RAFFLES INSTITUTION 2019 YEAR 6 PRELIMINARY EXAMINATION Higher 3

MATHEMATICS 9820/01

26 September 2019

3 hours

Additional materials: Answer Booklet List of Formulae (MF26)

READ THESE INSTRUCTIONS FIRST Write your name and CT 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 any 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. At the end of the examination, submit all answer booklets used. The number of marks is given in brackets [ ] at the end of each question or part question.

This document consists of 7 printed pages.

RAFFLES INSTITUTION

Mathematics Department

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1 A function F defined on ,a a is said to be odd if

F( ) F( ) for all ,x x x a a .

(i) Let f be an odd function defined on ,a a such that its inverse exists and the

Maclaurin series for f ( )x exists.

(a)

By writing down the Maclaurin series for f ( )x and f ( )x or otherwise,

show that the Maclaurin series for f ( )x contains only odd powers of x.

[3]

(b)

Show that its inverse function 1f is also an odd function. [2]

(ii) Let g( ) 2 tanx x x , 2

x

.

(a) Show that g is an odd function and its inverse exists. [2]

(b) Find the Maclaurin series for g( )x up to the term in 5x . [3]

(c) Hence find the Maclaurin series for 1g ( )x up to the term in 5x . [3]

2 Let nP denote the set of prime numbers less than or equal to n, 2n .

(i) Explain why for each

np P ,

21

1 1 1 11 ... .

1 n

pp p p

[1]

(ii) Hence deduce that

1

1

1 1

1n

n

jp P pj

. [2]

(iii) Justify the following statements:

(a) 1

1ln

n

j

nj

for all positive integers n, [3]

(b) 7 1

ln5 1

xx

on 1

0, .2

[3]

(iv) Hence show that 1 5

ln(ln )7

np P

np

and deduce that there are infinitely many prime

numbers. [3]

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3

(i)

Show that the number of ways to select 5 non-empty subsets 1 2 5, , ,A A A of

{1,2, ,9} such that

0, for all ,i jA A i j and

1 2 5 {1,2, ,9}A A A

is given by 9 9 9 9 9

5 5 5 55 4 3 2 1 .

1 2 3 4

[3]

Let S( , )r n denote the number of ways to distribute r distinct objects into n identical boxes

such that no box is empty.

(ii)

Using (i) or otherwise, find S(9,5) . [2]

(iii) Explain why for 3r , 1 11S( ,3) 3 1 2

2

r rr . [3]

(iv) Determine the number of ways to select two non-empty subsets of 1,2,...,n such

that one is not a subset of the other, and the intersection of these two sets is the null

set. [3]

4 (i) It is given that A, B and C are real numbers. Show that the quadratic function2h( ) 2t At Bt C is non-negative for all real values of t if and only if 0A

and 2.AC B [2]

(ii) Let f and g be continuous functions on the interval [ , ]a b . By considering

2

f ( ) g( ) db

at x x x , show that

2

2 2f ( ) d g( ) d f ( )g( )d

b b b

a a ax x x x x x x

and determine when equality holds. [3]

(iii) Hence show that 1

3

0

51 d

2x x . [3]

(iv) Let k be a continuous and differentiable function on the interval [0,1] satisfying

k(1) 0 . Show that

2

1 1 2

0 0

1k( ) d k ( ) d

3x x x x

and determine when equality holds. [4]

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5 For any positive integer n, the function f ( )n x is defined by

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

!n

x x x x nx

n

.

(a) By considering f ( 1) f ( )n nx x or otherwise, show that 1

f ( ) f ( 1) 1N

n N

n

x x

.

[3]

(b) Show that for all positive integers N, 1 1

0 f2 2 1

NN

. [5]

Hence or otherwise, show that the infinite series 1 1 3 1 3 5

...2 4 2 4 6 2 4 6 8

is

convergent and find its sum. [3]

6 A lattice point is a point such that its coordinates are all integers.

(i) Prove that among any set of five lattice points 1 1 2 2 5 5, , , ,..., ,x y x y x y in the

plane, there must be two of them whose midpoint is also a lattice point. [3]

(ii) Determine with proof, the least m such that among any set of m lattice points

1 1 1 2 2 2, , , , , ,..., , ,m m mx y z x y z x y z in space, there must be two of them whose

midpoint is also a lattice point. [3]

(iii) The centroid of 4 lattice points , , , , , ,a a b b c c d dx y x y x y x y is defined by

, .4 4

a b c d a b c dx x x x y y y y

Using (i) or otherwise, prove that among any set of 13 lattice points

1 1 2 2 13 13, , , ,..., ,x y x y x y in the plane, there must be four of them whose

centroid is also a lattice point. [3]

A lattice point ,x y is , modulo a b n if and only if

(modulo ) and (modulo ).x a n y b n

(iv) Let S be a set of twelve lattice points in the plane such that three of its points are

(0, 0) modulo 4, three are (0, 1) modulo 4, three are (1, 0) modulo 4 and three are

(1, 1) modulo 4. Show that there does not exist 4 points of S whose centroid is a

lattice point. [2]

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7 A string of 2n parenthesis with n open ‘(’ and n closed ‘)’ parentheses is said to be valid

if each open parenthesis has a matching closed parenthesis occurring after it. Otherwise, it

is invalid. For example, when 2n , (()) and ()() are the only two valid strings while ())(

is an invalid string.

