )( )xxx

xxxdxx

dxxxxxxdxx

xxxxxxdxx

dxxxxdxx

xxxxdxx

lnsinlncos2

lnsinlncoslncos 2

lncos lnsinlncoslncos

1lncoslnsinlncoslncos

lnsinlncoslncos

1lnsinlncoslncos

−−=

∫∫∫

∫∫

( ) ( ) ( )( )

0sin0cos2

lnsinlncos2

πππ

( ) ( ) ( )

( ) ( )( )2223

1432811123

811123

++++=+++

xxxAxxx

Comparing coefficient of 2

x , 2=A

( ) ( )

3 12 11 8

x x xdx

⌠⌡

( )( )

1 3 2 12 4

2 2 21

1 3 12 ln 2 4 tan

1 2 2 2

2 3 4ln 2 tan

1 2 2 2

xdx dx dx

x xx C

⌠⌡

⌠ ⌠ ⌠

⌡ ⌡⌡

= + ++ ++

+ = + + + + −

= − + + + ++

7 (SAJC 07 Prelim P1/Qn 14a)

∫ −

= θθ

da ∫

2cos33

= θθ

da ∫

12cos9

So a = 2.

Let θsin3=ax .

6 611 1

ydx dy

= ⇒ − =

− −

⇒ − =−

⇒ − =

∫ ∫

⇒ − − =

∴k = 2

Chapter 15: Areas and Volumes Solutions

1a [AJC 2007 Common Test]

Area = ln(1+x )

2ln(1 ) .

2(1 ) 2ln 2

2ln 2 2

ln 2 2 2 tan

ln 2 2 2 ln 2 24 2

xx x x dx

= + − +

+ −= −

= − −+

= − −

= − − = − +

Volume = (1) ln 2 1

ln 2 ln 2 1

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

π π π

− −

= − −

= − − − = − − = −

2 (SRJC 06FE Qn 10b)

When exy == ,1

(i) Area of R

( ) [ ]( ) ( )[ ]

units 42ln2

22ln2ln2

−−−−−=

−−−=

−−=

(ii) ln yy x x e= ⇒ =

When 2ln,2 == yx

Volume

( )( )

[ ] ( )

units 2ln462

2ln1442

2ln142

−−−=

−−

−−=

3 θcos4=x , θsin5=y

Using the GC, we can see that the curve is an ellipse.

sin4−=d

dx, θ

θcos5=

When 3

πθ = , 2

3cos4 ==

3cos5 ==

5−=−=

Equation of tangent:

( ) 32

+−−=

−=−

When 0=y ,

+−−=

The tangent cuts the x-axis at 8=x .

When 0=y ,

Area under curve

1 5 38 2

15 35sin 4sin

15 320 sin

15 3 1 cos 220

15 3 sin 210

15 3 1 210 0 0 sin

2 3 2 3

15 3 310

10 3 uni3

θ θ θ

= − −

−= +

= + −

= + − − −

= + −

Volume

1 5 38 6

75 25sin 4sin

75 100 sin sin

75 100 sin 1 cos

75 100 sin sin cos

cos75 100 cos

1 175 100 1 cos co

π π θ θ θ

π π θ θ θ θ

θπ π θ

ππ π

= − −

= + −

= + − +

= + − + − − +

2 1 175 100

3 2 24

575 100

325 units

= + − − − +

− = +

4 (NJ 06FE Qn 10)

a) Using the GC

dx+= 1

units 8

From GC

When 1=y

−=−

Volume⌡

−−=

Let ( )22 1 cosx θ= +

4sin cosdx

dθ θ

θ= −

When 2=x ,

cos122

When 3=x ,

cos123

( )( )

units 4

12cos4

cossin4sin

cossin4cos1

cossin4cos124

2cos12

πππ

ππππ

θθππ

θθθθ

θππ

θθθθ

θππ

θθθθ

θππ

⌡⌠ +

⌡⌠ −−=

⌠−

−−=

⌠−

−+−=

−−=

5 (CJ 06Prelim P1/Qn 14)

( ) ( )

2Area of

2 ln 1

2ln 2 1 2ln 0 1

= + − +

Volume ⌡

+==−

−∫0

(iii) When 5.0−=x , 515.0

+−+=y

−−

Volume

units 02.1

32ln42

454ln4

−−

−−=

+−−

−−=

−−

−−=

6i [ACJC 2007 Common Test]

( ) ( )( )( ) ( )( )

( ) ( )

( ) ( )( )( ) ( )( )

( ) ( ) ( )

( ) ( )

sin lncos ln cos ln

cos ln sin ln

cos ln cos ln sin ln

cos ln sin ln cos ln

2 cos ln cos ln sin ln

xx dx x x x dx

x x x dx

xx x x x x dx

x x x x x dx

x dx x x x

− = −

= + −

∴ = +

∫ ∫

∫∫ ( )

( ) ( ) ( ) cos ln

∴ =∫x x

cos ln x + sin ln x + K2 2

Turning point: (1, 1)

x-intercepts: ( )

2e ,0 = 0.208,0 and ( )

2e ,0 = 4.81,0

( ) ( )2

3 Required area cos ln cos ln

0.3871 0.0037

ex dx x dx

∫ ∫

1 secd d

1 tan 1x

+ +∫ ∫

ln sec tan

ln sec tan (Shown)

