Pre-Exam supplementary exercise - solution
Question A1
(
)
P ( A ) = 0.6, P ( A Ç B) = 0.45 and P A Ç B = 0.2 .
B
B
(
A
A
0.45
0.15
0.6
0.2
0.2
0.4
0.65
0.35
)
(a) P A Ç B = P ( A ) - P ( A Ç B) = 0.6 - 0.45 = 0.15
( ) (
) (
)
(b) P B = P A Ç B + P A Ç B = 0.15 + 0.2 = 0.35
(c) P ( A È B) = P ( A ) + P ( B) - P ( A Ç B) = 0.6 + (1 - 0.35) - 0.45 = 0.8
P ( A B) =
P ( A Ç B)
P ( B)
=
0.2
4
=
0.65 13
Question 2
(a) Consider the augmented matrix,
1
R ® R2
é 4 -4 14 ù - R2 + R1 ® R1 é1 -6 9 ù -3 R1 + R2 ® R2 é1 -6
9 ù 20 2
~
~
ê
ú
ê
ú
ê
ú ~
3
2
5
3
2
5
0
20
22
ë
û
ë
û
ë
û
Then, we have x2 = -1.1 and x1 = 9 + 6 ( -1.1) = 2.4
é1 -6
ê
ë0 1
(b) Consider the augmented matrix,
é 4 -4 14ù - R2 + R1 ®R1 é1 -4 - p 14 - q ù -3R1 + R2 ®R2 é1 -4 - q
~
ê
ú ~ ê
ú
ê
p
q û
ë3 p q û
ë3
ë0 12 + 4 p
(i)
Unique solution occurs when p ¹ -3 Ç q Î R
(ii)
Infinitely many solutions occur when p = -3 Ç q =
(iii)
No solution occurs when p = -3 Ç q ¹
21
2
21
2
14 - q ù
ú
4q - 42û
9 ù
ú
-1.1û
Question 3
(a) The value of the limits are
(
(i)
lim 1 + 6 x + 8 x
(ii)
lim-
x ®0
x ®3
(iii)
lim
x ®-¥
3
2 x
)
x -3
x - 7 x + 12
2
( (
= lim 1 + 6 x + 8 x
x ®0
= lim-
10 x - 68
9 x 2 - 3x + 5
x ®3
- ( x - 3)
2
( x - 3)( x - 4 )
= lim
x ®-¥
))
1
6 x +8 x
= limx ®3
2
×
(
3 6 x +8 x 2
x
)
= e18
-1
-1
=
=1
( x - 4 ) lim- x - 4
(
x ®3
)
10 x - 68
68
10 10
x
x
= lim
=3
9 x 2 - 3x + 5 x®-¥
9 x 2 - 3x + 5
x
x2
(b) The function continuous on all real numbers if
lim f ( x ) = f ( 0 ) and lim f ( x ) = f ( a ) .
x ®0
(
x ®a
)
lim x 2 + 2 = f ( 0 ) = ( 0 ) + b è b = 2
x ®0
lim ( x + 2 ) = a + 2 = f ( a ) = 2 ( a ) + 4 è a 2 - 4a = 0 Þ a = 4 or 0 (rejected)
2
x ®a
Question 4
(a) The derivative
dy
=
dx
(i)
(iii)
ò
8
0
(
)
cos( 3 x )
dy
= -6 x sin 3x 2 e
dx
dy
= 2 x cot ( 2 x ) - 2 x 2 csc2 ( 2 x )
dx
( )
(ii)
(b)
2x + 2
æ1ö
x 2 + 2 x ç ÷ - ( ln x )
è xø
2 x 2 + 2 x = ( x + 2 ) - ( ln x )( x + 1)
x2 + 2x
x2 + 2x x2 + 2x
2
3
5
8
0
3
5
f ( x ) dx = ò 10 xdx + ò 6 x 2 dx + ò 20 - 8 x3dx
3
5
8
é x2 ù
é x3 ù
= 10 ê ú + 6 ê ú + éë 20 x - 2 x 4 ùû
5
ë 2 û0
ë 3 û3
(
) (
æ 53 33 ö
4
4
= 45 + 6 ç - ÷ + 160 - 2 ( 8 ) - 100 - 2 ( 5 )
è3 3ø
= 45 + 196 + ( -8032 ) - ( -1150 )
= -6641
)
Question 5
x2 - 4
for x > 0
x
(i) The domain is {x | x ³ 2} and the range is {x | 0 £ x < 1} .