Let nC be the number of valid strings of n matching pairs of parentheses and define

0 1.C

(a) (i) Find 1C and

3C . [2]

(ii) Use the bijection principle to show that the number of invalid strings with

n open and n closed parentheses is the same as the number of strings with

1n open and 1n closed parentheses and find this number in terms of n.

[3]

(iii) Hence or otherwise, show that 21

1n

nC

nn

. [2]

(b) (i) For any positive integer n, explain why

1 0 2 1 1 2 0 1.n n n n nC C C C C C C C C

[3]

(ii) Hence show that if nC is odd, n and

1

2

nC are both odd. [3]

(iii) Hence determine all positive integers n such that nC is odd. [3]

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8 Taylor is at the origin O, lost in the middle of the woods in which the trunks of the trees

all have the same radius 0R and are represented by circles centered at the coordinates

( , )a b , where a and b are odd integers.

To get out of the woods, Taylor travels along ml , the half-line defined by y mx for

0x . For example, in the figure above, cl meets the tree centered at (1, 3) at P, and

Taylor is unable to move past P. However, Taylor is able to move past the first 2 columns

of trees if she moves along dl .

(i) Let m be a positive real number. Show that ml meets the tree with radius 0R

centered at ( , )a b if and only if 21b ma R m . [3]

It is known that any irrational number m can be well approximated by a rational number

b

a whose numerator and denominator are both odd integers. In other words, for any

0 , there exists two positive odd numbers a and b such that .b ma

(ii) Hence explain why if m is irrational, then ml meets a tree. [1]

Question 8 continues on the next page

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(iii) Suppose that p

mq

, where p and q are positive coprime integers.

(a) Suppose that p and q are odd. Show that ml meets a tree, explaining clearly

your reasoning. [1]

(b) Suppose that p and q are of different parity and that ml meets a tree. Show

that 2 21 R p q . [2]

(iv) Suppose 1

5R . Show, in any order, that the half-lines ml meets the tree

centered at

(a) 1, 1 if 1

22

m ,

(b) 1, , where is the closest odd integer to m if 2,m

(c) ,1 , where is the closest odd integer to m if 1

.2

m

[5]

We say that Taylor can get out of the woods if there exists a half-line ml which does not

meet any of the circles.

(v) Show that if Taylor can get past the first row or first column of trees, Taylor can

get out of the woods. [2]

[END OF PAPER]

1(i)

(a)

[3]

Method 1 2 2 1 2

0 1 2 2 1 2f ( ) ... n n

n nx = a a x+a x a x a x

2 12 2

0 1 2 2 1 2

2 2 1 2

0 1 2 2 1 2

f ( ) ...

= ...

n n

n n

n n

n n

x = a a x +a x a x a x

a a x+a x a x a x

Since f is odd and f ( ) f ( )x x

So 3 2 1

1 3 2 1

f ( ) f ( )f ( ) ...

2

n

n

x xx a x a x a x

(shown)

Method 2

Or you can compare coefficient of f ( ) and f ( )x x to arrive at the same conclusion.

2 2 1 2

0 1 2 2 1 2

2 2 1 2

0 1 2 2 1 2

2 2

0 2 2

f ( ) ...

f ( ) ...

2 ... 0

n n

n n

n n

n n

n

n

x = a a x+a x a x a x

x a a x a x a x a x

a a x a x

Hence the Maclaurin series of f will only contain the odd powers of x.

(i)

(b)

[2]

Method 1

If 1f exists, then 1f f ( )x x

1 1

1 1

1 1

f f ( ) f f ( )

f f ( ) f f ( ) since f ( ) f ( )

f f ( ) f f ( )

x x

x x x x

x x

therefore 1f is odd. We note that f(x) belongs to the range of f and thus is in the domain

of 1f .

Method 2

Let 1

1 1

1

1 1

f ( )

f ( ) f ( )

f ( ) f ( f ( ))

f (f ( )) since f is odd

=

f ( ) f ( )

y x

y x x y

x y

y

y

y x x

.