2 3y x= 2

1 122 31

xx= ⇒ =

( )( )

⇒ + − =

⇒ − + =

(ii) Area of Region R 2

31 1 1

d 32 21

3 3ln sec tan

3ln sec tan 3 3

3ln 2 3 (Ans)

x x−

= + −

(iii) Volume of solid

2y xπ

3 1 1d 0.513 (Ans)

= − =

8 [IJC 2007 Prelim]

ln 1dx

x t tdt t

= + ⇒ = +

Area required 2 2

1(2 ) 2 2 2 2 3

e e ett dt t dt t t e e

+ = = + = + = + −∫ ∫ units

(b) 2 4

−= − +

2 4 8 16d 1 1

2 2 2 (2 )

168ln 2

yy dy dy

y y y y

y y Cy

−= − + = − +

+ + + +

= − + − ++

∫ ∫ ∫

Volume required

2 2 21 1 0

2 | | 2 2

8ln | 2 | 16( 2)

161 8ln 3 8ln 2 8

14 28 ln

y y ydy dy dy

π π π

− −

− − += = +∫ ∫ ∫

= − + − + +

= − − + + +

9 [PJC/2007/Block Test]

area of PQRS < A < area of PQUT

Using GC, min point is (1, 2) 2 1.5 2.5 1.5A∴ × < < ×

3 < A < 3.75.

0.50.5

1 15( ) d ln 2 ln 2

xA x x x

= + = + = +

Thus 15

3 2ln 2 3.758

ln 216 16

⇒ < <

0.50.5

1 1 57( ) d 2

xV x x x

π= π + = + − =

10 [TPJC 2007 Common Test]

(a) Area of region required = 2

(ln )x e

When x = eu, 1 0 ; 1x u x e u= ⇒ = = ⇒ = and

Area of region required = 1 1

dxu du u e du

=∫ ∫ (shown)

Using integration by parts,

The exact area

1 112 2

2 2 2 2 2 2

u e du u e ue du

e ue e du e e e e

= −∫ ∫

= − − = − + − = −∫

(b) Using the graphic calculator, find the volume of the solid formed when the region bounded by

the curves cosy x= , the lines 3

y x= − + and 1y = is rotated 2π radians about the y-axis, giving

your answer correct to 3 significant figures.

When 3

y x= − + and cosy x= intersect, from GC, 0.9403x = and 0.5895y =

Volume required

0.5895

4 2cos

0.285 to 3sf

yx dxπ −−

= −∫

Chapter 16: Differential Equations

2sec tan

d yx x

Integrating both sides with respect to x: sec tan secdy

x x dx x Cdx

= = +∫

Integrating both sides with respect to x again: ( )

ln sec tan

y x C dx

x x Cx D

= + + +

The general solution is ( )ln sec tany x x Cx D= + + +

dy uu +=

2(ii) ∫ ∫= xdxdy 4

Cxy += 22

Cxxeu += 22

u += 2

Cxu ,2ln

+= is an arbitrary constant.

2 (iii)

3 0.4 (10 )

dx= −

− 1 x

2x = − 1

2x = y

( )( )

1 1 10.4

10 (10 )

1ln ln 10 0.4

dy dxy y

y y x C

+ =−

− − = +

∫ ∫

ln 4 1010

yAe A e

= =−

When x = −1, y = 5 ⇒ 4

4 44 4 4 4

10 (10 )

y ee y e y y

+= ⇒ = − ⇒ =

dy dzy z x

dx dx= − ⇒ = −

Substituting into 1 2

y dx y

−− = :

( ) ( )1 2

z x dx z x

dzz x x

− − − =

− −

− − + = −

Integrating w.r.t. x:

ln 1 , where is a constant

1, where

1 1 (Shown)

dz dxz

z x c c

z Ae A e

y x Ae y x Ae

− = +

+ = + ⇒ = − +

∫ ∫

When curve passes through ( )1,0 , A = 0 1y x⇒ = −

When curve passes through ( )0,2 , A = 1 1 xy x e⇒ = − +

(Notice that 1y x= − is an oblique asymptote for 1 xy x e= − + )

( )( )

2 41 4 2 4

xdy x dx y c y c

+= + ⇒ = + ⇒ = − +

− + ∫ ∫

At ( )2,7A = , 4

7 82 2

c = + =+

Thus, equation of the curve is 4

= − ++

Gradient of tangent at A

( ) ( )2

dx= = =

Gradient of normal at A 4= −

Equation of normal at A: ( )7 4 2 4 15y x y x− = − − ⇒ = − +

6 ( )1

dxkx x

dt= −

1y x= −

y x e= − +

( ) ( )

ln ln 1 ln

ln (shown)1

dx k dt dx k dtx x x x

x x C kt

= ⇒ + =− −

⇒ − − + =

∫ ∫ ∫ ∫

1When 0, ln 0 9

1When 1, ln ln 3

Ct x C

Ct x k k

= = ∴ = ⇒ =

9 9 9ln ln 3 ln ln 3 3

3 9 3 1

x x xt

∴ = ⇒ = ⇒ =

− − −

⇒ = − ⇒ =+

3 11 as (shown)