(a) Let f ( x ) =
(ii) Let y =
x2 - 4
. Then xy = x 2 - 4
x
è x2 y 2 = x2 - 4 è x2 - x2 y 2 = 4 è x =
(iii) As f -1 ( x ) =
4
1- y2
4
, then
1 - x2
g ( x ) = f -1 ( ( f  g )( x ) )
= f -1 ( tan x )
=
4
1 - tan 2 x
(b) (i) g (- x) = e- x sin ( - x ) - e ( ) = -e- x sin ( x ) - e x ¹ - g ( x ) ¹ g ( x )
Neither odd nor even function
- -x
(ii) h(- x) = - x tan 2 ( - x ) = x ( - tan x ) = x tan 2 x = h ( x )
Even function
2
(c) (i) 6 x 6 x 5 = 180
(ii) 2 x 6 x 5 = 60
Question 6
(a) The function value f ( 2) = 23 - 3 ´ 22 + 4 ´ 2 - 6 = -2
The derivative is f ' ( x ) = 3x 2 - 6 x + 4 .
The slope of the tangent line is f ' ( 2) = 3 ( 2) - 6 ( 2) + 4 = 4 .
2
The equation of the tangent line is
y - ( -2 )
= 4 è 4 x - y - 10 = 0
x-2
(b) By the limit definition, we have
sin ( 3 ( x + h ) ) - sin ( 3 x )
f '( x) = lim
h ®0
h
sin ( 3 x + 3h ) - sin ( 3 x )
= lim
h ®0
h
æ 3 x + 3h + 3 x ö æ 3 x + 3h - 3 x ö
2 cos ç
÷ sin ç
÷
2
2
è
ø è
ø
= lim
h ®0
h
æ 6 x + 3h ö æ 3h ö
2 cos ç
÷ sin ç ÷
2 ø è 2 ø
è
= lim
h ®0
h
æ 3h ö
sin ç ÷
æ 6 x + 3h ö
è 2 ø
= 2 lim cos ç
÷ lim
h ®0
h
®
0
h
è 2 ø
æ3ö
= 2 cos ( 3 x ) ç ÷
è2ø
= 3cos ( 3 x )
(c) F '( x) = f '( g ( x)) × g '( x)
(
)
Then F '(7) = f ' g ( 7 ) × g ' ( 7 ) = f ' ( -2 ) × 6 = 4 × 6 = 24
(d) Let f ( x ) = sin 2 (p x ) . Then f ( x + k ) = sin 2 (p x + p k )
Thus , sin 2 (p x ) = sin 2 (p x + p k ) when k is even number.
Hence, the period of the function is 2.
Question 7
(a)
6 x3 + 5 x - 2
( x - 1) ( x + 2 )
2
2
A
B
C
D
+
+
+
.
x - 1 x + 1 ( x + 2) ( x + 2) 2
º
(i) 6 x3 + 5 x - 2 = A ( x + 1)( x + 2 ) + B ( x - 1)( x + 2 ) + C ( x 2 - 1) ( x + 2 ) + D ( x 2 - 1)
2
2
(
)
When x = -2 , 6 ( -2 ) + 5 ( -2 ) - 2 = D ( -2 ) - 1 è D = -20
3
2
When x = -1 , 6 ( -1) + 5 ( -1) - 2 = B ( -2 ) è B =
3
When x = 1 , 6 (1) + 5 (1) - 2 = A ( 2)(3) è A =
3
When x = 0 , -2 =
(ii)
ò
3
ò
3
2
=
2
2
1
2
x -1
+
13
2
x +1
+
1
2
1
13
2
2
(1)( 2 ) + ( -1)( 2 ) + C ( -1)( 2 ) + ( -20 )( -1) è C = -1
2
2
6 x3 + 5 x - 2
( x 2 - 1) ( x + 2 )
13
2
2
dx.
