2008 PAPER 1 Solutions

1. 2 2

1dx x∫ =

23

13x⎡ ⎤

⎢ ⎥⎣ ⎦

=32

3 –

313

= 73

4

dya

y∫ =

4322

3a

y⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

= 3 32 22 (4) ( )

3a

⎛ ⎞−⎜ ⎟

⎝ ⎠=

322 8 ( )

3a

⎛ ⎞−⎜ ⎟

⎝ ⎠

322 8 ( )

3a

⎛ ⎞−⎜ ⎟

⎝ ⎠= 7

3

32a = 9

2⇒ a =

239

2⎛ ⎞⎜ ⎟⎝ ⎠

2.73 (to 3 sig fig)

2. [Actually we can write 1

1(2 1) ( 1)(4 5)6

n

rr r n n n

=

+ = + +∑ so that the context is more familiar to

you instead of Sn = 1 ( 1)(4 56

n n n+ + ) .]

Let Pn be the statement Sn = 1 ( 1)(4 56

n n n )+ + , n ∈Z+

For P1, LHS = u1= 1 (2(1) + 1) = 3 RHS = 1 (1)(1 1)(4(1) 5)6

+ + = 3 = LHS

∴P1 is true.

Assume Pk is true for some k∈Z+, i.e. Sk = 1 ( 1)(4 56

k k k )+ +

To prove that Pk + 1 is true, i.e. Sk +1=1 ( 1)[( 1) 1][4( 1) 56

k k k ]+ + + + + = 1 ( 1)( 2)(4 9)6

k k k+ + +

LHS = Sk + 1 = Sk + uk+1

= 1 k k + (k +1) (2(k +1) + 1) = ( 1)(4 5)6

k+ +1 ( 1)(4 56

k k k )+ + + (k +1) (2k +3)

= [ ]) (4 5) 6(2 3)6

k k k k+ + + +1 ( 1

= 21 ( 1 = ) 4 5 12 186

k k k k⎡ ⎤+ + + +⎣ ⎦21 ( 1) 4 17 18

6k k k⎡ ⎤+ + +⎣ ⎦

Factorize out (k+1)/6. Remember to insert the 6 for the second term! 

= 1 ( = RHS 1)( 2)(4 9)6

k k k+ + +

∴ Pk is true ⇒ Pk +1 is true Since P1 is true and Pk is true ⇒ Pk +1 is true, by Mathematical Induction, Pn is true for all

positive integers n.

B P

3. = , = OA143

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

OB51

0

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠ O A

(i) OP = OA + OB = ⎟⎜ ⎟ + ⎟

143

⎛ ⎞⎜

⎜ ⎟−⎝ ⎠

51

0

⎛ ⎞⎜−⎜ ⎟ = ⎜ ⎟⎝ ⎠

633

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

[by parallelogram law of addition]

(ii) i = OA OB OA OB cos AOB∠

cos AOB∠ =

1 54 13 01 54 13 0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

i

= 5 4 026 26− + = 1

26

AOB∠ = cos –1 ( 126

) = 87.8o (angle to 1 d.p.)

(iii) Area of parallelogram = OA OB× = 1 54 13 0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟× −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

= 3

1521

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

= 2 2 2( 3 = ) ( 15) ( 21)− + − + − 675 = 15 3 units2

4. (i) ddyx

= 2

31

xx +

⇒ dy∫ = 2

3 d1

x xx +∫ ⇒ dy∫ = 2

3 2 d2 1

x xx +∫

y = 23 ln | 1|2

x + + C = 23 ln( 1)2

x + + C

(ii) When x = 0, y = 2, 2 = 3 ln12

+ C ⇒ C = 2

Brackets or modulus of x2+1 are both acceptable.

The required particular solution: y = 23 ln( 1)2

x + + 2

(iii) When x→± , ∞ddyx→ 0. The gradient tends towards 0.

