99-AL

P MATHS HONG KONG EXAMINATIONS AUTHORITY

PAPER 2
HONG KONG ADVANCED LEVEL EXAMINATION 1999

PURE MATHEMATICS A-LEVEL PAPER 2

1.30 pm – 4.30 pm (3 hours)

This paper must be answered in English

1. This paper consists of Section A and Section B.

2. Answer ALL questions in Section A and any FOUR questions in Section B.

3. You are provided with one AL(C)1 answer book and four AL(D) answer books.
Section A : Write your answers in the AL(C)1 answer book.
Section B : Use a separate AL(D) answer book for each question and put the
question number on the front cover of each answer book.

4. The four AL(D) answer books should be tied together with the green tag
provided. The AL(C)1 answer book and the four AL(D) answer books must be
handed in separately at the end of the examination.


 

Hong Kong Examinations Authority

All Rights Reserved 1999

99-AL-PURE MATHS 2–1

 FORMULAS FOR REFERENCE

sin( A ± B) = sin A cos B ± cos A sin B

cos( A ± B ) = cos A cos B m sin A sin B
tan A ± tan B
tan( A ± B) =
1 m tan A tan B

A+ B A−B
sin A + sin B = 2 sin cos
2 2
A+ B A− B
sin A − sin B = 2 cos sin
2 2
A+ B A− B
cos A + cos B = 2 cos cos
2 2
A+ B A− B
cos A − cos B = −2 sin sin
2 2

2 sin A cos B = sin( A + B) + sin( A − B )

2 cos A cos B = cos( A + B ) + cos( A − B )
2 sin A sin B = cos( A − B ) − cos( A + B)

99-AL-PURE MATHS 2−2 −1−

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SECTION A (40 marks)
Answer ALL questions in this section.
Write your answers in the AL(C)1 answer book.

x sin x
1. (a) Evaluate lim .
x →0 1 − cos x

1
(b) Using (a) or otherwise, evaluate lim+ (1 − cos x) ln x .
x →0
(6 marks)

π π
2. (a) Let f be a continuous function. Show that
∫ 0
f( x) dx =
∫ 0
f(π − x ) dx .

π x sin x
(b) Evaluate
∫ 0 1 + cos 2 x
dx .

(6 marks)

1
3. Figure 1 shows the graph of r = + cos 2θ . Find the area of the shaded
2
region. π
θ=
2

Figure 1
(4 marks)

99-AL-PURE MATHS 2−3 −2− Go on to the next page

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 2x
4. Let f( x) = .
x −1
2

(a) Resolve f(x) into partial fractions.

(b) Find f (n ) (0) , where n = 1, 2, 3, K.

(5 marks)

 2 1
 x cos if x > 0 ,
5. Let f( x) =  x
ax + b if x ≤ 0
be differentiable at x = 0 . Find a and b .
(6 marks)

6. (a) Suppose f : R → R is a function satisfying f(a + x ) = f(a − x ) and

f(b + x ) = f(b − x) for all x , where a , b are constants and a > b . Let
w = 2(a − b) . Show that w is a period of f , i.e., f( x + w) = f( x ) for
all x ∈ R .

(b) Suppose g : R → R is a periodic function with period T > 0 satisfying

g( x) = g(− x ) for all x . Show that there exists c with 0 < c < T such
that g(c + x ) = g(c − x) for all x .
(6 marks)

7. Let L1 be the line of intersection of the planes x + y + z = 1 and

x − y − z = 5 , and L2 be the line passing through (1, 1, − 1) and intersecting
L1 at right angle.

(a) Find a parametric equation of L1 .

(b) Find the coordinates of the point of intersection of L1 and L2 , and a

parametric equation of L2 .
(7 marks)

99-AL-PURE MATHS 2−4 −3−

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SECTION B (60 marks)
Answer any FOUR questions in this section. Each question carries 15 marks.
Use a separate AL(D) answer book for each question.

1
8. Let f ( x ) = xe x for x ≠ 0 .

(a) Find lim− f( x ) and show that f(x) → ∞ as x → 0 + .

x →0
(3 marks)

(b) Find f ' ( x) and f '' ( x ) for x ≠ 0 .

(2 marks)

(c) Determine the values of x for each of the following cases:

(i) f ' ( x) > 0 ,

(ii) f '' ( x) > 0 .

