further mathematics 9649/01 - the learning space

NATIONAL JUNIOR COLLEGE SENIOR HIGH 2 PRELIMINARY EXAMINATION Higher 2

FURTHER MATHEMATICS 9649/01

Paper 1 30 August 2019

3 hours

Additional Materials: Answer Booklet List of Formulae (MF26)

READ THESE INSTRUCTIONS FIRST

Write your name, registration number, subject tutorial group, on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for diagrams or graphs. Do not use staples, paper clips, glue or correction fluid.

Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.

The number of marks is given in the brackets [ ] at the end of each question or part question.

This document consists of 6 printed pages.

www.KiasuExamPaper.com 131

2

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1 Use de Moivre’s theorem to prove that 5 3sin 5 16sin 20sin 5sinθ θ θ θ= − + . [3]

Hence find the five distinct roots of the equation 5 35 5 1 0x x x− + − = in the form sink φ ,

where k is a positive real constant and 0 2φ< < . [3]

2 The sequence 1 2 3, , ,...a a a is such that 1 5a > and 1

3 25

4 4n

nn

aa

a+ = + .

(i) Prove by mathematical induction that 5na > for every positive integer n. [4]

(ii) Prove that 1n na a +> for every positive integer n. [2]

(iii) Another sequence 1 2 3, , ,...b b b is such that 1 6b > and 1

3 25

4 4n

nn

bb

b+ = + . Is this

sequence convergent? Justify your answer. [2] 3 The hyperbola H has parametric equations

secx a θ= , tany b θ= ,

where a and b are positive constants.

It is given that H has a focus at F(c, 0) and a directrix x k= , where c and k are positive

constants. (i) State the values of c and k in terms of a and b. [2]

(ii) Show that for any point P ( )sec , tana bθ θ on H, 2 2 secPF a b aθ= + − . [3]

Hence show that the ratio of PF to the shortest distance from P to the directrix x k=

is constant. Express this ratio in terms of a and b in the simplest form. [2] 4 The population of snails in a park, P, can be modelled by the differential equation

d 1

1d 10 500

P PP

t= − ,

where t is time in years. A pesticide is introduced to weed out the snails at a constant rate.

(i) How many snails should be killed by the pesticide each year to maintain its population at 300, and what is the minimum initial population size that can permit this? [5]

(ii) Sketch on the same diagram a family of solution curves to show how the population of snail changes with time for different values of initial population size. [3]


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5 Solve the equation 6 3 1 0z z− + = , giving you answer in the exact form ier θ , where 0r >

and θ− < ≤ . [4]

Mark these solutions as points on an Argand diagram, and show that the convex hexagon

with these points as vertices have perimeter 6cos18

. [5]

6 (a) Using Simpson’s rule with 3 ordinates, obtain a value for0

0

4 dx h

x hx x

+

−, where h is a

positive constant, and show that the value you obtain is in error by 5kh , where k is a positive constant to be determined exactly. [5]

(b) The differential equation

2d2

d

yxy y

x= −

is to be solved numerically using the second order Taylor series

( ) ( ) ( ) ( )2

0 0 0 02!

hy x h y x hy x y x′ ′′+ ≈ + + .

Given that ( )0 1y = and taking 0.2h = , calculate successively the value of ( )0.2y

and ( )0.4y . [4]

7 (a) (i) Suppose that an invertible 3 3× matrix A has non-zero eigenvalues 1 2 3, ,λ λ λ

with corresponding eigenvectors 1 2 3, ,e e e . Show that the matrix 1−A has the

same eigenvectors and find the corresponding eigenvalues of 1−A . [2]

(ii) Another 3 3× matrix B has eigenvalues 1 2 3, ,μ μ μ with corresponding

eigenvectors 1 2 3, ,e e e . State the eigenvalues and the corresponding eigenvectors

of the matrix 1−A B . Hence, find the eigenvalue of the matrix C corresponding

to the eigenvector 1e , where

( ) ( )21 1 1...

n− − −= + + + +C I A B A B A B , n +∈ and 1 1μ λ≠ . [3]

(b) The matrix 1 0

a b=

−E , ,a b ∈ , has real eigenvalues 1 2,β β .

(i) If 1 2β β≠ , show that 2 4a b> . [2]

(ii) Assume that 0a ≠ . State the value of b when E is singular. Find a matrix P and

a diagonal matrix D such that 1−=E PDP when E is singular. [3]


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8 (a) The curve with equation 2

2e 3xy x= + − has exactly one stationary point in the

interval [ ]1, 0− . Use the Newton-Raphson method to find the x-coordinate of the

stationary point, correct to 4 decimal places. [5]

(b) (i) Show that the equation 3 7 2 0xx − + = has a root, α , in the interval [0, 1]. [1]

(ii) Student A uses the recurrence relation ( )31

114 2

7 n nnx x x+ = − − for finding α .

Explain why Student A will fail to find α . [2]

(iii) Student B uses the recurrence relation ( )31

12

7n nx x+ = + for finding α . Find the

approximate value of α to 5 decimal places. [2] 9

situation can be modelled using Cartesian coordinates shown in the diagram below. A point P(x, y) lies on the circumference of the coin and initially, P rolled in the direction of the positive x-direction, and it is always in contact with the x-axis.

In the diagram above, the dotted circle represents the position of the coin initially and the circle with a solid circle represents the position of the coin after it has rolled. Let C denote the centre of the coin, T denote the foot of perpendicular of C to the x-axis and θ denote the angle POT. (i) Show that the coordinates of P are given by ( )1.15( sin ), 1.15(1 cos )θ θ θ− − . [3]

(ii) Sketch the path traced out by P, labelling the x-intercepts exactly. [2] (iii) Calculate the exact length of the path. [5]

O

C

T

P

x

y

θ


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10 An entrepreneur started a sea-cage fish farm with an initial rearing of 0u silvery fish at the

start of year 2010. He intends to increase the total number of silvery fish on 2 January of each subsequent year by 20% of the number of surviving fishes in the previous year. The farm’s rearing technique and the conditions of sea water result in a death rate of 1% of the number of fishes at the end of each year.