99 3 13

x t= = → → ∞+ +

22death of rate birth of rate dt

x−=−=

At x = 10, 20101020 2 .a)(a)( =⇒−=

−=∴

ii) Selling away 1800 prawns daily, D.E. becomes 8.15

−−=x

( ) ( )( )195

−−−=+−−=

−−=

iii) Separating the variables, ( )( ) 5

−−

Integrating both sides w.r.t. t: ( )( )

⌡⌠ −=

⌠−−

( ) ( )

eA,Aex

xlnxln

+−=−−−

⌡⌠ −=

⌠−

constantarbitrary an is where,5

At 0=t , x = 13: 3

−=A 1

+=⇒=−

−∴

Chapter 10 and 17: Vectors

1 (SRJC/2009/Prelim/P1/Q2) 1. (a)

0 414 2

1 1 14

1 1 11

�� OB OAOP

+ λ − + λ + λ = = = λ + λ + λ + λ + λ

114 2 21 4

4 5 51 1

��CP

− + λ − + λ− + λ

λ = λ − − = + + λ + λ + λ −

− + λ − + λ

λ = + + λ

�� 2

2k⇒ =

4 14 25 4 8 2 2 4

k or kλ − + λ

+ = = − = − = −+ λ + λ

⇒ λ =

2cos150

µ− =

24 15 1µ µ⇒ = − +

2 216 15( 1)µ µ⇒ = +

2 15µ⇒ =

15µ = − (reject positive value of µ)

Equation of line l1 : ~

= + λ

λ ∈� .

Equation of line l2 : ~

= + β −

β∈� .

When the lines intersect,

βλ +=+ 131 ------ (1)

βλ =+ 22 ------ (2)

β+−= 15 ------ (3)

From (3), 6=β

Substitute into (2), 2=λ

Check (1) : LHS = 1 + 3(2) = 7

RHS = 1 + 6 = 7 = LHS

Therefore the lines intersect.

=+⇒=

------ (1)

=++⇒=

------ (2)

Solving, 3,2 −== ba .

Equation of line l3 : ,

= sr s ∈� .

Since V lies on l3,

sOV , for some s ∈� .

7 7 2 2

5 14 6 6 3 5 14 3 5 14

5 5 1 1

VX s s

= ⇒ − − − = ⇒ − =

��

14 5 14 5 5s s s⇒ = ⇒ = ⇒ = ±

Therefore V = ( )17, 9, 10− or ( )0 ,21 ,3− .

Alternative Solution:

15 14 3

= + −

�� or ( )

5 14 314

= − −

��

= + −

�� or

= − −

��

�� or

��

Therefore V = ( )17, 9, 10− or ( )0 ,21 ,3− .

: 4 15

⋅ − =

Distance from A to 1Π

6 3 15 4 15 11 11 50

5050 50 50 50 50

⋅ − − = − = − = =

15 4 and are on the same side of

50 50A O Π> ⇒

Vector parallel to 2Π

= − = − −

Normal vector of 2Π

1 1 4 2

0 2 12 2 6

2 10 2 1

− −

= × = = −

Acute angle between 1Π & 2Π

3 11 1

31cos cos

− ⋅ − −

46.8= ° (to 3 s.f.)

1 2 3: 5 4 3 15; : 2 6 6; : 8x y z x y z x y az bΠ Π Π− + = − + + = − + + =

For line of intersection of 1 2 and Π Π (also the intersection of the 3 planes):

5 4 3 15 1 0 1 3

2 6 1 6 0 1 0.5 0

− = ⇒ = − −

Let , z t t= ∈� .

Then 2

= − = − −

⇒ = + ∈ −

Take 0 and 1µ µ= = , two points lying on 3Π are ( )3,0,0 and ( )5,1, 2− .

Sub. into 8x y az b+ + = : ( ) ( )

( ) ( )

3 8 0 0 3

5 8 1 2 3 5

+ + = ⇒ =

+ + − = ⇒ =

(MJC/2009/Prelim/P1/Q5)

(i) Since AB = −b a��

and A, B and C are collinear,

2k∴ =

OC OB BC= +

b+ b a

��

(ii) 3

(iii) 2 2 2b 2 ( 2) 2 2 3= + − + =

In this case, the length of projection is ON��

and is equal to 4 in this question.

a.bcos

4 32 3

or 603

( ) ( )

DC DADE

− + −=

− −

��

(YJC/2009/Prelim/P1/Q9)

(a) 1l passes through A(0, 3, 1) and is parallel to

2l passes through B(1, 0, 2) and is parallel to

Since 12 2mm −= . 1l and 2l are parallel.

(b)(i) Since P lies on 1l , ie. =→

23 , for some λ∈ℜ.

If APB is a right angle, 0

−•→

−•

0)1()23(2)1( =−+−−− λλλ ⇒ 3

Hence the coordinates of P is

Shortest distance between 1l and 2l

4 1 13 2 units

= − = = −

(i) Let AB→

−2 . b =

k − 2

2k + 1

(k − 2) + 2(2k + 1) − 2(−2k) = 18

9k = 18 ⇒ k = 2

∴ b =

(ii) AB→

= 2 1 + 4 + 4 = 6.

