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HWA CHONG INSTITUTION MATHEMATICS DEPARTMENT PAGE 1 2011 GCE ‘A’ Level H2 Maths Solution Paper 2 SECTION A (PURE MATHEMATICS) 1(i) The shaded region above represents the region in which z lies. Note: Unless required in the question, it would be sufficient to just indicate the centre and radius of the circle in such a locus drawing. 1(ii) Minimum z is given by OA OC AC 2 2 2 5 3 29 3 Maximum z is given by OB OA AB 29 3 6 29 3 1(iii) Remark 1: The length DP in this case, is always the same regardless of which coordinates of P we are taking. This is because it happens that the line through CD is perpendicular to the line representing 1 arg 4 z . This can be easily verified by finding the gradient of the line through CD . Re Im 3 Re Im C Re Im C P P D

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MATHEMATICS DEPARTMENT PAGE 1

2011 GCE ‘A’ Level H2 Maths Solution Paper 2 SECTION A (PURE MATHEMATICS)

1(i)

The shaded region above represents the region in which

z

lies.

Note: Unless required in the

question, it would be sufficient to

just indicate the centre and radius

of the circle in such a locus

drawing.

1(ii)

Minimum z is given by OA

OC AC

2 22 5 3 29 3

Maximum z is given by OB

OA AB 29 3 6 29 3

1(iii)

Remark 1: The length DP in

this case, is always the same

regardless of which coordinates

of P we are taking. This is

because it happens that the line

through CD is perpendicular to

the line representing

1

arg4

z . This can be easily

verified by finding the gradient of

the line through CD .

Re

Im

3

Re

Im

C

Re

Im

C P

P D HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 2

Combining the restriction given by 1

0 arg4

z , the

resulting locus of z

is given as the shaded region above

(see diagram).

The point D

represents the complex number 6 i .

The maximum value of 6 iz

is given by DP (there

are two possible such positions that give this maximum

value).

Observe that 2, 2

and 5, 5 both lie on the circle

which has Cartesian equation 2 2 22 5 3x y .

Thus, these two points are the possible positions of P .

This means the maximum value required is DP

2 2

6 2 1 2 17 .

Remark 2: This part can also be

solved by changing the equations

of the circle and the half-line to

Cartesian form and do

simultaneous equations solving.

The equation of the circle is

of the half-line is y x .

2(i) From the question’s description, the box will have base

area PS SR and height x .

Thus the volume is given by

2 2 2V n x n x x

2 22 2 2x n nx xn x 2 2 32 6 4n x nx x . (Shown!)

2(ii) Find 2 2d

2 12 12d

Vn nx x

x and let

d0

d

V

x

2 26 6 4 6

12

n n nx

6 12 1 3

12 2 6

nx n

From the diagram (for example, from the side of AD ),

we deduce that 2x n , and thus 1 3

2 6x n

is the

3(i) Write f ln 2 1 3x x .

Then, f 3

2 1 ex

x

f 31

e 12

xx

.

Thus, 1 31f e 1

2

xx .

The domain of 1f

is ,

or we can write as .

The range of 1f

is 1

,2

.

Remark 1: Domain of 1f

equals

to the range of f and range of 1f

equals to the domain of f .

Remark 2: Generally, to find the

range of a function, we will need

a sketch of the graph. HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 3

3(ii)

The coordinates where the curves intersect the axes are:

0, 3A , 31e 1 , 0

2B

, 310, e 1

2C

and

3, 0D .

3(iii) When the two curves intersect, they also intersect at the

line y x . That is, 1f fx x is equivalent to

f x x . Thus we have ln 2 1 3x x

ln 2 1 3x x .

Using the graphic calculator, the values of x are

0.4847 and 5.482 , correct to 4 s.f.

Note: TI84 Plus screenshots to

get these values (based on the

intersections of the two graphs

3y x and ln 2 1y x ) are

shown:

4(a)(i) Use integration by parts twice:

2 2 2 2 2

0 00

1 1e d e 2 e d

2 2

nn nx x xx x x x x

22 2

0

e e d2

nn xn

x x

22 2 2

00

1 1e e e d

2 2 2

n nn x xn

x x

22 2 2

0

1e e e

2 2 4

n

n n xn n

22 2 21 1

e e e2 2 4 4

n n nn n

22e 1

2 2 14 4

n

n n

.

Note: Usually when the integrand

consists of two different families

of terms, we should be prepared

to apply integration by parts. This

is only a guideline, as there are

exceptions such as ln

dx

xx

where it is not necessary to

employ the technique of

integration by parts. Another

familiar exception is 3sec dx x

where the usual method is to use

integration by parts to solve it.

x

y HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 4

4(a)(ii)

2

2 2 2

0

e 1 1e d lim 2 2 1

4 4 4

nx

nx x n n

.

Note that 2

2e2 2 1 0

4

n

n n

as n .

Note 1: Integrals which have

limits ending with infinity such as

2 2

0

e dxx x

are called improper

integrals.