Let nu denotes the total number of surviving silvery fish in the farm at the end of year n.

(i) Write down expressions for 1,u 2u and 3u in terms of 0u . [2]

(ii) By considering ,nu or otherwise, find the year when the total number of surviving

silvery fish in the farm will first exceed 5 times its initial number. [2] On 1 January 2017, a mysterious disease wiped out 40% of the silvery fish in the farm. Changes were made to the rearing technique that enhanced the survival rate of the fishes. As a result, the death rate of the fishes has maintained at a constant value of 10 per year since 2017.

Let 0 500u = and nv be the total number of surviving silvery fish in the farm at the end of

year n, starting from January 2017.

(iii) Write down a recurrence relation for nv , stating the value of 0v correct to the nearest

whole number. [2]

(iv) Hence find nv in terms of n. [3]

(v) Comment on the behaviour and validity of the model obtained in part (iv). [2]


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11 Two 50-litre tanks, Tank A and Tank B (as shown in diagram below), containing salt solution are connected by two horizontal pipes. Both tanks have inlets and outlets where salt solution flows in and out of the tanks. The rates of flow in the inlets, outlets and pipes are managed in such a way that both tanks will be full at all times.

Tank A is receiving salt solution at a concentration of 1 gram per litre at a rate of 1 litre per minute and Tank B is receiving pure water at the same rate. Salt solution is flowing out from Tank A from the bottom outlet at a rate of 2 litres per minute. Salt solution flows from Tank A to Tank B through one of the horizontal pipes at a rate of 2 litres per minute and flows in the reverse direction through the other horizontal pipe at a rate of 3 litres per minute.

(i) Suppose that at time t minutes, the amount of salt in Tank A and Tank B are x grams

and y grams respectively. Show that

d 2 31

d 25 50

xx y

t= − + + and

d 1 3

d 25 50

yx y

t= − . [2]

(ii) Prove that 2

2

d d,

d d

y yp qy r

t t+ + =

where p, q and r are constants to be determined. [3] (iii) Find the general solution of the differential equation in (ii). [4] (iv) By expressing x in terms of t, find the ratio of the amount of salt in Tank A to the

amount of salt in Tank B in the long run. [3]

--- END OF PAPER ---

Pure water Salt solution

Tank A Tank B

Salt solution


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2019 SH2 FM Prelim Exam Marker Comments Q1 Solution Comments

5

25 4 3

3 4 52

cos5 sin 5 cos sin

cos 5cos sin 10cos sin

10 cos sin 5cos sin sin

i i

i i

i i i

Comparing imaginary parts,

4 2 3 5

22 2 3 5

3 5

sin 5 5cos sin 10cos sin sin

5 1 sin sin 10 1 sin sin sin

5sin 20sin 16sin

This part is well done.

5 35 5 1 0x x x Let sinx k ,

5 5 3 3

4 5 2 3

sin 5 sin 5 sin 1 01sin 5 sin 5sin

k k k

k kk

Let 2k , 5 3 116sin 20sin 5sin

21sin 52

π 13π 25π 37π 49π π5 , , , , or 2 π, 0,1,2,3,46 6 6 6 6 6π 13π 5π 37π 49π π 2 π, , , , or , 0,1,2,3,4

30 30 6 30 30 30 5

k k

k k

Hence the roots are π 13π 5π 37π 49π2sin , 2sin ,2sin , 2sin , 2sin

30 30 6 30 30x

A lot of students were not able to find k. They were not able to produce five distinct solutions as well.

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Q2 Solution Comments (i) Let nP be the statement 5na for every positive integer

n. By given condition, 1 5a so 1P is true. Assume that kP is true for some positive integer k.

1

2

2 22

2

3 255 54 43 20 25

4 3 3

3 20 10 10 254 3 3 3 3

3 10 254 3 9

kk

k

k kk

k kk

kk

aaa

a aa

a aa

aa

2

2

2

10 5When 5,3 3

10 253 9

10 25 03 9

3 10 25 04 3 9

k k

k

k

kk

a a

a

a

aa

So 1 5ka and 1kP is true. Since 1P is true and kP is true implies 1kP is true, nP is true for all positive integers n.

This part was well done. Some students need to revise this topic.

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(ii)

1

2

3 254 4254 41 25

40 since 5

nn n n

n

n

n

nn

n

aa a aa

aa

aa

a

Although most students were able to get full credit, the proofs used were often tedious.

(iii) Yes. This is because 5nb and the sequence is decreasing. So it is a convergent sequence.

Most students failed to recognise that this sequence is a subcase of part (i).

Q3 Solutions Comments (i) 2 2

2 2 1x ya b

2 2c a b and 2

2 2

aka b

This part is well done in general.

(ii) Consider any point P sec , tana b on E.

2 22 2

2 2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 2 2 2 2 2

22 2

2 2

sec tan

sec 2 sec tan

sec 2 sec sec

sec 2 sec

sec

sec

PF a a b b

a a a b a b b

a a a b a b

a b a a b a

a b a

a b a

This part is well done in general. A few students intended to remove the modulus signs which is not necessary.

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Q3 Solutions Comments

Distance P to the directrix is 2

2 2seca a

a b

2 22 2

2 2

2

2 2 2 2

2 2

secsec

sec sec

aa ba b a a ba aa a

a b a b

a ba

This part is well done in general.

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Q4 Solution Comments (i) Let P denotes the population of the snail at time t and k

denotes the constant rate of pesticide. The appropriate logistic model is

d 1 1d 10 500P P P kt

For stable population values, d 0dPt .