(3 marks)

(d) Find all relative extrema of f(x) .

(2 marks)

(e) Find all asymptotes of the graph of f(x) .

(3 marks)

(f) Sketch the graph of f(x) .

(2 marks)

99-AL-PURE MATHS 2−5 −4− Go on to the next page

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9. Let n be a positive integer.

( L + t )+ 1t− t
2n
1 2 n −2
(a) Show that = 1+ t2 + for t 2 ≠ 1 .
1− t2 2

(2 marks)

(b) For −1 < x < 1 , show that

x 1
∫
t
(i) dt = ln ,
0 1− t 2
1− x 2

x t 2 n +1  x2 x4
L + x2n  .
2n
1
(ii)
∫ 0 1− t 2
dt = ln −

1− x 2  2
+
4
+

(7 marks)

k n +1
8 9 8
∑
n 1
(c) Show that 0 ≤ ln 3 −   ≤   .
=
k 1 2k 9 2n + 2  9 
k
1 8
∑
∞
Hence evaluate   .
k =1 2k  9 

(6 marks)

99-AL-PURE MATHS 2−6 −5−

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10. (a) Let f : R → R be a strictly increasing bijective function.

(i) Show that the inverse function f −1 is also strictly increasing.

(ii) Let a < b and t 1 , t 2 , K, t n ∈ [a, b] , n ≥ 2 .

1 
∑
n
Show that a ≤ f −1  f(t i )  ≤ b .
n i =1 

Find a necessary and sufficient condition on t 1 , t 2 , K, tn

1 
∑
n
such that a < f −1  f(t i )  < b .
n i =1 
(8 marks)

1
(b) Let g : R → R be defined by g( x) = x3 .

(i) Show that g is bijective and strictly increasing.

3
 
L
1 1
1 + 23 + + n3 
Hence show that 1 <   < n for n ≥ 2 .
 n 
 
 

(ii) Find the area enclosed by the graphs of y = g(x) and

−
y = g 1 ( x) .
(7 marks)

99-AL-PURE MATHS 2−7 −6− Go on to the next page

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11. (a) For n = 0, 1, 2, K and y ≥ 0 , define I n ( y) = ∫ 0
y
−
t n e t dt .

Prove that I n ( y ) = − y n e − y + n I n −1 ( y ) for n ≥ 1 and y ≥ 0 .

Hence deduce that I n ( y ) ≤ n! for n ≥ 0 and y ≥ 0 .

(5 marks)

(b) Let n be a positive integer.

(i) By considering g( x) = n ln(n + x ) − n ln(n − x ) − 2 x for

0 ≤ x < n , show that (n + x) n e − ( n + x) ≥ (n − x ) n e −( n − x ) for
0≤ x<n .

2n n
∫ ∫
− −
(ii) Use (b)(i) to show that u n e u du ≥ u n e u du .
n 0

2n n
∫ ∫
− −
Hence deduce that t n e t dt ≥ 2e n (n − t ) n e t dt .
0 0
(8 marks)

n en
∫
1
(c) Using the above results or otherwise, show that (n − t ) n e t dt ≤
n! 0 2
for all positive integers n .
(2 marks)

99-AL-PURE MATHS 2−8 −7−

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 ( x − h) 2 y2
12. (a) The equation of the ellipse E is 2
+ = 1 , where a, b, h ∈ R
a b2
and a, b > 0 .
(i) By integration, find the area enclosed by E .
(ii) If the straight line y = mx is tangent to E , show that
b2
m2 = .
h − a2
2

(6 marks)

(b) For n = 1, 2, 3, K , let E n be the ellipse given by

( x − hn ) 2
y2
2 2
=1 ,+
an p 2 an
where p > 0 , hn > hn +1 and hn > a n > 0 . Suppose for all n , E n
and E n +1 touch each other externally and the straight line y = mx is a
common tangent to all E n as shown in Figure 2.

y y = mx

x
O
E3
E2
E1
Figure 2
(i) Express hn − hn +1 in terms of a n and a n +1 .
(ii) Using (a)(ii) and the result of (b)(i), or otherwise, show that
a n +1 h1 − a1
= .
an h1 + a1

(iii) Let S n be the area enclosed by the ellipse E n . Evaluate

∑
∞
S n in terms of a1 , h1 and p .
n =1
(9 marks)

99-AL-PURE MATHS 2−9 −8− Go on to the next page

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13. Let f(x) be a differentiable function on R such that f ′( x ) ≤ f( x) for all
x∈R.