(ii)

(a)

[2]

g( ) 2 tan( ) ( )

2 tan since tan is an odd function

(2 tan ) g( )

x x x

x x

x x x

Hence g is an odd function. 2 2 2g '( ) 2sec 1 2(1 tan ) 1 1 tan 0x x x x

Therefore g is strictly increasing. Hence its inverse exists.

Alternatively,

2 20 cos 1 sec 1 sec 1 2sec 1 1 02

x x x x x

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Or using the horizontal line test, since every line y = k, k cuts the graph of

g( ) 2 tanx x x at one and only one point, g is 1-1 and thus the inverse exists.

(ii)

(b)

[3]

Method 1

1

13 5 2 4

23 5 2 4 2

3 5 2 4

3 5 3 5

g( ) 2 tan

2sin (cos )

2 + 13! 5! 2! 4!

( 1)( 2)2 + 1 ( 1)

3! 5! 2! 4! 2! 2!

52 + 1

6 120 2 24

52

2 24 6

x x x

x x x

x x x xx + x

x x x x xx + x

x x x xx x

x x x xx

5

3 5

+12 120

2 4

3 15

xx

x xx

Method 2

2

2

4 2 2

(4) 4 4 2 3

6 4 2 4 2

(5)

6 2 4 4 2

g( ) 2 tan

g ( ) 2sec 1

g ( ) 4sec tan

g ( ) 4 sec 2sec tan

g ( ) 4 4sec tan 4 sec tan sec tan

4sec 16sec tan 16sec tang ( ) 4

4sec 8sec tan 12sec tan

x x x

x x

x x x

x x x x

x x x x x x x

x x x x xx

x x x x x

Hence (5)g (0) 1, g (0) 4, g (0) 32

and thus 3 52 4

g( ) ...3 15

x xx x

(ii)

(c)

[3]

Since g is odd then 1g is also odd.

Let 1 3 5g ( ) ...x ax bx cx

3 51

3 51 1

1

3 53 5 3 5

3 5

2 4g g ( ) and g( ) ...

3 15

2 g ( ) 4 g ( )g ( )

3 15

2 ... 4 ......

3 15

x xx x x x

x xx x

ax bx cx ax bx cxax bx cx x

Compare the coefficient of x: 1a

Compare the coefficient of x3: 32 2

03 3

ab b

Compare the coefficient of x5: 5

22 4 16(3 ) 0

3 15 15

ac a b c

Therefore 1 3 52 16

g ( ) ...3 15

x x x x

2(i)

[1] 2 1 21

1 1 1 1 1 1 1 11 ... ... 1 ...

1 n n n

pp p p p p p p

since all the terms that are omitted are positive.

(ii)

[2]

Hence

21

1 1 1 11 ...

1n n

np P p Pp

p p p

and we need to explain why

21

1 1 1 11 ...

n

n

njp P p p p j

In the product on the left hand side, since nP denote the set of prime numbers less than or

equal to n, we must have the reciprocals of all the integers less than or equal to n by the

Fundamental Theorem of Arithmetic.

In fact, all integers less than n can be uniquely expressed as a product of prime powers,

where each prime is less than n. 1

1 2 ... , n ka aa

k i nn p p p p P . In addition each of the exponents cannot exceed n since 2n n

for all positive integers n.

There are additional terms after expanding, but they are all positive. Hence

11

1 1

1n

n

jp P pj

.

(iii)

(a)

[3]

Consider the graph of 1

.yx

1 2 3 4 …. n n + 1

y

x O

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1

1n

j j

represents the sum of all the areas of the rectangles shown. This is clearly greater

than the area under the graph of 1

yx

from 1 to 1n . Therefore,

1

1 1

1 1 d ln( 1) ln

nn

j

x n nj x

.

(iii)

(b)

[3]

Note that 1

ln ln(1 )1

xx

. If

2

1f ( ) ln(1 ) f ( ) 0

(1 )x x x

x

and thus the

curve is convex in the interval 1

0, .2

Hence the chord joining the points 0,0 and

1, ln 2

2

must lie above the curve.

Thus, 12

ln 2 1 1ln 2ln 2 ln

1 1x x

x x

.

But 7

2ln 2 1.386 1.45

and thus 7 1

ln5 1

xx

on 1

0, .2

OR

Consider 7 1 7

f ( ) ln ln 15 1 5

x x x xx

.

