Po Leung Kuk Celine Ho Yam Tong College
Form Seven Pure Mathematics
2010
HKALE
Paper One
Marking Scheme
Solution
Marks
1M
n
1. (a) Note that (1  2 x) n   ak x k .
k 0
n
Putting x = 1, we have
a
k 0
1
 3n .
k
n
(b) Differentiating both sides of (1  2 x) n   ak x k w.r.t. x,
k 0
1M
n
We have 2n(1  2 x) n1   kak x k 1 .
k 1
n
Putting x = 1, we have
 ka
k 1
k
1
 2n3n1 .
n
(c)
 3k  1a
k 0
k
n
  3kak  ak 
k 0
n
n
k 0
k 0
n
n
 3 kak   ak
1M
 3 kak   a k
k 1
k 0
n1
 3(2n3 )  3n
1M
(by (b) and (a))
 2n3  3
 (2n  1)3n
n
2. (a) Let
n
C
C
C
C
x
 1  2  3  4 .
2
( x  1)( x  4) x  1 x  1 x  2 x  2
1
(7)
1M
2
Solving, we have C1 
Thus, we have
(b) Note that
1
1
1
1
, C2 
, C3 
and C 4 
.
6
6
6
6
1A for all correct
x
1  1
1
1
1 
 




2
( x  1)( x  4) 6  x  1 x  1 x  2 x  2 
2

d 
x
 3x 4  5 x 2  4
 2


.
dx  ( x  1)( x 2  4)  ( x 2  1) 2 ( x 2  4) 2
1A
x
1  1
1
1
1 
 




2
( x  1)( x  4) 6  x  1 x  1 x  2 x  2 
Differentiating both sides with respect to x, we have
By (a), we hve
2
 3x 4  5 x 2  4
1
1
1
1
.




2
2
2
2
2
2
2
( x  1) ( x  4)
6( x  1)
6( x  1)
6( x  2)
6( x  2) 2
 3x 4  5 x 2  4
1
1
1
1




2
2
2
2
2
2
2
( x  1) ( x  4)
6( x  1)
6( x  1)
6( x  2)
6( x  2) 2
1/10
1A
Solution
Marks
Remarks
3k 4  5k 2  4

2
2
2
2
k 3 ( k  1) ( k  4)
n
(c)

1 n  1
1   1
1 
 







2
2
2


6 k 3   (k  2) (k  1)   (k  2) (k  1) 2  

1 n  1
1  n  1
1 
  
   



2
2 
2

6  k 3  (k  2)
(k  1)  k 3  (k  2)
(k  1) 2  
(by (b))
1M
1 1
1
1 

=   
1
2
6  16 (n  2)
(n  1) 2 
5
1
1



2
32 6(n  2)
6(n  1) 2
 3k 4  5k 2  4

2
2
2
2
k 3 (k  1) (k  4)
 5

1
1

 lim  

2
2
n 32
6(n  2)
6(n  1) 


1A
(6)
3. (a) Note that f ( x)  ( x  1)( x  4)r ( x)  x  k , where r (x ) is a polynomial.
Also note that f ( x)  ( x  4) 2 s( x)  kx  10 , where s (x) is a polynomial. 1M
So, we have f (4)  4  k and f (4)  4k  10 .
Hence, we have  4  k  4k 10 .
Thus, we have k  2 .
1M
1A
(b) As f (x) is a cubic polynomial, we have f ( x)  ( x  4) 2 ( x  A)  2 x  10 ,
where A is a constant.
By (a), we have f (1)  1 .
So, we have 9(1  A)  8  1.
1M
Solving, we have A  0 .
f (x)
 ( x  4) 2 x  2 x  10
 x 3  8 x 2  18 x  10
Thus, we have g ( x)  8x 2  18x  10 .
(c) When
g x 3
1A
is divided by x  1, the remainder is  g  1 .
3
As g (1)  36 , we have
g  13  46656 .
Thus, the required remainder is -46656.
4. (a) The required matrix


 cos
2


 sin
2

 sin
1M
1A
(7)