(iii) BA→

= 2 AC→

⇒ a − b = 2( c − a)

2c = 3a − b

2c = 3

−4 ⇒ c =

(iv) CD→

−3 −

−4 −

−6 = 3

∴ CD→

×××× CB→

−5 ×××× 3

−2 = 3

Therefore equation of plane is

5 5 5 9

− −

⋅ = ⋅ = −

Since 0

⋅BD , the length of projection of line BD on the plane = 5122 =+ units

��,

��

AB OB OA

= − = −

��

(i) 1l :

= + − ∈

3 1cos

35 3θ

− • − =

28.56θ = °

Acute angle between 1l and 1n is 28.56° .

Therefore, angle between 1Π and 1l is 61.4° .

(iii) Since A lies on 1Π , perpendicular distance required is 1

•��

− • − =

Perpendicular distance = 1.52 units

(iv) 2 1n AB n= ×��

= − × −

= − = − − −

2Π : ( ) ( ) ( )2 2 4 2− − = − + + − −r i j k i j k i j ki i = – 11

Cartesian Eqn: x – y – 2z = – 11

(v) Direction vector of 2 1 2l n n= ×

= − × − −

= + ∈

8. (i) Since P lies on l ,

− + → = − +

for some λ ∈R .

, 2 0.

OP l OP

→ → ⊥ ⋅ − =

− ⋅ − = +

12 9 4 5 0

λ λ λ

∴ − + + + + =

1 12 2

− → ∴ = − = −

3 5 3 20 101 1

2 2 2 38 192 2

1 11 1 16 8

− → × − = − × − = =

∴ ⋅ =

The Cartesian equation is 10 19 8 0x y z+ + = .

Remark: Better to use

instead of OP→

− ⋅ = − +

8 6 2 5 3kλ λ λ− + − + + = −

3 (7 2 ) 3k λ− + − = −

7 2 0k⇒ − =

2k⇒ =

OR Use point P (or any point on l).

− ⋅ = −

2k⇒ =

Alt. Since l lies in Π2,

⋅ − =

2k⇒ =

(iv) Let θ be the angle between planes 1Π and 2Π .

8 1cos

525 17.25θ

6.8θ∴ =

9. (a) The augmented matrix

2 2 1 4

2 3 4 1

4 3 1 2

− − = − −

The RREF of the augmented matrix

1 0 0.5 0

0 1 1 0

0 0 0 1

− = −

The final row of the RREF shows that 0 0 0 1x y z+ + = , which implies that the

system of equations is inconsistent. ∴ The equations do not have a solution.

The 3 planes represented by the 3 equations do not intersect in a point or line.

Furthermore, since neither of the planes is parallel to any other (∵ neither of the

normals is parallel to any other), the 3 planes form a triangular prism.

2 3 10

1 4 10

− × =

A vector to both normals of and is 2

π π ∴ ⊥

( ) 1 2

2 2 4 ---------(1)

2 3 4 1 ---------(2)

When 0, 2 2 4 -------(4)

and 2 3 1 --------(5)

(5) (4) gives 1 and 1.

1, 1, 0 is a point on both and .

12 is para2

− + = −

+ − =

= − = −

− = = −

∴ −

1 2llel to both and .

1 1a vector equation of is 1 + 2 , .

λ λ− ∴ = ∈

(ii) A vector normal to π is

A vector equation of π is

1 1 11

2 1 232 2 3 2

= − =

10 10.

: 2 0l r

� and 1

: 2 2r

Π ⋅ = �

Since 1l lies on 1Π ⇒

2 1 2 2 (shown)

⋅ = ⇒ = −

Also since 1

0 1 0 1

⋅ = ⇒ = −

Therefore, 1

l r λ

= + − −

3, 2 (verified)

r OAλ→

= − = =

2 1 2 1

Normal of is 2 0 0 2 0

2 1 2 1

Π = × = = − − − −

Thus 2

Π ⋅ = �

: 5 1 5 10

Π ⋅ = ⋅ = �

AB OB OA

− −

= − = − =

→ → →

Thus length of projection of 3 on AB Π→

→�

x + y + z = 2, x + z = 0 and x + 5y + z = 10, intersects at 1l .

Otherwise method

For , 2 5 10 2 lies on

Π ⋅ = ⇒ Π − −

When 3

3 1 3 3

1, 2 5 10 2 also l 2 ies on

3 1 3 3

= ⋅ = ⇒ Π − − −

lies both on 1 3 and l Π

⇒ 1 2 3, and Π Π Π all intersects on the line 1.l

Solving method

1 1 1 2

1 0 1 0

1 5 1 10

1 1 1 2 1 0 1 0

1 0 1 0 0 1 0 2

1 5 1 10 0 0 0 0

⇒ →

Thus y = 2, x + z = 0.

Let ,z xµ µ= = −

� which is line 1.l

⇒ 1 2 3, and Π Π Π all intersects on the line 1.l

OA and

=∴→

Equation of AV is

λ , ℜ∈λ or r

λ , ℜ∈λ

Equation of l is r

If AV and l intersect, let

−−

⇒ )1.....(..........2−=− µλ and 25 −=+ µλ ……….(2)

Solving eq (1) & (2) we have 0=µ and 2−=λ .