Then 7 1

f '( )5 1

xx

and 7 1 5 2

f '( ) 0 15 1 7 7

x xx

Also with

2

1f "( ) 0

1x

x

,

on the interval 2

0,7

, 7 1

f '( )5 1

xx

> 0 (increasing) and f (0) 0

so 2 2 7 2 1

0< f ( ) f f ln 0.6353 027 7 5 7

17

x

y

x O 1

and on the interval 2 1

,7 2

,

7 1f '( )

5 1x

x

< 0 (decreasing) and

1 7 1 1f ln 0.00685 0

12 5 21

2

so 0.00685<f ( ) 0.6353x

Therefore 7 1 7 1

f ( ) ln 0 ln5 1 5 1

x x xx x

on

10, .

2

(iv)

[3] Taking ln on both sides of

11

1 1

1n

n

jp P pj

, we have

1

1

1

7 1ln using (iii)(b)

5 1

1 ln

1

1 ln using (ii)

ln(ln )

n n

n

p P p P p

p P p

n

j

p

j

n

using (iii)(a)

Rearranging, we get 1 5

ln(ln )7

np P

np

. This shows that 1

np P p

diverges as

5ln(ln )

7n as n and thus the number of primes has to be infinite.

(i)

[3]

This is the same as distributing 9 distinct objects into 5 distinct boxes where no box is

empty.

Let iB denote the event that box i is empty.

No. of ways to distribute 9 distinct objects into 5 distinct boxes such that no boxes are

empty = 1 2 5B B B =

95 1 2 5...B B B

First, iB = No. of ways of distribution such that box i is empty = 94 for each i from 1

to 5. 5

9

1

54

1i

i

B

.

Similarly,

i jB B = No. of ways of distribution such that boxes i and j are empty = 93 for

1 5i j .

95

32

i j

i j

B B

.

i j kB B B = No. of ways of distribution such that boxes i, j and k are empty = 92

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95

23

i j k

i j k

B B B

.

i j k mB B B B = No. of ways of distribution such that boxes i, j, k and m are empty

= 1

5

4i j k m

i j k m

B B B B

.

1 2 5...B B B = No. of ways of distribution such that all 5 boxes are empty is zero.

By the principle of inclusion and exclusion,

1 2 5...B B B = 9 9 9 95 5 5 5

4 3 2 1 01 2 3 4

Hence, no. of ways to distribute 9 distinct objects into 5 distinct boxes such that no

boxes are empty = 9

1 2 55 ...B B B

9 9 9 9 95 5 5 5

5 4 3 2 1 .1 2 3 4

(ii)

[2] From GC, 9 9 9 9 95 5 5 5

5 4 3 2 1 834120 5! (9,5)1 2 3 4

S

.

Hence S(9,5) 6951 .

(iii)

[3]

Method 1

We will first consider the case where the 3 boxes are distinct. Then the

total number of ways to distribute r distinct objects into 3 distinct boxes

with no box empty is using the PIE as in (i)

(the total number of ways to distribute r distinct objects into 3 distinct

boxes with no restriction) 3

1

(the number of ways to distribute r

distinct objects into 3 distinct boxes with Box 1 empty) + 3

1

(the

number of ways to distribute r distinct objects into 3 distinct boxes with

Boxes 1 and 2 empty) = 3 3 2 3 1r rr .

The desired number of ways is thus

1 1

3 3 2 3 1 13 1 2

3! 2

r rr

r r

.

Method 2

Alternatively, we can do this:

(the total number of ways to distribute r distinct objects into 3 distinct boxes with no

restriction) (the number of ways to distribute r distinct objects into 3 distinct boxes

with 1 or 2 boxes empty).

Total number of ways to distribute r distinct objects into 3 distinct boxes with no

restriction is 3r since each of the r objects has a choice of 3 boxes.

Case 1: Exactly 1 box is empty.

There are 3 possibilities for 1 box to be empty. To distribute the r distinct objects into the

remaining 2 boxes with no box empty, the number of ways is 2S(r,2) =12(2 1)r .

Thus, by MP, the total number of ways with exactly 1 box empty is 6(2r1 1) .

Case 2: Exactly 2 boxes are empty.

There are 3

2C possibilities for 2 boxes to be empty and only 1 way to distribute the r

distinct objects into the remaining box.

Thus, by MP, the total number of ways with exactly 2 boxes empty is 3.

Hence, the total number of ways to distribute r distinct objects into 3 distinct boxes, with

no box empty is 3r 16(2 1) 3r .

The total number of ways to distribute r distinct objects into 3 distinct boxes with no box

empty is 3! S(r, 3).

6S(r, 3) = 13 3 6 2 1r r S(r, 3) = 11213

2

1

rr .

(iv)

[3]

We are looking for unordered pairs ( , )X Y where X and Y are distinct subsets of

{1,2, , }n fulfilling the condition in the question. (i.e. ( , )X Y and ( , )Y X are

considered identical). Since their intersection is empty this means that X and Y must be

disjoint.

