2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
St. Mark’s School
2003/04 Half-yearly Examination
Pure Mathematics
Time allowed: 3hr.
S.6B
Section A:
1. (a)
Jan. 2004
A16 + 2R
SHK
a n 1  a n b  ab n  b n 1
=
a n a  b   b n a  b 
=
a  ba n  b n 
=
a  b 2 a n1  a n2 b  a n3b 2   ab n2  b n1 
 a, b  0, a  b
0
1M
1
(b) Let S (n ) be the statement “ a n1  b n n  1a  nb  0 ”
For n  1,
a 2  b2a  b  a 2  2ab  b 2  a  b   0 .
2
 a  b
 S (1) is true.
Assume S (k ) is true for some k N.
1
i.e. a k 1  b k k  1a  kb  0
For n  k  1,
L.H.S. = a k  2  b k 1 k  2a  k  1b
= a k  2  b k 1 k  1a  kb  ab k 1  b k  2
 a k  2  a k 1b  ab k 1  b k  2
0
2.
(by assumption)
(by (a))
1
1
(5)
1 3
2n  1
 
2 4
2n
2 4
2n
Let y n     
.
3 5
2n  1
Consider,
xn 
2n
2n  1
1
4n 2  2n  12n  1

=
=
>0
2n  1
2n
2n2n  1
2n2n  1




1M
2n
2n  1

2n  1
2n
y n  xn
 x n 2  x n y n
xn 

1 2 3 4
2n  1 2n
1
   


2 3 4 5
2n 2n  1 2n  1
2M
1
2n  1 .
So, 0  xn 
1
1
2n  1
, and lim
n 
1
2n  1
0
By sandwich rule, lim x n  0 .
1
n 
(5)
p.1 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
3.
(a)
P (x )
Q (x)
1
2 x 4  3x 3  x 2  x  1
x 1
2
1
2x 4  x3  0x 2  x
2
1
 4x3  x 2  x  1
2
3
2
 4x  2x  0x  1
1
x2  x
2
4x3  2x 2  0x  1
4x  4
4x3  2x 2
4x 2  0x  1
4x 2  2x
2x  1
2M
1A
 the greatest common divisor of P (x ) and Q (x) is 2x  1 .
1
(b) Let r1 ( x)  x 2  x ,
2
d ( x)  2 x  1
1

 P ( x)   x  1Q( x)  r1 ( x)
2

Q( x)  4x  4r1 ( x)  d ( x)
1M
 d ( x)  Q( x)  4x  4r1 ( x)


1

 Q( x)  4 x  4 P( x)   x  1Q( x)
2




1

 4 x  4P( x)  1  4 x  4 x  1Q( x)
2




 4 x  4P( x)  2 x 2  2 x  3 Q( x)
 S ( x)  4 x  4
T ( x)  2 x 2  2 x  3
2A
(6)
4.
(a)
1  xn  C0n  C1n x  C2n x 2  C3n x 3    Cnn x n
1M
Integrate both sides w.r.t. x from 0 to 1.
1
1 n 2 1 n 3 1 n 4
1
 n
n n 1 
0 1  x dx  C0 x  2 C1 x  3 C2 x  4 C3 x    n  1 Cn x  0
1
n
1M
 1  x n 1 
1 n 1 n 1 n
1
n
C nn

  C0  C1  C 2  C3   
2
3
4
n 1
 n 1 0
1
1 n 1 n 1 n
1
2 n1  1
n
.
 C  C1  C 2  C3   
Cn 
2
3
4
n 1
n 1
n
0
(b)
2x 2  x  4

x x2  2

2


A Bx  C Dx  E


2
x x2  2
x2  2


1M




A x 2  2  xBx  C  x 2  2  xDx  E 
2

x x 2



2

2

 2 x 2  x  4  A x 2  2  x x 2  2 Bx  C   xDx  E 
2
1
p.2 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
  A  B x 2  Cx 3  4 A  2B  Dx 2  2C  E x  4 A
Comparing cofficients, we get
 A B

C

4 A  2 B  D
 2C  E


4A

A
B

 C
D

 E

 0
 2
1M
 1
 4
1
 1

0

0

1
2x 2  x  4

 0
x x 2
2

2

1
x
1
.
 2

2
x x 2 x 2 2

1

(6)
5.
(a)  r  r is a root of x 3  ax  b  0

 r r

3


a r  r b  0
1M
r 3  3r 2 r  3r 2  r r  ar  a r  b  0
r
3


1A

 3r 2  ar  b  r 3r 2  r  a  0
 r is not a rational
 r 3  3r 2  ar  b  0 and 3r 2  r  a  0
(b) (i)
1
Sub. r  r into x 3  ax  b  0