23 ln 1 22

y x= + + (iv) y

x

23 ln 12

y x= +

23 ln 1 22

y x=

2

2−0 + −

The gradient at x = 0 is 0 when you sub x = 0 into dy/dx. So turning points (min)  should be drawn rather than pointed cusps. Graphs should also show that the gradient at the two ends tends towards 0. 

5. (i) 13

20

1 d1 9

xx+∫

= 13

202

1 1 d9 1

3

xx⎛ ⎞ +⎜ ⎟

⎝ ⎠

∫

= 131

0

1 1 tan9 1/ 3 1/ 3

x−⎡ ⎤⎢ ⎥⎣ ⎦

= 1

1 30

1 tan 39

x−⎡ ⎤⎣ ⎦

= 1 11 tan ( 3) tan (0)3

− −⎡ ⎤−⎣ ⎦ = 1 03 3

π⎡ ⎤−⎢ ⎥⎣ ⎦ =

9π

(ii) u = ln x , ' = xn v

='u 1x

, v = 11

1nx

n+

+

1

ln de nx x x∫

= 1

1

1 (ln )1

ne

x xn

+⎡ ⎤⎢ ⎥+⎣ ⎦ – ( )1

1

1 1 d1

nex x

n x+

+ ∫

= 11 (ln )

1ne e

n+

+ – 11 1 (ln1)

1n

n+

+ – 1

1 d1

e nx xn + ∫

Remember to sub the limitsinto the first term! 

=11 0

1ne

n+ −

+ – 1

1

11 1

enxn n

+⎡ ⎤⎢ ⎥+ +⎣ ⎦

=11

1ne

n+

+ – ( )

12

1 11

nen

+⎡ ⎤−⎣ ⎦+

=( )

1 12

1 ( 1) 11

n nn e en

+ +⎡ ⎤+ − +⎣ ⎦+

=( )

1

21

1

nne n

+ +

+ 6(a) AC2 = AB2 + BC2 – 2(AB)(BC) cos θ COSINE RULE. PLEASE REVISE! AC2 = 12 + 32 – 2(1)(3) cos θ AC2 = 10 – 6 cos θ

Since θ is a small angle, cosθ = 1 – 2

2θ

1

3

AC2 10 – 6(1 – ≈2

2θ ) = 4 + 3 2θ θ

Exact form for tan−1  3  required by Q and 

should be evaluated in terms of radians. 

Remember make the coefficient of x2 1 by pulling out 1/9. 

AC (4 + 3≈ 2θ )12 (Shown)

AC 4≈12 (1 +

234θ )

12 = 2 [ 1 +

21 32 4

θ⎛ ⎞⎜⎝ ⎠

⎟ + …]

= 2 + 23

4θ + …

∴ a = 2, b = 34

(b) f(x) = tan (2x +4π )

f ( )x′ = 2 sec2 (2x + 4π ) or f ( )x′ = ( )22 1 (f ( ))x+

f ( )x′′ = 4 sec (2x +4π )[2 tan (2x +

4π )sec (2x +

4π )]

= 8 tan (2x + 4π ) sec 2(2x +

4π ) or f ''( )x = ( )2 2f ( )f '( )x x

f(0) = 1 f = 4 f(0)′ (0)′′ = 16

f(x) = 1 + 4x +162!

x2 + …

= 1 + 4x + 8x2 + …

7. Total hours = 180 = 3(x + 2y) + 9( 12

)(22xπ ⎛ ⎞

⎜ ⎟⎝ ⎠

)

180 = 3x + 6y + 92

xπ

60 = x + 2y + 32

xπ

2y = 60 – x – 32

xπ

y = 30 – 2x – 3

4xπ

Let A be the area of the flower – bed.

A = xy +

2

22

xπ ⎛ ⎞⎜ ⎟⎝ ⎠ = x[30 –

2x – 3

4xπ ] +

2

8xπ Replace y in terms of x so that A is in 

terms of one variable, that is x only.  = 30x –

2

2x –

234xπ +

2

8xπ = 30x –

2

2x –

258xπ

ddAx

= 30 – x – 54

xπ

Let ddAx

= 0 .

