Form 6 Pure Mathematics Paper II
Final Examination 2006-2007
Marking Scheme
1. (a) lim x ln x x

0

= lim x

0
 ln x x

1
1
= lim x

0
  x x

2
 lim (
 
0 x

0

(b) f ( x ) is differentiable at x = 0
 f ( x ) is continuous at x = 0
 lim ( )
 lim f x x

0
 x

0


 k lim x

0

 h sin x
 k cos 2 x
 lim x

0
 x
2 ln x
1
1 lim x

0
 ln x x

2
 lim x

0
 
2 x x

3
 x lim

0
 x
2

2

0
 k = 1
2.
1A
1M+1A
(ii) t lim

0
 f (0
  f (0)
= t lim t

0
 t
2 ln t

0 t
 lim ln t

0 t

0

1A lim h

0
 f (0
  f (0)
= h lim t

0
 h sin t
 cos 2 t

1 t
 t lim

0
 h cos t

2sin 2 t
1
 h 1A f ( x ) is differentiable at x = 0
 f  '(0)
 f
+
'(0)
 h = 0 1M
6

0 x
F u
2
) du =

(
2
)

0

0 x uF u
2 u du
[ Let t = u 2 ]
1M+1A
1M = xF x
2
)
 
0 x
2
=
 x
2 x f u du
0
( )
 
0 x
2
=

0 x
2
( x

) ( )
1A
1A
5

3. (a) ( )
 

1 x
0

 x
1 when 0 when 1 otherwise
1
2
y
1
1
x
O 2
1M+1A
(b) f is not an surjective function as for any y > 1 or y < 0, there doesn’t exists an x in R such that y = f ( x ).
1
(c)(i) g(

1) = f (

2) f (0) + 1 = 1 g (1) = f (0) f (2) + 1 = (1)(1) + 1 = 2

g (

1)
 g ( x ) is not an even function. 1
(ii) ( 1)
( 1)


|
0 x

2 |

 0
 g ( x ) =

2
 1 when 1 otherwise otherwise x
3 otherwise
1 when x

1
1A
1A
y
2
1
O 1
x
1A
7

4. and f ( x ) = sin x cos
4 x = sin (
  x ) cos
4
(
  x )
= f (
  x )
By (a) ,

0
 x sin cos xdx =
=
=
1
2
 
0

4 xdx


1
2

5

0
 cos
4 xd x
5.
1 x
3 
1
=
1
( x

1)( x
2  
1)
=
1 1
(
3 x

1
 x

2 x
2  
1
)

1 1
3

( x

1
 x

2 x
2  
1
) dx
=
1
3 ln x
 
1
3
 x

2 x
2  
1 dx
=
1
3 ln x
 
1
3

( x 2 x
 x
1
2
 
1

3 x 2
2 x 1
) dx
1
1M for attempt to solve by partial fraction
2
For
 dx x

1
 ln | x

1|
.
1M
1M
1A
6
1M+1A 1
1A 2
1M
1M
1M
1A

=
=
=
1
3 ln x
 
1
6 ln
2 x x
1
2
 1 x
2  
1 dx
1
3 ln x
 
1
6 ln
2 x x
1  1
2
1 ( 1)
6 ln x
2 x

 
2

1
1 3

1 tan (
2 x

3
1
)
( x

1
)
2 
3
2 4

C dx
1A
5
6. (a) f ( x ) = x x ln f ( x ) = x ln x
 f ' ( x )

1
 ln x f ( x )
 f '( x ) = (1 + ln x ) f ( x ) = x x
(1 + ln x )
=
= x lim

1 ln x x x

 x x

1


0
0 type

 x lim

1 x x
( 1

1 ln

1 x x )