Note 2: A way to understand why

2

2e2 2 1 0

4

n

n n

as n

is by manipulating the expression

2 2

2

2

e 2 2 12 2 1

4 4e

n

n

n nn n

and think of

2 3

2 2 2e 1 2 ...

2! 3!

n n nn

(according to MF15) which can

be considered an infinite sum

consisting of terms in n of much

higher degrees.

4(b) From the diagram, the shaded region will to undergo a

rotation of one full round about the -axis.

Volume of solid obtained is given by . 1

2

0

dV y x

21

220

16d

1

xx

x

.

Using the substitution tanx :

2dsec

d

x

.

When 0 0x and when 14

x

.

Thus,

24 2

220

tan16 sec d

tan 1

V

x

y

1 0 HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 5

24

20

tan16 d

tan 1

(since 2 2tan 1 sec )

24

20

tan16 d

sec

.

24 4 2

2 20 0

sin 116 d 16 sin d

cos sec

. (Shown!)

Thus, 4 2

0

16 sin dV

4

0

1 cos 216 d

2

(by cosine double angle formula)

4

0

18 sin 2

2

1

8 2 24 2

.

2011 GCE ‘A’ Level H2 Maths Solution Paper 2 SECTION B (STATISTICS)

5 From P 40.0 0.05X , we standardise X :

40.0P 0.05

X

40.0P 0.05Z

40.01.64485

----- (1)

Similarly, for P 70.0 0.975X , we have

70.0P 0.975Z

70.01.95996

----- (2)

Solving (1) and (2): 53.7 , 8.32 , correct to 3 s.f.

Note 1: To use the InvNorm

function in the graphic calculator,

we must ensure the probability is

of the form P ... ...X or

P ... ...X , with the

inequality sign being “less than”

or “less than or equal”.

Note 2: The method of

standardisation is very efficient in

solving normal distribution

problems involving unknown

mean and/or unknown variance

(or standard deviation).

6(i) Quota sampling might be carried out by

First divide the range of ages into several groups (12 –

16, 17 – 21, 22 – 26 etc).

Next, assign a quota (required number of residents to

be interviewed) for each age group.

Interviewer will then pick residents at his/her own

discretion for interviews, until all the quotas are

fulfilled. HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 6

6(ii) One disadvantage is that the interviewer may likely have

collected a biased sample of residents due to his/her non-

random selection process of residents (selection is based

solely on his/her judgement).

Remark: Despite the obvious

it is still used widely nowadays

mainly due to its big advantage of

not needing to have a sampling

frame to carry out the sampling

process.

6(iii) Stratified sampling.

This method is not realistic in this context because it will

our sampling frame. For example, it will be difficult to

get a whole list of residents with information such as

their ages from a government or public agency.

such as randomly selected

residents may not be willing to

participate in the interview.

7(i) Assumption 1: All the friends should not have any

knowledge of this experiment to be carried out, before or

during the experiment, to ensure that each trial of

contacting a friend is independent of one another.

Assumption 2: The experiment should be conducted in a

realistic span of time in the evening to ensure that time

does not become a factor to affect the constant success

probability of contacting a friend.

7(ii) No matter how quickly the experiment can be carried

out, all the calls are bound to be done at different times in

the evening, and in turn will make it difficult for the

probability of success in contacting a friend to stay

constant.

7(iii) For 8n , ~ B 8, 0.7R and P 6R

1 P 5R

0.552 correct to 3 s.f.

TI84 Plus screenshots: HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 7

7(iv) For 40n , ~ B 40, 0.7R .

Since 40n is considered large, 28 5np and

1 12 5n p , thus, ~ N 28, 8.4R approximately.

The parameters in this new distribution are the mean (28)

and variance (8.4).

Therefore P 25R .P 24.5

c cR

0.114 correct to 3 s.f.

Note: c.c stands for continuity correction.

TI84 Plus screenshots:

8(i)

Note: The points should follow a

quadratic curve of the form (in

8(ii) From the graphic calculator, the value of the product

moment correlation coefficient, r is 0.992 correct to 3

s.f.

If we observe the differences of consecutive y values

such as 18.8 16.9 1.9 etc:

D 1.9 2.4 2.8 3.1 3.7 4.1

We would notice that differences follow an increasing

trend, which would not be the case if the data have

followed a linear model (the differences would be more

or less kept constant).

TI84 Plus screenshots to

calculating the value of r :

Note:

Below is a screenshot from TI84

Plus showing the seven observed

points and the best fit line:

y

x HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 8

Though the linear model is not

the best, it is not obvious from the

diagram to conclude that the

linear model.

8(iii) Let 1r and 2r denote the product moment correlation

coefficients based on the linear model and the quadratic

model respectively. If 2 1r r , then we would deem the

quadratic model to be the better model. If 2 1r r , then

the linear model would be better.

Note: We can of course find the

value of r for the quadratic

model and do a comparison with

that obtained for the linear model.

However, since the question

asked us to “explain how to use”,

it will be more ideal to just

propose a general way to help

decide which model is better.