(a) Suppose a ≥ 0 and f(a) = 0 . Let x ∈ ( a , a + 1) .

(i) Using Mean Value Theorem or otherwise, show that there exists
ξ 1 ∈ (a, x) such that f( x) ≤ f(ξ 1 ) ( x − a) .

(ii) Using (a)(i) or otherwise, show that for each n = 1, 2, 3, K ,

there exists ξ n ∈ (a, x ) such that f( x) ≤ f(ξ n ) ( x − a) .
n

(iii) Using (a)(ii) or otherwise, show that f(x) = 0 for all

x ∈ [ a , a + 1] .

You may use the fact that there is M > 0 such that f( x) ≤ M
for all x ∈ [ a , a + 1] .

(9 marks)

(b) Suppose f(0) = 0 .

(i) Using (a) or otherwise, show that f(x) = 0 for all x ∈ [0, ∞) .

(ii) Show that f(x) = 0 for all x ∈ R .

(6 marks)

END OF PAPER

99-AL-PURE MATHS 2−10 −9−

 All Rights Reserved 1999
1999 Paper 2
Section A

1. (a) 2

(b) e2

π2
2. (b)
4

π 3 3
3. +
4 16

1 1
4. (a) +
x −1 x +1

− 2n! when n is odd

(b) 
0 when n is even

5. a=b=0

x = 3

7. (a) y = t ,t∈R.
z = −t − 2


(b) (3, 0, −2) ,

x = 2s + 1

 y = −s + 1 , s ∈ R .
z = − s − 1


 All Rights Reserved 1999

1999 Paper 2
Section B

1
8. (a) lim− f( x ) = lim− xe x = 0
x →0 x →0
1 ′
1 1
1 ex  1
lim+ f( x) = lim+ xe x
ex  x
= lim+ = lim+ = lim+ e x → ∞
x →0 x →0 x →0 1 x →0 ′ x →0
1
x  
 x

(b) For x ≠ 0 ,
1 1 1
 1   1
f ' ( x) = xe x  − 2  + e x = 1 − e x
 x   x
1 1 1
 1  1   1  1
f '' ( x ) = 1 − e x  − 2  + e x  2  = 3 e x
 x  x  x  x

x −1
(c) (i) f ' ( x) > 0 ⇔ > 0 ⇔ x < 0 or x > 1 .
x
i.e. f ' ( x) > 0 on (−∞, 0)  (1, ∞) .

1
(ii) f '' ( x) > 0 ⇔ >0 ⇔ x > 0.
x3
i.e. f '' ( x) > 0 on (0, ∞) .

(d) x (−∞, 0) 0 (0, 1) 1 (1, ∞)

f(x) ↑ Undefined ↓ e ↑
f ' ( x) + Undefined − 0 +
f '' ( x ) − Undefined + + +

(1, e) is a relative minimum point.

 All Rights Reserved 1999

1999 Paper 2
Section B

(e) From (a)(i),  f(x) → ∞ as x → 0 +

∴ x = 0 is a vertical asymptote.

1
f( x)
 lim = lim e x = 1
x →∞ x x →∞
1 ′
1 1
1 ex 
ex−1  x
lim (f( x) − x ) = lim ( xe x − x) = lim = lim = 1
x →∞ x →∞ x →∞ 1 x →∞ ′
1
x  
 x
∴ y = x + 1 is an oblique asymptote.

(f) The graph of f(x):

y

y=x+

(1, e)
1

O x

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1999 Paper 2
Section B

9. (a) For t 2 ≠ 1 ,

(1 + t )+ 1t− t 1 − (t 2 ) n
2n
t 2n
+ + t
2n− 2
= +
2
2
1− t 2 1− t 2
1
=
1− t 2

x t 1 x2 1
(b) (i)
∫ 01− t2
dt =
2 ∫ 0 1− u
du (putting u = t 2 )

