mathematics 9820/01 · 2019 jc2 h3 mathematics preliminary examination paper 1 3 (a) let x, y, z be...

2019 JC2 H3 Mathematics Preliminary Examination Paper 1 [Turn over

EUNOIA JUNIOR COLLEGE

JC2 Preliminary Examination 2019

General Certificate of Education Advanced Level

Higher 3

MATHEMATICS

Paper 1 [100 marks]

9820/01

24 September 2019

3 hours

Additional Materials: Answer Booklet

List of Formulae (MF26)

READ THESE INSTRUCTIONS FIRST

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Answer all 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.

At the end of the examination, fasten all your work securely together.

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

This document consists of 7 printed pages.

2

2019 JC2 H3 Mathematics Preliminary Examination Paper 1

1 (i) Find the general solution of the differential equation

2

d ( ),

d 1

x c t x

t t

(A)

where c is a constant. [5]

(ii) Make the substitution y tx in the differential equation

d

,d

y ax by

x bx ay

(B)

where a and b are constants and 0,a and show that the resulting equation is

equivalent to equation (A). [3]

(iii) Hence show that equation (B) has solutions of the form

,b a b a

x y D x y

where D is an arbitrary constant. [3]

2 (i) Show that f ( ) d f ( ) db b

a a

x x a b x x for any function f which is continuous on

the domain , .a b [1]

(ii) Using the result in (i), show that

π π

4 4

0 0

2ln(1 tan ) d ln d .

1 tan

[2]

(iii) Hence, by using an appropriate substitution, find the exact value

of 1

20

ln(1 ) d

1

xx

x

. [3]

(iv) Use the result in (i) to also find the exact value of

π

4

π0 4

cos d

cos( )

xx

x

. [3]

3


3 (a) Let x, y, z be real numbers such that 0 .x y z Show that 2 2 2, , x y z are three

consecutive terms of an arithmetic sequence if and only if 1 1 1

, , y z x z x y

are

three consecutive terms of another arithmetic sequence. [3]

(b) An infinite sequence of integers, , ,na n is defined by 1 2 1a a and

2

1

1

2, 2.n

n

n

aa n

a

(i) Show that 2 1 1

1

n n n n

n n

a a a a

a a

for 2.n [2]

(ii) Hence or otherwise, show that 1 14n n na a a and explain why na is an odd

integer for all .n [2]

(c) Prove that there is no arithmetic progression which has 1, 2, 3 among its

terms. [5]

4 Let n be a positive integer greater than 2 and 0 1 2, , , ..., nx x x x be real numbers with

0 1.x

(i) By using the Cauchy-Schwarz Inequality or otherwise, show that

2 2

1

1

1( ) .

1

n

i i n

i

x x xn

[3]

(ii) Given that 2 2

1

1

1( ) ,

1

n

i i n

i

x x xn

find kx for 0 , ,k n k leaving your

answer in terms of n and k. [3]

4


5 A factory supervisor intends to distribute work to his three subordinates. He has 24

indistinguishable cards each representing one unit of work, and he intends to distribute

them into three different colored boxes, red, green and blue, representing each of his

subordinates.

(i) Find the number of ways he can distribute the cards into the boxes without

restriction. [1]

The supervisor wishes to split the work relatively evenly amongst his three subordinates

so that each subordinate has at most 10 units of work.

(ii) Find the number of ways the supervisor can distribute the workload. [3]

The red box represents the workload of Sue, who is pregnant, and the supervisor will like

to reduce her workload such that she will only have a workload at most half of any of her

other two colleagues. For example, if Sue is given 3 units of work, then her colleagues

must have at least 6 units of work each.

(iii) Find the number of ways the supervisor can distribute the workload. [3]

Peter is given r units of work. He places the r indistinguishable cards into n

indistinguishable boxes for him to randomly choose one box for each of the subsequent

days.

