Page 37 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

Example 30

1
x
1 If x + 8 < x – 1 , find the set of values of x which satisfy the inequality.

Solution: Given that x < 1
x + 8 x – 1
x – 1 < 0
x + 8 x – 1
x(x – 1) – (x + 8) < 0
(x + 8)(x – 1)
x – 2x – 8 < 0
2
(x + 8)(x – 1)
(x + 2)(x – 4) < 0
(x + 8)(x – 1)

By considering each of the four factors as positive, we get
x + 2 . 0, i.e. x . –2
x – 4 . 0, i.e. x . 4
x + 8 . 0, i.e. x . –8
x – 1 . 0, i.e. x . 1
These ranges can be represented on the real number line as follows.

x + 2 > 0
x – 4 > 0
x
– 8 –2 0 1 4
x – 1 > 0
x + 8 > 0
The numerator (x + 2)(x – 4)  0 in {x : x < –2 or x  4}
The denominator (x + 8)(x – 1) . 0 in {x : x , –8 or x . 1}
Hence, (x + 2)(x – 4) < 0 if the numerator and denominator are of opposite signs,
(x + 8)(x – 1)
i.e. when –8 , x < –2 or 1 , x < 4.
Hence, the set of values of x such that x < 1
x + 8 x – 1
is {x : –8 , x < –2 or 1 , x < 4}.

Inequalities involving modulus signs

Inequalities involving absolute value or modulus signs may be solved using the analytical or graphical method.

Example 31

Find the values of x such that |2x + 1| , 3.

Solution: Given that |2x + 1| , 3
This means –3 , 2x + 1 , 3
–4 , 2x , 2

–2 , x , 1
Hence, the inequality is valid if –2 , x , 1, i.e. x  (–2, 1).

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