1
[ln | 1 − u |] 0x
2
= −
2
1
= ln for −1 < x < 1
1 − x2

(ii) From (a),

1− t
t
2
(
= t + t ++ t − +
3 2n 1
) t 2 n +1
1− t2
for t 2 ≠ 1

∫ (t + t )dt + ∫ 1t − t
x t x x 2 n +1
++ t
∫
2 n −1
dt = 3
dt for −1 < x < 1
01−t2 0 0 2

= (t + t ) ∫
1 x x t 2 n +1
+  + t 2 n −1 dt +
∫
3
ln dt ( by (b)(i) )
0 1− t2
1− x2 0

x
1 t 2 t 4 t 2n  x t 2 n +1
++
ln
1− x2
= +
 2 4
 +
2n 
0
∫ 1− t
0 2
dt

1  x2 x4 x 2n  x t 2 n +1
++
ln −

1− x 2  2
+
4 2n 
=
∫ 0 1− t 2
dt

 All Rights Reserved 1999

1999 Paper 2
Section B

8
(c) Putting x 2 = in (b)(ii), we have
9
8
t 2 n +1
k
1 8
∑ ∫
n
ln 3 −   = 9 dt
k =1 2k 9
  0 1− t 2
8
For 0 ≤ t ≤ ,
9
t 2 n +1
 ≥0
1− t 2
8
t 2 n +1
∴
∫ 0
9
1− t 2
dt ≥ 0

t 2 n +1 t 2 n +1
 ≤ = 9t +
2n 1
1− t 2 1 8
−
9
8 8
t 2 n +1
∴
∫ 0
9
1− t 2
dt ≤ 9
∫ 0
9 t 2 n +1 dt

8
 9  9
=  t 2n+ 2 
 2n + 2 0
n +1
9 8
=  
2n + 2  9 
k n +1
8 9 8
∑
n 1
Hence 0 ≤ ln 3 −   ≤  
k =1 2 k
 
9 2 n +2 9

n +1
9 8
 lim   =0
n →∞ 2 n + 2  9 
k
1 8
∑
n
∴ lim   = ln 3
k =1 2 k 9
n →∞  

 All Rights Reserved 1999

1999 Paper 2
Section B

10. (a) (i) Suppose on the contrary that f −1 is not strictly increasing.
Then there exist y1 and y2 such that y1 < y2 and
f −1 ( y1 ) ≥ f −1 ( y2 )
( ) (
⇒ f f −1 ( y1 ) ≥ f f −1 ( y2 ) ) ( Q f is strictly increasing)
⇒ y1 ≥ y2
which contradicts that y1 < y2 .

(ii) Q t1 , t 2 , K, t n ∈ [a, b] and f is strictly increasing

∴ f(a) ≤ f(t i ) ≤ f(b) for i = 1, 2, K, n
∑
n
⇒ n f(a) ≤ f(t i ) ≤ n f(b)
i =1

f(a) ≤ ∑
1 n
⇒ f(t i ) ≤ f(b)
n i =1

⇒
1
a ≤ f −1  ∑
n
i =1

f(t i )  ≤ b ( Q f −1 is strictly increasing)
n 

Q f is strictly increasing
∴ For t 1 , t 2 , K , t n ∈ [a, b] ,

f(a) =
1
n ∑
n
i =1
L
f(t i ) iff t1 = t 2 = = t n = a and
1
n ∑
n
i =1
L
f(t i ) = f(b) iff t1 = t 2 = = t n = b .

Again since f −1 is strictly increasing,

1 
∑
n
a < f −1  f(t i )  < b
=
n i 1

iff t 1 , t 2 , K , t n not all equal to a and not all equal to b .

 All Rights Reserved 1999

1999 Paper 2
Section B

(b) (i) For any x1, x2 ∈ R ,

1 1
g( x1 ) = g( x 2 ) ⇒ x1 3 = x 2 3
⇒ x1 = x 2
∴ g is injective.

For any y ∈ R , let x = y 3 , then g( x) = y .

∴ g is surjective.

Q g ′( x ) =
1
2
> 0 for x ∈ R \ {0} and g is continuous on R
3x 3
∴ g is strictly increasing.