Set 300P and we have 1 31 300 0

10 512

k

k

To find the equilibrium population, we solve

2

0

50

1

00

1 12 010 500

500 12

500 60000 0200 300

0

PP

P P

P P

PP

200 or 300P P Thus, minimum initial population size is 201.

This question is well done for students who chose the appropriate logistic model. The pesticide information is the hint to the harvesting condition. The common mistake is to take the minimum as 200. However, based on the graph in part (ii), we know that taking 200 as the initial population will cause the population to be stagnant. Then combined with the knowledge that number of snails is considered to be discrete, we should take the answer to be 201.

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Q4 Solution Comments (ii)

Generally, students have no issues with the behaviour above 300 and between 200 and 300. Students tend to think that the behaviour below 200 can be concave upwards, as well as tending towards zero (t-axis).

We can check this by observing d 1 1 12d 10 500P P Pt

for

different values of P for 200P

d 0.22dPt

when 190P

d 4.6dPt

when 90P

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Q5 Solution Comments

6 3 3 1 31 02

z z z

π πi i 2 π

3 3 31 1 3i e e2

kz

or

π πi i 2 π

3 3 31 1 3i e e2

kz

5π π 7π 7π π 5πi i i i i i9 9 9 9 9 9e , e , e ,e , e , ez

All the students were able to carry out the correct steps to find the cubic roots of a complex number after finding the two quadratic roots. However, a few made calculation errors in the process.

π2sin9

CD FE AB

This part is generally done well. A few students did not manage to show the correct relative positions of these points.

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Q5 Solution Comments 7π 2π2sin π 2sin9 9

AF BC DE

The perimeter of ABCDEF is π 2π 2π π2sin 3 2sin 3 6 sin sin9 9 9 9

1 2π π 1 2π π6 2sin cos2 9 9 2 9 9

π π12sin cos6 18π6cos

18

This part is generally done well. A few students took a long way to prove the results. They are encouraged to devise a plan before writing.

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Q6 Solution Comments (a)

00

00

4 5

5 50 0

5 4 3 2 2 3 4 50 0 0 0 0

5 4 3 2 2 3 4 50 0 0 0 0

4 2 3 50 0

4 2 3 50 0

1d5

1 15 51 5 10 10 551 5 10 10 551 10 20 25

22 45

x hx h

x hx h

x x x

x h x h

x x h x h x h x h h

x x h x h x h x h h

x h x h h

x h x h h

By Simpson’s rule,

0

0

4 44 40 0 0

4 3 2 2 3 4 40 0 0 0 0

4 3 2 2 3 40 0 0 0

4 2 2 40 0

4 2 3 50 0

d 43

( 4 6 4 43

4 6 4 )

6 12 23

22 43

x h

x h

hx x x h x x h

h x x h x h x h h x

x x h x h x h hh x x h h

x h x h h

Error term is 5 5 52 2 43 5 15

h h h where 415

k

This part is not done in general. Common Mistakes: 1. Some students hesitated to carry out the binomial expansion with power 5, and similar problems occurred in simplifying the results from Simpson’s rule. 2. Calculation errors are seen commonly. Students need to be prepared to perform some calculations that may look tedious in a Maths Paper.

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Q6 Solution Comments (b) 2d 2

dy xy yx

2

2

d d d2 2 2d d d

y y yy x yx x x

By letting 0 0x and 0.2h ,

20.20.2 0 0.2 0 0

2!y y y y

0 1y 20 2 0 1 1 1y 0 2 1 2 0 1 2 1 1 4y

0.2 1 0.2 1 0.02 4 0.88y

By letting 0 0.2x and 0.2h ,

20.20.4 0.2 0.2 0.2 0.2

2!y y y y

0.2 0.88y 20.2 2 0.2 0.88 0.88 0.4224y

0.2

2 0.88 2 0.2 0.4224 2 0.88 0.42242.334464

y

0.4 0.88 0.2 0.4224 0.02 2.334464

0.842209280.842

y

This part is not one well in general. Though most of the students managed to find the derivatives correctly, many were still lacking of accuracy in evaluating the expressions.

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Q7 Solution Comments (a)(i)

, 1, 2,3i i i i Ae λ e 1 1

1 1i i i

i i i

A Ae A eA e e

Hence the matrix 1A has the same eigenvectors 1 2 3, ,e e e and the corresponding eigenvalues are 1 1 1

1 2 3, , .

This part was well answered. Students have to remember to state the eigenvectors with the respective eigenvalues.

(a)(ii)

The eigenvalues of 1A B are 1 1 11 1 2 2 3 3, , and the

corresponding eigenvectors are 1 2 3, ,e e e .

21 1 11 1

21 1 11 1 1 1 1 1 1

111 1

111 1

...

1 ...

11

n

n

n

Ce I A B A B A B e

e

e

The eigenvalue of the matrix C corresponding to the

eigenvector 1e is 11

1 11

1 1

11

n

.

This part was well done.

(b)(i)

2

01

0

0

a b

a b

a b

2 42

a a b

Since 1 2 , 2 24 0 4 (Shown)a b a b

Most students were able to show clear working.

(b)(ii)

0b The eigenvalues of E are 0 and a.

This part was well done, except that some students made careless mistakes in finding the eigenvectors.

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Let 1 0 , then solve 0 0

1 0 0a x

y

and

01

is an

eigenvector.

Let 2 a , then solve 0 0 01 0

xa y

and 1a

is

an eigenvector.

So 01 1

a

P and 0 00 a

D .

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Q8 Solution Comments (a) 2d 4 e 1 0

dxy x

x

The Newton-Raphson formula as follows: 2

2 21 2

4 e 18 e 4e

n

n n

xn

n n x xn

xx xx

1

3

4

0

2

0.50.296470.239060.236420.23641

xxxxx

Hence, 0.2364x . Checking,

2

2

0.23635 4

0.23645 4

4 0.23635 e 1 2.9 10 0

4 0.23645 e 1 1.8 10 0

This question is well done for students who noticed that the

f(x) is dd

yx

, rather than y.

