2011 H2 Math 9740 Paper 2 Solution - WordPress.com H2 Math 9740 Paper 2 Solution 2011 H2 9740 Paper...

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2011 H2 Math 9740 Paper 2 Solution

2011 H2 9740 Paper 2 Solution Affinity Education Place

1.

2 5 29i+ =

Minimum | z | = 29 3−

Maximum | z | = 29 3+

The half line ( )4

arg zπ

= represents portion of line y = x in Cartesian coordinates and it passes through

2 + 2i and 5 + 5i

This means possible P is 2 + 2i or 5 + 5i.

Maximum distance = 17 .

2. (i) V = ( ) ( )2 2 2n x n x x− −

= 2 2 32 6 4n x nx x− +

(ii) 2 22 12 12

dVn nx x

dx= − +

At stationary of V, 0dV

dx= .

⇒ 2 22 12 12 0n nx x− + =

⇒ 2 26 6 0n nx x− + =

⇒2 26 36 24

12

n n nx

± −=

= ( )3 3

6

n ±

However, 2

nx < and ( )3 3

6 2

n n+> .

Therefore, ( )3 3

6

nx

−= .

Re

Im

O

2+5i

4π

z

P

6 + i P




3. Given ( ) ( )2 1 3f x ln x= + + , 1

2x > −

(i) Let ( )2 1 3y ln x= + +

( )2 1 3ln x y+ = −

31 1

2 2

yx e −= −

Range of ( )1f x−is

1

2,

− ∞

Domain of ( )1f x−is ( ),−∞ ∞

( )1f x− = 31 1

2 2

xe − − , x∈ℝ .

(ii) Refer to the graph

(iii)The graphs of ( )y f x=

and ( )1y f x−= intersect along the line y = x.

Hence the points of intersection satisfy

( )f x x=

⇒ ( )2 1 3ln x x+ + =

⇒ ( )2 1 3ln x x+ = −

Solve the equations by GC, 2the x-coordinates

are -0.4847and 5.482 (4 s.f.)

4. (a) (i) 2 2

0

n xx e dx−∫

= 2 2 2

00

1

2

nnx xx e xe dx− − − + ∫

=2 2 2 2

00

1 1 1

2 2 2

nnn x xn e xe e dx− − − − + − + ∫

=2 2 2 2

0

1 1 1

2 2 4

n

n n xn e ne e− − − − − + −

=2 2 2 21 1 1 1

2 2 4 4

n n nn e ne e− − −− − − +

(ii) As n →∞ , 2nne−

and 2 2nn e−

0→ and 2 0ne− → .

2 2

0

1

4

xx e dx∞ − =∫

x

y

O

1

2x = −

( )3 0,

( )y f x=

1

2y = −

( )3 12

0e ,− −

( )3 12

0 e,− −

( )0 3,

( )1y f x−=




(b) 2

4

1

xy

x=

+

Volume of the solid formed =

( )2

1

20 2

16

1

xdx

xπ

+∫

x tanθ= , 2 21

dxsec x

dθ

θ= = +

( )2

1

20 2

16

1

xdx

xπ

+∫

= ( )

2

420

16tan

dsec

π θπ θ

θ∫

= 2

40

16 sin dπ

π θ θ∫

Volume of the solid formed =2

40

16 sin dπ

π θ θ∫

= 40

8 1 2cos dπ

π θ θ−∫

= 4

0

28

2

sinπ

θπ θ −

=1

84 2

ππ −

= ( )2 2π π − unit3.

5. ( )2X ~ � ,µ σ

( ) 4040 0 05 1 64485P X . .

µσ−

< = ⇒ = −

( ) 7070 0 975 1 95996P X . .

µσ−

< = ⇒ =

Solving 1 64485 40.µ σ− = and 1 95996 70.µ σ+ =

53 7.µ = (3 s.f.) and 8 32.σ = (3 s.f.)

6. (i) Consider different age groups of residents, such as

6 to 16 years old, more than 16 to 25 years old, more than 25 years old to 35 years old, more than 35 to 45

years old and above 45 years old.

The surveyors interview a given number of residents base on their preference.

(ii) The selection of interviewees is biased and there is lack of fair representation of residents from

different age groups.

(iii) Using stratify sampling method. The number of interviewees from each age group is proportional to

the total number of residents within the age group.

However, this method will not be realistic because it may be difficult to determine the exact number of

residents in each age group.




7. (i) Any two assumptions: (1) The probability of successfully contact a friend remain constant at 0.7.

(2) The n calls are independent. In another words, when the outcome of a call is known this will not affect

the outcome of another call.

(3) There are only 2 possible outcomes of a call. That is, successfully contact a friend or fail to contact a

friend.

(ii) The probability of success in contacting a friend may not be constant for different people.

Alternatively, the calls may not be independent. For example, knowing one of his friends cannot be contacted

will lead to knowing another friend is not contactable.

(iii) Given n = 8. R ~ B(8, 0.7)

( )6P R ≥

= ( )1 5P R− ≤ = 0.552 (3 s.f.)