. Also, 2262 =⇒+=+ tt λµ

Acute angle between AV and line l = 1

cos 1 . 1 / 9 105 60.8

Perpendicular distance required = 079.60sin

AM = 2.62 units

12 or OA

OZ OA AZ OB= + +

= +i k

��

By ratio Theorem

8 1 8 11 1 1

0 4 41 1 1 1 1

OP OZ OC

λλ λ

λ λ λ λ λλ

= + = + = + + + + +

��

Since OP ⊥⊥⊥⊥ CZ,

CZ OZ OC

= − = −

�� i

��

8 1 71

4 4 0 56 7 16 16 0 8

λ λ λ

− = ⇒ + − + = ⇒ =

1 1 41 4

4 81 1 / 8 9

1 / 2 1

��

2 2 244 8 1 4

9OP = + + =��

7 / 9 71

32 / 9 329

-32 / 9 -32

BP OP OB

= − = =

��

The direction vector of line BP is

Cartesian Equation of line BP:

7 32 32

x y z− −= =

− or equivalent

Write down the position vector of Z in terms of i, j and k.

The point P divides CZ such that 1

λ==== . Given that OP

�� is perpendicular to CZ

��, find the

value of λ and evaluate OP��

Find a cartesian equation of the line BP.

Ans: (i) 8 4OZ = +i k��

; (ii) 1

8λ = ; 4OP =

��; (iii)

7 32 32

x y z− −= =

4 2 , for some

OP s s

= + ∈

→� .

Since OP is perpendicular to l,

4 2 2 0 3 8 4 0

s s s•

− − ⇒ + = ⇒ − + + + =

1s⇒ = −

= + − =

3 1 6 1

3 6 (1)

4 2 3 (2)

− = + − − − − − − − − − − −

⇒ + = + − − − − − − − − − − − = − − − − − − − − − − −

From (1) : 3s t= − − ------------------- (4)

Sub (4) into (2) : 4 2( 3 ) 3t a t+ − − = +

5 2t a⇒ = − − ----------- (5)

Sub (5) into (3) : 5 ( 2 )a a= − −

2 2 5 0a a⇒ + + =

2( 1) 4 0a⇒ + + =

Since, there are no real solutions for the equation 2( 1) 4 0a + + = , therefore there does not exist

real values of a such that the two lines l and m intersect. Hence, the 2 lines do not have a common

point.

1 1 10cos60 4 10

2 310 .1 10

a aa a a

= ⇒ = ⇒ = + ⇒ = ±

Since AB ⊥��

1Π , AB��

��

2 1 2 1

OB AB OA k k

− −

= + = + = + − −

��

B lies on 1Π , so

2 18 2 1 2 18 9 18 2

OB k k k

⋅ = ⇒ + ⋅ = ⇒ = ⇒ = − − −

��

∴ = −

��

Perpendicular distance from A to 1Π ( )22 22 4 4 6AB= = + + − =

��

By Ratio Theorem,

( )2 13

2 0 31

3 1 5 12

OC OBOA OC OA OB

+= ⇒ = −

− −

∴ = − = − −

��

2 3 5 0 3 3

5 1 6 , 5 1 6

3 2 5 4 2 6

− −

= − − = = − − = − − − −

��

5 3 6 2

6 6 15 3 5

5 6 12 4

− −

× = × = = − −

��

Therefore equation of plane DBC is

5 5 5 9

− −

⋅ = ⋅ = −

BD OD OB

= − =

��

Length of projection of line BD on 1Π

1 2 4 4 25 165

1 1 4 4 9

× − + +

= = = =+ +

15 [HCI 2007 Prelims]

Sub. equation of line into plane:

1 1 . 0 3

λ µ and get 3 2λ µ= − .

Sub. 3 2λ µ= − into 2Π :

1 0 3 2

(3 2 ) 1 1 3 1

0 1 0 1

µ µ µ

= − + = + −

To show the two lines are parallel:

The direction vector of 2l is scalar multiple of direction vector of 1l , thus 1l and 2l are parallel.

Note that B lies on 2l , thus

Shortest distance

2 0 2 21 1 1 30

1 2 1 036 6 6

1 0 1 4

= × = × = = − − −

��

= + −

�� for some α ∈�

0 2 2 1 0 2

BA BC α α

⋅ = ⇒ − ⋅ − + = ⇒ = −

�� .

∴ = −

��

Since , 3 2 . 5

CD BA OD OD

= − = ∴ = − −

��

Observe that O lies on 2Π . Midpoint of OA which is ( )3 12 2, , 0 then lies on 3Π .

= × = −

1 1 1 1

1 0 1 1

⋅ − = ⋅ − ⇒ ⋅ − =

r r (shown)

Let F be foot of perpendicular from A to 3Π and pick a point E (1, 0, 0) on 3Π .

1 1 11 1 1

1 1 133 3

= ⋅ − − = − −

��

OF OA AF

∴ = + = −

��

[NJC 2007 Prelims]

BN ⊥ 1Π ⇒ BN // n ⇒ BN k= n

Since N lies on 1Π ,

⋅ − =

��

ON OB BN k

= + = + − −

�� , so

18 2 2 4 8

− ⋅ − = ⇒ = − +

18 16 2

∴ = − = − +

��

⊥ distance from B to 1Π ( )( )22 2

8 2 8 1 2 1 8 6

= = − = + − + =

��

To verify that point A lies on 1Π :

0 2 2 2 4

= ⋅ − = + = =

RHS (Verified)

Let B′ be the reflection of B in 1Π .