54

xπ 30 – x – = 0

30 = 5(1 )4

xπ+

x = 30≈5(1 )

4π

+ 6.09 (to 3 sig fig)

y = 3 0 – 6.082

– 89 3 (6.0889)4

π≈ 12.6 (to 3 sig fig)

2 A

2

ddx

= – 1 – 54π < 0

∴ x = 6.09 and

y =12.6 when area is maximum.

STION SAYS GC CANNOT BE US D.

(i) (1 +

You can also use  derivative test or GC  drawing the curve of 

 first by

y = 30x – 2

2x

– 25

8xπ

 to show that  

maximum is attained. 

8 NOTE: YOU CAN USE GC TO CHECK ANSWERS WHEN QUE

E3 i)3 (1)= 1 + 33 2( 3 i) + 3(1) ( 3 i) + (2 3 i) 3

= 1 + 3 3 i – 9 – 3 3 i – 8 =

(ii) Since 1 + 3 i is a ro f 2z + az + + 4 = 0, it satisfies the equation. ot o bz

Sub z = 1 +

3 2

3 i into the equation,

2(1

+ 3 i )3 + a(1 + 3 i )2 + b(1 + 3 i) + 4 = 0

2( 8) ( 2 2 3) (1 3) 4 0a b− + − + + + + =

( )16 2 4 3(2 ) 0a b a b i− − + + + + =

12 2 0 nd 2 0a b a b− − + = + = ⇒ 6, 3b a= = − a (iii) Since z1 = 1 + 3 i is a r of oot 2z3 + az2 + bz + 4 = 0, z1

* = 1 – 3 i is another root of the equation.

[z – (1 + 3 i)] [z – (1 – 3 i)] Im

Re1

3 = (z – 1)2 – ( 3 i )2 = z2 – 2z + 4 2z + az + bz + 4 = 2(z – 2 4)(z + Comparing the constant, 8c ⇒ c = 1/2

3 2 2 z + c) = 4

he roo s are 1 + T t 3 i, 1 – 3 i and 12

− . 12

−

3−

1z

2z

3z

x

y

32

y =

12

x = −

73

32

y = −

f(x) = ax b9. cx d

++

(i) f ( )x′ = ( )( )2

( )cx d b c+=

a ax

cx d

+ −

+ ( )2acx ad acx bc

cx d+ − − =

( )2ad bccx d

−

+

+

Since ad – bc ≠ 0, f ( )x′ fo values of x, the graph of y = f(x) has no turning poin

≠ 0 r all real ts.

(ii) When ad – bc = 0, f ( )x′ =( )2ad −bccx d+

= 0. i.e. The gradient of the curve at all points is 0.

(iii)

The graph is a horizontal line.

For f(x) = 3 7x2 1x− , a = 3, b = – 7, +

c = 2, d = 1

ad – bc = (3)(1) – (– 7)(2) = 17

f ( )x′ =)

( 2d+

= ( )

ad bccx

−22 1x +

> 0 for all real values of x. 17

3 7 ∴ y = 2 1xx−+

has a positive gradient at all poin the graph.

v)

ts of

(i

y

x

32

y =

12

x = −

73

7−

3 72 1xy −

=x +

The gradient at x = 7/3 (which is originally an x‐intercept in y origin

2y 3 72 1xx−+

=

al graph) is // to the y‐axis. You will be penalized if you draw it pointed. 

10 ) 10, 13, 16, 19, … = 3

(i a = 10, d

[ ]2 ( 1)2n a n d+ − > 2000 ⇒ [ ]2(10) ( 1)3

2n n+ − > 2000

[ ]172

3n n+ > 2000 ⇒ 3 17 4000n n+ − > 0 2

33.7939.46− n < – 39.46 or n > 33.79 Least n = 34 months or 2 years 10 months

∴ She will first saved over $2000 on 1st Oct 2011.

