Chapter 2 Equation Inequalities and Absolute Values

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

Q1. 2003/2004 (b) 2x 1  1 2x 1  1
By substituting a  3x , solve the equation x2 x2
9x  3  28(3x1) . 2x 1  1 or 2x 1 1 0
x2 x2
9x  3  28(3x1) 2x 1 1  0 2x 1 (x  2)  0
x2
32x  3  28(3x ) 2x 1 (x  2)  0 x2
3 x2 3x 1  0
x3 0 x2
Let U  3x x2

U 2  3  28U ( multiply 3 both sides) x  2  x  3  2  x  1
3 Final sulotion 3

3U 2  28U  9  0

3U 1U  9  0

U 1 @U 9
3

3x  31 @ 3x  32
x  1 x2

Q2. 2003/2004 Interval form : (,2)  (2, 1)  (3,)
Solve the following inequalities 3

(a) x 2  x  12  0 Q3. 2003/2004
(b) 2x 1  1 Solve 3ln 2x  3  ln 27

x2

3ln 2x  3  ln 27

(a) x 2  x  12  0 3ln 2x  3  ln 33

(x  3)(x  4)  0 3ln 2x  3  3ln 3
4 3
ln 2x  1 ln 3

x-3   + ln 2x  ln 3  1

x+4 ++ + ln 2x   1
(x – 3)(x + 4)  + 3

Solution : (,4)  (3, ) : 2x  e1  x  3e
Edited by limhweecheng 3 2

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

Q4. 2004/2005
Solve x5e3ln x  4x  21 .

x5e 3ln x  4x  21 Solution : (0,)
x e5 lnx3  4x  21
x5  x 3  4x  21 (b) 1 log 2 x  6log x 2  0
x2  4x  21  0
(x  7)(x  3)  0 1 log 2 x  6 log 2 2   0
x  7 or x  3 log 2 x

Q5. 2004/2005 1  log 2 x   6 x   0
Solve the following inequalities: log 2
(a) 4x  9  12
Let y  log 2 x
x 1 y  6  0
(b) 1 log 2 x  6log x 2  0
(c) x  5  1 y

2x  4 y2  y 6  0
y
(a) 4x  9  12
x ( y  2)( y  3)  0
y
4x  9 12  0
x 3 y 0 or y2
 3  log 2 x  0 log 2 x  2
4x2 12x  9  0 23  x  20 x  22
x 1  x 1 x4
8
(2x  3)2  0
x

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

c) x  5  1 (x2  6)(x2  2)  0
2x  4

x2  6 @ x2  2
x 6
x5 1 or x  5  1
2x  4 2x  4
x 5 1 0
x 5 1 0 2x  4 Q7. 2005/2006
2x  4 3x 1  0
2x  4 Solve the following inequalities
9x 0
2x  4 a) 7x 2  x  6  x 2  4
b) x  3  3

x 1

(2,9] [ 1 ,2) a) 7x 2  x  6  x 2  4
Final Solution : 3
6x2  x  2  0
(2x 1)(3x  2)  0

Solution :  2 , 1
3 2 
[ 1 ,2)  (2,9]
3 x3 3
b) x  1
Q6. 2005/2006
Find the values of x satisfying the equation x  3  3 or x  3  3
log 4 (x4  4)  1 log 4 (x4  4) x 1 x 1

log 4 (x4  4)  log 4 4  log 4 (x2  4) x 3 3  0 x330
log 4 (x4  4)  log 4 4(x2  4) x 1 x 1
 x4  4  4(x2  4)
x  3  3(x 1)  0 x  3  3(x 1)  0
x4  4x2 12  0 x 1 x 1

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

6  2x  0  4x  0 Q9. 2006/2007
x 1 x 1 Obtain the solution set for 2x 1  x2  4

 2x 1  x2  4 or 2x 1    x2  4

(1,3) (0,1) x2  2x 3 0 x2  2x 5  0

(x  3)(x 1)  0 x   (2)  (2)2  4(1)(5)
2(1)

 1.45 @ 3.45

Final solution = (0,1)  (1,3)

x  3 x 1 x  1.45  x  3.45
Final Solution set:
Q8.

By substituting a  2x , solve the equation
4x  3  2x2 .