By midpoint theorem,

( )1 2

2AN AB AB AB AN AB′ ′= + ⇒ = −��

1 2 7 2 7

2 2 0 18 0 14

7 2 1 2 13

− − − ′∴ = − − − = − − − −

�� , which is parallel to

− −

2 1Π Π⊥ ⇒ 2Π //

Direction vector of line l =

7 2 9 3

18 0 18 3 6

1 2 3 1

− − −

= − = = − − − − −

��

Normal vector of 2Π

= − × − =

Equation of 2Π :

4 2 4 4

2 0 2 2 8

0 2 0 0

⋅ = ⋅ ⇒ ⋅ =

Note that the triangle ABN is a right-angle triangle with 90ANB∠ = ° .

( )2 2 2

3 6 3 3 6 1 3 46

= − − = + + =

��

By Pythagoras’ Theorem,

Distance of A from 3Π ( ) ( )( )2 2 9 46 64 6 30AN AB BN= = − = − =

Acute angle between l and 3Π

= 1 1 8 6

cos cos 15.63 46

− − ∠ = = = °

(to 3 s.f.)

[MI 2007 Prelims]

xz yµ

+= = + = . Then 3 1, 3, 2x z yµ µ= − = − = .

: 3 1 ,

l µ µ

= − + ∈

To check if lines intersect:

1 2 1 3

+ − +

= − + −

2 3 2 -----(1)

3 -----(2)

1 -----(3)

− = −

⇒ − = −− =

Solving equations (2) & (3): 1λ = − and 1 3 2µ = − + =

Substitute into equation (1): LHS ( ) ( )2 1 3 2 8 RHS= − − = − ≠

Therefore the lines do not intersect.

Let the angle between a and b be θ .

( )( )

3 2 3 cos5

4.15 (to 3 s.f.)

− ⋅ = ⋅ − ⋅

a b a a a b a

Chapter 18: Complex Numbers 1

(a) 2 3 4w i= +

Let w x iy= +

2( ) 3 4x iy i+ = +

2 22 3 4x xyi y i+ − = +

2 2 3x y− = ----- (1)

2 4xy = ----- (2)

From eq (2): 2

Sub into eq (1):

22 4 22

3 3 4 0x x xx

− = ⇒ − − =

Solving, we get 2x = ± , 1y = ±

Hence (2 )w i= ± +

(b) Let 4 16z = −

4 ( 2 )

2 , 2, 1,0,1

= = − −

3 34 4 4 42 , 2 , 2 ,2

i i i iz e e e e

π π π π− −=

2 [MI/2010/Prelim/P2/Q2]

(i) 5 (2 )

2 where 2, 1,0

= = ± ±

5 32 1 02

− − =

Re (z)

Im (z)

k ki i

k k ki i i

w we e

π π π

cos sin5 5

1 cot (shown)5

* 2 3 6

π π π− = − − − =

1 1 3 1 3 1cos sin

* 3 6 6 3 2 2 6 6

ii i i

π π − = + = + = +

is purely imaginary, ( )2 1 , 6 2

n k kπ π

= + ∈� ,

( )3 2 1 ,n k k∴ = + ∈� .

24 4 20 ,

z i z i e k

− = ⇒ = = ∈�

8 8 8 8

, 2, 1,0,1

i i i i

z e e e e

π π π π

− −

= = − −

( )1 2

1 4 3arg

8 2 8 8z z

π π π + = + =

5 [RI/2010/Prelim/P1/Q1]

Let z = x + iy. Substitute the second equation into the first.

( * (1 i)) 2 0

( i ) ( )(1 i) 2 0

2 2i ( )i 2 0

x y x y

x xy x y

+ + − =+ + − =+ + − =+ + − =

+ + + + − =+ + + + − =+ + + + − =+ + + + − =

+ + + − =+ + + − =+ + + − =+ + + − =

On comparing real and imaginary parts, 2 2 2 22 2 0, 2 ( ) 0x xy x y x y− = + + = + =− = + + = + =− = + + = + =− = + + = + =

1, 1x y= ± == ± == ± == ± = ∓ .

When 1 iz = −= −= −= − , w = 2i

When 1 iz = − += − += − += − + , 2iw = −= −= −= −

6 [NYJC/2010/Prelim/P1/Q2]

(1+ai)(b+2i) = 8i

=> (b-2a)+(ab+2)i = 8i

Comparing real/imaginary parts, b-2a = 0 and ab+2 = 8

i.e. b = 2a and a(2a) + 2 = 8

i.e. 2

i.e. a = - 3 (since a < 0) and b = - 2 3

w = - 2 3 - 2i => arg(w) = 5

z2 + z2

Hence, arg(wn) = 5

nπ−

∴Least n = 3

(then argument is 5

π−, so that wn is of form – ki with k > 0).

7 [DH/2010/Prelim/P1/Q7]

(i) No. The statement is not always true. It applies only for (polynomial) equation in z with real

coefficients.

