St. Mark’s School
Mid-Year Examination (2005  2006)
Form 6 Pure Mathematics
Suggested Answer
Section A (40 marks)
1.
(a)
1  xn  C0n  C1n x  C2n x 2  ...  Crn x r  ...  Cnn x n
… (*)
Diff. (*) w.r.t. x,
n1  x
n 1
 C1n  2C2n x  3C3n x 2  ...  nCnn x n1 … (**)
1M
Put x  1, we have
C1n  2C 2n  3C3n  ...  nC nn  n  2 n 1 .
1A
(b) From (**), we have
nx1  x
n 1
 C1n x  2C2n x 2  3C3n x 3  ...  nCnn x n … (***)
Diff. (***) w.r.t. x,
nn  1x1  x
n 2
 n1  x
n 1
 C1n  2 2 C2n x  32 C3n x 2  ...  n 2Cnn x n1
1M
Put x  1, we have
C1n  2 2 C 2n  3 2 C3n  ...  n 2 C nn  nn  1  2 n  2  n  2 n 1
 n  2 n2 n  1
1A
(4)
2.
Let S n be the statement “ a n  3 n 1  2 ”.
For n = 0,
LHS = a 0  1 , RHS  3 01  2  1
 S 0 is true.
For n  1,
LHS = a1  7 , RHS  311  2  7
 S 1 is true.
1M
Assume S k  and S k  1 are true for some non-negative integer k.
i.e. a k  3 k 1  2 and a k 1  3 k  2  2
1M
For n  k  2 , ak  2  4ak 1  3ak  0

 
a k  2  4 3 k  2  2  3 3 k 1  2

 12  3 k 1  8  3  3 k 1  6
 9  3 k 1  2
 3 k 3  2
1M
1M
1A
 S k  2 is true.
 By mathematical induction, S n is true for all non-negative integers n.
(5)
Page 1 of 10
3.
(a) Tr 1  Tr
Crkn x r  Crkn1 x r 1
kn! x r 
kn!
x r 1
r  1!kn  r  1!
r !kn  r !
correct
kn  r  1
x 1
r
1A for both C rkn and C rkn1
( x  0 )
kn  r  1  2 
  1
r
k
1M
2kn  2r  2  rk
2kn  1
r
k2
(b) Putting n  17 and k  3 in (a), we have
251  1
r
3 2
r  20.8
1A
1M
1A
So, we have T1  T2  ...  T21 and T21  T22  ...  T52 .
20
2
The greatest term is T21  C   .
3
51
20
1A
(6)
4.
(a)
1
A
B
C
 

xx  1x  2 x x  1 x  2
1M
Ax  1x  2  Bx x  2  Cxx  2  1
1
2
Put x  1 , B  1
1
Put x  2 , C 
2
1
1
1
1
.




xx  1x  2 2 x x  1 2x  2
Put x  0 , A 
n
(b)
1A
1
 k k  1k  2
k 1
n
 1
1
1 

  


k  1 2k  2 
k 1  2k

1M
1 1 1
 
2 2 6
1 1 1
  
4 3 8
1 1 1
  
6 4 10
1 1 1
  
8 5 12
Page 2 of 10

1
1
1


2n  2 n  1 2n
1
1
1

 
2n  1 n 2n  1
1
1
1



2n n  1 2n  2
1
1
1
1
 


4 2n  1 n  1 2n  2
1
1
1
 

4 2n  1 2n  2

n
lim
n 
1M
1A
1
1
1

1
 k k  1k  2  lim  4  2n  1  2n  2 
n 
k 1



1
4
1A
(6)
5.
(a)
2x 2  1 2x 5
2x 5
+1
 x 3  3x 2
3
2
 2x  2x
1
 x3  x2
3
 x 1
x
2
x 2
x
x3
x3
 Px  and Qx  are relative prime.
x
 x2
 x2
2
 x +1
 2x
 x 1
x 2
3
x 1
1M+1A
(b) Let r1 x   x 2  x  2 .


Px   2 x 2  1 Qx   r1 x  … (1)
Qx  x  1r1 x  3 … (2)
1M
(1)  x  1 : x  1Px   2 x 2  1x  1Qx   x  1r1 x 
1M
x  1Px   2 x 2  1x  1Qx   Qx   3
1  xPx  2 x 3  2 x 2  x  2Qx  3


1
1  x Px   1 2 x 3  2 x 2  x  2 Qx   1 .
3
3
1
1
mx   1  x  , nx   2 x 3  2 x 2  x  2 .
3
3


1A
1A
(6)
6.
 2 cos x, x  c
For f x    2
,
ax  b, x  c


lim f x   lim ax 2  b  ac 2  b
xc 
xc
1M+1A
lim f x   lim 2 cos x  2 cos c
x c 
1A
x c
The necessary condition f x  to be continuous at x  c is
Page 3 of 10
ac 2  b  2 cos c
2 cos c  b
a
, c  0.
c2
For c = 0,
b = 2.
aR.
7.
(a)
1
n

1
n 1
1M
1A
1A
(6)
1
 ... 