8(iv) From the graphic calculator, the equation of the least-

squares regression line of y on 2x is

20.85621 22.23049y x .

When 3.2x , the estimate value of y is

2

0.85621 3.2 22.23049 13.5y correct to 3 s.f.

Some TI84 Plus screenshots: HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 9

9 A probability tree is drawn for the situation:

Remark: A probability tree is for

the purpose of clearer illustration.

If the question did not ask, it is

not necessary to draw a

probability tree and use it to solve

probability problems.

9(i)(a) P a lens is faulty 0.6 0.05 0.4 0.07

0.058 .

9(i)(b) P a lens is made by given that the lens is faultyA

P lens is made by | lens is faultyA

P lens is made by lens is faulty

P lens is faulty

A

and

0.6 0.05

0.058

0.517 correct to 3 s.f.

9(ii)(a) P exactly one of the lenses is faulty

P 1st lens faulty and 2nd lens not faulty

P 1st lens not faulty and 2nd lens faulty

0.058 1 0.058 1 0.058 0.058

0.109272

Note 1: We can also use the

complementary method to obtain

the probability:

1 P both faulty P both not faulty

Note 2: The answer in this case is

exact, thus it is not necessary to

round it off to 3 significant

figures. 9(ii)(b) P both were made by , given that exactly one is faultyA

P both were made by | exactly one is faultyA

P both were made by exactly one is faulty

P exactly one is faulty

A

and

0.6 0.05 0.6 0.95 2

0.109272

0.313 correct to 3 s.f.

10(i) Let denote the mean time taken to install a

component.

Null hypothesis 0H is 38.0 .

Alternative hypothesis 1H is 38.0 .

0.4

A

B

0.05

0.95

0.07

0.93

Faulty

Faulty

0.6 Not faulty

Not faulty HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 10

10(ii) To reject null hypothesis, -value level of significancep .

That is P 0.05T t

38.0

P 0.055.0

50

tZ

38 501.64485

5

t

36.837t .

Since t denotes the sample mean time, therefore the set

of values required is : 0 36.8t t .

10(iii) If the null hypothesis is not rejected,

-value level of significancep .

That is, P 0.05T t

37.1 38.0P 0.05

5.0Z

n

0.91.64485

5

n

83.5n

Since n is a positive integer, the set of values required

can be written as :1 83n n .

11(i)

18 12

4 6P 4 0.0941

30

10

R

correct to 3 s.f.

11(ii) Starting with P P 1R r R r , we have

18 12 18 12

10 1 9

30 30

10 10

r r r r

18 12 18 12

10 1 9r r r r

18! 12! 18! 12!

! 18 ! 10 ! 2 ! 1 ! 17 ! 9 ! 3 !r r r r r r r r

1 ! 17 ! 10 ! 2 !

! 18 ! 9 ! 3 !

r r r r

r r r r

1 ! 17 ! 9 ! 3 ! ! 18 ! 10 ! 2 !r r r r r r r r

(Shown!)

Note 1:

!

! !

n n

r r n r

Note 2: Most probable number

here refers to the mode of R .

Note 3: We can use TI84 Plus to

verify the correctness of r : HWA CHONG INSTITUTION

MATHEMATICS DEPARTMENT PAGE 11

1 10

18 3

r r

r r

1 3 18 10r r r r

5.53r

Thus, the required value of r is 6 .

12(i) Let X denote the number of people joining the queue in

a period of 4 minutes. Then ~ Po 4.8X .

Required probability P 8 1 P 7X X

0.113 correct to 3 s.f.

12(ii) Let Y denote the number of people joining the queue in

a period of t seconds.

Then 1.2

~ Po60

Y t

~ Po 0.02Y t

Given P 1 0.7Y

P 0 P 1 0.7Y Y .

Thus

0 1

0.02 0.020.02 0.02e e 0.7

0! 1!

t tt t .

0.02 0.02e 0.02 e 0.7t tt .

From the graphic calculator, 54.87t .

Thus 55t seconds (correct to the nearest whole

number).

TI84 Plus screenshots that display

the solving of the equation:

12(iii) Let J denote the number of people joining the queue in

a period of 15 minutes and let L denote the number of

people joining the queue in a period of 15 minutes.

Then ~ Po 15 1.2J and ~ Po 15 1.8L

~ Po 18J and ~ Po 27L .

Since in both cases, the mean number is larger than 10.

Thus, ~ N 18,18J and ~ N 27, 27L approximately.

This means ~ N 18 27,18 27J L approximately.

~ N 9, 45J L

By graphic calculator, the required probability is

P 11J L 0.617 correct to 3 s.f.

Remark: In the process of

computing P 11J L , we

did not do a continuity correction

even though J L is a discrete

random variable (it follows a

Skellam distribution).

12(iv) If the time period is in terms of several hours, the rate of

people joining the check-in queue would probably not be

constant throughout, which violates one of the conditions

for a Poisson model to be valid.

Remark: The rate of people

joining the queue would certainly

be higher at a certain period of

time just before the departure of a

flight. HWA CHONG INSTITUTION

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