2

cos 
2 
1A
 0  1

 
1
0


2/10
Solution
Marks
Remarks
(b) (i) Let (x,y) be the coordinates of Q.
 x
 
 y
 0  1 1 
 
 
 1 0  3 
  3
  
 1 
1A
Thus, the coordinates of Q are (-3,1).
 0  1

(ii) (1) M  
1 0 
1A
 0  1 x    y 
   
 for any point (x,y).
(2) Note that 
1
0
y

x

  

So, T transforms any point (x,y) to the point (-y,-x).
Thus, T is the reflection in the straight line x  y  0 .
(3) The area of O' P ' Q'
= The area of OPQ


1 2
1  32
2

1
10
2
1A for reflection +
1A for x  y  0
1M provided T is a
reflection

2
1A
=5
(7)
5 (a) zz  (12  16i ) z  (12  16i ) z  375
zz  (12  16i ) z  (12  16i ) z  375
zz  (12  16i )z  (12  16i ) z  (12  16i )(12  16i )  375  (12  16i )(12  16i )
1M
z  12  16i z  (12  16i)  375  400
z  12  16i z  (12  16i)  25
z  12  16i   5 2
2
1A
z  12  16i   5
Thus, the centre of the circle is 12  16i .
The radius of the circle is 5.
1A
1A
(b) The distance between the centre of the circle and the pole

12  02  16  02
= 20
z1
 20  5
= 15
1M
3/10
Solution
Marks
Remarks
z1

15
12  16i 
20
1M
1A
(7)
 9 12i
6 (a) By Cauchy-Schwarz’s inequality, we have
          
 

2
2
2
2
2
2
2
2

  1    1    1    1  
             



1
1
1
1 
 
 
 
 




 

2
1M
        1  1  1  1   1  1  1  12
    
        1  1  1  1   16
    
1
(b) Since  ,  ,  and  are positive real numbers,
     ,      ,      and      are positive.
                          
1
1
1
1



                

  16

1M


1
1
1
1
  16
3       



                 
        3  3  3  3   16
                 

3
3
3
3




                

16
 
        
1


3
3
3
3
16
(c) 



 
                             


1
1
1
1
16




        
3
                 
                            16




  
  
   
   
3
1




16
1
1
1

  
  
   
    3
1M




16



 4
                3




4




                3
7 (a) (i)
(E) has a unique solution
 0
1 1
  a 0
1
(6)
1M
1
4  0
3 4 a4
  12  4a  16  a(a  4)  0
 4  a2  0
4/10
Solution
Marks
 a 4
 a  2 and a  2
 a  2 ,  2  a  2 or a  2
2
Remarks
1A
1A
When (E) has a unique solution,
x
2 1
1
2 0
4
b 4 a4
4a
1 2
1
2
2(16  2b  a)
4  a2

 22  6a  4b  ab  2a 2
4  a2
1M for Cramer’s
Rule
4
a 2
y

3 b a4
4  a2
1 1 2
a 0 2
z
3 4 b
4a
2

 2  8a  ab
4  a2
1A+1A(1A for any
one, 1A for all)
(ii) When a  2 , the augment matrix of (E) becomes
1 1 1 2
1 1
1 1 1 2 
1 2 






 2 0  4 2 ~ 0  2  6  2  ~ 0 1 3 1  .
3 4 6 b
0 1
0 0 0 b  7
3 b  6 





So, (E) is consistent when b  7 .
Thus, the solution set is (1  2s,1  3s, s) : s  R.
1A
1A or equivalent
(8)
(b) Note that the augmented matrix of the first three equations of (F) is
1 1 1 2 
1 1 1 2 