L.H.S. = r  r
  ar  r   b
3
= r 3  3r 2  ar  b   r 3r 2  r  a 
=
0
1
(by (a))
 r  r is also a root of the equation.
(ii) From (a), 3r 2  r  a  0  r 2  
ra
3
Sub. into r 3  3r 2  ar  b  0

r 2  ar
 3r 2  ar  b  0
3
1M+1A
8r 2  2ar  3b  0
 r a
8 
  2ar  3b  0
3 

6a  8r  9b  8a  0
r
8a  9b
23a  4
for a 
4
3
1
p.3 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
Alternatively,
 r 3  3r 2  ar  b  (1)
From (a), 
2
 (2)
  3r  r  a
r 3r 2  ar  b
(1)

:
3
ra
(2)
2
8r  2ar  3b  0
r  a 2ar  3b

Hence,
3
8
8r  8a  6ar  9b
8a  9b
4
for a 
r
3
2(3a  4)
6.
(a) [Differentibility]
f (0  x)  f (0)
f ' (0)  lim
x 0
x
x 3 sin 1  0
x
= lim
x  0
x
1
2
= lim x  sin
x 0
x
(b)
1
1M
1
2
is bounded and lim x   0 )

x

0
x
 f is differentiable at x  0 and f ' (0)  0 .
[Continuouity]
 f is differentiable at x  0 .
 f is continuous at x  0 .
Alternatively,
1
lim f ( x)  lim x 3 sin  0
x 0
x 0
x
f (0)  0
 f is continuous at x  0 .
= 0.
1M+1A
( sin
1
1
 2
3x sin  x cos
, x0
f ' ( x)  
x
x

0
, x0

f ' (0  x)  f ' (0)
f " 0  lim 
x 0
x
1
1
2
3x  s i n  x  c o s  0
x
x
 lim 
x  0
x
1
1
1
1A
1
1 

 cos 
= lim   3x sin
x 0 
x
x 
does not exist.
 the second derivative of f at x  0 does not exist
1
(5)
p.4 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
7.
(a)
y' 
2 sin 1 x
1 x2
1 x2 
y" 

2
1 x
 2 sin 1 x 
2
 2x
2 1 x2
1 x2
2 1  x 2  2 x s i n1 x
1  x  1  x
2

1A
2

 1  x 2 y" xy'

2 1  x 2  2 x sin 1 x

1 x2
2 x sin 1 x
1 x2
 2.


 1  x 2 y" xy'  2 …(*)
1
Differentiate (*) w.r.t. x n-times.
( n 2)
 n 1 x2
1  x y
( n 2)
 2nxy( n1)  nn  1y ( n)  xy( n1)  ny ( n )  0
1  x y
( n 2)
 2n  1xy( n1)  n 2 y ( n )  0 … (**)
2
2
2


nn  1
1 x2
2
1  x y
(1)
y ( n 1) 


( 2)
y ( n )  xy ( n 1)  nx (1) y ( n )  0
1M
1
(b) Put x  0 into (**), y ( n 2)  n 2 y ( n ) (0) .
1
For n is even, let n  2m  2 , where m  N.
y ( 2 m ) 0  2m  2 y ( 2 m2) 0
2
 2m  2 2m  4 y 2 m  4  0
2
2
 2m  2 2m  4  2 2  y ( 2) (0)
2
2
 22m  22m  4 4  2 .
2
 y   0  2
 y ( 2 n ) 0  22n  22n  4 4  2 .
2
2
1
For n is odd, let n  2m  1 , where m  N.
y ( 2 m 1) 0  2m  1 y 2 m1 0
2
 2m  1 2m  3 y 2 m 3  0
2
2
 2m  1 2m  3 1  y (1) 0
2
 0.
2
 y
(1)

 0
 y ( 2 n1) 0  0 .
1
(7)
p.5 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
Section B:
8.
(a) (i)
Let f ( x)  x 3  3 px  2q
If (*) has a repeated root, then f ( x)  0 and f ' ( x)  0 has a common root.
1M
f ' ( x)  3x 2  3 p

f ' ( x)  0
x2  p
1A
x p
If x 
p is the common root, then
 p   3 p p  2q  0
3
 p
3
q
p3  q2
If x   p is the common root, then
 p   3 p p  2q  0
3
 p
3
 q
p3  q2
Alternatively,
f ' ( x)  0
1

x2  p

 (1)

2
f ( x)  0
 ( 2)
 x x  3 p  4q
(*) has a repeated root iff f ( x)  0 and f ' ( x)  0 has a common root.
2
2
Sub. (1) into (2), p p  3 p   4q
2
p3 , f ( p) 
 p   3p
3
1M
2
p3  q2
(ii) If q 
1A
2
1
(3)
 
p  2 p3  0
1A
p satisfies both f ( x)  0 and f ' ( x)  0

p is a repeated root of (*).