4x  3  2x2 {x : x  1.45  x  1}

   2x 2  3  2x 22 Q10. 2006/2007
   2x 2  3  4 2x (a) Find the solution et of the inequality

a2  4a  3  0 2 1 1
3 2x x  4
(a 1)(a  3)  0
a 1 or a  3 (b) Solve the following inequalities equation for all
x is real numbers. Write your answer in set
2x 1 or 2x  3x form .
2x  20 x  log 2 3 4  3 2x 1
x0 x  log 3 1 x

log 2

x  1.59

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(a) 1 1 Final solution:
3 2x x  4
Set solution : {x : x  6  x  0}
1  1 0 Q11. 2007/2008
3 2x x  4 Given that 81y  3(2y3)x and 218y6x  64xy .
(x  4)  (3  2x)  0 Find the values of x and y.
(3  2x)(x  4)

3x 1  0
(3  2x)(x  4)

Solution set : {x : 4  x   1  x  32} 81y  3(2y3)x
3
34 y  3(2 y3)x
4 y  (2 y  3)x

(b) 4  3  2x  1 x  4 y      (1)
1 x 2y 3

3 2x  3 218y6x  64xy
1 x

3 2x  3 or 3  2x  3 218y6x  26xy
1 x 1 x 18 y  6x  6xy
3  2x  3(1 x)  0 3y  x  xy
3  2x  3(1 x)  0
1 x 1 x 3y  x( y 1)

 5x  0  6x 0 x  3y      (2)
1 x 1 x y 1

x  1 x  0 x  6  x  1 (1) = (2)
4y  3y

2y 3 y 1
4y2  4y  6y2 9y

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

2 y25y  0 x  2x 1  0
y(2 y  5)  0
(x  4)(2x 1)
y0 @ y5
2

From (1), y  0, x  0

y 5, x5
2

Q13. 2007/2008 (4,1]  (1 ,2]
Solve the following inequalities: 2
(a) x  1
(b) x  2
x  4 2x 1 x4
(b) x  2
x  2 and x  2
x4 x4 x4

(a) x  1 x 20 x 20
x  4 2x 1 x4 x4
x  1 0 x8 0  3x  8  0
x  4 2x 1
x4 x4
x(2x 1) 1(x  4)  0
(x  4)(2x 1)
2x2  2x  4  0
(x  4)(2x 1)

2x  2x 1  0

(x  4)(2x 1)

Final solution: {x : x  8  x   8}
3

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

Q13. 2008/2009 x  2  10  x2

Solve the equation 3log x 3  log 3 3 x  10 x  2  10  x2 or x  2  10  x2
3 x  2  10  x2 x2  x 12  0
x2  x 8  0 (x  3)(x  4)  0
3log x 3  log 3 3 x  10
3

3 log x 3  1 log 3 x  10 x  1 1 4(1)(8)
3 3 2

3 log 3 3   1 log 3 x  10 x  2.37,3.37
log 3 x 3 3

3  log 3 x  10
log 3 x 3 3

Let U  log 3 x

3  U  10 multiply 3U both sides
U3 3

9  U 2  10U

U 2 10U  9  0 Final solution: (,3)  (2.37,) interval form

U  9U 1  0

U 9 @ U 1 Q15. 2009/2010

log 3 x  9  x  39  Solve the equation 32x 10 3x1 1  0
log 3 x  1  x  3
 3x2 10 3x  1  0
 3 Let u  3x

Q14. 2008/2009 u2  10 u 1  0
Determine the interval of x satisfying the 3
inequality x  2  10  x2
3u 2 10u  3  0

3u 1u 3  0

u1 @ 3
3

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

3x  31 3x  3 Q18. 2010/2011
 x  1  x 1
Solve the equation ln x  3  2
Q16. 2009/2010 ln x

Determine the solution set for 2x  3  5 ln x  3  2
x ln x

2x  3  5 Let u  ln x
x
u  3  2
2x2  5x  3  0 u
x
u2  2u  3  0
2x  3x 1  0
u 1u  3  0
x
u  1 or u  3

u  1, u  3

 ln x  1 or  ln x  3

xe  x  e3

Q19. 2010/2011

Solution set : {x : x  0 1  x  3} Solve the following inequalities
2
(a) 3x2  x  4  0 (b) x 1  2
Q17. 2009/2010 2x2  3x  2 x3
Solve 2 5  x  x
(a) 3x2  x  4  0
25  x 2   x 2 2x2  3x  2
 4 25 10x  x2  x2
3x  4x 1  0
3x2  40x 100  0 2x 1x  2

