Page 38 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

Note: Some inequalities involving moduli may be solved by squaring both sides of the inequality. However, it
is only valid if both sides of the inequality are positive or zero for all x  R.
1
Example 32

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

Solution: Given that 2|x – 1| < |x + 3|
2
4(x – 1) < (x + 3) 2 Both LHS and RHS are  0.
Squaring both sides of inequality.
4x – 8x + 4 < x + 6x + 9
2
2
3x – 14x – 5 < 0
2
f(x )
(3x + 1)(x – 5) < 0
2
f(x) = 3x – 14x – 5
From the sketch graph of
f(x) = 3x – 14x – 5,
2
1
we see that f(x) < 0 if – < x < 5, 1 _ 0 5 x
3 – 3
1
3
4
i.e. in the closed interval – , 5 .
3
Example 33
Using graphical method, find the range of values of x for which the inequality |2x – 1| < |x| + 3 is valid.
2
Solution: We first sketch the graphs of y = |2x – 1| and y = |x| + 3 as follows:
2
y
y = –2x + 1
3
3
y = –x + — y = x + — A
2
2
B
3
—
1 2 y = 2x – 1
x
1 O 1 5
– — — —
2 2 2
The graphs of y = |2x – 1| and y = |x| + 3 intersect at A and B.
2
For A, we solve y = 2x – 1 and y = x + 3 , i.e. 2x – 1 = x + 3 ⇒ x = 5 .
2 2 2
1
For B, we solve y = –2x + 1 and y = –x + 3 , i.e. –2x + 1 = –x + 3 ⇒ x = – .
2 2 2
From the graphs, we notice that |2x – 1| < |x| + 3 is valid in the range
2
1
– < x < 5 .
2 2

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