1

3

  p  0
(iii) If q   p 3 , f ( p )   p  3 p  p  2
3
(2)
1A+1A
 p satisfies both f ( x)  0 and f ' ( x)  0
 (*) has a repeated root and the root is  p .
1
(3)
p.6 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
(b) (i)
Let x  y  h , then (**) becomes
2 y  h   3 y  h    y  h   c  0
3
2

 

2 y 3   6h  3y 2  6h 2  6h  1 y   2h 3  3h 2  h  c  0
1A
1
, (**) becomes
2
1
2y3  y  c  0
2
Put h 
1
c
y 3  3  y  2   0 … (3)
 12 
4
1A
(2)
(ii) Using the results of (a)(ii) and (iii), (3) and hence (**) has a repeated root if
3
1
c
   
 12 
4
c2 
1
108
c
1
6 3
2
1M
( c  0 )
Consider (3) when c 
1
6 3
1A
,
 1 
1
y 3  3  y  2
  0 … (4)
 12 
 24 3 
Taking p 
q 
p3 ,
1
1
and q 
in (a),
12
24 3
p
1
is a repeated root of (4).
12
1M
2

1 
1 
 (4) becomes  y 
 y 
  0
3 
2 3

y
1
1
(repeated)
1A
1
1
1 1
or  
(repeated)
x 
2 2 3
2
3
1A
3
or
2 3
For the roots of (**),
1
x y
2
(5)
(15)
p.7 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
9.
(a) [Existence]
Let S (n ) be the statement “ there exist a positive integers a n and bn such that
1  3 
2n
 an  bn 3 , (i) an  3b 2  2 2n and (ii) a n and bn are both divisible by
2
2 n .”
For n  1,

L.H.S. = 1  3

2
 42 3
1A
Put a1  4 , b1  2 , then
(i)
a1  3b1  16  12  4  2 2
2
2
(ii) a1 ,b1 are both divisible by 2
1
Assume S (k ) is true for some integer k, k  1 .

i.e. 1  3

2k
ak  3bk  2 2k and (ii)
 ak  bk 3 , (i)
2
2
a k , bk are both divisible by
2k .
For n  k  1,
1  3  
2 k 1
=
1  3  1  3 
a  b 3 4  2 3
=
4ak  6bk   2ak  4bk 
=
2k
k
2
2M
k
3
Put a k 1  4a k  6bk , bk  2ak  4bk , then
ak 1  3bk 1
2
(i)
2
=
4ak  6bk 2  32ak  4bk 2
=
16ak  48ak bk  36bk  3 4ak  16ak bk  16bk
=
4ak  12bk
=
2 2  2 2k
2

2
2
=
2
=

2
4 ak  3bk
2
2
2


2 2k 1
1
(ii)  ak , bk are divisible by 2 k
 ak 1  22ak  3bk  and bk 1  2ak  2bk  are divisible by 2 k 1 .
1
 S (k  1) is true.
By the principle of mathematical induction, S (n ) is true for all positive integers n.
[Uniqueness] Suppose a  b 3  c  d 3 for some a, b, c, d  Z+, then
(a  c)  (b  d ) 3
 3 is irrational,
 a  c  b  d  0  a  c and b  d
p.8 of 13
2
(8)
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
(b) By (a), an  3bn  2 2n
2
2
2 2n
 an  bn 3 =
1M
a n  bn 3
1  3 

3  1  3 
2n
2 2n
1 
=
2n
2 2n
=
 2
2n
1  3 
2n
2n
1  3 
2n
=
Alternatively,

For n  1, 1  3

Assume 1  3
For n  k  1,
1  3  
2 k 1


2
2k
1
 4  2 3  a1  b1 3
 ak  bk 3 , where k  1.
1M
1  3  1  3 
a  b 3 4  2 3
2k
=
=
k
2
k
4ak  6bk   2ak  4bk 
=
3
= ak 1  bk 1 3
By the principle of mathematical induction, it is true for all positive integer n.
Alternatively,
1  3 
2n
2n
  C r2 n
r 0
 3
r
n
=
 C22rn
r 0
n
 a n   C 22rn
1  3 
r 0
2n
 3
2r

C  3
n
=
r 0



2r
1A
r
2r
2n
2r
r 0

2r
 
n 1
, bn   C 22rn1 3
  C r2 n  3
r 0
 
 n 1
 3  C 22rn1 3
 r 0
2r
 3
2n
1
 
 n 1 2 n
 3  C 2 r 1 3
 r 0
2r



= an  bn 3
1
(2)

a
1 3
(c) From (b), n  3 
bn
bn

2n
1M
1
 0.
n  b
n
 bn is divisible by 2 n
 lim
 1 3  1
 lim 1  3
n 

a

Hence lim  n  3   lim 1  3
n b
 n
 n


2n

2n
0
1 For either
1
0
n b
n
 lim
an
 3
n  b
n
 lim
1
(3)
p.9 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)