Method 1:

We can distribute the n distinct balls to 3 identical boxes, but note that one box can be

empty (as it is not X and not Y). So we have two mutually exclusive cases of this last

box empty or not empty. This gives 3S( , 3) S( , 2)n n , because the last box is non-

empty in the first term, and can be any of the 3 identical boxes (X, Y takes the other two

identical boxes), or it could be empty in the second term.

1

1 13 1 13S( , 3) S( , 2) 3 2 2 1 3 1 2 .

2 2

nn n n nn n

Method 2:

This is similar to putting n distinct balls into 3 identical boxes, where one non-empty

box is X, one nonempty box is Y, and the last box can be empty. To use Stirling number

of the second kind (as the condition is that each box must be non-empty), we need to

add one more distinct object, so that whichever box has this distinct object is neither X

nor Y. Hence the answer is 1

S( 1, 3) 3 1 2 .2

n nn

Method 3:

Consider X Y k where 2 ,k n since they cannot be empty. There are n

k

ways

to select the k distinct elements of {1,2, , }n in the sets X and Y. If ( , )X Y is an

ordered pair, there are 2 21 2 1

kk k k

k

ways to partition these k distinct

elements between X and Y. Hence number of ways to partition these k distinct elements

between unordered pairs ( , )X Y is 12 1k . By MP, 12 1kn

k

and summing over

2 ,k n the answer is

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1

2

2 2 0 0

0 0

2 1

1 12 2 2 1 1

2 2

1 1 1 12 1 2 (1 1)

2 2 2 2

3 12 .

2

nk

k

n n n nk k

k k k k

n nnk n

k k

nn

n

k

n n n nn n

k k k k

n n

k k

4(i)

[2]

Method 1

2

2

2

2

h( ) 2 0

0 and 2 4 0 (using discriminant)

0 and

0 and

t At Bt C

A B AC

A B AC

A AC B

Method 2

2

2 2

2 2

h( ) 2t At Bt C

B BA t A C

A A

B AC BA t

A A

If 0A and 2 2. 0AC B AC B , then

2

0B

A tA

and

2

0AC B

A

Therefore,

2 2

h( ) 0B AC B

t A tA A

for all real values of t.

If

2 2

h( ) 0B AC B

t A tA A

for all real values of t, then

clearly A > 0. (If A < 0, h( ) 0t for sufficiently large t)

Also

22 2h 0 0

B AC BAC B AC B

A A

Therefore 0A and 2.AC B

(ii)

[3] For all real values of t,

2f ( ) g( ) 0t x x

Thus we have

2

f ( ) g( ) d 0b

at x x x

2 22 f ( ) 2 f ( )g( ) g( ) d 0

b

at x t x x x x

2 22 f ( ) d 2 f ( )g( ) d g( ) d 0

b b b

a a at x x t x x x x x

Let 2

f ( ) db

aA x x , f ( )g( ) d

b

aB x x x and

2g( ) d

b

aC x x

From (i), we have

2

2

22 2

f ( ) g( ) d 0

f ( ) d g( ) d f ( )g( ) d

b

a

b b b

a a a

A C B

t x x x

x x x x x x x

Equality holds when 2

f ( ) g( ) d 0b

at x x x . Since f and g are continuous this means

that we must have f ( ) g( ) 0t x x for all real x, i.e. f is a scalar multiple of g.

(iii)

[3] Let

3f ( ) 1 and g( ) 1x x x

Since f ( ) g( )x c x , therefore equality does not hold, so we have

221 1 12 3 3

0 0 01 d 1 d 1 dx x x x x

2

1 13 3

0 01 1 d 1 dx x x x

1

4 21

3

00

1 d4

xx x x

2

13

0

51 d

4x x

1 13 3

0 0

51 d since 1 d 0

2x x x x

(iv)

[4]

1 11

00 0

1

0

2 21 1

0 0

1 1 22

0 0

1 2

0

k( ) d k( ) k ( ) d

k ( ) d since k(1) 0

k( ) d k ( ) d

d k ( ) d

1 = k ( ) d

3

x x x x x x x

x x x

x x x x x

x x x x

x x

Equality holds if and only if 2k ( ) k( ) since k(1) 0.x cx x ax a

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5

(a)

[3]

Method 1

1

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

! !

( 1)( 2)...( 1)[( ) ]

!

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

( 1)!

f ( 1)

n n

n

x x x n x x x x nx x

n n

x x x nx n x

n

x x x n

n

x

Therefore,

1 1

1 2

1

1

2 1

1 1

f ( ) f ( ) f ( 1) f ( 1)

f ( ) f ( 1) f ( 1)

f ( ) f ( 1) f ( 1)

f ( 1) ( 1)

f ( 1) 1

N N

n n n

n n

N N

n n

n n

N

N

N

x x x x

x x x

x x x

x x x

x

Method 2

Let PN denote the statement that 1

f ( ) f ( 1) 1N

n N

n

x x

.