Let the number of ways for him to do so be denoted by ( , ).P r n

(iv) State the recurrence relation between ( , ),P r n ( 1, 1)P r n and ( , ).P r n n [1]

(v) Using the result in (iv) or otherwise, show that 1

( , ) ( , )n

k

P r n P r n k

for 1.r n [2]

(vi) Using (iv) or (v), find the value of (10, 3).P [2]

5


6 You have an unlimited supply of 1 1 , 1 2 and 2 2 tiles. Tiles of the same size are

indistinguishable.

(i) Let nT is the number of ways of tiling a 1 n path.

(a) State the value of 1T and 2 .T [1]

(b) Write down an appropriate recurrence relation between 2nT , 1nT and .nT [1]

Consider the tilings of a 2 n path. (The 1 2 tiles can be rotated in the tilings.)

Let nP be the number of tilings of

Let nQ be the number of tilings of

(ii) Show that 1n n nP P Q for 1.n Explain your reasoning clearly. [2]

(iii) Show that 1 1 12 2n n nQ P Q for 2.n Explain your reasoning clearly. [4]

(iv) Use (ii) and (iii) to show that 2 1 12 3 2n n n nP P P P for 2.n [2]

It is given that the solution to the above recurrence relation is

1 2 2 1 2 2

(2 .(

2) (2 2)7 14

1)

14

nn n

nP

(v) Find the number of distinct ways of tiling a 2 n path. [2]

…

…

n

…

…

n

6


7 (a) Show that if 2 27 ( )a b then 7 a and 7 b where a and b are positive integers. [3]

(b) Given that n is an integer greater than 1, by considering the cases when n is even

and n is odd, show that 4 4nn cannot be a prime number. [4]

(c) Show that for every positive integer n, 16 10 1n n is divisible by 25. [5]

8 Let 2 31

1 2 1 3

( )( )( ) .

( )( )

x x x xL x

x x x x

(i) State the values of 1 1( ),L x 1 2( )L x and 1 3( ).L x [2]

(ii) Hence or otherwise, obtain expressions of quadratic polynomials 2L and 3L such

that ( ) 1,i iL x and ( ) 0,i jL x ,i j 2, 3,i and 1, 2, 3.j [2]

It is known that if p is a polynomial of degree m, and ( ) 0,ip x for 1, 2, , 1,i m

where 1 2 3 1n mx x x x x , then ( ) 0,p x for all .x

(iii) Suppose there are 3 distinct values 1,x 2 ,x 3x and any real values 1,y 2 ,y 3.y

Using the above result, explain why there is at most one quadratic polynomial P

such that ( )i iP x y for 1, 2, 3.i [3]

(iv) Obtain the quadratic polynomial P as a combination of iy and iL such that

( )i iP x y for 1, 2, 3,i and justify your answer. [2]

(v) Let ix i for 1, 2, 3,i and1 2 3{ , , } {1,2,3}.y y y Find all possible quadratic

polynomials Q satisfying ( )i iQ x y , ,iy i for 1, 2, 3,i with appropriate

justification to be provided. [You need not simplify your answer] [4]

7


9 Let ( )S n denote the sum of all divisors of the number n. For example,

(6) 1 2 3 6 12S .

(i) Find

(a) ( )S p in terms of p, [1]

(b) ( )S pq in terms of p and q, [1]

(c) ( )m nS p q in terms of p, q, m and n, [2]

where p, q are distinct prime numbers and m, n are positive integers.

Aliquot divisors of a number are divisors of the number excluding the number itself

(i.e. the sum of all proper divisors of the number). For example, the aliquot divisors of

6 are 1, 2 and 3 respectively.

Two numbers are called amicable if each equals the sum of the aliquot divisors of the

other, and the two number form an amicable pair.

(ii) Verify that 220 and 284 is an amicable pair. [2]

Assume that two amicable numbers, M and N, has the form and ,M apq N ar where

p, q, r are distinct prime numbers, a (not necessarily prime) is the common divisor

of M and N, and a is co-prime to p, q and r.