 1 0 2  1 ~  0 1  1 3  .
3 4 2  
0 0 0   9




So, the first three equations of (F) are solvable when   9 .
Under this case, the solution set for the first three equations is
(1  2t, t  3, t ) : t  R.
Hence, (F) is consistent when   9 and   7(1  2t )  17(3  t )  3t .
1A
1A or equivalent
1M
1A
(4)
Thus, we have   9 and   44 .
2
and b  5 in (a), the first three equations of (G) have the
3
1M
unique solution x  3 , y  1 and z  0 .
1A
Note that 5(3)  2(1)  18(0)  17  16 .
(c) Putting a 
Thus, (G) is inconsistent.
1A f.t.
(3)
5/10
Solution
Marks
Remarks
8 (a) ‘  ’
Since  is a repeated root of the equation p( x)  0 ,
p( x)  ( x   ) 2 q1 ( x) , where q1 ( x) is a polynomial.
p' ( x)  2( x   )q1 ( x)  ( x   ) 2 q1 ' ( x)
Thus, we have p( )  p' ( )  0 .
1M
1
‘’
As p ( )  0 , we have p( x)  ( x   )q2 ( x) , where q2 ( x) is a polynomial.
p' ( x)  q21 ( x)  ( x   )q2 ' ( x)
As p ' ( )  0 , we have q2 ( x)  ( x   )q3 ( x) , where q3 ( x) is a polynomial.
1M
Therefore, we have p( x)  ( x   ) 2 q3 ( x) .
Thus,  is a repeated root of the equation p( x)  0 .
1
(4)
(b) (i) Since f (  )  0 and f (0)  1 , we have f (  )  f (0) .
Thus, we have   0 .
1
(ii) Since f (  )  0 , we have  6  a 5  b 4  c 3  b 2  a  1  0 .
2
3
4
5
6
1 1
1
1
1 1
So, we have 1  a   b   c   b   a      0 .
 


 
1M
1
Hence, we have f    0 .

Note that  is a repeated real root of f ( x)  0 .
1
By (a), we have  6  a 5  b 4  c 3  b 2  a  1  0 and
6 5  5a 4  4b 3  3c 2  2b  a  0 .
Therefore, we have a 5  2b 4  3c 3  4b 2  5a  6  0 .
2
3
4
1M
1M
5
1
1
1
1
1
So, we have a  2b   3c   4b   5a   6   0 .





1
Hence, we have f '    0 .

1
By (a),
is a repeated root of the equation f ( x)  0 .
1

(6)
1M
(c) (i) Note that g (2)  0 .
Also note that g ' ( x)  24 x 5  80 x 4  68x 3  21x 2  34 x  16 .
As g ' (2)  0 , 2 is a repeated root of the equation g ( x)  0 .
(ii) By (c)(i) and (b)(ii),
1
is a repeated root of the equation g ( x)  0 .
2
1A f.t.
1M
1M
Therefore, we have g ( x)  ( x  2) (2 x  1) ( x  x  1) .
Note that 12  4(1)(1)  3  0 .
Thus, g (x ) cannot be factorized as a product of linear polynomials with 1A f.t.
2
2
real coefficients.
9. (a) (i) xn  yn

2
(5)
5
1
7
2

xn1  y n1   xn1  y n1 
6
6
9
9

6/10
Solution

Marks
11
( xn1  y n1 )
18
 11 
 
 18 
Remarks
1M
n 1
( x1  y1 )
>0
Thus, we have xn  yn .
1
(ii) xn1  xn

5
1
xn  y n  xn
6
6
1
 ( y n  xn )
6
<0
Thus, xn  is a strictly decreasing sequence.
yn1  yn

2
7
xn  y n  y n
9
9

2
( xn  y n )
9
>0
Thus,
yn 
is a strictly increasing sequence.
(iii) By (a)(i) and (a)(ii), we have y1  yn  xn  x1 .
xn  is a strictly decreasing sequence and is bounded below by y1 .
yn  is a strictly increasing sequence and is bounded below by x1 .
Thus, xn  and y n  are convergent sequences.
1
1
1A
1A
(iv) Let lim xn  A and lim y n  B .
n 
n 
Then, we have A 
5
1
A B.
6
6
So, we have A  B .
1M
1
Thus, we have lim xn  lim y n .
n 
n 
(v) 4 xn1  3 yn1
10
2
2
7
xn  y n  xn  y n
3
3
3
3
 4 xn  3 y n