(a) herefore, compound interest = 10(1.02)24 – 10 = $6.084

(ii)

Her original $10 will get her 2410(1.02) after 24 months [that is 2 years] T ≈$6.08 (to 3 sig fig)

(b) [10(1.02) + 10](1.02) = 10(1.02)2 + 10(1.02)

10(1.02) nd o 4th month =

2)24 = 10 ((1.02) + (1.02)2 + … (1.02)23 + (1.02)24)

10

End of 1st month = 10(1.02) End of 2nd month = End of 3rd month = [10(1.02)2 + 10(1.02) + 10](1.02) = 10(1.02)3 + 10(1.02)2 +E f 2 10(1.02)24 + 10(1.02)23 + … + 10(1.02) = 10(1.02) + 10(1.02)2 + … 10(1.02)23 + 10(1.0

= 241.02(1.02 1)

1.02 1−⎣ ⎦

⎡ ⎤−⎢ ⎥

$310.30 ≈ ≈$310 (to 3 sig fig)

(c) End of nth month = 10 ((1.02) + (1.02)2 + … + (1.02) )

= 10

n

1.02(1.02 1)1.02 1−⎣

n −⎢ ⎥

⎦

0 (1.02n – 1) n – 1) > 2000

1.02n >

⎡ ⎤

= 51510(1.02

200 151

⎛ ⎞+⎜ ⎟

n ln 1.02 > ln

⎝ ⎠200 151

⎛ ⎞+⎜ ⎟⎝ ⎠

n > 80.5 Least n = 81 m hs

Inequality sign does not change as ln 1.02 is positive. 

ont

11. p1 : 2x – 5y + 3z = 3

0.9 y z = 1

Using GC, x =

p2 : 3x + 2y – 5z = – 5 p3 : 5x – 2 + 17 6.6

4 4 711

− , y = 11

− , z = 11

Coordinates of the intersection point = 4 4 7, ,11 11 11

⎛ ⎞− −⎜ ⎟⎝ ⎠

(i) p1 : 2x – 5y + 3z = 3 p2 : 3x + 2y – 5z = – 5

GC, Using z = α , α ∈ R y = – 1 + α x = – 1 + α

xyz

⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎛ =

11

αα

α

− +⎜ ⎟− +⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞ r = =

⎞ +

Vector equation of the line l: r =

11

0

−⎛⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠

111

α⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

11

0

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠

+ 111

α⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

, α ∈Eqn should be written as r = …

NOT line = … or line = r = … 

R

(ii) p3 : 5x + λ y + 17z = μ or r i 5

17λ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= μ

Must know the concepts we Since l lies in p3, ll for 

1. the direction vector of l is perpendicular to the normal of p3:

(ii) and (iii).

111

⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎛.

5

17λ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= 0 ⇒ λ = – 22

2. a point on the line (−1, −1, 0) should also lie on the plane p3:

=

11

−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝

. 0 ⎠ 17⎜ ⎟

⎝ ⎠

522⎜ ⎟−⎜ ⎟

⎛ ⎞μ ⇒ μ = 17

herefore,T μ = 17 and λ = – 22.

(iii) Since there are no point in common, l is just // to p3 but does not lie on p3. 1. the direction vector of l is perpendicular to the normal of p3:

1⎜ ⎟⎜ ⎟

1⎛ ⎞⎜ ⎟ .

5

1⎝ ⎠ 17λ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟

= 0 ⇒ λ

⎝ ⎠

= – 22

2. a point on the line (−1, −1, 0) should NOT lie on the plane p3:

11

−⎛ ⎞⎜ ⎟−⎜ ⎟ .

0⎜ ⎟⎝ ⎠ 17⎜ ⎟

⎝ ⎠

522−⎜ ⎟

⎛ ⎞⎜ ⎟ ≠ μ ⇒ μ ≠ 17

Therefore, μ ≠ 17 and λ = – 22.

iv)

    Another direction vector =

(

1 11 1

3 0

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

n = =

r . ⎞⎟ = .