(iii) Show that ( ) ( 1)( 1) ( ).a pq r p q S a [2]

(iv) Express 1r in terms of p and q only, and use (iii) to show that

( 1)( 1) 2 ( ) ( 2).p q a S a a p q [3]

The equation in (iv) can be further manipulated to the following equality:

2

1 12 ( ) 2 ( ) 2 ( )

a a ap q

a S a a S a a S a

(v) Find the only possible solution for p and q for 4,a and hence find M and N. [2]

End of Paper

2019 JC2 H3 Mathematics Preliminary Examination Paper 1 [Turn over

2019 Year 6 H3 Math Prelim Exam Solutions

Qn Solution Remarks

1(i)

2

2

2 2

2

122

1

2

1

2

1

2

1

2

d

d 1

1 d

d 11 1 2

d d1 2 1

1 1ln ln ln 1

2 1 2

1ln ln ln 1 1

1

1

1

1, where is an arbitrary constant.

1

c

c

c

c

c

c t xx

t tc tx

x t tc t

x tx t t

c tx d t

t

x tt t

k t

tx

kt

p tx p

t

(ii)

d d

d d

y tx

y tt x

x x

2

2

2

d

d

d

d

d

dd

d1 d

d 1

d

d 1

ax b txtt x

x bx a tx

a b ttt x

x b ata b tt

x tx b att a at

xx b atx b at

x t a t

bt x

x a

t t

and the resulting equation is equivalent to equation

(A) with .b

ca

2


1(iii) 1

2

1

2

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

2 2

2

1

1

b

a

b

a

b b b b

a a a a

b b b b

a a a a

b a b a

a a

b a a b a

b a b a

y

x x

k

y

x

x x y x k x y x

x x y k x y x

x x y k x y x

x x y k x y x

x y D x y

Total Marks: 11

3


Qn Solution Remarks

2(i) Using the substitution ,y a b x we have

f ( ) d f ( ) d

f ( ) d

f ( ) d (shown)

b a

a bb

ab

a

x x a b y y

a b y y

a b x x

(ii)

π

4

0π

4

0π

4

0π

4

0π

4

0

ln 1 tan d

πln 1 tan d

4

πtan tan

4ln 1 d

π1 tan tan

4

1 tanln 1 d

1 tan

2ln d (shown)

1 tan

(iii)

1 π

42 0

0

ln 1 d ln 1 tan d

1

xx

x

by using the

substitution tan .x

Also, we have

π1

4

20 0

ln 1 2 d ln d .

1 1 tan

xx

x

Summing up both equalities, we have

π1 π

44

2 00 0

π

4

0π

4

0

ln 1 22 d ln 1 tan d ln d

1 1 tan

ln 1 tan ln 2 ln 1 tan d

πln 2 d ln 2

4

xx

x

1

20

ln 1 π d ln 2

1 8

xx

x

4


2

(iv)

π

π 4

4

0

0

π

4

0

π

4

0

πcos

cos 4 d d

π π πcos cos

4 4 4

π π πcos cos sin sin

4 4 4 d

cos

π π πcos cos sin sin

4 4 4 d

cos

1 1cos sin

2 2

cos

xx

x x

x x

x

xx

x

xx

x x

π

4

0π

4

0π

4

0

d

1 1tan d

2 2

1ln sec

2 2

π 1 πln sec 0

44 2 2

π 1ln 2

4 2 2π 1

ln 24 2 2 2

xx

x x

xx

Total Marks: 9

5


Qn Solution Remarks

3(a)

2 2 2 2

2 2 2 2

1 1 1, ,

are three consecutive numbers of an AP

1 1 1 1

y z x z x y

x z y z x y x z

y z x z x z x y

x z y z x y x z

y x z y

x z y z x y x z

y x x y z y y z

x y x z y z x y x z y z

y x z y

x y x z y z x y x z y z

y x z y

x

2 2 2, , are three consecutive numbers of an APy z

(b)

(i)

2

1

2

1 12 2

1

12

1

1

1 1

2

2

2

, 2

nn

n n n

n n

n n

n n

nn

n

n

n n

n

aa

a a a

a a

a a

a a

aa

a

a

a an

a

(b)

(ii)

2 2

23

1

2 1 23

1

aa

a

1 1 3 1

2

1 1

3 1... 4

1

4 , 2.

n n

n

n n n

a a a a

a a

a a a n

Therefore 1na has the same parity as 1na since 4 na is

even. Since both 1a and 2a are both odd, by induction,

na is an odd integer for all .n

Alternatively, let P(n) be the statement that na is odd for

.n

P(1) and P(2) are both true trivially by definition.