3x 10x 10  0

 ,10   10 , Solution :   , 4     1 ,1  2,

3   3  2 
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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(b) x 1  2 Q21. 2011/2012
x3 (a) Solve the equation

x 1  2 and x 1  2 logx  4  2 log 3  1  log x 
x3 x3
2
x 1  2  0 x 1  2  0 (b) find solution set for x  3  2
x3 x3
3x  5  0 x 1
x3
x7 0  (a)
x3

log x  4 2log 3  1 log x 

2

log x  4 log 32  log x   1

2


 9x  4
log   1
 x 

2

9x  4  10

x

Final solution:  ,7  5 ,   2
3  9x  36  10x  5x

2
4x  36

Q20. 2011/2012 x 9
Solve 32x1  28(3x )  9  0

 32x1  28 3x  9  0 x3 2
(b) x  1

   3x 2  3  28 3x  9  0 x3 2 and x  3  2
x 1 x 1

Let m = 3x x 3  2  0
x 1
3m2  28m  9  0 x 3 2  0 x  3  2x  2  0
x 1
m 93m 1  0 x  3  2x  2  0 x 1
3x 1  0
m  9 or m1 x 1 x 1
3x  9  32 3 x5 0
x 1
3x  3-1

x  2 x  1

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

1 1
2x 1 x  2

x  2  2x  1  0
2x 1x  2

 x3 2  0

2x  1x 

Solution set: {x : x  5  x  1} Solution set : {x : 2  x  1  x  3}
3 2

Q22. 2012/2013 Q24. 2012/2013
Find the value of x which satisfies the equation

log 25  x log 2x  2  3 log 21 x

log 25  x log 2x  2  3 log 21 x

log  5  x1 x  3 (a) Solve x2  x  3  3
 x  2
2  


5  x1  x  23 (b) Find the solution set of the inequality
x  2 2x2  9x  4  4
x2
x2  4x  21  0

(x  7)(x  3)  0

x  3 @ x  7 (a) x2  x  3  3

x 3 x2  x  3  3 or x2  x  3  3

Q23. 2012/2013 x2  x  0 x2  x  6  0

x  0,1 x  2,3

Determine the solution set of the inequality  x   2,0,1,3
1 1

2x 1 x  2

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(b) 2x2  9x  4  4 U2U 1  0
x2
U 0 @ U1
2x2  5x  12  0 2  1
2
x  2 log 3 x  0 @ log 3 x

2x  3x  4  0 x  30 1
x  2 x  1, x  1
x  32

x 3

Q26. 2013/2014
Solve the equation 22x2  2x1  2x  23

22x2  2x1  2x  23

Solution set : {x : x  4  2  x  3}  2x 2  2x 2  2x 8
2
22

Q25. 2013/2014 Let U  2x

Find the value of x which satisfies the equation U 2  2U  U  8
4
log 9 x  log 3 x2 , x >1
U 2  3U  8  0
log 9 x  log 3 x2 4

U 2 12U  32  0

U  4U  8  0

log 3 x  log 3 x2 U 4 @ U 8
log 3 9
2x  4 2x  8

log 3 x log 3 x2 2x  22 @ 2x  23
log 3 32
 x2 x3

log 3 x  log 3 x2
2
Q27. 2013/2014
Let U  log 3 x Find the solution set of 2  3x  x  3

U U2 2  3x  x  3
2
2  3x2  x  32

2U 2 U  0

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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

Squaring both sides Q29. 2014/2015

2  3x2  x  32 Solve the inequality 1  1
6 x x 1
4 12x  9x2  x2  6x  9
8x2 18x  5  0 1 1
6 x x 1
2x  54x 1  0 1  1 0
6 x x 1
Solution set : {x :  1  x  5} (x 1)  (6  x)  0
42 (6  x)(x 1)
2x  7  0
(6  x)(x 1)