⎞⎟ = – 2

3x – y – 2z = – 2

1 21 01 3

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

312

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

31

⎛⎜ −⎜ ⎟

2⎜ ⎟−⎝ ⎠ 0⎜ ⎟⎝ ⎠

11

−−⎜ ⎟

⎛ ⎞⎜ ⎟

⎛⎜

312−⎜ ⎟

⎜ ⎟−⎝ ⎠

2008 Paper 2 Solutions Section A: Pure Mathematics [40 marks] 1

(i)

Sketch f for the required domain only! 

Note: There is only 1 intersection point at (0, 0) between f(x) and g(x).

(ii) f(x) = e x sin x

= (1 + x + 2

2x +

3

6x + …)(x –

3

6x + …)

= x –3

6x + x2 +

3

2x + … = x + x2 +

3

3x + …

Standard series used here is faster! 

Your graph of g is a cubic curve and you should exhibit understanding that they are close near x = 0 since g is an approx to f. 

(iii) See above diagram for 3

2 3xy x x= + + = g(x)

(iv) | f(x) – g(x) | < 0.5 is the same as | f(x) – g(x) | − 0.5 < 0.

Sketch y = |ex sin x – ( x + x2 +3

3x )| − 0.5.

y

0.5y =

x1.561.96−

From the graph, – 1.96 < x < 1.56 for | f(x) – g(x) | − 0.5 < 0.

2 (i) y2 = x (1 – x)12

y =± x12 (1 – x)

14

Area of R = 21 112 4

0(1 ) dx x−∫ x = 2

1 112 4

0(1 ) dx x−∫ x = 0.9988 = 0.999 (to 3 sig fig)

(ii) Let u = 1 – x ⇒ x = 1 – u When x = 0, u = 1 – 0 = 1 When x = 1, u = 1 – 1 = 0 ddux

= – 1 ⇒ dx du= −

Volume = 1 2

0dy xπ ∫

= 112

0(1 ) dx x xπ −∫

= 102

1(1 ) ( d )u u u− −∫π

= −1 302 2

1du u−∫π

Limits change correspondingly.  Upper limit is x = 1 which changes to u = 0. 

u = −

03 52 2

1

3 52 2

u u⎡ ⎤⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

π

= 2 23 5

π ⎡ ⎤−⎢ ⎥⎣ ⎦= 4

15π units3

(iii) y2 = x (1 – x)12

2y ddyx

= (1 – x)12 + x 1

2⎛ ⎞⎜ ⎟⎝ ⎠

(1 – x)12

−(– 1) = (1 – x)

12 − 1

2 1x

x−

ddyx

= [(1 – x)12 − 1

2x (1 – x)

12

−] / 2y

When ddyx

= 0, (1 – x)12 − 1

2 1x

x−= 0

1 x− = 12 1

xx−

⇒ (1 – x) = 12

x

x = 23

Exact form means you must use the analytical method which is by differentiation! You cannot use GC to find 0.66666 which is 2/3 in exact form. You can use GC to check answers though. ☺ 

3 (a) 1*

w rpw r

= = = arg( ) arg arg arg * ( ) 2*

wp w ww

= = − = − − =θ θ θ

( )55 2 1i 0ip e e= =θ θ = cos (10θ ) + i sin(10θ )

For p5 to be real, sin(10θ ) = 0, i.e. 10θ = π , 2π , 3π , 4π , …

3 2, , ,10 5 10 5

=π π π πθ [Since 0

2< <

πθ ]

For p5 to be positive ⇒ cos (10θ ) > 0 . Therefore 3 2(rejected), , (rejected),10 5 10 5

=π π π πθ .