Using (a) and putting t i = i where i = 1, 2, K, n ,

Q t1 , t 2 , K, t n not all equal to 1 and not all equal to n
1 
∑
n
∴ 1 < g −1  g(i )  < n
i =1
n 
3
 1 1 
⇒
1 + 23 + + n3
1< 
L 
 <n
 n 
 
 

∫ (g− g )dx
1
(ii) Area enclosed by g and g −1 = 2 −1
0

1 
1
 x 3 − x 3  dx
= 2
∫
0



1
3 4 1 
= 2  x 3 − x4 
 4 4 
0
= 1

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1999 Paper 2
Section B

y
11. (a) I n ( y) =
∫ 0
t n e − t dt


= −  t n e −t [ ] −∫ y
0
y 
nt n −1 e −t dt 
 0 
yn y
= −
e y
+n ∫ 0
t n −1 e −t dt

yn
= − + n I n −1 ( y ) for n ≥ 1 and y ≥ 0
ey

I0 ( y) =
∫
y
e −t dt = − e −t [
≤1 ] y
0 = 1−
1
0 ey
Suppose I k ( y ) ≤ k! for some non-negative integer k.
yn
I k +1 ( y ) = − + (k + 1) I k ( y )
ey
≤ (k + 1) I k ( y )
≤ (k + 1)!
By the principle of mathematical induction, the result follows.

(b) (i)  n
g ′( x ) =+
n+ x n−x
n
−2 = 2
2x 2
≥ 0 for 0 ≤ x < n .
n − x2
∴ g is increasing for 0 ≤ x < n
i.e. g( x) ≥ g( 0) for 0 ≤ x < n

Hence for 0 ≤ x < n ,

n ln(n + x ) − n ln(n − x ) − 2 x ≥ g(0) = 0
n ln(n + x) ≥ n ln(n − x ) + 2 x
(n + x ) n ≥ (n − x ) n e 2 x
( n + x ) n e − ( n + x ) ≥ ( n − x) n e − ( n − x )

 All Rights Reserved 1999

1999 Paper 2
Section B

2n n
(ii)
∫ n
u n e −u du =
∫ 0
(n + x) n e −( n + x ) dx ( putting u = n + x )
n
≥
∫ 0
( n − x ) n e − ( n − x ) dx ( by (b)(i) )
0
=
∫ n
y n e − y (−dy ) ( putting y = n − x )
n
=
∫ 0
u n e −u du
2n n 2n
Hence
∫ 0
t n e −t dt =
∫ 0
t n e − t dt +
∫ n
t n e − t dt
n n
≥
∫ 0
t n e −t dt +
∫ 0
t n e −t dt
0
= 2
∫ n
(n − x ) n e −n + x (−dx) ( putting t = n − x )
n
= 2e −
∫ (n − t ) n e t dt
n
0

(c) For n > 0 ,

1 n 1 2n

n! 0 ∫
(n − t ) n e t dt ≤ − n
2e (n!) 0
t n e −t dt
∫ ( by (b)(ii) )

1
= −n
I n ( 2n )
2e (n!)
1
≤ −n
(n!) ( by (a) )
2e (n!)
en
=
2

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1999 Paper 2
Section B

( x − h) 2 y2  ( x − h) 2 
12. (a) (i) + =1 ⇒ y 2 = b 2 1 − 
2 2  a2 
a b  
∴ Area enclosed by E
h+ a ( x − h) 2
= 2 ∫ h− a
b 1−
a2
dx

a y2
= 2 ∫ −a
b 1−
a2
dy ( by putting y = x − h )

4b a
=
a ∫ 0
a 2 − y 2 dy


4b
=
a ∫ 0
2 a 2 cos 2 θ dθ ( by putting y = a sin θ )


1 + cos2θ
= 4ab ∫ 0
2
2
dx


 sin 2θ  2
= 2ab θ +
 2  0
= abπ

(ii) Sub. y = mx into the equation of E , then

( x − h) 2 m 2 x 2
+ 2 =1
a2 b
(a 2 m 2 + b 2 ) x 2 − 2b 2 hx + b 2 h 2 − a 2 b 2 = 0
Q y = mx is tangent to E
∴ 4b 4 h 2 − 4(a 2 m 2 + b 2 )b 2 (h 2 − a 2 ) = 0
b 2 h 2 − (a 2 m 2 + b 2 )(h 2 − a 2 ) = 0
b 2 − m 2 (h 2 − a 2 ) = 0
b2
m2 =
h2 − a2

