First slide

Introduction to real valued functions

Question

The solution set of the inequality $\frac{| x + 2 | - x}{x} < 2$ is

Moderate

A

(0, 1)
B

[0, 2]
C

$(- \infty, 0) \cup (1, \infty)$
D

none of these

Solution

We have $\frac{| x + 2 | - x}{x} < 2$
$\Rightarrow \frac{| x + 2 | - 3 x}{x} < 0$
Let $| x + 2 | - 3 x = 0$
$\begin{array}{l} \Rightarrow x + 2 = \pm 3 x \\ \Rightarrow x = 1 \end{array}$
Sign scheme of the expression $\frac{| x + 2 | - 3 x}{x}$ is as shown in the
following figure:

From the sign scheme, $x \in (- \infty, 0) \cup (1, \infty)$

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