6


Suppose ka and 1ka are both odd for some positive

integer k. Then

22

12

odd 22 oddeven

odd odd

kk

k

aa

a

Hence P , P 1 both true P 2 is true.k k k

Therefore, since P(1) and P(2) are both true,

and P , P 1 both true P 2 is true,k k k by the

Principle of Mathematical Induction, na is an odd

integer for all .n

(c) Suppose there is an arithmetic progression 1 2, , ...a a

with common difference d that has 1, 2, 3 among its

terms.

Then there exist distinct positive integers m, n and p such

that 1, 2, 3.m n pa a a

Thus we have 2 1 n ma a n m d and

3 2 ,p na a p n d so 3 2

2 1

n m

p n

is

rational.

Since 3 23 2 2 1 6 2 3 2

2 1

thus 6 3 2a is a rational number.

Then,

2 2

2

2

2

2 6 3

2 6 3

2 2 2 6 2 18 3

2 2 6 2 7

2 6 2 7

a

a

a a

a a

a a

Since 3a does not satisfy the equality, we can divide

throughout by 2 6a so

272

2 6

a

a

which is a

contradiction. Hence the supposition does not hold and

the result required is shown.

Total Marks: 12

7


Qn Solution Remarks

4(i) Applying Cauchy-Schwarz Inequality 2

1 1 12 2

1 1 1

n n n

i i i i

i i i

a b a b

with

1i i ia x x for 1, 2, ..., ,i n 1n na x and

1ib for 1, 2, ..., 1,i n we have

21

2 2 2

1 1

1 1 1

2 2

1

1

1 1

1

1

n n n

i i n i i n

i i i

n

i i n

i

x x x x x x

x x xn

(ii) Since equality holds, we have (constant).i ia mb m

Thus, 1 2 1... .na a a

Since 1

1

1,n

i

i

a

therefore we have

1

1ia

n

for 1, 2, ..., 1,i n

Thus,

1

2 1 1

1,

12

,1

3,

1

n

n n n

n n n

xn

x a xn

x a xn

and so on so forth. Generalizing, we have

1,

1k

n kx

n

0, 1, 2, ..., .k n

Total Marks: 6

8


Qn Solution Mark Scheme

5(i) Number of ways 24 3 1

3253 1

B1

(ii) The problem is

1 2 3 24

0 10, 1, 2, 3i

x x x

x i

with 1 2 3, , x x x being the number of cards in the red,

green and blue boxes respectively.

Method 1

Let iA denote the event where 0 10.ix

Number of ways

1 2 3

1 2 3

3

1 2 3

1, 1,2,3

13 3 1 2 3 1325 3 3 0

3 1 3 1

325 3 105 3 6 0 28

i i j

i i ji j

A A A

S A A A

S A A A A A A

Method 2

4ix since 3ix for any i will result in

1 2 3 24x x x for 0 10, 1, 2, 3.ix i

1x 2 3x x

2 3, x x or

corresponding

cases for 2x

Number

of Ways

4 20 10, 10 1

5 19 9, 10 , 10, 9 2

6 18 28 10x 3

7 17 27 10x 4

8 16 26 10x 5

9 15 25 10x 6

10 14 24 10x 7

Therefore number of ways = 1 + 2 + … + 7 = 28

9


(ii) Method 3

Number of ways

* **

3 3

1 2

(8,8,8) (5,9) ... (9,5)(6,9,9) (7,8,9) (4,10,10)

(none with 10 units) + (one with 10 units)

(two with 10 units)

1 3 6 5

10 15 3 28

C C

(iii) Sue can take either no work, or up to 4 units of work,

since it is not possible to have Sue take 5 or more units

of work as the total amount of work is 24 units

(5 + 10 + 10 =25 > 24).

If Sue takes k units of work where k = 0 to 4, then the

other 2 must take at least 2k units of work from the

remaining 24 k units of work,

no. of ways 24 4 2 125 5

2 1k k

k

since k units of work is assigned to Sue, and we place 2k

units of work each into the other two boxes.

