QS015
1.
MODEL 3
a) Simplify
2 5
3 2
.
[3 marks]
b) Solve the equation log 2 ( x 2  2)  1  log 2 ( x  5) .
[4 marks]
2.
Find the solution set of the inequalities x  3x  4  6
[5 marks]
3.
Given the complex number z and its conjugate z satisfy the equation z z  2 z i  12  6i .
Find z .
[7 marks]
4.
Solve the equations of 8 x  17  2 x  2 x  9 .
[7 marks]
5.
a)
If the second term and the fifth term of the geometric progression are
1
respectively, find the first term and the common ratio.
32
1
and
4
[5 marks]
1
b)
Expand 1  3x  3 in ascending powers of x , up to and including the term in x 3 .
State the range of values of x for which the expansion is valid. Hence, evaluate
3
6.
1.03 correct to four decimal places.
 4 1 2


Given that A    2 3 1  and B 
 1 3 1


[7 marks]
 0 5 5 


1  6 8 
 3 13  14 


a) Find AB as a single matrix.
b) By using the result in (a), solve the following system linear equations.
4 x  y  2 z  13
 2 x  3 y  z  9
 x  3 y  z  7
[12 marks]
5
QS015
MODEL 3
Final Answers
1.
a)
b)
2 15  2 10
x  4 or -2
2.
1
{ x : 5  x  }
2
3.
z  3  3i , z  3  i
4.
x 8
5.
a)
r
b)
1 x  x2 
a)
b)
5I
x  2 , y  3 , z  4
6.
1
1
, a
2
2
5 3
1
1
x  ..... ,the expansion is valid if   x  , 1.0099 (4 dp)
3
3
3
6
QS015
MODEL 3
Answer Scheme
2 5
1. a)
3 2
=
2 5

3 2
3 2
3 2
2 5 3 2
=
3 2
= 2 5( 3  2)


 2 15  2 10
b) log 2 ( x 2  2)  log 2 ( x  5)  1
 x2  2 
  1
log 2 
x

5


2
x 2
 21
x5
x 2  2x  8  0
x  4x  2  0
x  4 or  2
2. x  3x  4  6
3x  4  6  x
Using basic definition,
3 x  4  6  x AND 3 x  4  (6  x )
4x  2
2 x  10
1
x
x  5
2
5
Solution set
1
2
1
{ x : 5  x  }
2
7
QS015
MODEL 3
3. Let z  x  iy , z  x  iy
z z  2 zi  12  6i
( x  iy )( x  iy )  2i ( x  iy )  12  6i
x 2  y 2 i 2  2 xi  2i 2 y  12  6i
x 2  y 2  2 y  2 xi  12  6i
Comparing real parts
x 2  y 2  2 y  12 -----------(1)
Substitute (2) into (1)
32  y 2  2 y  12
compare imaginary parts
2 x  6 , x  3 ------(2)
y2  2y  3  0
 y  3 y  1  0
y  3 or  1
z  3  3i , z  3  i
4.
8 x  17  2 x  2 x  9
Squaring both sides
8 x  17  2 8 x  17 2 x  2 x  2 x  9

 
8 x  8  2 (8 x  17)(2 x)
4 x  4  16 x 2  34 x
16 x 2  32 x  16  16 x 2  34 x
2 x  16 , x  8 .
5.
1
1
………(1) , T5  ar 4 
……….(2)
4
32
ar 4
1 1
(2)  (1) ,


ar
32 4
1
r3 
8
1
r
2
1
By substituting r  into (1),
2
1
1
a
2
4
1
a
2
a)
T2  ar 
8
QS015
MODEL 3
11 
1  1  1

  1
  1  2 
3
1
3 3 
3x 2  3  3  3  3x   .....
b) 1  3 x   1  (3 x)  
3
2!
3!
5
= 1  x  x 2  x 3  .....
3
The expansion is valid if 3x  1
1
3
x

3
1
3
1
1
x
3
3
1
1.03  1  0.03 3
1
 1  3(0.01)  3
So, we can use the above expansion by substituting x  0.01 .
5
2
3
3
1.03  1  0.01  0.01  0.01  ....
3
= 1  0.01  0.0001  0.00000167
 1.00990167
= 1.0099 ( 4 decimal places)
6.
a)
 4 1 2  0  5 5 



AB    2 3 1    1  6 8 
  1 3 1   3 13  14 



1 0 0
 5 1 0 0 = 5 I
1 0 0
b)
AB  5 I
1
A 1 = B
5
 0 5 5 

1
 1  6 8 
5

 3 13  14 
 4 1 2   x   13 

   
 2 3 1  y =  9
 1 3 1  z   7

   
9
QS015
MODEL 3
AX  B
X  A 1 B
 0  5 5   13 
 
1
= 1  6 8    9
5
 
 3 13  14    7 
 0  45  35 

1
   13  54  56 
5

 39  117  98 
 10 

1
   15 
5

 20 
 2 
 
=   3
 4 
 
x  2 , y  3 , z  4
10