Check P1: L.H.S.1

1

1

f ( )n

x x

R.H.S. 1

1f ( 1) 1 1

1!

xx x

L.H.S.

Therefore P1 is true.

Assume that Pk is true for some k , i.e. 1

f ( ) f ( 1) 1k

n k

n

x x

, then for Pk+1:

1

1

1

1

1

1

L.H.S. f ( )

f ( ) f ( )

f ( 1) 1 f ( )

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

! ( 1)!

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

! 1

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

( 1)!

f ( 1) 1 R.H.S.

k

n

n

k

n k

n

k k

k

x

x x

x x

x x x k x x x x k

k k

x x x k x

k k

x x x k x k

k

x

Therefore 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

N ,

i.e. 1

f ( ) f ( 1) 1N

n N

n

x x N

. (Proven)

(b)

[5]

Method 1

Since 1 1 3 5 2 1

f2 2 4 6 2

n

n

n

and all the terms are positive, it is clear that

1f 0

2n

.

It suffices to show 1 1 3 5 2 1 1

f2 2 4 6 2 2 1

n

n

n n

.

When n = 1, LHS 1 1 1

2 3 2 1

= RHS.

Suppose 1 3 5 2 1 1

2 4 6 2 2 1

n

n n

for some positive integer n.

Then

2

2

1 3 5 2 1 2 1 1 2 1 1

2 4 6 2 2 2 2 22 1 2 3

2 3 2 2

2 1 2 1

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

3 4, which is true.

n n n

n n nn n

n n

n n

n n n

Therefore, by mathematical induction, the result we want to prove is true for all positive

integers n.

Method 2

Let QN denote the statement that 1 1

0 f2 2 1

NN

.

Check Q1: 1

1 1 1 10 f

2 2 3 2 1

Therefore Q1 is true.

Assume that Qk is true for some k , i.e. 1 1

0 f2 2 1

kk

, then for Qk+1:

1

11

1 1ff ( ) 2 1 1 2 1 12 2 f f

1f ( ) 1 1 2( 1) 2 2( 1) 2f

2

k

kk k

kk

kx x k k k

x k k k k

Since 1 1

0 f2 2 1

kk

, 1

1 2 1 1 2 1f

2 2( 1) 2( 1)2 1k

k k

k kk

Since A.M > G.M. (and equality does not hold since 2 1 2 3k k )

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

2

k kk k k

i.e. 1

1 2 1 2 1 1 1f

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

k k

k k k k k

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and it is obvious that 1

1f 0

2k

, so 1

1 10 f

2 2( 1) 1k

k

.

Therefore Qk is true Qk+1 is true

Since Q1 is true and Qk is true Qk+1 is true, by mathematical induction, QN is true for

all N ,

i.e. 1 1

0 f ( )2 2 1

N NN

. (Proven)

[3] From (a),

1

f ( ) f ( 1) 1N

n N

n

x x N

Sub 1

2x ,

1

1 1f f 1 1

2 2

N

n N

n

1 1 1 1 1 1 1 11 1 1 2 2 32 2 2 2 2 2 2 22

1! 2! 3! 4!

1 1 1 11 2 ... 1

12 2 2 2... f 1

! 2N

N

N

1 1 1 3 1 3 5 1 3 ... (2 3) 1... f 1

2 2 4 2 4 6 2 4 6 8 2 4 6 ... 2 2N

N

N

Taking the limit as N on both sides,

1 1 1 3 1 3 5 1

... lim f 12 2 4 2 4 6 2 4 6 8 2

NN

From (b): 1 1

0 f2 2 1

NN

, and

1lim 0

2 1N N

1lim f 0

2N

N

Then

1 1 1 3 1 3 5... 0 1

2 2 4 2 4 6 2 4 6 8

1 1 3 1 3 5 1...

2 4 2 4 6 2 4 6 8 2

Therefore the infinite series 1 1 3 1 3 5

...2 4 2 4 6 2 4 6 8

is convergent series and its

sum is 1

2.

6(i)

[3]

Consider the parity of each coordinate, which is either odd or even. Hence there are only

four ‘holes’ (odd, odd), (odd, even), (even, odd) or (even, even). Since there are five lattice

points, there must be two of them whose parity for each coordinate are the same. The sum

of the two coordinates must thus be even and thus their midpoint 1 2 1 2,2 2

x x y y

is an

integer.

(ii)

[3] Similarly, least

32 1 9m .