Q28. 2014/2015

Solve the equation 3y  31 y  12

3y  31 y  12

3y  3  12 {x :1  x  7  x  6} set solution
3y 2

Let U  3y , 1, 7   6, Interval form

U  3  12  2
U
Q30. 2014/2015
U 212U  27  0
(a) Solve the following equation
U  3U  9  0 6x2  x 11  4

U 3 @ U 9 (b) Find the solution set for the ineqaulities
3y  9 2   x  2   5
3y  3  x  4
y 1 @ 3y  32
y2 (a) 6x2  x 11  4

6x2  x 11  4 or 6x2  x 11  4

6x2  x 15  0 6x2  x  7  0

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(2x  3)(3x  5)  0 (6x  7)(x  1)  0 y log 4  y log 3  2 log 4
y  2 log 4  9.638 ( 3 dp)
x  3 or x   5 x   7 or x  1
23 6 log 4  log 3

The set solution x  { 5 ,  7 ,1, 3} Q32. 2015/2016
36 2
Solve the equation 2  log 2 x  15 log x 2

(b) 2   x  2   5 2  log 2 x  15 log x 2
 x  4

 x  2  2  log 2 x  15 log 2 2 
 x  4  log 2 x
 3  0

4x 10  0 2  log 2 x  15 x
x4 log 2

Let U  log 2 x

2  U  15
U

U 2  2U 15  0

U  3U  5  0

U  5 @ U  3

The solution set: {x : x  5  x  4} log 2 x  5 log 2 x  3
2
x  25 @ x  23

Q31. 2015/2016 x 1 x8
32
4y2 1
Evaluate the solution of  3 y up to

three decimal places. Q33. 2015/2016

4y2  1 (a) Solve the inequalities x  1  2
3 y x3

(b) Show that . Hence, find the
interval for x so that

4y2  3y

 2x  42x
8x
log 4 y2  log 3y  13 2x  36  0

( y  2) log 4  y log 3
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Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(a) x  1  2 Thus, 2x  4 or 2x  9
x3
x 1  2 or x 1  2 2x 22 or x  log 2 9
x3 x3 x2 x  3.17
x 1  2  0
x 1  2  0 x3 The interval of the solution is
x3 (,2]  [3.17,)

x 1 2(x  3)  0 x 1 2(x  3)  0 Q34 : 2016/2017
x3 x3
3x  5  0 Determine the values of x which satisfy the
x7 0  x3 equation 32x1  4(3x )  9 .
x3

32x1  4(3x )  9

 3x 2  4(3x )  9
3
U 2  4U  9
3

U 2 12U  27  0

U  3U  9  0

U 3 @ U 9

Solution :  7,3    3, 5  3x  3 3x  9
x 1 @ 3x  32
 3
x2

(b) 2x  42x  2x  24x  2x4x3x  22x shown Q35. 2016/2017
8x 23x

 2x  42x (a) If 7  3 5  x  y
8x (b) Solve the equation log 2 x  log 4(3x  4)  0
 13 2x  36  0

 22x 13 2x  36  0

(a) 7  3 5  x  y

Let U = 2x

Squaring both sides

U 2 13U  36  0   7  3 5 2 x 2

U  4U  9  0 y

From sign table , we get U  4 or U  9 7  3 5  x  2 xy  y

 x  y 2 xy

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

By comparing, By comparing, x2  (3x  4)
x y7
y  7  x      (1) x2  3x  4  0

2 xy  3 5 x 1x  4  0
4xy  9(5)
x  1(ignore) @ x  4
4xy  45      (2)
Q36. 2016/2017
4x(7  x)  45
28x  4x2  45 (a) Solve the following equation 3  7, x  4
4x2  28x  45  0 x4
(2x  5)(2x  9)  0
(b) Find the solution set for the inequality
x5 @ x9 4x  x4 , x3
22 x3

(a) 3  7
x4

When x  5 , y  7  5  9 3 7 or 3  7
2 22 x4 x4

When x  9 , y  7  9  5 3  7(x  4) 3  7(x  4)
2 22 7x  31 7x  25
x  31 x  25
Since x  y  0
Thus x  9 and y  5 is the solution only. 7 7