 All Rights Reserved 1999

1999 Paper 2
Section B

(b) (i) Q E n and E n +1 touch each other externally

∴ hn − a n = hn +1 + a n +1
hn − hn +1 = a n + a n +1

p 2 an 2
(ii) By (a)(ii), m 2 =
hn − a n
2 2

p 2 an 2 p 2 a1
2
∴ =
hn − a n h1 − a1
2 2 2 2

a n 2 h1 2 = a1 2 hn 2
h
hn = 1 a n
a1
h1 h
Hence a n − 1 a n +1 = a n + a n +1 ( by (b)(i) )
a1 a1
 h1  h 
 − 1a n =  1 + 1a n +1
 1 
a  1 
a
a n +1 h1 − a1
=
an h1 + a1

 All Rights Reserved 1999

1999 Paper 2
Section B

∑ ∑
∞ ∞
(iii) Sn = a ( pa n )π ( by (a)(i) )
n =1 n =1 n

∑
∞
= pπ a 2
n =1 n
2
a n +1 2  h −a 
Q =  1 1  ∈ (0, 1)
 h1 + a1 
2
an
∴ The summation is an infinite sum of a geometric sequence
2
 h −a 
with common ratio  1 1  .
 h1 + a1 
2

∑
∞ a1
Hence S n = pπ ⋅
n =1 2
h −a 
1 −  1 1 
 h1 + a1 
a1 (h1 + a1 ) 2
2
= pπ ⋅
(h1 + a1 ) 2 − (h1 − a1 ) 2
pa1 (h1 + a1 ) 2 π
=
4h1

 All Rights Reserved 1999

1999 Paper 2
Section B

13. (a) (i) By Mean Value Theorem, there exists ξ 1 ∈ (a, x) such that
f(x) − f(a) = f ′(ξ 1 )( x − a) for x ∈ ( a, a + 1)
 f(a) = 0 and x − a > 0
∴ f( x ) = f ′(ξ 1 ) ( x − a) ≤ f(ξ 1 ) ( x − a )

(ii) Using (a)(i), the statement holds for n = 1.

Assume that there exists ξ k ∈ (a, x) such that f( x ) ≤ f(ξ k ) ( x − a) k .
By Mean Value Theorem, there exists ξ k +1 ∈ (a, ξ k ) such that
f(ξ k ) − f(a ) = f ′(ξ k +1 )(ξ k − a)
f(ξ k ) = f ′(ξ k +1 ) (ξ k − a) ≤ f(ξ k +1 ) ( x − a)
∴ f( x) ≤ f(ξ k ) ( x − a) k ≤ f(ξ k +1 ) ( x − a ) k +1
By the principle of mathematical induction, the result follows.

(iii)  there exists M > 0 such that

f( x) ≤ M for all x ∈ [ a , a + 1]
∴ 0 ≤ f( x ) ≤ M ( x − a) n for all n = 1, 2, 3,  and x ∈ (a , a + 1)
x ∈ (a , a + 1) ⇒ lim M ( x − a) = 0 n
n→ ∞
Hence f(x) = 0 for all x ∈ (a , a + 1)
 f is continuous on [ a , a + 1]
∴ f(x) = 0 for all x ∈ [ a , a + 1]

 All Rights Reserved 1999

1999 Paper 2
Section B

(b) (i) f(0) = 0 ⇒ f(x) = 0 for all x ∈ [0, 1] ( by (a)(iii) )

Assume f(x) = 0 for all x ∈ [k, k+1] where k is a non-negative integer.
Hence f(k+1) = 0 ⇒ f(x) = 0 for all x ∈ [k+1, k+2] .
By the principle of mathematical induction,
f(x) = 0 for all x ∈ [n, n+1] where n = 0, 1, 2, . K
i.e. f(x) = 0 for all x ∈ [0, 1] ∪ [1, 2] ∪ [2, 3] ∪ L
f(x) = 0 for all x ∈ [0, ∞)

(ii) Now define g(x) = f(−x) for x ∈ R , then

g is a differentiable function on R with
g(0) = f(0) = 0 and
g ′( x) = − f ′(− x) = f ′(− x) ≤ f(− x) = g( x )
Using (b)(i),
g(x) = 0 for all x ∈ [0, ∞)
⇒ f(x) = 0 for all x ∈ (−∞, 0]
Hence f(x) = 0 for all x∈ R

 All Rights Reserved 1999