Since there are 8 ‘holes’ (2 choices for each coordinate) and 9 lattice points, there must be

two lattice points whose parity for each coordinate is the same. The midpoint is an integer

using the same argument in (i).

We will show 8 points is insufficient by a counter-example. Consider the 8 vertices of the

unit cube. It is clear that the midpoint of each edge is not a lattice point. In fact, along the

edge which the x, y or z coordinate increases by 1, the midpoint along that coordinate is not

an integer.

(iii)

[3]

Using the result in (i) to any five of the 13 lattice points we have a pair of lattice points such

that their midpoint is a lattice point. Call these two points 1 2,P P .

Now with 11 remaining lattice points, apply the result again to get another two lattice points

3 4,P P whose midpoint is also a lattice point.

Continue in this manner until we have 5 remaining lattice points where we apply the result

one more time to get another 2 lattice points 9 10,P P .

Now consider the midpoints of these 5 pairs of lattice points. Call them

1 2 3 4 5, , , , .M M M M M They are all lattice points.

Now applying the result in (i) one last time to these 5 points, we can find two of them whose

midpoint is a lattice point. However, the midpoint of these two lattice points is the centroid

of the original 4 points which the 2 pairs came from.

WLOG, if 1 2,M M are the two lattice points, then we already know they are of the form

1 ,2 2

a b a bx x y yM

and 2 ,2 2

c d c dx x y yM

. Then the midpoint of 1 2,M M is

2 2 2 2, ,2 2 4 4

a b c d a b c d

a b c d a b c d

x x x x y y y yx x x x y y y y

,

which is the centroid of the original 4 points.

(iv)

[2]

Suppose there are 4 such points. Then the sum of the points of S modulo 4 must be (0, 0).

Thus all 4 points must have the same first coordinate modulo 4 and the same second

coordinate modulo 4.

But this means that the four points must be the same modulo 4, which we have chosen S to

have no more than three of each.

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7

(a)

(i)

[2]

1

3

1 ()

5 ()()(), ()(()), (())(), (()()), ((()))

C

C

(ii)

[3]

Let D be the set of invalid strings with n open and n closed parentheses, and E be the set

of strings with 1n open and 1n closed parentheses.

Define a mapping f : D E where for the first “)” not matching a “(” in a valid way

before it in a string in D, we flip all subsequent “(” to “)” and vice versa in that string. It is

clear that there are k matching pairs before this first non-matching “)”, where 0 1.k n

After the map, the image string has n k closed parentheses and 1n k open

parentheses after the first non-matching “)”, making it a total of ( 1) 1k n k n open

and 1 ( ) 1k n k n closed parentheses. Hence the image lies in set E.

first non-matching “)”

Pre-image k matching pairs ) n k “(” and 1n k “)”

Image k matching pairs ) n k “)” and 1n k “(”

To show that f is injective, we see that two different invalid strings in set D will be

mapped to different strings in E because the ordering of the open and closed parentheses

will be different.

To show that f is surjective, take any string with 1n open and 1n closed parentheses

in set E, and start reading from the left for the first “)” not matching a “(” before it in a

valid way. There will be k pairs before this first non-matching “)”, where 0 1.k n

This is always possible as there are more closed than open parentheses, and that this first

non-matching “)” will not be at the rightmost position. Flip all subsequent parentheses

after it, and we have a total of ( )k n k n open parentheses, and 1 ( 1)k n k n

closed parentheses. This is an invalid string with n open and n closed parentheses in set D.

So every element in E has a preimage in D.

Hence f is a bijection and 2

1

nD E

n

.

(iii)

[2]

To find the number of valid strings, we subtract the number of invalid strings from the

total possible arrangements of n open and n closed parentheses.

2 2

1

2 2 2 (2 )! (2 )! since

11 ( 1)!( 1)! ( 1) ! !

21

1

n

n nC

n n

n n nn n n n

n n nn n n n n n

n

nn

(b)

(i)

[3]

In any valid string of n open and n closed parentheses, the first character (from the left)

has to be “(”. Somewhere on the right there is a matching “)”. In between there is a valid

string let’s call it A, and on the right of “)” there is another valid string B. We have ( )A B

where A or B can contain zero pairs, where there is exactly one way for zero pairs, defined

as 0 1C . Either A or B can contain 0 to 1n pairs of parentheses, but if A contains k

pairs, B contains 1n k pairs so that there is a total of n pairs. This is computed by MP

using 1k n kC C .

Hence we sum all the mutually exclusive cases where A contains 1n pairs, B contains 0

pairs, where A contains 2n pairs, B contains 1 pair, all the way till where A contains 0

pairs, B contains 1n pairs. This gives the recurrence formula as kC is the number of

valid strings for 0 1.k n

(ii)

From 1 0 2 1 1 2 0 1n n n n nC C C C C C C C C , we know that if n were even, there is an

even number of terms which can be paired up, i.e. 2 2

1 0 2 1 12 n nn n nC C C C C C C

,

contradicting the assumption that nC is odd.