1
(vi) By (a)(v), we have 4 xn  3 yn  4 xn1  3 yn1  ...  4 x1  3 y1 .
So, we have lim (4 xn  3 yn )  4 x1  3 y1 .
n
1M
By (a)(iii), we have 4 lim xn  3 lim y n  4 x1  3 y1 .
n 
n 
By (a)(iv), we have 7 lim xn  4 x1  3 y1 .
n 
1M
1A
4 x  3 y1
Thus, we have lim xn  1
.
n 
7
(12)
(b) If x1  y1 , then  x1   y1 .
7/10
Solution
and bn  yn for n  1,2,3,... .
Define an   xn
Marks
Remarks
1M
5
1
2
7
an  bn and bn 1  a n  bn .
6
6
9
9
are convergent sequences.
Then, we have an  b1 , an1 
By (a)(iii), a n  and bn 
Thus, xn  and y n  are convergent sequences.
1A f.t.
1A f.t.
10. (a) ’  ’
a b 
 , where a, b, c and d are real numbers.
Let M  
c
d


 a b  a c   a 2  b 2

  
Then, we have 
 c d  b d   ac  bd
ac  bd   1 0 

.
c 2  d 2   0 1 
Hence, we have a 2  b 2  c 2  d 2  1 and ac  bd  0 .
Therefore, the points ( a, b) and (c, d ) lie on the circle x 2  y 2  1.
a  cos 
 c  sin 
So, there exist  ,   R such that 
and 
.
d  cos 
 b  sin 
 cos  sin  
 .
Then, we have M  
 sin  cos  
Since ac  bd  0 , we have cos sin   sin  cos   0 .
So, we have sin(    )  0 .
Therefore, we have     n , where n Z .
Hence, we have   n   , where n Z .
sin( n   )   sin 
Note that when n is an even number, we have 
.
 cos(n   )  cos 
 sin( n   )  sin 
Alos note that when n is an odd number, we have 
.
cos( n   )   cos
sin    sin 
 sin   sin 
So, we have 
and 
.
 cos   cos
cos    cos
 cos  sin  
 cos  sin  
 or M  
 .
Thus, we have M  
  sin  cos  
 sin   cos  
1M
1M
1M
1M
1
‘’
 cos 
If M  
  sin 
sin  
 cos 
 , then MM T  
cos  
  sin 
 cos 2   sin 2 
So, we have MM T  
0

 cos 
If M  
 sin 
 cos 2   sin 2 
So, we have MM  
0

 sin  
.
cos  
 1 0

.
cos   sin    0 1 
0
2
2
sin  
 cos 
 , then MM T  
 cos  
 sin 
T
sin   cos 

cos   sin 
sin   cos 

 cos   sin 
sin  
.
 cos  
 1 0

.
cos   sin    0 1 
0
2
2
1
(6)
Thus, M is orthogonal.
 cos 
(b) (i) Since M  
  sin 
sin  
 , the statement is true for n  1.
cos  
 cos k sin k 
 , where k is a positive integer.
Assume that M k  
  sin k cos k 
M k 1
 M kM
8/10
Solution
Marks
 cos k sin k  cos  sin  

 (by induction assumption)
 

sin
k

cos
k


sin

cos




cos
k

cos


sin
k

sin

cos
k
 sin   sin k cos 


 

sin
k

cos


cos
k

sin


sin
k

sin


cos
k

cos



cos(
k

1
)

sin(
k

1
)




 

sin(
k

1
)

cos(
k

1
)



 cos n sin n 
 .
By mathematical induction, we have M n  

sin
n

cos
n



1 0

(ii) Note that M 2  
0 1
M
Thus, we have M n  
I
Remarks
1
1A
If n = 1, 3, 5, …
If n = 2, 4, 6, …
1A
 cos sin  
 .
(c) By (a) and (b), we have X  

sin

cos



cos
400

sin
400


1
0

 
  
 .
So, we have 

sin
400

cos
400

0

1

 

Hence, we have cos 400  1 and sin 400  0 .
Then, we have 400  (2n  1) , where n Z .
2n  1

 cos
400
Therefore, we have X  

2
  sin n  1
400

2n  1

 cos
400
Thus, we have X  

2
  sin n  1
400

sin
sin
1M
1A
2n  1 
400

2n  1
cos
400

 , where n Z .