θ =5π , 2

5π

OR Shorter Method: For p5 to be real and positive, arg (p) = 10θ must be multiples of 2π . 10θ = 2π , 4π , …

θ =5π , 2

5π [Since 0

2< <

πθ ] Im

Re

(8,6)

(4,3)

8 6z z i= − −

6z ≤

2 2 36x y+ =

(b)

(i) See diagram. (ii) 10, 5, 6 radiusOA OB OQ= = = =

Let PO∠ = B QOB= ∠ θ = 1 1 5cos cos6

OBOQ

− −=

Arg (8+6i) = 1 16 3tan tan8 4

− −= Do follow the accuracy specified by Q which is 3 decimal places. 

Least value of arg z = 1 3tan4

− − 1 5cos6

− = 0.0058 (3 dp)

Greatest value of arg z = 1 3tan4

− + 1 5cos6

− = 1.229 (3 dp)

4 (i)

41

1

4

y

x

y x=

f( )y x=

1f ( )y x−=

(ii) Let y = f(x)

y = (x – 4)2 + 1, x > 4 y – 1 = (x – 4)2, x > 4 x – 4 = 1y± −

x = 4 1y− − (rejected ∵ x > 4) or x = 4 1y+ −

f – 1(x) = 4 1x+ − , x > 1 1fD − = = (1, fR ∞ )

(iii) Refer to diagram

(iv) y = f(x) is reflected about the line y = x to obtain y = f –1 (x) .

To find the intersection between f (x) = f –1 (x), it is equivalent to find the intersection between f(x) = x. Note x > 4 (domain of f). (x – 4)2 + 1 = x x2 – 8x + 17 = x x2 – 9x + 17 = 0

2( 9) ( 9) 4(1)(17)2(1)

x− − ± − −

= = 9 132

±

x = 9 132

− (rejected ∵x > 4)

Exact value of x = 9 132

+

Exact value required. Numerical solution gains no credit. 

Section B: Statistics [60 marks]

5 Number the students from 1 to 950. 95050

= 19. Randomly select a student from 1 to 19 and

every 19th student thereafter until a sample of 50 students is obtained. Important to mention the random start point and numbering of students. 

Reason Stratified sampling (eg different level as stratas) gives a sample that is more representative of

the student population in the school. Students from different levels may have different responses in terms of sports facilities.

Important to mention the context by giving example on what the strata might be. Key thing is that it is more representative sample. Do not accept “allows the opinions of diff stratas to be considered separately”

6 Let X be the random variable “the mass of calcium in a 1 litre bottle”

x = x

n∑ = 1026

15 = 68.4

Unbiased estimate of population variance 2s

=( )2

211

xx

n n

⎡ ⎤⎢ ⎥−⎢ ⎥−⎣ ⎦

∑∑ = ( )21026.01 77265.9014 15

⎡ ⎤−⎢ ⎥

⎢ ⎥⎣ ⎦= 506.25

H0 : μ = 78 H1 : μ ≠ 78 Get some method mark from 

this step if your p‐value happens to be wrong due to carelessness. 

2

68.4 78506.25 15

XTs n

− −= =

μ = −1.652

p – value = 0.121 Since p – value = 0.121 > 0.05, we do not reject Ho and conclude that there is insufficient

evidence to conclude that the mean mass of calcium in a bottle has changed at 5% significance level.

B

A

B

B

A

A

B

A

B

A

0.6

0.4

0.7

0.3

0.2

0.8

0.3

0.7

0.8

0.2

st1 set nd2 set rd3 set

7

Realize that A wins two sets or B wins two sets, the third set is not played. 

(i) P(A wins the second set) = 0.6(0.7) + 0.4(0.2) = 0.5 (ii) P(A wins the match) = 0.6(0.7) + 0.6(0.3)(0.2) + 0.4(0.2)(0.7) = 0.512 (iii) P(B won the first set | A wins the match)

= ( won the first set AND wins the match)( wins the match)

P B AP A

= 0.4(0.2)(0.7)0.512

= 0.109 or 764

8 (i) From GC, r = 0.970. (to 3 sig fig)

Since r is close to 1, this indicates that there is a strong positive linear correlation between x and t, thus a linear model is appropriate.