(iv) n = 40. R ~ B(40, 0.7), p = 0.7

Since np = 28 > 5 and n(1 – p) = 12 > 5, R can be approximated by Normal distribution with mean 28 and

variance 8.4. That is, R ~ �(28, 8.4) approximately.

( )25P R <

= ( )24 5P R .< (by c.c.)

= 0.114 ( 3 s.f)

8. (i)

(ii) For y = c + dx, r = -0.99231 (5 d.p.)

Although the | r | is close to 1, the scatter diagram of y on x shows quadratic correlation (or non-linear

correlation). Hence, linear correlation model may not be the best to describe the correlation between x and y.

(iii) For 2y a bx= + , r = -0.99998 (5 d.p) (a = 22.23029, b = -0.85621)

The new | r | value is closer to 1, hence, it shows that 2y a bx= + is a better model.

y

x O

y

x O




(iv) Equation of the least-squares regression line of y on x2 is y = 22.23029 – 0.85621x

2.

When x = 3.2, y = 13.5 (3 s.f)

9. (i) (a) P(a selected lens is faulty)

= 0 6 0 05 0 4 0 07. . . .× + ×

= 0.0580 or 29

500

(b) P(a selected lens from A | it is faulty)

= 0 6 0 05

0 058

. .

.

×

= 0.517 (3 s.f) or 15

29

(ii) (a) P(exactly one of 2 lens selected is faulty)

= ( )( )2

0 058 1 0 0581

. .

−

= 0.109272

= 0.109 (3 s.f)

(or state that X ~ B(2, 0.058) where X is the number of faulty lens out of 2 lens selected)

(b) P(Two selected lens are both from A | exactly one is faulty)

=( )

( )Two lens from A are selected and exactly 1 is faculty

exactly 1 out of 2 lens is faulty

P

P

= 0 6 0 05 0 6 0 95 20 109272

. . . ..

× × × ×

= 0 03420 109272

..

= 0.313 (3 s.f.)

10. T ~ �(38, 25)

(i) Let µ be the mean time taken to install a component.

Null hypotheses, Ho: 38µ =

Alternate hypotheses, H1: 38µ <

(ii) Level of significance = 5%. Since T follows normal distribution and population variance is known, using z-

test.

Under Ho, 25

3850

T ~ � ,

.

To reject Ho, ( ) 0 05P T t .< ≤ .

38

0 0525

50

tP Z .

− ⇒ < ≤

.

381 6449

2550

t.

−⇒ ≤ − 36 83688t .⇒ ≤ .

Therefore, the set of values of t is ( ){ }36 8 3s.f.t | t .∈ ≤ℝ .

0.6

0.4

A

B

Faulty

Not

Faulty

Faulty

Not

Faulty

0.05

0.95

0.07

0.93




(iii) Given 37 1t .= . When the null hypothesis is not rejected, 37 1 38

0 0525

.P Z .

n

− ⇒ < >

.

37 1 381 6449

25

..

n

−⇒ > − . (no equality sign because ( )1 6449 0 05P Z . .< − = )

0 91 6449

5

. n.

−⇒ > − 83 5n .⇒ <

(3 s.f)

Therefore, the set of values of n is { }0 83n | n∈ < ≤ℤ .

11. (i) P (R = 4) =

18 12

4 6

30

10

C C

C=

816

8671 or 0.0941 ( 3 s.f.)

(ii) Given P (R = r) > P (R = r + 1) where r is the most probable number of women to be selected.

⇒18 12 18 12

10 1 9

30 30

10 10

r r r rC C C C

C C

− + −>

⇒( ) ( ) ( ) ( ) ( ) ( ) ( )

18 12 18 12

18 10 2 1 17 9 3

! ! ! !

r ! r ! r ! r ! r ! r ! r ! r !>

− − + + − − +

⇒ ( ) ( ) ( ) ( ) ( ) ( ) ( )1 17 9 3 18 10 2r ! r ! r ! r ! r ! r ! r ! r !+ − − + > − − + (shown)

Further simplify above inequality, we get ( )( ) ( )( )1 3 18 10r r r r+ + > − − .

32 177r⇒ >

5 53r .⇒ >

Consider

r ( > 5.5) P (X = r)

6 0.306

7 0.233

8 0.096

Therefore, r = 6.

12. (i) Let Y be the number of people joining an airport check-in queue in 4 minutes. Y ~ Po(4.8)

P (Y ≥8)

= 1 – P (Y≤7)

= 0.113 (3 s.f.)

(ii) Let 1 2

60t

.Y ~ Po t

.

( )1 0 7tP Y .≤ =

⇒ 0 02 0 020 02 0 7. t . te . te .− −+ =

From GC, t = 54.867

Therefore t = 55 seconds. (to the nearest whole number)

(iii) Let X be the number of people leaving the queue in 15 minutes. X ~ Po(27)

Since E(X) = 27 > 10, X ~ N(27, 27) approximately.

P(X≤ 11)

= P(X < 11.5) (by c.c.)

= 0.00143 (3 s.f.)

(iv) Poisson distribution may not be valid because number of people check-in is not constant over several hours

period.

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