2n  1 
400

2n  1
cos
400

 , where n  1,2,...,400 .


 cos  sin  
 or
(d) By (a), there exists   R such that M  
  sin  cos  
 cos sin  
 .
M  
 sin   cos 
 cos 401 sin 401 
 or
By (b), we have M 401  
  sin 401 cos 401 
 cos  sin  
 .
M 401  
 sin   cos  
Thus, M 401 is orthogonal ( by(a) ).
(4)
1M for either one
1A f.t.
(2)
11. (a) (i) Let f ( x)  ln x  x  1 for all x  0 .
 0
1 x 
Note that f ' ( x ) 
 0
x 
 0
if 0 < x < 1,
if x = 1
if x > 1
1M for testing
Therefore, f (x) attains its absolute maximum at x  1.
So, we have f ( x)  f (1) for all x  0 .
Since f (1)  0 , we have ln x  x  1  0 for all x  0 .
Thus, we have for all ln x  x  1 . for all x  0 .
9/10
1
Solution
(ii) By (a)(i), we have ln bk  bk  1 for k =1, 2, …, n.
Hence, we have ak ln bk  ak bk  ak for k =1, 2, …, n.
Then, we have ln bk
ak
n
So, we have
 ln b
Remarks
 ak bk  ak for k =1, 2, …, n.
n
n
k 1
k 1
  ak bk   ak .
ak
k
k 1
Marks
1M
Therefore, we have ln( b1 1 b2 2 ...bn n )  0 .
a
a
a
1
(4)
Thus, we have b1 1 b2 2 ...bn n  1 .
a
a
a
(b) (i) Note that
ln x x  ln x
 x ln x  ln x
 ( x  1) ln x
 0

 0
 0

when 0 < x < 1
when x = 1
when x > 1
1M
Therefore, we have ln x x  ln x for all x  0 .
Since ln x is strictly increasing, we have x x  x for all x  0 .
1
(ii) By (b)(i), we have ck k  ck  0 for k =1, 2, …, n.
c
Multiplying, we have c1 1 c2 2 ...cn n  c1c2 ...cn .
c
c
c
1
(3)
Thus, we have c1 1 c2 2 ...cn n  1.
c
c
c
n
n
r 1
r 1
(c) (i) Let S1   xr and S 2   xr .
2
For each k =1, 2, …, n, define a k 
n
Then, we have
 ak 
k 1
xk
x S
and bk  k 1 .
S2
S1
S1
 1 and
S1
n
a b
k k
k 1

1M for either one
S2
 1.
S2
1M withhold 1M if
checking is omitted
By (a)(ii), we have b1 1 b2 2 ...bn n  1 .
a
a
a
x1
xn
x2
 x S  S1  x S  S1  x S  S 1
So, we have  1 1   2 1  ... n 1   1 .
 S2   S2   S2 
xS 
Hence, we have  1 1 
 S2 
x1
x1
x2
x2
xn
 x2 S1   xn S1 

 ...
  1
 S2   S2 
x2
Therefore, we have x1 x2 ...xn
x1
1M for using (a)(ii)
Thus, we have x1 x2 ...xn
xn
xn
S 
  2 
 S1 
x1  x 2  ...  x n
.
 x 2  x2 2  ...  xn 2 

  1

 x1  x2  ...  xn 
(ii) Let G  n x1 x2 ...xn .
10/10
x1  x 2  ...  x n
.
1
Solution
Marks
For each for k =1, 2, …, n, define ck 
xk
.
G
1M
x1 x2 ...xn
1
Gn
Then, we have c1c2 ...cn 
1M withhold 1M if
checking is omitted
By (b)(ii), we have c1 1 c2 2 ...cn n  1.
c
x1
c
Remarks
c
xn
x2
 x G  x  G  x  G
So, we have  1   2  ... n   1 .
G  G   G 
x
x
1M for using (b)(ii)
x
1
2
n
 x1   x2   xn 
Therefore, we have     ...   1 .
G  G   G 
Hence, we have x1 1 x2 2 ...xn n  G x1  x2 ...  xn .
x
x
x
Thus, we have x1 1 x2 2 ...xn n  n x1  x2  ...  xn
x
x
x
x1  x 2  ...  x n
.
END OF MARKING SCHEME
11/10
1
(8)