Some students are still not doing the checking of sign change.

(b) (i)

3 7 2

f 0 2

f 1 4

f xx x

Since f 0 f 1 0 and f is continuous, there is a root in [0, 1].

Well done.

(b) (ii) 3

1 21 147n n nx x x

1 2d 32d 7

nn

n

xx

x

For 0 1nx ,

1d11 27 d

n

n

xx .

Note has to be taken to write the argument clearly. The question has already reveal that it would not converge to .

Thus, there must be justification about why

1d1

dn

n

xx .

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Since

1d1

dn

n

xx , the sequence does not converge to .

Alternatively, use graph:

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(b) (iii)

0

1

2

3

4

5

6

10.4285714290.29695960.2894553410.2891788340.2891689150.28916856

xxxxxxx

0.289 17 Checking,

3 5

3 5

0.289165 7(0.289165) 2 2.4 10 00.289175 7(0.289175) 2 4.4 10 0

This question is well done , except some students are still not doing the checking of sign change.

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Q9 Solution Comments (i) As the coin is rolling without losing contact with x -axis,

we note that arc length PTOT . Consequently,

1.15 1.15sin1.15 s

sin

in

x OT PC

Similarly,

1.15 1.15cos1.15 1

cos

cos

y PT PC

This question is well done.

(ii)

Students confuse the x-intercepts as the number of revolutions.

(iii) Length of path traced Students confuse the limits as the x-intercepts of the graph in part (ii).

2.3 4.6 O

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2 2

2 2

2 2

2

0

2

0

2

02

0

2

0

2 2

d d2 dd d

2 1.15 1 1.15 d

2 1.15 1 d

cos sin

cos sin

cos cos s2.3 1 2 in

c

d

2.3 2 s1 do

x y

2

022

0 0

22.3 2 2 d2

4.6 d 4.

sin

sin cos

cm

6 22 2

18.4

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Q10 Solutions Comments (i) Year n nu

0 0u 1 00.99u 2

0 0

20

1.2 0.99 0.01 1.2 0.99

1.2 0.99

u u

u

3

2 2 2 20 0

2 30

1.2 0.99 0.01 1.2 0.99

1.2 0.99

u u

u

This part is very poorly done. Most of the students were not able to interpret the context correctly and translate it into mathematical language. Some students simply assumed a wrong pattern by merely observing the first two or three terms.

(ii) From (i), we observe that 101.2 0.99n n

nu u for

1.n For 05 ,nu u

1 10 01.2 0.99 5 1.2 0.99 5n n n nu u

From GC, 10

11

4.675.54

uu

Thus, the total number of surviving silvery fish in the farm will first exceed 5 times its initial number in 2020.

Those who successfully obtained the answer for part (i) were able to get the correct answer for this part.

(iii) 6 70 70.6 0.6 1.2 0.99 500

834.94835 (nearest whole number)

v u

11.2 10, 1n nv v n

Similar issue is observed in the answers. Many students were not able to interpret the context correctly. As a result, many obtained a wrong initial value.

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Q10 Solutions Comments (iv) 1.2 n

nv A B

11.2 1.2 1.2 10

1.2 1.2 1.2 100.2 10

50

n n

n n

A B A B

A B A BB

B

0835 1.2 50 785A A

785 1.2 50, 0nnv n

This part is generally well done. Most of the students were able to recall and apply the approach to find the general solution to a first order recurrence relation.

(v) The model is not appropriate as it suggests that the total number of fishes in the farm increases exponentially in the long run but the capacity of the farm has a limited size.

This part is well answered in general.

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Q11 Solution Comments (i) d rate of salt flowing in rate of salt flowing out

d3 2 2150 50 50

2 3 125 50

xt

y x x

x y

d rate of salt flowing in rate of salt flowing outd

2 350 501 325 50

yt

x y

x y

Most students were able to show clear and detailed working.

(ii) d 1 3d 25 50y x yt

2

2

d 1 d 3 dd 25 d 50 d

1 2 3 3 d125 25 50 50 d

2 3 1 3 d625 1250 25 50 d2 d 3 3 1 3 d25

625 d 2 1250 25 50 d7 d 3 1

50 d 1250 25

y x yt t t

yx ytyx yt

y yy yt t

y yt

2

2

d 7 d 3 1d 50 d 1250 25

y y yt t

Some students were careless in their arithmetic and could not get the correct second order DE.

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Q11 Solution Comments (iii) 2 7 3 0

50 12501 3 or

50 25

m m

m

Complimentary solution is 1 350 25e e

t t

cy A B

Particular solution is 503py

The general solution is 1 350 25 50e e

3t t

y A B

Some students were careless in solving the characteristic equation, leading to the wrong complimentary solution. Some students were not aware that a constant is sufficient to guess the particular solution.

(iv)

1 3 1 350 25 50 25

1 350 25

d 325d 2

1 3 3 5025 e e e e50 25 2 3

3e e 252

t t t t

t t

yx yt

A B A B

A B

As t , 25 350 23

xy

This was well done.

[Turn over

NATIONAL JUNIOR COLLEGE SENIOR HIGH 2 PRELIMINARY EXAMINATION Higher 2

FURTHER MATHEMATICS 9649/02 Paper 2 18 September 2019 3 hours Additional Materials: Answer Booklets List of Formulae (MF26) READ THESE INSTRUCTIONS FIRST Write your name, registration number, subject tutorial group, on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in the brackets [ ] at the end of each question or part question.

This document consists of 7 printed pages and 1 blank page.