1
 x lim

1 x x
( 1
 ln x )
2

1 x x

1


0
0
= type

 x lim

1
 x
 x

2
( 1
 ln x )
2  x x

1

 
2 x
2
(b) lim x

0
=
=

0 x t sin(sin x
3 t ) dt lim x

0 x sin(sin
3 x
2 x ) lim x

0
1
3 sin(sin sin x x ) sin x x


0
0 type



1
3
1A
1M
2A
4
1M+1A
2
7. (a) f '( x ) = x
3 
45 x
2 
332 x

288

( x

8)( x

1)( x

36)
( x

15)
3
( x

15)
3 f "( x ) =
686( x

6)
( x

15)
4
1A+1A
1A

(b) x <1 1 (1, 6) 6 (6, 8) 8 (8, 15) 15 (15,36) 36 >36 f ( x )
 max
 inf
 min
   min
 f '( x ) + f "( x )

0



0

+
0
+
+
+



+
0
+
+
+
(i) f '( x ) > 0 when x

(

, 1)

(8, 15)

(36,

)
(ii) f "( x ) > 0 when x

(6, 15)

(15,

)
1A+1A
1A (c) Maximum point = (1,
7
4
)
Minimum point = (8, 0) , (36,
224
)
3
Inflexional point = (6,
16
27
)
1A+1A
1A
4
(d) x = 15 is the equation of a vertical asymptote.

( x

6)( x

8)
2
( x

15)
2
 x
3 
10 x
2 
32 x

384 x
2 
30 x

225
 
20)

343( x

12) x
2 
30 x

225
1A y = x + 20 is the equation of a horizontal asymptote. 1M+1A
(e)
y y = x + 20
75
(0, )
128
(0, 20)
O
(6,
16
27
)
(8, 0)
15
(1,
7
4
)
(36,
224
3
) x = 15
x
1A for the shape
1A for the special points
1A for the asymptotes

8.
1
(a) F '( x ) =
1
 

 g '
 x
  a
1
  


If x = a , then x
  a
1
 
 a .

(i)
 x

a , x
1



 a
 x
1



 x
 x
F '(a) = 0
(as 0

1 +

) g " ( x ) < 0
 x > 0
 g ' (x) is a strictly decreasing function on R
+
 g ' x
  a
1
 
 


(ii)

x

a, x
1



 a
 x
1



 x
 x
F ' ( x ) < 0
(as 0

1 +

)
 g ' x
  a
1
 
 


F ' ( x ) > 0

F( x ) attains its minimum value at x = a .
1M
1A
Since F ( x ) has only one turning point, F ( x ) attains its least value at x = a.
1M+1A
4
(b) (i) By (a) , F( x )

F(a)
=
For m = 2 , let a = g x
2 a
1



 a
 

( )
1




and

=


2
1
= 0
. Then we have from (a) g x
1
1



 x
2
 
 g x
1



 g x
2
)

0 i.e. g



 x
1
1



2

1

2

1 x
2


 
 g x
1



2

1

2

1 g x
2
) i.e. g


  

2




1 g x



2 g x
2
)
2

The statement holds for m = 2.
1A
1M+1A

Suppose the given statement is true for 2

m

k .
Let

1
+

2
+...+
 k

1
= p and x =
1 p
(

1 x
1
+

2 x
2
+...+
 k

1 x k

1
)

  
2 2

2
...
 k

1 x k

1
 k

1


 k

= px
 p


 k
= x

1

 k p
 k x k p

By (a), we have g (
  

2 2
2

...
...
 k

1 x k

1
 k

1


 k

= g ( x
1


 k p
 k x k p
)
)
1M
1A

1

1
 k p



 k p g x k
)



1

1
 k p





 k

1
  r g x r
) r

1

2
...
 k

1

 k p g x k
)






= p
 p
 k


1 p k r

1


1
 r g x r
)

 k p
[ By assumption ] g x k
)




1 1
 
2
(
2
 

2
...
 k
 k g x k
)

By principle of induction, the statement holds for m

2.
(b) (ii) Let g ( x ) = ln x which is differentiable and g '( x ) =
1 x
, g " ( x ) =
 x
1
2
< 0.
For any positive numbers a
1
, a
2
, ..., a m
, let x i
= a i
(1

i

m)
 i
= 1 (1

i

m) and
1A
1A
1M+1A
1M

9.
Then by (i) , we have i.e. i.e. i.e. g ln a
1
 a
2
...
m a
1
 a
2
...
m a
1
 a
2
...
m a
1
 a
2
...
m a m a m a m
 

 

1 m k m


1 g a k
)
1 m ln( a a ...
a m
)
1
 e ln( a a ...
a m
) m a m  m a a ...
a m
1A
11
(a)(i) For 0
  

4
, 0

tan n+1  
tan n  
1 .