Similarly, since n is odd, we know that we can write

1 3 1 12 2 2 2

1 0 2 12 n n n nn n nC C C C C C C C C

which implies that 1 12 2

n nC C is odd and thus 12

nC is odd.

(iii) From the above, we see that if nC is odd, then n and 1

2nC are odd. This means 1

20n and

1 32 4

1

2

n nC C is odd. Continuing this, we will eventually get

2 10

2

k

k

n for some

nonnegative integer k. This means that we have 2 1kn for some nonnegative integer k.

Conversely, since 0 1 3, ,C C C are odd, using

1 3 1 12 2 2 2

1 0 2 12 n n n nn n nC C C C C C C C C ,

We have 12 1 2 1 2 12 ...k k kC C C

, which means that we must have 2 1kC

being odd for all

nonnegative integers k.

8(i)

[3]

Method 1

The distance between the point (a, b) and the half-line ml is given by

2 2

1

0 0 | |

1 1

a

b m

am b

m m

.

Hence the half-line meets the circle with radius R centered at (a, b) if and only if the

distance d between the point (a, b) and the half-line ml satisfies d R , i.e.

21b ma R m .

Method 2

ml meets a tree centered at (a, b) if and only if the equation

2 2 2( ) ( )x a mx b R has a real solution. Grouping terms we have 2 2 2 2 2(1 ) 2( ) 0m x a bm x a b R . Since there is a real solution, the discriminant

is nonnegative and thus

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2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2

4( ) 4(1 )( ) 0

2

(1 ) 2 ( )

a bm m a b R

a abm b m a b R a m b m R m

R m b abm a m b am

Taking square root on both sides of the inequality yields the desired result.

Other methods

One may also use similar triangles to find the distance between the center of the circle

(a, b) and the line y = mx which is tangential to the circle. In that case, the triangles

need to be clearly labelled.

(ii)

[1]

Suppose that m is irrational. From the result given to be true, there exists odd positive

integers a and b such that 21b ma R m . (It suffices to choose 21R m ).

However this means the half-line ml will meet the circle with radius R centered at (a,

b).

(iii)

(a)

[1]

Since 20 1p qm R m , from (i), we know the half-line ml will meet the circle

with radius R centered at (q, p).

(iii)

(b)

[2]

Suppose that p and q are of different parities and that ml meets a tree centered at (a, b)

where a and b are both odd integers. Then from (i), we have

2

1p p

b a Rq q

.

Multiplying by q > 0 on both sides of the inequality, we have 2 2bq pa R p q .

Since p and q are of different parities and a and b are both odd, ap bq is odd and thus

2 21 ap bq R p q .

(iv)

[5] Suppose that

1

5R and let us show that all half-lines ml where 0m will meet a

tree planted at ,1 or 1, .

It suffices to show this is true for 1

5R (since a tree meeting ml for this value of R

will also meet ml for any value of R greater than this).

A natural approach is to first consider for what values of m will the line y = mx intersect

the circle centered at (1, 1). This is equivalent to determining the range of m such that 2 2 1

5( 1) ( 1)x mx has a solution. Using the fact that this means we have a

nonnegative discriminant, we have 2 2 95

4(1 ) 4(1 ) 0.m m This of course reduces

to 1

(2 1)( 2) 0 2.2

m m m

If we consider symmetry, then it suffices to consider m > 2. We need to show the line

now meets a circle centered at (1, ), where is an odd integer depending on m.

We claim that if is the closest odd integer to m, 21

5

mm

and we are done. In

fact, if is chosen as such, then 1m . Therefore 2 21 2 1

15 5

mm

since 2m .

Alternatively, if no symmetry is considered, the case 12

2m is the same as 1 2m .

We need to just provide the construction for the case 1

2m .

Suppose 1

2m . Then

12m

m and thus there is an odd positive integer such that

2

1

5

mm

. Since

1m

m , this implies that

211

5

mm

.

(v)

[2]

We prove this statement using the contrapositive. If Taylor cannot get out of the woods,

then all the half-lines ml will meet a tree and therefore using (iii)(b) with 2

21

m ,

(since 2 and 1 have different parities), 2 2 11 1 2

5R R . But from (iv), if

1

5R , Taylor cannot get past the first row or first column of trees.

Note that the result is mathematically true in the following more restricted sense: If there

is a half-line that does not intersect any circle centered at 1, or ,1 , then there is a

half-line (not necessarily the same as the previous) that does not intersect any of the

circles.