-1
-20
+
dx
( x + 2) ( x + 2) 2
3
13
20 ù
é1
= ê ln x - 1 + ln x + 1 - ln x + 2 +
2
x + 2 úû 2
ë2
13
13
æ1
ö æ1
ö
= ç ln 2 + ln 4 - ln 5 + 4 ÷ - ç ln1 + ln 3 - ln 4 + 5 ÷
2
2
è2
ø è2
ø
13
31
= - ln 3 + ln 2 - ln 5 - 1
2
2
(b) Since P ( A È B ) = 0.85, P ( A ) = 0.32, P ( B ) = 0.48., then
P ( A | B) =
P ( A Ç B)
P ( B)
=
P ( A) + P ( B ) - P ( A È B )
P ( B)
=
(1 - 0.32 ) + (1 - 0.48) - ( 0.85)
(1 - 0.48)
0.35
0.52
= 0.6731
=
(c) For any integer n ³ 1 , Let Pn be the statement that
n
å ( 3 j - 2) = 1 + 4 + 7 +  + (3n - 2 ) =
j =1
For n = 1 , the statement P1 says that 1 =
n ( 3n - 1)
2
1( 3 - 1)
which is true.
2
Assume n = k , and the statement Pk is true, i.e.
k
k ( 3k - 1)
j =1
2
å ( 3 j - 2 ) = 1 + 4 + 7 +  + (3k - 2 ) =
For n = k + 1 , we have
k +1
å ( 3 j - 2 ) = 1 + 4 + 7 +  + ( 3 ( k + 1) - 2 )
j =1
k ( 3k - 1)
+ 3k + 1
2
k ( 3k - 1) + 2 ( 3k + 1)
=
2
2
3k - k + 6k + 2
=
2
2
3k + 5k + 2
=
2
( k + 1)( 3k + 2 )
=
2
( k + 1) ( 3 ( k + 1) - 1)
=
2
=
Therefore, the statement Pk+1 holds.
Thus, by the principle of mathematical induction, Pn holds for all integer n ³ 1 .
Question 8
(a) Since the denominator, lim ( x + 4 )( x - 4 ) = 0 , we also need the limit of the
x ®4
numerator to be zero as x ® 4 for limit to exist.
lim ( x - 1)( x - a ) = 0 è ( 4 - a ) = 0 è a = 4
x ®4
( x - 1) 4 - 1 3
( x - 1)( x - a ) = lim ( x - 1)( x - 4 ) = lim ( x - 1) = lim
x ®4
=
=
x ® 4 ( x + 4 )( x - 4 )
x ® 4 ( x + 4 )( x - 4 )
x ®4 ( x + 4)
lim ( x + 4 ) 4 + 4 8
x ®4
lim
(1) × ( 3) × ( 5) = 5 and is obviously true.
(b) For n = 1, the statement reduces to 12 + 22 =
3
Assuming the statement is true for n = k,
12 + 22 + 32 +  + ( 2k ) =
2
k ( 2k + 1)( 4k + 1)
3
For n = k + 1, the left-hand side of the statement becomes
12 + 22 + 32 +  + ( 2k ) + ( 2k + 1) + ( 2k + 2) .