2n

lim
n 1
n 
2n
1
2n
n 1

1
2n
 ... 
1
2n
1A
2n

1A
 1
1
1 
 lim 

 ... 
   . i.e. limit does not exist.
n 
n 1
2n 
 n
 x2 1

(b) lim  2
x  x  1 


x2
1M

1
 lim 
x 
1

1
x2
1
x2






1A
x2
x2
1 

1  2 
x 
 lim 
x2
x 
1 

1  2 
x 

1M
x
x

1  
1 
 lim 1  2  1  2 
x 
 x   x 
2
2



 e e  e 2 .
1M
1A
Alternatively,
 x2 1

lim  2
x  x  1 


x2
2 

 lim 1  2

x 
 x 1
x2

2 

 lim 1  2

x  
 x 1

1M
2 x2
x 2 1
2
 x 2 1



1M
 e2
1A
(7)
Page 4 of 10
Section B (60 marks)
8.
(a) Let f x   x 4  3x 2  k
If    , then  is a root of f x   0 with multiplicity  2 .
1
  is a root of f ' x   0
1
i.e. 4 x 3  6 x  0
6
2
  0 , 
1
     2  2
which contradicts that     2
Hence,    .
Alternatively,
If    , then   1 .
Sub. Into f x   0 gives k  2 .
1
1
1
1
Solving x 4  3x 2  2  0 , we have x  1 or  2
or no two roots are equal
 sum of any two roots  2
which contradicts that     2
which contradicts that   
Hence,    .
1+1
1
(5)
(b)    3  k  0
4
 
2 2
2
 
 32  k  0
1
  2 is a root of y 2  3 y  k  0
Similarly,  2 is a root of y 2  3 y  k  0 .
    and     (     2 )
 2   2
1
      4
1A
2
 2   2  2  4
3  2  4
 
1
2
1A
k    
2
1
4
1A
(5)
(c)
1
0
4
4 x 4  12 x 2  1  0
x 4  3x 2 
x2 

12  144  16
8
1A
64 2
4
1A
Page 5 of 10
2  2 

2
22

2

x  1 
2 

1A+1A
    2
 the values of  and  are 1 
2
.
2
1A
(5)
(15)
9.
(a) Let
x 3  x 2  3x  2
x 2 x  1
2

A B
C
D
.
 2 

x x
x  1 x  12
1M
Axx  1  Bx  1  Cx 2 x  1  Dx 2  x 3  x 2  3x  2
2
2
 A  C x 3   2 A  B  C  Dx 2   A  2Bx  B  x 3  3x  2
1M
A  1 , B  2 , C  0 , D  1
i.e.
x 3  x 2  3x  2 1 2
1
.
  2
2
2
x x
x  12
x x  1
1A
(3)
(b) (i)
Px   mx  1 x   2 x   3 x   4 
P' x   mx   2 x   3 x   4   x  1 x   3 x   4 
 x  1 x   2 x   4   x  1 x   2 x   3 

P'  x  4
1

Px  i 1 x   i
4
(ii) From (b)(i),
1
1
 x 
i 1
1A

i
P'  x 
.
P x 
Diff. both sides w.r.t. x,
4
1
 x   
2
i 1
4
2

Px 2
i
1
 x   
i 1
Px P' ' x   P' x 
2
i

P' x 2  Px P' ' x 
Px 2
1
(3)
(c) (i)
Solving f x   0 , we have
b  b 2  4a 2
x 
2a
2
 b 2  4a 2 and a  0 , | b | b 2  4a 2  0
Page 6 of 10
1A
 ab  0  ( a  0 and b  0 ) or ( a  0 and b  0 )
If b  0 , then b  b 2  4a 2  0 and a  0 .
If b  0 , then  b  b 2  4a 2  0 and a  0 .
Both cases imply that x 2  0 .
Hence all the four roots of f ( x)  0 are real.
1
Besides, f 0  a  0 and
1
f 1  2a  b  0  b 2  4a 2  2a  b2a  b  0
1
 Both 0 and 1 are not the roots of f x   0 .
 i 3   i 2  3 i  2

 i 2  i  12
i 1
4
(ii)
4
1
i 1
i

4
 
i 1
4
 2
i 1
4
1
i
2

i 1
1
 i  12
(by (a))
1M
4
4
1
1
1
 2


2
0   i  i 1 0   i  i 1  i  12
 f ' 0  f 0 f ' ' 0   f ' 1  f 1 f ' ' 1
f ' 0

2
f 0
 f 02
 f 12
2
2
(by (b)(i)&(ii))
 f x   ax 4  bx 2  a , f ' x   4ax 3  2bx and f ' ' ( x)  12ax 2  2b
1M
1A
 f 0  a , f ' 0  0 , f ' ' 0  2b and
f 1  2a  b , f ' 1  22a  b , f ' ' 1  26a  b .
Hence,
 i 3   i 2  3 i  2