Answer to the Q whether it is appropriate or not. Do not just say strong correlation! 

x

t

2 4 6 8 10

2

4

6

8

10

(4.8,7.6)P

(ii)

Scatter diagram should be drawn carefully, indicating the scale and axes. x is the independent variable so it must be plotted on the horizontal axis. 

(iii) With P (4.8, 7.6) removed, the remaining points lies on a ln-type curve. You may also draw in the ln curve onto the same scatter diagram to illustrate as shown below.

x

t

2 4 6 8 10

2

4

6

8

10

(4.8,7.6)P

(iv) Using GC, t = 1.42 + 4.40 ln x. a = 1.42 and b = 4.40 (iv) When x = 4.8, t = 1.42 + 4.40 ln (4.8) = 8.32 (to 3 sig fig) (vi) x = 8 is out of the data range, the value of t obtained by extrapolation may not be

accurate even though the model is a good fit.

9 Let X be the random variable “the number of grand pianos sold in a week” Be careful with P(X ≥  4) = 1 – P(X ≤  3) not 1 – P(X ≤  4) 

X ~ Po (1.8) P(X ≥ 4) = 1 – P(X 3) = 1 –0.89129 = 0.10871 = 0.109 (to 3 sig fig) ≤ Let Y be the random variable “the number of upright pianos sold in a week” Y ~ Po (2.6) X +Y ~ Po (1.8 + 2.6 = 4.4) P(X +Y = 4) = 0.191736 = 0.192 (to 3 sig fig) Let W be the random variable “the number of grand pianos sold in 50 weeks” W ~ Po (50 × 1.8)

Remember to do cc for this approx from poisson to normal. 

W ~ Po (90) Since λ = 90 > 10, W ~ N(90, 90) approximately P(W < 80) .c c⎯⎯→ P(W < 79.5) = 0.13419 = 0.134 (to 3 sig fig)

The rate at which grand pianos are sold over the period of a year may not be constant. This is because the demand for piano may be higher during certain months eg year end when bonuses are given or sales.

10 (i) Required number of ways = 3C2× 4C3× 5C3 = 120

(ii) Required number of ways = 9C8 = 9 (iii) Required number of ways = 5C4× 7C4 + 5C5× 7C3 = 210 (iiii) Required number of ways = 12C8 – (K and M only) – (L and M only which is part (ii))

= 12C8 – 8C8 – 9 = 485

11 X~ N(50, 82) X1 + X2 ~ N(2×50,2×82) X1 + X2 ~ N(100, 128) P(X1 + X2 > 120) = 0.0385 (to 3 sig fig) X1 – X2 ~ N(50 – 50, 82 + 82) X1 – X2 ~ N(0, 128)

(i) P(X1 > X2 + 15) = P(X1 – X2 > 15) = 0.0924 (to 3 sig fig)

74 146 Let Y ~ N(μ , 2σ )

74 146 1102

μ += =

0.0668

Note:  Complementary method is faster and also less complicated. 

Delegates from K and L = 7 so it is not possible to choose 8 from solely K and L.  

It is easier if you notice that we can combine K and M together in a single group and choose 8. 

Your answers should relate to the rate within the period not constant and give a reason in the context of the Q why it is not constant in order to score a full 2 marks. 

E(Y) = μ = 110 (by symmetry)

P(Y < 74) = 0.0668

P(Z < 74 110σ− ) = 0.0668 ⇒ 74 110

σ− = −1.5000

σ = 24

A common mistake is writeVar (aX + b) = aVar (X) + b. 

Var (Y) = 242 = 576 E(Y) = aE(X )+ b = 50a + b Var(Y) = Var (aX + b) = Var (aX) + Var(b) = a2 Var (X) + 0 = 64 a2 50a + b = 110 ------------------- (1) 64 a2 = 576 ------------------- (2) Solving (2), a = 3 (since a > 0) Solving (1) by substituting a = 3, E(Y) = 50a + b 110 = 50(3)+ b b = – 40