2
Then, we have
LHS =
k ( 2k + 1)( 4k + 1)
3
2
+ ( 2k + 1) + ( 2k + 2 )
2
2
3
(8k + 6k + k ) + (12k 2 + 12k + 3) + (12k 2 + 24k + 12 )
3
=
2
k ( 2k + 1)( 4k + 1) + 3 ( 2k + 1) + 3 ( 2k + 2 )
2
=
2
2
3
8k + 30k + 37k + 15
3
( 2k + 2 + 1)( 4k + 4 + 1)
= ( k + 1) ×
3
= RHS
=
3
2
Therefore, by the principle of mathematical induction, the given statement is
true for all positive integers n.
Question 9
(a) (i) The augmented matrix of the system is
é1 1
1
ê
t
ê2 3
2
êë 3 4 t - 3t + 4
1 ù -2 R + R ® R é1 1
1
ú 1 2 2ê
t -1 ú
~ ê0 1 2 t - 2
-3 R1 + R3 ® R3
êë0 1 t - 3t + 1
2t + m úû
é1 1
1
ê
~ ê0 1 2 t - 2
êë0 0 t - 4t + 3
Therefore a = t 2 - 4t + 3 = ( t - 1)( t - 3) and b = t + m
(ii) By the result of (a)(i)
- R2 + R3 ® R3
1
ù
ú
t -3 ú
2t + m - 3úû
1 ù
ú
t -3ú
t + m úû
1. Unique solution: ( t - 1)( t - 3) ¹ 0 Ç m Î R or t ¹ 1 Ç t ¹ 3 Ç m Î R .
2. No solution: ( t - 1)( t - 3) = 0 Ç m ¹ -t .
è When t = 1 , m ¹ -1 . When t = 3 , m ¹ -3
3. Infinitely many solutions: ( t - 1)( t - 3) = 0 Ç m = -t .
è When t = 1 , m = -1 . è When t = 3 , m = -3 .
1
1
f ( x + h) - f ( x)
x +1
(b) f ¢ ( x ) = lim
= lim x + h + 1
h ®0
h
®
0
h
h
æ
ö
1 ç x +1- x + h +1 ÷
= lim
h®0 h ç
x + h +1
x +1 ÷
è
ø
(
(
1
= lim ×
h®0 h
1
= lim ×
h®0 h
1
= lim ×
h®0 h
= - lim
h®0
=-
(
)
)
)(
x - x+h
(
)(
x + h +1
)
x +1
x - x+h
(
)(
x + h +1
)
x +1
x + x+h
×
x + x+h
x - ( x + h)
(
)(
x + h +1
)(
x + h +1
1
( 2 x )(
)
x +1
2
)(
x +1
1
)(
x +1
x + x+h
x + x+h
)
)
Question 10
(a) Since
x 2 + 3x + 5
( x + 1)( x + 3)
2
A
B
C
+
+
, we have
x + 1 x + 3 ( x + 3) 2
=
x2 + 3x + 5 = A( x + 3) + B ( x + 1)( x + 3) + C ( x + 1)
2
Put x = -1 , we have ( -1 + 3) A = ( -1) + 3( -1) + 5 è A =
2
2
3
4
Put x = -3 , we have ( -3 + 1) C = ( -3) + 3( -3) + 5 è C = 2
Put x = 0 , we have
5
2
3 2
5
( 3) + 3B + æç - ö÷ = 5 è B = 1
4
4
è 2ø
(b) By the result of (a), we have
1
x 2 + 3x + 5
ò ( x + 1)( x + 3)
0
dx = ò
2
1
0
3
4
x +1
+
1
4
x+3
-
5
2
( x + 3)
2
dx
1
1
5 -1 ù
é3
= ê ln x + 1 + ln x + 3 - ×
4
2 x + 3 úû 0
ë4
1
5 1ö æ3
1
5 1ö
æ3
= ç ln 2 + ln 4 + × ÷ - ç ln1 + ln 3 + × ÷
4
2 4ø è4
4
2 3ø
è4
5
1
5
= ln 2 - ln 3 2
4
24
Question 11
(a) Using augmented matrix and performing elementary row operations, we have
é7 5 4
ê
ë3 3 6
120 ù -2 R2 + R1 ® R1 é1 -1 -8
~
ú
ê
90 û
ë3 3 6
-60ù
ú
90 û
é 1 -1 -8
ê
ë 0 6 30
-60ù
ú
270 û
-3 R1 + R2 ® R2
~
1
R2 ® R2
6
é1
ê
ë0
R2 + R1 ® R1 é1
~ ê
ë0
~
-1 -8
1
5
0 -3
1
5
-60 ù
ú
45 û
-15ù
ú
45 û
(b) Let z = t , then x = -15 + 3t and y = 45 - 5t .