2

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Section A: Pure Mathematics [50 marks]

1 A sequence of real numbers { }1 2 3, , , ..., , ...nu u u u is defined by the second-order

recurrence relation

1 24,u u k= = and 2 1 1 2n n nu k u k u+ += + for 1,n ≥

where 1 2, and k k k are real constants.

(i) If the general term nu is of the form n nA Bα β+ , where A, B, α and β are real

constants and α β≠ , prove that 21 24 0k k+ > . [2]

(ii) Solve the recurrence relation when 1 2k = and 2 10k = − . You should leave your

answer in a form involving complex numbers. [4]

2 (a) Let A be a nonzero square matrix such that 2 =A A . Determine all possible values of ( )det A . [2]

Determine if the following statements are true. Justify your answer. (i) −I A is always invertible. (ii) +I A is always invertible. [3]

(b) Let

a b c

d e f

g h i

=B . Given that B is the inverse of a matrix C, and D is the matrix

obtained from C by adding to the second row of C twice the first row of C, find 1−D in a similar form to B. [3]

3 Find the general solution of d

3 2 0d

yx y x

x− − = , expressing y in terms of x. [3]

(i) Obtain the particular solution of the solution curve C that passes through the point

( )2, 4 . [2]

(ii) Sketch and label clearly on the same diagram, C and two other members of the family of solution curves, of which one is a straight line and the other has a different number of stationary points from C. [3]

4 The curve D has polar equation 1

6sin2

r θ= , where 0 2θ≤ < .

(i) Sketch D, indicating clearly all key features and symmetries of the curve. [2] (ii) Find the arc length of D. [2]

The locus of points ( ),r θ satisfying 1

6sin 32

rθ ≤ ≤ forms a region R.

(iii) Find the exact area of R. [5]


3

[Turn over

5 Prove that the locus described by the equation ( ) ( )1 i 2 3 iw w− + = − + , where w is a

variable complex number, is a circle and find its radius and the coordinates of its centre. [3]

The variable complex number z satisfies the following inequalities:

(I) ( ) ( )1 i 2 3 iz z− + ≥ − + ;

(II) ( )4 6 2iz z− ≥ − + .

Sketch, on a single Argand diagram, the region in which the point representing z can lie in. [3]

The range of ( )arg z is ( )1 1tan arg tana z b− −≤ ≤ . Determine the exact values of a and b.

[3] 6 The linear transformation 4 3T : → is represented by the matrix M, where

1 5 1 2

1 3 4 3

1 11 8 1

− −

= − −M .

(a) (i) Find a basis for ( )TR , the range space of T. Give a precise geometrical

description of ( )TR . [3]

(ii) Find a basis for ( )TK , the null space of T. [2]

(iii) Hence find the general solution of the equation

( )6 5

T 4 3

12

α β

α β

α β

−

= − +

−11

x ,

where α β, ∈ , leaving your answer in terms of α and β . [2]

(b) Let 3V ⊆ be the set that satisfies the following properties:

(A) ( ) { }TV R∩ = 0 ;

(B) ( ) 3TV R∪ = .

Determine whether V is a subspace of 3 . [3]


4

[Turn over

Section B: Probability and Statistics [50 marks]

7 In an election in which there were a large number of voters, a sample of 2000 voters at certain polling stations were interviewed just after they had voted, and 1612 of them said they had voted for candidate Y. The overall proportion of voters who voted for candidate Y is p. (i) Assuming that the figures above can be used to estimate p, the symmetrical

( )100 %α− confidence interval for p is found to be ( )0.78118, 0.83082 . Find the

value of α , giving your answer to 1 decimal place. [3] (ii) After the votes were all counted, candidate Y had in fact obtained 89% of the votes.

Give one possible reason why this figure does not lie within the confidence interval in part (i). [1]

8 X and Y are independent random variable having Poisson distributions with mean λ and

μ respectively.

(i) Show that ( )P |x n x

nX x X Y n

x

λ μ

λ μ λ μ

−

+= + = =

+, where n, x are non-

negative integers and n x≥ . [3] (ii) Name the probability distribution given in part (i) with its parameter(s), justifying

your answer. [2] 9 A bag contains 12 white and 18 black balls. Balls are drawn, one at a time, in which any

black ball drawn is returned to the bag before the next ball is drawn and any white ball drawn is not replaced.

Let 1, 0,1, 2, ,11,rT r+ = denote the number of draws required to obtain the ( )1 thr +

white ball after the rth white ball has been drawn.

(i) Show that ( )( )

( )

1

1

12 18P , 1, 2, 3,

30

s

r s

rT s s

r

−

+

−= = =

−. [2]

(ii) Find the expected number of draws needed until all the white balls have been selected, giving your answer correct to four significant figures. [3]


5

[Turn over

10 Part of a research study of identical twins who had been separated at birth involved a random sample of 9 pairs, in which one twin had been raised by the natural parents and the other by adoptive parents. The IQ scores of these twins were measured when they were 7 years old, with the following results.

Twin Pair 1 2 3 4 5 6 7 8 9

IQ score of twin raised by natural parents

86 88 115 132 104 94 113 88 125

IQ score of twin raised by adoptive parents

82 92 115 124 97 95 111 83 120

(a) One researcher claimed that there was no evidence of a difference between the two

population IQ scores. Use a suitable t-test to test this claim at 10% level of significance and state a necessary assumption for this test. [5]

(b) Another researcher wanted to investigate the same claim by using non-parametric tests at the same level of significance. Carry out both (i) a signed test, and (ii) a Wilcoxon matched-pairs signed rank test. Explain why part (b)(ii) is a better non-parametric test than part (b)(i). [7]


6

[Turn over

11 It can be assumed that the lifetime in months of a smartwatch from Brand K follows an exponential distribution and has an expected value of 20.

(i) Peter bought a smartwatch from Brand K and started using it on 1 January 2018.