Number of ways supervisor can distribute the workload

4

0

25 5 75k

k

(iv) ( , ) ( 1, 1) ( , )P r n P r n P r n n

(v)

( , )

( 1, 1) ( , )

(applying on ( , ))

( 2, 2) 1 1 , 1 1

( , )

(applying on ( 1, 1))

( 2, 2) ( , 1) ( , )

(rearrangement)

1 , 1

, 2

P r n

P r n P r n n

P r n

P r n P r n n

P r n n

P r n

P r n P r n n P r n n

P r n

P r n n n

(iv)

(iv)

1

( , )

(applying on ( 2, 2), , until ( 1, 1))

1, 1 , 2 ( , )

, 1 , 2 ( , )

( 1, 1 , 1 1)

( , ) (shown)n

k

P r n n

P r n P r n

P r n P r n P r n n

P r n P r n P r n n

P r n P r n

P r n k

(iv)

10


(vi) Method 1 (Using (v)) 3

1

2 3

1 1

2 2

1 1

(10, 3) (7, )

(7, 1) (7, 2) (7, 3)

1 (5, ) (4, )

1 (5, 1) (5, 2) (4, 1) (4, 2) (4, 3)

1 1 (5, 2) 1 (4, 2) 1

4 (3, ) (2, )

4 (3, 1) (3, 2) (2, 1) (2, 2

k

k k

k k

P P k

P P P

P k P k

P P P P P

P P

P k P k

P P P P

)

4 1 1 1 1 8

Method 2 (Using (iv))

(10, 3)

(9, 2) (7, 3)

(8, 1) (7, 2) (6, 2) (4, 3)

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

1 1 (4, 1) (3, 2) 1 (3, 1) (2, 2) 1

8

P

P P

P P P P

P P P P

P P P P

Total Marks: 12

11


Qn Solution Remarks

6(i)

(a) 1 1T , 2 2.T

(i)

(b) 2 1n n nT T T for 1.n

(ii) Consider the “odd” tile / last tile in a tiling of 1nP . It

can only be covered by a 1 1 or a 1 2 tile.

If it is covered by a 1 1 tile, the rest form a tiling of

nQ .

If it is covered by a 1 2 tile, the rest form a tiling of

nP .

Thus 1n n nP P Q .

(iii) Consider the last column of 2 tiles in a tiling of 1nQ .

The following cases are possible:

2 2 tile: The rest form a tiling of 1nQ .

1 2 tile (vertical): The rest form a tiling of nQ

Two 1 1 tiles: The rest form a tiling of nQ .

Two 1 2 tiles (horizontal): The rest form a

tiling of 1nQ .

One 1 1 tile and one 1 2 tile (horizontal):

The rest form a tiling of nP . Note that this case

counts twice (depending on which tile covers

the top line and which tile covers the bottom

line).

Thus 1 12 2 2n n n nQ Q Q P

1 12 2n nP Q using the result from (ii).

Alternative

Tiling of 1nQ can be obtained from

1nQ using a 2 2 tile or two horizontal 1 2

tiles.

nQ using one vertical 1 2 tile

1nP using one 1 1 tile

The first two cases involve non-1 1 tiles on the last

column, and the last case involves 1 1 tiles on the last

column.

However, the last case we must consider the double

counting of the subcase where tiles from the last

column are both 1 1 tiles, and the number of ways to

do so is .nQ

Thus 1 1 12 2n n n n nQ Q Q P Q 1 12 2n nP Q

12


(iv) Add 1 12n nP P to both sides of (iii):

1 1 1 1 1 1 12 2 2 2n n n n n n nP P Q P P P Q

2 1 12 3 2n n n nP P P P (using result from (ii))

(v) Number of tilings of 2 n path is nQ

So

1n n nQ P P

2 1 2 22 2 2 1

7 11 2

4

nn

1 2 22 2 2 1

142

n

2 5 3 2 5 3 21 2 2 2 2

7 14 14

n nn

Total Marks: 12

13


Qn Solution Remarks

7(a) We show by proving the contrapositive statement:

if 7 | a or 7 | b then 2 27 | ( ).a b

For positive integers n not divisible by 7,

(mod 7)n 2 (mod 7)n

1 1

2 4

3 2

4 2

5 4

6 1

Therefore 2 2( ) mod 7a b can only take values

1 (from 4 + 4), 2 (from 1 + 1), 3 (from 1 + 2),

4 (from 2 + 2), 5 (from 1 + 4) 6 (from 2 + 4) but never

0.