As x, y and z are positive integers, we have -15 + 3t > 0 and 45 - 5t > 0 .
Thus, 5 < t < 9 or 6 £ t £ 8 .
t
x
y
z
6
3
15
6
( x, y, z ) = (3,15,6), ( 6,10,7 ) and (9,5,8)
7
6
10
7
8
9
5
8
Question 12
-2 x + 1
-2 x + 1
=
¹ ± f ( x) ,
9 + 3 ( - x ) 9 - 3x
f ( x ) is neither odd nor even.
(a) f ( - x ) =
2x + 1
. Then y ( 9 + 3x ) = 2 x + 1 è 9 y + 3xy = 2 x + 1
9 + 3x
1- 9y
è 3xy - 2 x = 1 - 9 y è x =
3y - 2
1 - 9x
2
Then f -1 ( x) =
for x ¹
3x - 2
3
(b) Let y =
(
)
3
(c) ( g  f )( x ) = ( x + 1) è g f ( x ) = ( x + 1)
3
3
æ -1 - 6 x ö
æ 1 + 6x ö
Then g ( x ) = ç
÷ = -ç
÷
è 3x - 2 ø
è 3x - 2 ø
(d)
3
3
( f  g )( x ) = ( x + 2) è f ( g ( x )) = ( x + 2)
2
è g ( x) =
1 - 9 ( x + 2)
2
3( x + 2) - 2
2
æ 1- 9y
ö
è g ( y) = ç
+ 1÷ .
è 3y - 2 ø
2
(
)
è f -1 f ( g ( x ) ) = g ( x ) .
Question 13
(a)
lim
x ®4
5 - x2 + 9
20 - x 2 - 2
= lim
x ®4
5 - x2 + 9
×
5 + x2 + 9
20 - x 2 - 2 5 + x 2 + 9
×
20 - x 2 + 2
20 - x 2 + 2
25 - x 2 - 9 20 - x 2 + 2
20 - x 2 + 2 2 + 2 4 2
×
=
lim
=
=
=
x ® 4 20 - x 2 - 4
x ®4
5 + 5 10 5
5 + x2 + 9
5 + x2 + 9
= lim
x
x +10 -27 x
×
x +10
x
-27 ö
-27 ö -27
æ x - 17 ö
æ
æ
=
lim
1
+
=
lim
1
+
(b) lim ç
÷ x ®¥ ç
÷
ç
÷
x ®¥ x + 10
è
ø
è x + 10 ø x ®¥ è x + 10 ø
=e
lim
x ®¥
-27 x
x +10
= e -27
(c) By Chain rule, we have
(
)
d
ln sin ( e3 x )
dx
1
=
× cos ( e3 x ) × ( e3 x ) × (3x)
3x
sin ( e )
f ¢( x) =
3e3 x
=
× cos ( e3 x )
3x
sin ( e )
(d) Let u = 2 + x
Therefore,
2
Þ du = 2 xdx . i.e.
Þ ò x csc2 (2 + x 2 )dx
1
= ò csc2 (u ) × du
2
1
= (- cot(u )) + C
2
1
= - cot(2 + x 2 ) + C
2