Given that this smartwatch was still working on 1 January 2019, find the probability that this smartwatch will break down in the year 2019. [3]

(ii) Brand K had a promotion drive on 1 August 2019 and a large quantity of its

smartwatches were sold on that day. Assuming all the customers who bought the smartwatches on this date started using them on the same day, estimate the date before which half of these smartwatches will break down. [3]

A smartwatch from Brand K is priced at 450 dollars with a default warranty period of 12 months. At the point of purchase, the customer has a choice to extend the warranty period to 24 months for an additional k dollars. The customer may replace the smartwatch for free if it breaks down within the warranty period but there will be no more warranty for the replacement. Mary has the following plan in mind before purchasing a smartwatch from Brand K. (1): If the smartwatch breaks down within the warranty period, she will get a

replacement for free, and she will not buy from Brand K again if the replacement breaks down.

(2): If the smartwatch breaks down within three years but the warranty has expired, she will buy a new one from Brand K at the same price without extending the warranty, but she will not buy from Brand K again if this new smartwatch breaks down.

By considering the expected amount of money that Mary will spend on Brand K within three years, discuss how the value of k, to the nearest dollar, affects Mary’s decision on whether to extend the warranty. [5] The probability that Mary still has a working smartwatch from Brand K at the end of three years is denoted by p. Explain how the value of p is affected by Mary’s decision. [1]


7

[Turn over

12 (a) The number of attempts at dialling needed to get through to a service centre’s hotline was investigated, and the results for 120 occasions are shown in the table below.

Number of attempts

1 2 3 4 5 6 7≥

Frequency 45 41 16 12 4 2 0

It is assumed that the distribution of the number of attempts is geometric. (i) Use the sample mean to estimate the value of p, the probability of getting

through to the hotline. By using this estimate of p, carry out an appropriate test, at 5% significance level, to assess whether or not the assumption is reasonable. [6]

(ii) It is now given that 0.5p = . Determine the change in the degree of freedom

from the test in part (i). [1] (b) A local newspaper conducted a survey of the level of satisfaction with a particular

shopping centre. A random sample of adults were asked to rate their levels of satisfaction with the restaurants and the supermarket in the shopping centre, using a scale of ‘low’, ‘medium’ and ‘high’ for each.

The table below, showing the responses in percentages, was published in a newspaper article. The article did not state the number of respondents to the survey.

Level of satisfaction with restaurants Low Medium High Level of satisfaction with supermarket

Low 3% 11% 6%

Medium 7% 32.5% 4.5%

High 10% 19% 7%

(i) A reader used the figures in the table to investigate if the levels of satisfaction

with the restaurants and the supermarket in the shopping centre are independent. He carried out a chi-square test at 5% level of significance with the sample size assumed to be 200n ≥ . He concluded that they are not independent. If he had instead assumed that the sample size was 2n, without any calculation, state the change in the value of the test statistic and explain whether or not the conclusion of the test would be changed. [2]

(ii) What is the minimum sample size required to show, at 0.1% level of significance, that the two categories are not independent? [3]

--- END OF PAPER ---


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2019 SH2 FM Paper 2 Markers Comments Q1 Solution Comments (i) Characteristic equation is 2

1 2 0m k m k .

Since , we have 21 24 1 0k k .

Hence 21 24 0k k .

Most students were able to write down the characteristic equation. Students are reminded to show sufficient working for “show” questions, even if the steps are short and simple.

(ii) Characteristic equation is 2 2 10 0m m . Solving, we get 1 3im . Hence the general solution is

1 3i 1 3in nnu A B .

1 4 1 3i 1 3i 44 and 3 3 02

u A BA B A BA B

The general solution is 2 1 3i 2 1 3in nnu

Mistakes usually occur when solving the characteristic equation. Some students used values of 1u and 2u to solve for A and B, thus overcomplicating their solution.


2

2

det det

det det

det det 1 0

A A

A A

A A

Hence, det 0 or 1A .

Some students think that for 2 A A , then A 0 or A I . This stems from the working

2

2

A AA A 0

A A I 0, which wrongly leads to A 0 or A I 0 .

Counterexamples can be 3 61 2

and 2 2 41 3 4

1 2 3

.

Matrices that satisfy 2 A A are called idempotent matrices.

of 20

Q2 Solution Comments (a) (i)

No. Counterexample: A = I. Then I A 0 .

Usually if we think that the statement is false, we prove it by providing a counterexample, rather than a direct proof.

(a) (ii)

True. Suppose not, then there exists a nonzero x such that

2

(1)

I A x 0x Ax 0

A x Ax A0

Ax A x 0Ax Ax 0

Ax 0

Sub Ax 0 into (1), we get x 0 , a contradiction. Alternatively,

Consider 1 22

I A . Then we have

2 21 12 2 22 2

1 221 22

I A I A I A A A

I A A

I

I

The first approach is using proof by contradiction. The second approach is to come up with a general form of the inverse matrix, which implies that it is invertible.

(b) 1C B

Let 1 0 02 1 00 0 1

E , which defines the row operation.

Students who can define the elementary row operations would be able to solve the problem.

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1

1 1

1 0 02 1 0

0 0 1

222

a b cd e fg h i

a b b cd e e fg h h i

D ECEB

D BE

Q3 Solution Comments d 3 2 0

dyx y xx

d 3 2dy yx x

3 d 3IF ex

x x

3 4 3

3 3

3 2

3

d 3 2d

d 2d

yx x y xx

x y xx

x y x cy x cx

Most students were able to obtain the correct integrating factor and solve the DE correctly.

(i) When 2x , 4y , so 34 2 84

c c .

Particular solution is 334

y x x .

This is well done.

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Q3 Solution Comments (ii)

Two common mistakes: 1. Students did not sketch the case when 0c .

2. Students did not sketch 334

y x x correctly.