(ii) 4 3+ i 0z + = ⇒

4 3 iz = − −

4 62ez

π − =

1 5 (12 5)i ( 2 ) i4 6 242 e 2 e

π ππ

−− +

= = , 0,1, 2,3k =

1 1 1 1

4 4 4 4

5 7 19 17-i i i -i

24 24 24 242 e or 2 e or 2 e or 2 ezπ π π π

(iv) The quadrilateral is a square.

Let the length of each side be L

Pythagoras Theorem:

12 2 242| | =2(2 ) 2 2L z= =

8 [YJC/2010/Prelim/P1/Q9]

(a) z = 1 + ip , w = 1 + iq

zw = (1 + ip)(1 + iq)

3 – 4i = 1 – pq + i(p + q)

∴ 3 = 1 − pq … … (1)

& � 4 = p + q ⇒ q = – 4 – p

Substitute into (1) ⇒ 3 = 1 − p(– 4 – p)

p2 + 4p – 2 = 0

p = 62)1(2

)2)(1(4164±−=

−−±−

Since p > 0 ∴ p = 62 +− //

q = – 4 – ( 62 +− ) = 62 −− //

(b) a = 1 + i 3 =

∴ 1 + a + a2 + a3 + … + a9 = a

= )3i1(1

)i(2123

−−−

i 3512513

= i 3171512 +− //

(i) z5 = 32eπ + 2kπ( )i

, k = 0,±1,±2

52 , 0, 1, 2

π π+ = = ± ±

, 2 ie

(ii) w5 = 32i ( )

55 32 32iw iw⇒ = − ⇒ = −

Let z iw= ⇒ w iz= −

Area of triangle OW1W21 2

2 2 sin 1.902 5

π= × × × = units2.

(iii) ( )3 3

5 5 5 52 2 2 2 2i i i i

iz e z e z e z e z e

π π π ππ

− − − − − − −

Re( )z

Im( )z

–1 ××××

××××

( )3 3

2 25 5 5 52 4 2 4 2i i i i

z z e e z z e e z

π π π π− −

− + + − + + +

( )2 2 34cos 4 4cos 4 2

5 5z z z z z

π π − + − + +

p = 4,α =π

5,β =

5,k = 2

10 [NJC/2010/Prelim/P1/Q10]

(a) 2 *( i) 16 i 0z a z b+ − + + =

( ) ( )

22 3i ( i) 2 3i 16 i 0

5 12i 2 3 i 2i 3 16 i 0

8 2 10 3 i 0

+ + − − + + =

⇒ − + + − − − + + =

⇒ + + − + =

By comparing real and imaginary coefficient,

Real: 8 2 0 4 (ans)

Im :10 3 0 22 (ans)

+ = ⇒ = −

− + = ⇒ = −

(b) 5 1 0z + =

i 2 1 π5

i 2 1 π

e , 0, 1, 2

= = ± ±

iπ iπ i3π i3π

5 5 5 51, e , e ,e ,ez− −

i i ii

1 e e e e

e cos isin + cos i sin2 2 2 2

2cos e (shown)2

θ θ θ

θ θ θ θ

− + = +

= − + − +

(ii) Replace complex number z with 1w − ,

i 2 1 π

⇒ − =

∴ = +

From (i),

( ) ( ) ( )i 2 1 π

2 1 πi

102 1 π

1 e 2cos e 10

∴ = + =

for 0, 1, 2k = ± ± .

Chapter 19: Complex Numbers II

1 [TPJC/2010/Prelim/P2/Q2]

( ) ( ) ( ) ( )π

arg i 0 arg arg i 0 arg2

z z z= ⇒ + = ⇒ = −

(i) maximum value of 14 3 8 11z AP+ = = + =

(ii) 2 2

8 4 3 4 3 3tan

4 4OAP

− − −= =

( )1 4 3 3tan arg 4 π

− −

∴− ≤ + ≤

or ( )0.776 rad arg 4 πz− ≤ + ≤

2 [YJC/2010/Prelim/P2/Q3(b)]

C (–1, 1)

51 −=− zz

2i1 =−+w

(i) Min value of 2=− wz

(ii) Intersection point at B = w

22 −=−== OCCBOBw

−=∴ 4 i

( )( )i112 −−=

( ) ( )i1212 −−−=

(ii) Least value of z w− = 1

(iii) Greatest ( )arg 3z + = 1 11 3

tan sin 0.826 (3 dp)5 26

− − + =

Least ( )arg 3z + = 1 11 3

tan sin 0.432 (3 dp)5 26

− − − = −

4 [CJC/2010/Prelim/P1/Q11]

(a)(i)

[k= –1]

[k= –2]

[k= –3]

i (2,1)

O −3

(5,−3)

Equation of circle:

(b)(i) iiz +−≤+− 3223

( ) 11223 +≤−− iz

( ) 1323 ≤−− iz

Circle, center (3, -2), radius

half-line from point (6, -5), arg =

Area of shaded region = = =

(a) 13 13pq i= +

( )( )2 13 13

2 2 13 13

( 2 ) ( 2) 13 13

ia b i i

b i abi a i

a b i ab i

+ − = +

− + + = +

+ + − = +

Comparing real and imaginary parts,

2 13 - (1)

2 13 - (2)