(d)  1  3
  1  3 
2n
2n

 2an  Z+ and 0  1  3

 2a n is the smallest integer greater than 1  3


2n
2n
1
1M
.
 a n is divisible by 2 n
 2a n is divisible by 2 n 1 .
1
(2)
(15)
10. (a) (i)
Case 1: for n is a positive integer
Let S (n ) be the statement “ g (nx)  g ( x) ”.
n
For n  1,
L.H.S. = g (x )
R.H.S. = g (x ) .
 S (1) is true.
Assume S (k ) is true for some integer k  1.
i.e. g (kx)  g ( x)
k
1M
For n  k  1,
gk  1x
=
=
g (kx  x)
g (kx)  g ( x)
=
g ( x)k 1
 S (k  1) is true.
By the principle of mathematical induction, S (n ) is true for all positive integer n. 1M
Case 2: for n = 0
g (0)  g (0  0)  g (0)  g (0)
 g (0)1  g (0)  0
 g (0)  0 or g (0)  1
If g (0)  0 , then for any x R
g ( x)  g ( x  0)  g ( x)  g (0)  0
which contradicts that g (x ) is a non-zero function.
 g (0)  1
 g (0  x)  g (0)  1
and g ( x)  1
0
1M
1M
1M
Case 3: for n is negative integer
For any negative integer m, let m  n ,
 g (mx  nx)  g (mx)  g (nx)
1  g (0)  gm  nx
= g (mx)  g ( x)
n
 g (mx) 
1M
(since n is a positive integer)
1
g ( x)n
 g ( x ) 
n
p.10 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
= g ( x ) 
m
1M
 g (nx)  g ( x) for all integer n.
n
(ii) Let M be such that g ( x)  M for all x R.
(7)
1M
If g (x ) is not congruent to 1, then there exists   R such that
g ( )  1 or g ( )  1 .
Case 1: g ( )  1
If g ( )  1 , then there exists n N such that
g ( )]n
M
 g ( n )  M
which contradicts that g (x ) is bounded.
1M
Case 2: g ( )  1
If g ( )  1 , then
1
 1.
g ( )
 there exists a negative integer m such that
g ( )
m
M
 g (m )  M
which contradicts that g (x ) is bounded.
 g ( x)  1 .
(b) (i)
f (1) 
g (1)
g (1)1
1 0 .
 f is a non-zero function.
f ( x  y) 
=
1M
g ( x  y)
g (1)x y
g ( x)  g ( y )
g (1)x  g (1)y
= f ( x)  f ( y )
(ii)
1M
(3)
1M
(2)
f ( x  1)  f ( x)  f (1)  f ( x)
 f is a periodic function of period 1.
1M
Since g (x ) is bounded on 0,1 , f (x) is bounded on 0,1 .
And f is periodic with period 1.
 f is bounded on R.
1M
(2)
(iii) By (a) (ii), we have f ( x)  1 .
 g ( x)  g (1)
x
for any x R.
1M
(1)
(15)
p.11 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
11. (a)
F ( x, y ) 
1
f  y  x
2x
F 1, y  
1
f  y  1
21

1A
1
y2
f  y  1 
 y5
2
2
1M
f  y  1  y 2  2 y  10   y  1  9
2
1A
 f  y  x   y  x  9
2
i.e. F  x, y  


1
 y  x 2  9 .
2x
1
(4)
(b) (i)
x0  0 , xn1  F xn ,2 x  , for n  0,1,2, 




x1  F x0 ,2 x0  
1
2 x0  x0 2  9  1 x0 2  9
2 x0
2 x0
x 2  F x1 ,2 x1  
1
1
2
2
 2 x1  x1   9 
x1  9
2 x1
2 x2






1M


x n 1  F x n ,2 x n  

1
2 xn  xn 2  9  1 xn 2  9 .
2 xn
2 xn
1A
(ii) As x0  0 , xn  0 for all n N.
x n 1 

1
2
xn  9
2 xn



1
xn  32  6 xn
2 xn

1
6 x n 
2 xn

1M
1M
3
{xn } is bounded below by 3.
x n 1  x n 


1
2
xn  9  xn
2 xn

xn  9  2 xn
2 xn

9  xn
2 xn
2

1A
1M
2
2
1A
3  xn 3  xn 
1M
2 xn
p.12 of 13
2003/04 S.6 Pure Maths. Half-yearly Exam. (Solution)
0
1
 xn1  xn for all n N.
 {x n } is monotonic decreasing.
Let lim x n  l .
n
n 

1 2
l 9
2l
l2  9
or
l 3
l

1
2
xn  9
n  2 x
n
 lim x n 1  lim


1M
l  3
(rejected)
1A
 lim x n  3 .
n 
(11)
(15)
p.13 of 13