Some students did not check which values of c will give a

different number of stationary points from 334

y x x , which

they should do since it is easy to check for this question.

y x

334

y x x

3, 0y x cx c

of 20

Q4 Solution Comments (i)

This part is well answered. Students are reminded to label the key features in polar system (e.g. coordinates and equations)

(ii) Arc length of C 2π 2

2

02π 2 2

0

d dd

1 16sin 3cos d2 2

29.06529.1

rr

This part is well answered.

6, π 0

of 20

Q4 Solution Comments (iii)

R is one of the regions bounded by C and the circle with polar equation 3r , as shown in the diagram. To find the point of intersection

16sin 321 1sin2 21 π 1 5π or 2 6 2 6

π 5π or 3 3

This part is not answered well by some students. In the process of evaluating the area of the region, some wrongly applied addition instead of subtraction.

of 20


π23

0

π32 2

0

π3

0

π30

Area of

π 12 area of sector with angle 6sin d3 2 2

1 π2 3 18sin d2 3 2

3π2 9 1 cos d2

3π 18 sin

π π3π 18 sin3 3

9 3 3π

R

of 20

Q5 Solution Comments 1 i) 2 3 iz z

2 2 2 2

2 2

2 2

1 1 2 3 1

10 18 2 0

5 1 8

x y x y

x x y y

x y

Hence the radius is 2 2 and the centre is (5, 1).

Most students were able to obtain the Cartesian equation of a circle, except for those who made careless mistakes.

For the students who obtained the correct Cartesian equation of the circle in the previous part, they were able to sketch the locus properly. Most graphs were neat enough and the required information was indicated clearly.

Cartesian equation of 4 6 2iz z is

2 2 224 6 28 16 12 36 4 4

6

x y x yx x y

y x

Solving for intersection points, we get

This part was well done.

Re(z)

Im(z)

(3, 3)

O

(5, 1)

(7, –1) 2 2

of 20

Q5 Solution Comments 3, 3x y or 7, 1x y .

Hence 17

a and 1b .

of 20


rref1 5 1 2 1 0 8.5 4.51 3 4 3 0 1 1.5 0.5

1 11 8 1 0 0 0 0

Since the first two columns are pivot columns, a basis for

TR is 1 51 , 3

1 11

.

TR is a plane passing through the origin and

perpendicular to the vector1 5 81 3 6

1 11 2

.

The first part is well done. The second part requires a geometrical description, thus using mathematical expressions to describe is not accepted. For e.g. TR is a plane with equation

1 51

1,3 ,

1 1

r � is not allowed.

(a)(ii)

Consider the augmented matrix 1 5 1 2 01 3 4 3 0

1 11 8 1 0

for the linear system Mx 0 . By the GC, we can obtain

8.5 4.51.5 0.5

T span ,1 00 1

K

.

Hence,

8.5 4.51.5 0.5

,1 00 1

is a basis for TK .

This question is well done.

of 20

Q6 Solution Comments (a) (iii) Since

6 5 1 54 3 1 3

12 1

, the general

solution is 1 2

8.5 4.51.5 0.5

0 1 00 0 1

k k

x ,

1 2,k k .

Students who can express 6 54 3

12

as a linear

combination of the basis of TR would be able to score in this question. Note to be taken to state that 1 2,k k are real scalars.

(b)

Let 1 01 , 0

0 1V

. To show that these vectors belong to

V, we need to show that they do not belong to the range space of T. Consider the matrix rref

rref1 5 1 0 1 0 0 11 3 1 0 0 1 0 0

1 11 0 1 0 0 1 1

.

This implies that the 1st, 2nd and 3rd columns, as well as the 1st, 2nd and 4th columns are linearly independent.

Hence, 1 01 , 0

0 1V

.

Now, 1 0 11 0 1 range space of T

0 1 1

The issue mainly is perhaps students did not have a sense of whether they want to prove or disprove.

of 20


11

1V

.

So, V is not closed under vector addition and hence not a subspace of 3 .

of 20

Q7 Solution Comments (i) 1612ˆ 0.806

2000p (or 0.78118 0.83082ˆ 0.806

2p )

1 0.005

1 0.005

1 0.005

0.806 1 0.8060.83082 0.806

20000.0088420586 0.02482

2.8070386231 0.005 0.997500

0.005 0.0025000.5 (1 d.p.)

z

zz

This part is well answered.

(ii) The voters sampled were from the same polling stations. This is not a random sample from the entire population, so it is not valid to compute the confidence interval with these figures, as candidate Y may be more popular at other polling stations.

The key concept in this answer is that the sample is not random. Students are expected to illustrate further why a 99.5% confidence interval cannot capture the mean.

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Q8 Solution Comments (i)

P |

PP

e e !! ! e

!! !

x n x

n

n x

n

n x

n x

x n x

x

x

x

X x X Y n

X x Y n xX Y n

nx n x

nx n x

nx

nx


(ii)

P |

PP

1

x n x

x n x

X x X Y n

X x Y n xX Y n

nx

nx

It is a binomial distribution, B ,n

.

Some students missed out the instruction that the distribution is to be named with its parameters.

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Q9 Solution Comments (i) After the rth white ball has been drawn,

probability of drawing a black ball 1830 r

and

probability of drawing a white ball 1230

rr

.

1

1

1

18 12P30 3012 18

30

s

r

s

s

rT sr rr

r


(ii) When 0r , 1230

p . Thus expected number of draws is 3012

.

When 1r , 1129

p . Thus, expected number of draws is 2911

.

When 2r , 1028

p . Thus, expected number of draws is 2810

.

…

Thus, total number of draw 11

0

3012

67.86r

rr

Students who realised that the drawing of rth white ball follows geometric distribution would be able to use the expectation of geometric distribution.

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Q10 Solution Comments (a) To conduct a paired-sample t-test.