Hence the contrapositive statement is shown.

(b) When n is even, 4 4nn is even and hence not a

prime.

When n is odd, i.e. 2 1n k with integer 1,k

(since 1n )

24 2 2

22 2 2 2

2 22 1

2 1 2 1

4 2 2 2

2 2

2 2

2 2 2 2

n n n

n k

n k

n k n k

n n n

n n

n n

n n n n

The smaller factor is 2 12 2 ,n kn n and

2 1 2 1

2 2 2 2 1

22

2 2 2 2

2 2 2 2 2

2 2 1

n k k n

k k k k

k k

n n n n

n n

n

so 4 4nn cannot be prime.

14


7(c) Method 1: Consider the values of

16 , 10 and 1 (mod 25)n n

n 16 (mod 25)n 10 (mod 25)n 1

Sum

(mod

25)

1 16 10 1 0

2 6 20 1 0

3 21 5 1 0

4 11 15 1 0

5 1 0 1 0

6 16 10 1 0

7 6 20 1 0

Since the table repeats cyclically for every 5 values

of n, 16 10 1n n is divisible by 25, for .n

Method 2: Mathematical Induction

Let P(n) be the statement that16 10 1n n is

divisible by 25, for .n

P(1) is true since 116 10 1 1 25 which is

divisible by 25.

Suppose P(k) is true for some positive integer k.

Then for P 1 ,k

116 10 1 1

16 16 10 1 160 16 10 1 1

16 16 10 1 150 25

k

k

k

k

k k k

k k

which is divisible by 25.

Hence P is true P 1 is true.k k

Therefore, since P(1) is true, and

P is true P 1 is true,k k by the Principle of

Mathematical Induction, 16 10 1n n is divisible by

25, for .n

Total Marks: 12

15


Qn Solution Remarks

8(i) 1 1( ) 1,L x

1 2( ) 0,L x 1 3( ) 0.L x

(ii) 1

2

2 1 2

3

3

( )( )( )

( )( )

x x x xL x

x x x x

1 23

3 1 3 2

( )( )( )

( )( )

x x x xL x

x x x x

(iii) Suppose there exist two quadratic polynomials 1p , 2p

(of degree less or equal to 1 3 1 2)n , with

1 2 ,( ) ( )i i ip x p x y for 1, 2, 3.i

Then the difference polynomial 1 2q p p is a

polynomial of degree less or equal to 1 3 1 2n

that satisfy ( ) 0,ih x for 1, 2, 3.i

Since the number of zeroes of a nonzero polynomial

is equal to its degree, it follows that

1 2( ) ( ) ( ) 0,q x p x p x ,x

i.e. 1 2( ) ( ),p x p x for all .x

(iv) 1 1 2 2 3 3( ) ( ) ( ) ( ),P x y L x y L x y L x .x

This is because since ( ) 1,i iL x ( ) 0i jL x for ,i j

, 1, 2, 3i j from the earlier parts, we have

,( )i iP x y for 1, 2, 3.i

(v) Let {1, 2, 3}X Y and a function f : ,X X

satisfying

,f ( )i ix y iy i for 1, 2, 3,i

where ,ix i and ,iy Y X for 1, 2, 3.i

Note that the above translates to a derangement

problem, where the number of such possible

mappings is given by 3

1 1 13! 1 2.

1! 2! 3!D

In fact, the 2 derangements are given by

Derangement 1: f (1) 2, f (2) 3, f (3) 1,

Derangement 2: f (1) 3, f (2) 1, f (3) 2.