 i 2  i  12
i 1
4
 a 2b  42a  b   22a  b 6a  b 
 02

a2
2a  b 2
2

4b  4a  2b

a
2a  b

4a 2  10ab  4b 2
a2a  b 
1A
1A
(9)
(15)
Page 7 of 10
10. (a) (i)


x 4  2ax 2  4bx  c  x 2  2tx  r x 2  2tx  s

Comparing the coefficient of x , 2ts  2tr  4b .
1M
t s  r   2b
 b  0 , t  0 .
(ii) Comparing the coefficient of x 2 , s  r  4t 2  2a
2b
By (a)(i), we hve s  r 
.
t
Thus, we have
b
b
r  a  2t 2 
, s  a  2t 2  .
t
t
1A+1A
(iii) Comparing the constant terms, rs  c .
By (a)(ii), we have
1
b 
b

2
2
 a  2t   a  2t    c
t 
t

a  2t 
2 2


1M
b2
c
t2

2
t 2 2t 2  a  b 2  ct 2  0


4t 6  4at 4  a 2  c t 2  b 2  0 .
1
(6)
(b) (i)
Putting y  x  h in y 4  4 y 3  2 y 2  52 y  9  0 ,
x  h 4  4x  h 3  2x  h 2  52x  h   9  0



1M

x 4  4h  4h 3  6h 2  12h  2 x 2  4h 3  12h 2  4h  52 x  h 4  4h 3  2h 2  52h  9 =0
 4h  4  0
h  1
1M
1A
Thus, when y  x  1 , (*) can be written as x 4  8 x 2  64 x  48  0 .
(ii) Put a  4 , b  16 and c  48 in (a), we have



x 4  8x 2  64 x  48  x 2  2tx  r x 2  2tx  s , where
4t 6  16t 4  64t 2  256  0 .
i.e. f (t) = t 6  4t 4  16t 2  64  0
 
f 2  2 6  46  16 t 2  64  0
4

1A
1M

f t   t  2t  2 t 4  16  0
Take t  2 .
By (a)(ii), r  4 and s  12 . So, we have
x 4  8 x 2  64 x  48  0
x
2
x


 4 x  4 x 2  4 x  12  0
1A (accept t  2 )
1M
 4  16  16
4  16  48
or x 
2
2
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x  2  2 2 or x  2  2 2i
1A
Thus, all the roots of (*) are
 3  2 2 ,  3  2 2 , 1  2 2 i and 1  2 2 i .
1M
(9)
(15)
11. (a) Let S n be the statement “ an  n and bn  n ”
For n = 1, a1  b1  1
 S 1 is true.
Assume S k  is true for some integer k.
i.e. a k  k and bk  k
For n  k  1,
a k 1  a k  2bk
 n  2n
 n 1
1
bk 1  ak  bk
 nn
 n 1
1
 S k  1 is true.
By mathematical induction, S n is true for all positive integers n.
an  2bn  an1  2bn1   2an  bn 
2
2
2

  an1  2bn1
2
2
2

1M

  1
n 1
  1
a
1
n 1
  1
2
 2b1
2

1  2
n
1
(4)
(b) (i)
If n is odd,
an  2bn  1
2
2
an  2bn
2
1M
2
an
 2
bn
1
If n is even,
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an  2bn  1
2
2
an  2bn
2
2
an
 2
bn
1
an 2 an an1  2bn1 an



bn 2 bn
an1  bn1 bn
1M

an  2bn   2an  bn  an

an  2bn   an  bn  bn

1M

2 a n  2bn

bn 2a n  3bn 

2
2
 1n1 2
bn 2a n  3bn 
1
(ii) From the above results,
a 2 k 1
a
a
 2 and 2 k 1  2 k 1 , i.e. strictly increasing and bounded above by
b2 k 1
b2 k 1 b2 k 1
a2k
a
a
 2 and 2 k 2  2 k , i.e. strictly decreasing and bounded below by
b2 k
b2 k 2 b2 k
a 
a 
By monotone convergence theorem,  2 k 1  and  2 k  converge.
 b2 k 1 
 b2 k 
Let lim
n 
2
2.
1M
1
a2k
a
 l1 and lim 2 k 1  l 2 .
n


b2 k
b2 k 1
an1
a  2bn
 lim n
n b
n a  b
n 1
n
n
lim
an
2
bn
 lim
n a
n
1
bn
If n is odd, l1 
1M
l2  2
, i.e. l1l 2  l1  l 2  2
l2  1
If n is even, l 2 
l1  2
, i.e. l1l 2  l 2  l1  2
l1  1
Comparing these two results, l1  l 2  2 .
1M
1A
(9)
(15)
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