Differences X are: 4, 4, 0, 8, 7, 1, 2, 5, 5 . To test: H0: 0 Against H1: 0 at 10% level of significance.

Under H0, 0 ~ 8

9

XT tS

.

By GC, p-value = 0.0596 < 0.10. We reject H0. There is sufficient evidence at 10% level of significance to conclude that the two population means differ. Therefore, the researcher’s claim is not valid. Assumption: The difference between the IQ scores follows a normal distribution.

This part is well answered in general, though a few students made a critical mistake in choosing the appropriate test. A common error is that the students failed to include the degree of freedom or t(8) in the presentation.

(b) (i)

To test H0: 0Dm H1: 0Dm at 10% level of significance.

D 4 4 0 8 7 1 2 5 5 S + - + + - + + +

Under H0, the number of positive signs ~ B 8, 0.5K

p-value = 2P 6 0.289 0.10K . We do not reject H0. There is insufficient evidence at 10% level of significance to conclude that the differences between the populations have a median differing from 0.

This part is fairly scored. Some students failed to compute the correct p-value for a two-tail test.

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Q10 Solution Comments (b) (ii)

D 4 4 0 8 7 1 2 5 5 S + - + + - + + + R 3.5 3.5 8 7 1 2 5.5 5.5

Assume the differences are symmetric. To test H0: 0Dm H1: 0Dm at 10% level of significance. Perform Wilcoxon signed rank test where 8n . Critical Region: 5W

Calculation: 3.5 8 7 2 5.5 5.5 31.51 3.5 4.5

PQ

min( , ) 4.5 5W P Q . Reject H0. There is sufficient evidence at 10% level of significance to conclude that the differences between the populations have a median differing from 0.

This part is well-answered.

(b)(ii) is better as it takes the differences from the assumed median into consideration in the test in the form of ranks.

This part is well-answered.

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Q11 Solution Comments (i) Let X be the lifetime in months of the smartwatch from

Brank K. Then 1~ exp20

X

. Its cdf,

0.05

0 if 0F

1 e if 0x

xx

x

0.05 12

P 12 24 | 12

P 0 12

1 e0.451180.451

X X

X

This part is fairly scored by students. Students who applied the memoryless property scored well while students chose to solve it in other approach were often seen careless in their working.

(ii) Let the median be m.

0.05

P 0.5

1 e 0.50.05 ln 0.5

13.863 months 13 months and 26 days

m

X m

mm

The date is estimated to be 27 September 2020.

This part is fairly scored by students. Mistakes occurred are mainly due to students missed out the key word ‘date’ or ‘before’.

(1) If Mary decides not to extend the warranty, the expected amount she needs to spend is

0.05(12) 0.05(36) 0.05(36)

450P 12 900P 12 36 450P 36

450 0.45118 900 e e 450 e

622.58

X X X

(2) If Mary decides to extend the warranty the expected amount she needs to spend is

This part is poorly answered. Only a few students managed to compute the expected amount under different decisions. Among those who successfully computed the two expectations, majority did not make use of the condition ‘integer’ to discuss the cases.

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Q11 Solution Comments 0.05(24) 0.05(36) 0.05(24) 0.05(36)

450 P 24 P 36 900P 24 36

450 1 e e 900 e e

511.15

X X X

k

k

k

Let 622.58 511.15 k , then 111.43k . If 111k , Mary should extend the warranty. If 112k , Mary should not extend the warranty.

The value of p will not be affected as Mary will always get a new smartwatch if the first smartwatch brought breaks down within 3 years, independent of whether it is within the warranty period.

This part is fairly scored by the students.


Let X be the number of calls needed to get through to the hotline.

From GC, 172.1258

x so 80.47117

p .

Number of attempts 1 2 3 4 5 6 7

Frequency 45 41 16 12 4 2 0 Expected Frequency

56.471

29.896

15.827

8.3692

4.4361

2.3455

2.6421

We need to combine the last 3 columns. Number of attempts 1 2 3 4 5

Frequency 45 41 16 12 6 Expected Frequency 56.471 29.896 15.827 8.3792 9.4266

H0: ~ Geo(0.471)X H1: ~ Geo(0.471)X By GC, value 0.025924 0.05p (degree of freedom is 3)

This part was well done. Students must remember to give their intermediate working to at least 2 more decimal places or significant figures than the required accuracy. Some students used the wrong degree of freedom. Students must remember that as they had to estimate the value of the parameter p, the degree of freedom will be reduced.

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Q12 Solution Comments We have sufficient evidence to reject the null hypothesis, so the distribution of the number of attempts is not geometric.

(a)(ii)

The degree of freedom will increase by 1. OR The degree of freedom changes from 3 to 4.

The degree of freedom should increase since the parameter p is now given.

(b)(i)

The test statistic will double. The conclusion of the test will not change because if the original test statistic is larger than the critical value, two times the original test statistic will also be larger than the critical value.

This part was well answered.

(b)(ii)

The observed and expected frequencies are 0.03n

(0.04n) 0.11n

(0.125n) 0.06n

(0.035n) 0.07n

(0.088n) 0.325n

(0.275n) 0.045n

(0.077n) 0.1n

(0.072n) 0.19n

(0.225n) 0.07n

(0.063n)

2 2 2

2

2 2 2

2 2 2

0.03 0.04 0.11 0.125 0.06 0.035

0.04 0.125 0.035

0.07 0.088 0.325 0.275 0.045 0.077

0.088 0.275 0.077

0.1 0.072 0.19 0.225 0.07 0.063

0.072 0.225 0.0630.065340

n n n n n n

n n n

n n n n n n

n n n

n n n n n n

n n nn

We need 0.065340 18.47 282.675n n . Minimum 283n .

This part was poorly done. Main mistakes: 1. Students did not bother to calculate the test statistic properly. 2. Some students did not know that the critical value is obtained from MF26.

further mathematics 9649/01 - the learning space

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