0

0
 
0

4 tan n

1  
0

4 tan n

0

I n+1

I n for n

0
1M+1A
(ii) For n

2 ,
I n
+ I n - 2
=

0

4
 tan n   tan n

2

=

0

4 tan n

2  sec
2  
=

 tan n

1 n

1
 

 
0

4
=
1 n

1
1M+2A
(iii) For n

2 ,
1 n

1
= I n
+ I n

2

2 I n
[ by (a)(i) ]

1
2( n

1)

I n
1M+1A
Also


1 n

1
= I n+2
+ I n

2I n
1
2( n

1)

I n
1
2( n

1)

I n

1
2( n

1)
[ by (a)(i) ]
1
8

(b) For n

1, I
2n+1
=
1
2n

I
2n

1
=
2
1 n

2 n
1

2
+I
2n

3
= ...
=
2
1 n

2 n
1

2

1
2 n

4
  
1) n-1
1
2
+(

1) n
I
1
1M+1A

I
2n+1
=
(

1 ) n

1
(a n

2 I
1
)
2
Now I
1
=

0

4 tan
 dx ln cos



4
0

1 ln 2
2

I
2n+1
=
(

1 ) n

1
(a n
 ln 2)
2
From (a)(iii)
1
2 2 n

2 )

I
2n+1

1
2 2 n )
,
Since both lim n

1
2(2 n

2)
and lim n


 lim n

I
2 n

1

0. lim n
 a n
 ln 2
1
equal zero
1A
1A
1A
1M+1A
7

10. (a) Let have
( )
 tan

1 x , x

R . Apply Mean Value Theorem to f ( x ) on [a, b], we
( )

( ) b
 a
 f for some a <

< b.
 tan

1 b
 tan

1 a

1
1
  2
1M+1A
In addition, 1 + a
2
< 1 +
 2
< 1 + b
2

1
1
 a
2

1
1
  2

1
1
 b
2

1
1
 b
2
 tan

1 b
 tan

1 a

1

1 a
2
1M+1A
4
(b)(i) Let t = tan
3

8
. Then
1 tan
3

4

3
 tan 2( )
8

3

2 tan
2
8
3

8

2 t
1
 t
2
 t
2 
2 t

1 = 0
1M+1A



(ii) tan
24 t
 tan
2

8
3

2
 
2
8
 
2
= tan

3
 
8 3


(1
 tan

3

8
3


8 tan tan

3

3
2 is rejected as

(1

2)

3
1 (1 2) 3 t > 0)
=
=
=
=
1
1


2
3


3
6

6
6


3
3


2
2


2
2

1

2

3

6

3

2

2

(6 3) ( 3

6)( 2
 
6

3

2

2

1

2

3

6

3

2

2


3

6

2 3

12

2 6

6

3

2

2
1

2

3

6

3

2

2

1

3

2
= 6

2

3

2 .
1A
1A
1A
1M+1A
7

(c) Put a = 0 and b = 6

2

3

2 in (a), we have
1
1


6

2

3

2
2

 tan

1

6

2

3 2
6

2

3

2


1
24



6
6


2
2


3
3


2
2


2
24

6

2

3

2
1
24



6
6


2
2


3
3


2
2
=
=


2
24
 
6 6


6

2

3

2

1 6 2 3 4 2 12

2 18

4 6

2 6

4 2

4 3
24


6

2

3 2



1 tan 0

=
=
12


6


2


3

2
3 6


6
6


2
2
12

6

2

3 2

6

2

8 6
      
=
12

6

2

3 2

6

2

4 6

4 2
 
1
 2
1M+1A
= 3( 6

2)