We now show that we can extend this derangement

property of function f defined on {1, 2, 3} to a

quadratic polynomial Q, which is defined on the real

line by finding explicitly 2 quadratic polynomials 1Q ,

2Q such that they satisfy

1(1) 2,Q 1(2) 3,Q 1(3) 1,Q

2 (1) 3,Q 2 (2) 1,Q 2(3) 2.Q

16


To do so, we can use either of the following two

methods:

Method 1

We can find two quadratic polynomials 1,Q

2Q such

that they satisfy

1(1) 2,Q 1(2) 3,Q 1(3) 1,Q

2 (1) 3,Q 2 (2) 1,Q 2(3) 2.Q

from parts (iii) and the interpolatory property (iv), we

get

1 1 2 3( ) 2 ( ) 3 ( ) 1 ( ),Q x L x L x L x

2 1 2 3( ) 3 ( ) 1 ( ) 2 ( ),Q x L x L x L x

where

1

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

(1 2)(1 3) 2

x xL x x x

2

( 1)( 3)( ) ( 1)( 3),

(2 1)(2 3)

x xL x x x

3

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

(3 1)(3 2) 2

x xL x x x

Method 2

Let 2

1 2 1 0( )Q x a x a x a , and

2

2 2 1 0.( )Q x b x b x b

We solve for coefficients 2 1 0, ,a a a and 2 1 0, ,b b b such

that

1(1) 2,Q 1(2) 3,Q 1(3) 1,Q

or 2 (1) 3,Q 2 (2) 1,Q 2(3) 2.Q

We obtain the resulting system of linear equations 2

2 1 02

2 1 02

2 1 0

(1) (1) 2,

(2) (2) 3,

(3) (3) 1.

a a a

a a a

a a a

2

2 1 02

2 1 02

2 1 0

(1) (1) 3,

(2) (2) 1,

(3) (3) 2.

b

b

b

b b

b b

b b

2 1 0

3 112,, ,

2 2a a a

2 1 0, ,3 13

8,2

2

b b b

Therefore, 2

1

3 11

2) 2,

2(Q x x x

2

2

3 138.

2(

2)Q x x x

Total Marks: 13

17


Qn Solution Remarks

9(i)

(a)

( )S p 1p

(i)

(b) ( ) ( ) ( ) 1 1S pq S p S q p q

(i)

(c)

1 1

( ) ( ) ( )

1 1

1 1

1 1

m n m n

m n

m n

S p q S p S q

p p q q

p q

p q

(ii) 2220 2 5 11

The proper divisors of 220 are

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110, and

Sum of all proper divisors of 220

1 2 4 5 10 11 20 22 44 55 110

284

2284 2 71

The proper divisors of 2924 are 1, 2, 4, 71 and 142, and

Sum of all proper divisors of 284

1 2 4 71 142 220

(iv)

( )

( ) 1 1

a pq r M N

S M

S apq

S a S p S q

S a p q

(v) Since M , N form a pair of amicable numbers,

( )S M M N , and ( )S N N M .

Simplifying, we get ( ) ( )S M S N , and thus

1 1 1

1 1 1

2 ( ) 1 1

2 1 1 ( ) 1 1

2 1 1 ( )

2 1 1 1 1 1

2 1 (2 )

2 2 2 2 (2 )

2 2 (shown)

S apq S ar

p q S a r S a

r p q

a S a p q

a p q S a p q

a p q a pq r

a p q a pq p q

a pq p q a pq p q

apq ap aq a apq ap aq

ap aq a a p q

18


(vi) 4, (4) 1 2 4 7.a S 2

1 12 ( ) 2 ( ) 2 ( )

a a ap q

a S a a S a a S a

becomes

2

4 4 41 1

2 4 7 2 4 7 2 4 7

3 3 16

p q

p q

WLOG let ,p q

3p 3q p q r

1 16 4 19 99

2 8 5 11 71

4 4 7 7 63

Since p , q , r are distinct primes,

so the only solution is 5, 11, 71,p q r and

4 5 11 220, 4 71 284.M N

Total Marks: 13

mathematics 9820/01 · 2019 jc2 h3 mathematics preliminary examination paper 1 3 (a) let x, y, z be...

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