22 Thus, x   25 , 31
 
 7 7 

(b) log 2 x  log 4(3x  4)  0 (b)  4  x  x  4
x3
log x  log 2 (3x  4)  0 4x x40
log 2 4 x3
2  4  x  x(x  3)  4(x  3)  0
x3
log x  log 2(3x  4)  0  4  x  x2  3x  4x  12  0
2 x3
2

log x  log 2 (3x  4)
2
2

2log 2 x  log 2(3x  4)

log 2 x2  log 2 (3x  4)  x2  2x  8  0
Edited by limhweecheng x3

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

x2  2x  8  0 Solve the equation 3 log 9 x  log 3 x2
x3
3 log 9 x  log 3 x2
(x  4)(x  2)  0
x3

3 log 3 x   log 3 x 2
log 3 9

3log 3 x  log 3 x2
2

Let U  log 3 x

The solution set is x : x  4  2  x  3 3U  U 2
2

3U  2U 2

Q37. 2017/2018 U (2U  3)  0

 Solve the equation 32x1 16 3x  5  0 U 0 @ U 3
2
 32x1 16 3x  5  0
 32x 3 16 3x  5  0 Therefore,

log 3 x  0 @ log 3 x  3
2
x  30
Let U  3x x 1 3

x3 2

3U 2 16U  5  0 x  5.20

3U 1U 5  0

U 1 @ U 5 Q39. 2017/2018
3 Find the interval of x for which the following
inequalities are true.
3x  1 3x  5
3 (a) 5 1  0
x3
3x  31 @ x  log 3 5
(b) 3x  2  2
x  1 x  log 5  1.46 2x  3
log 3

Q38. 2017/2018

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(a) 5 1  0 Final solution :
x3   8, 3     3 ,  4 
5  (x  3)  0  2  2 7 
x3
2x 0
x3

Solution is (,3)  [2,) interval form

(b) 3x  2  2
2x  3

3x  2  2 or 3x  2  2
2x  3 2x  3

3x  2  2  0 3x  2  2  0
2x  3 2x  3

3x  2  2(2x  3)  0 3x  2  2(2x  3)  0
2x  3 2x  3

x8 0  7x  4  0
2x  3 2x  3

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

(53)8 ((35)4)2 0 32 2
(53)6−2 ((35)3)2
17

=

5 8 −(6−2 ) 5 8−(−6)
(3) = (3)

5 10 −6 5 14 41/ 2018/2019
(3) = (3)
. Solve
Comparing : (a) 6x 1  x  3
(b) Determine the solution set of x which
10 − 6 = 14 satisfies the inequality
= 2 2 x
x 1 x 3

(b) 1  8 (a) √ + − √ =
4 2x x
(√ + = (3 + √ )2
18
4 − 2 − ≥ 0 )
− 8(4 − 2 )
6 + 1 = 9 + 6√ +
(4 − 2 ) ≥ 0
17 − 32 5 − 8 = 6√
(4 − 2 ) ≥ 0
(5 − 8)2 = (6√ )2

25 2 − 80 + 64 = 36

25 2 − 116 + 64 = 0

(25 − 16)( − 4) = 0

= 16 @ = 4

25

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

Checking the answer

When = 16

25

√ + − √ = √ + − √ = ≠ 3
( )

Therefore ≠ 16

25

When = 4,
√ + − √ = √ ( ) + − =

Therefore = 4 is the solution.

(b) 2 x
x 1 x 3
2 42. 2019/2020
+ 1 − + 3 < 0
Solve the following:

2( + 3) − ( + 1)  (a) 3 52x 1 x1
( + 1)( + 3) < 0
 252  200

− 2 + + 6 (b) x  4  x2  x  12
( + 1)( + 3) < 0

2 − − 6
( + 1)( + 3) > 0

( + 2)( − 3)
( + 1)( + 3) > 0

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

43. 2019/2020
Solve the following

(a) log2 2x  2log4(x  4)
(b) 2 x  3  1

2x 1

Edited by limhweecheng

Chapter 2 : Equations, Inequalities & Absolute Mathematic
Values sm015

1 1 < ≪ 7
< 2 2 4

Final Solution : ∪


{ : < ∪ < ≪ }

Edited by limhweecheng