(2): 15

Subst. 15

= into (1):

(3, -2)

(6, -5)

( )( )

152 13

13 30 0

3 10 0

3 or 10

− + =

− − =

Greatest

Max arg( )

z i PAR

− = ∠

6 [MI/2010/Prelim/P1/Q10]

(a) Given that 1 iz k= + is a root, so substitute into the given equation

( ) ( ) ( ) ( )4 3 2

1 i 1 i 9 1 i 29 1 i 60 0k k k k+ − + − + + + − =

( ) ( )2 3 4 2 3 21 4 i 6 4 i 1 3 i 3 i 9 18 i 9 29 29 i 60 0k k k k k k k k k k+ − − + − + − − − + − + + − =

(4, 3)

Locus of z

P (4, 1)

A(0, 1)

R (4,5)

Comparing the real or imaginary parts on both sides,

3 3 34 4 3 18 29 3 12 0

2 or 0 (N.A.)

k k k k k k k k

− − + − + = − + =

⇒ = ± =

OR, 2 4 2 2 4 26 3 9 9 31 6 40 0

By GC, 2

k k k k k k

− + + − + − = + − =

⇒ = ±

Hence, ( )( ) ( )( ) 21 2i 1 2i 2 5z z z z− − − + = − + is a factor of the given equation

( )( )4 3 2 2 29 29 60 2 5 12 0z z z z z z z z− − + − = − + + − =

( ) ( )2 22 5 0 or 12 0

1 2i , 4 or 3

z z z z

− + = + − =

∴ = ± = − =

(b) (i)

(ii) min. (arg z) = argument of any complex numbers along CD

1 2tan

At A, 2 2 and 2 2x y= − = + . -----(1)

max. (arg z) = argument of a, where A ≡ a

tan -----(2)

2 2tan

+= −

Hence, ( ) 1 2 2arg tan

4 2 2z

π − +

≤ ≤ −

(Shown)

7 [PJC/2010/Prelim/P2/Q5]

i 2i 2 2z − − ≤ and ( )Re 1 3iz > +

arg 2 24

− − =

( )arg 2 24

− − =

C α α Note : α = π/4

i 2+2i 2z − ≤ and ( )Re 2z >

( )2 2i 2z − − ≤ and ( )Re 2z >

( )arg 2 2i2 2

zπ π

− < − − ≤ − −

( ) 1 2arg 2 2i sin

2 2 4z

π π − − < − − ≤ − −

( )arg 2 2i2 3

zπ π

− < − − ≤ −

At maximum ( )arg 2 2iz − − ,

( )2 2cos ( 2 2sin )iz α α= + + − +

2 2cos ( 2 2sin )i6 6

zπ π

= + + − +

( )2 3 iz = + −

1 2d k d≤ <

2 21 2 3k≤ < +

1 13k≤ <

•(5,−2)

Im (z)

(2,−2)×

Re (z)

locus of z

πα−

(a) 0 arg( 1 )2

≤ + − ≤z iπ

and 2z i− =

Let P represent the complex number z x iy= + .

2cos 24

a x= = =π

2sin 24

b = =π

2 1y = +

( )2 2 1z i∴ = + +

(b) (i)

2 cos sin cos sin4 4 6 6

w i iπ π π π

= + −

Let 13 3

cos sin4 4

w iπ π

2 cos sin6 6

w iπ π

( )( )31 22∴ =w w w

1 22 2(1)(1) 2= = =w w w

( )( )( )3

arg( ) arg 2

arg(2) arg( ) 3arg( )

= + + −

2 cos sin4 4

w iπ π

∴ = +

(ii) 2 cos sin4 4

n n n nw i

π π = +

P(x,y)

arg( 1 )z i+ − = 0

arg( 1 )z i+ − = 2π

Since 4 , n k k= ∈�

4 42 cos sin

( 1) 2

k kw i

(a)(i) 64 3 16 0ww i iw∗ + + =

( )( ) ( )64 3 16 0x yi x yi i i x yi+ − + + + =

2 2 64 3 16 16 0x y i ix y+ + + − =

( ) ( )2 2 16 16 64 3 0x y y x i+ − + + =

Comparing coefficients,

( )16 64 3 0x + = and ( )2 2 16 0x y y+ − =

4 3x = − 2 16 48 0y y− + =

( )( )4 12 0y y− − =

y = 4 since y < 5

4 3 4w i∴ = − +

(ii) 8w = , ( )5

= , thus 5 5

8 cos sin6 6

n n n nw i

π π = +

Since n

w is to real, Im ( ) 0nw = , 5

ππ= ∈�

kn k⇒ = ∈�

(b) 1 3 1z i iz− − = +

Let z x iy= + , then ( )1 3 1x iy i i x iy+ − − = + +

( )1 1 3 1x i y ix y− + − = + −

( ) ( ) ( )1 1 3 1x i y y ix− + − = − +

( ) ( ) ( )2 2 221 1 3 1x y x y− + − = + −

( )2 2 2 22 1 2 1 9 1 2x x y y x y y− + + − + = + − +

2 28 2 8 16 7x x y y+ + − = −

8 64x y

+ + − =

The locus is a circle of centre 1

of radius 3

8units.

10 [VJC/2010/Prelim/P2/Q1]

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