3( 6

2) 24( 6

2

3

2) .
2A
4

Section A
Question
Number
1
2
3
4
5
6
Section B
Performance in General
(a) Good. Only one or two students were not aware that they have to apply
L’ Hospital rule in finding the limit.
(b) Poor. Obviously, the students confused about the definitions of differentiability and continuity.
Poor. Only one student did this problem correctly. Some of them were not aware that they have to apply the fundamental theorem of calculus, some of them failed to apply the method of integration by parts and some of them did not realize the use of the substitution t = u 2 .
(a) Good. Some students were not aware the importance of

and

in graph sketching.
(b) Poor. The students mixed up the definition of injective and surjective.
(c) Poor. The students overlooked the domain of the functions f ( x

1) and f ( x + 1).
(a) Fair. Some students were not aware the method of substitution.
(b) Poor. Most of the students did not get the functions f ( x ) and g ( x ) correctly. Even if they got the correct functions, they forgot to check whether the functions satisfied the given properties.
Good. Some students simply did not know how to tackle this problem.
Good. Apart from the careless computational mistakes, the students knew how to tackle this question.
Question
Number
7
8
9
10
Performance in General
(a) Poor. Most of the students could not find the derivatives correctly. Their performance in this part revealed that the students are weak at the simplification of complicated expressions.
(b)(c)
For those students who got the correct answers in (a) could finish these two parts satisfactory.
(d) Fair. Some students did not know how to find the vertical asymptote and some of them have difficulties in finding the constant b of the slant asymptote y = ax + b.
(e) Poor. Only two students could sketch the graph of f ( x ) correctly.
Poor. Only one student attempted this question and his performance is fair.
(a) (i) Fair. The students did not realize that 0

tan
 
1 for 0
  

4
.
(ii) Good.
(iii) Poor. The students failed to set up a connection between part (i), (ii) and (iii).
(b) Good.
(c) Fair. Some students did not know how to find I
1
, some could not set up the relation between I
2n+1
and a n
and some of them were not aware the sandwich theorem.
(a) Good. Most of the students could finish this part correctly.
(b)(c) Poor. A lot of computation errors were found.

2.
3.
4.
1. Let f ( x )



 ax

 e bx sin
If

is continuous at x if if x
 x


2

.
,
2

2
, show that a


2 e b

2 .
Furthermore, if

is differentiable at

2
, find the values of a and b.
(a) Prove that x lim

0 x n ln | x |

0 for any positive rational number n .
(b) Let f ( x ) = 
0 x
3 ln | x |
(i) Find f ‘( x ) for all x

0.
(ii) Prove that f ‘(0) = 0. when x

0
. otherwise
(iii) Is f '( x ) continuous at x = 0. Justify your answer.
If
Let f : R

R be a continuous function. Show that
1
0 z

0 x z for all x

R .
0 z
1 f xt dt = 0 for all x

R , show that f (x) = 0 for all x

R .
Let f : R

R be a real valued function defined by f ( x ) =
|

0 x

1 |
(a) (i) when

1
 x

1 otherwise
Sketch the graph of y = f ( x ) for

2
 x

2.
(ii) Is f an even function? Explain your answer.
(b) Let g : R

R be defined by g ( x ) = f ( x ) f ( x + 1).
(i) Sketch the graph of y = g ( x ) for

2
 x

2.
(ii) Is g an even function? Explain your answer.

5. (a) Suppose f (x) is continuous on [ 0 , a ].
Show that
0 z a
( )

0 z a
(

) .
Further, if f (x) + f (a

x) = K for all x

[ 0 , a ], where K is a constant, prove that
(i)
(ii)
K = 2 f ( a
2
) ;
0 a z f x dx
 af a
( )
2
.
(b) Hence, or otherwise, evaluate
0
2 z  1
. e sin x 
1 dx
- The End -