First slide

Introduction to sets

Question

Number of integral values of x satisfying the inequation $\frac{x}{x + 2} \leq \frac{1}{| x |} is$

Moderate

A

2
B

8
C

4
D

3

Solution

$\frac{x | x |}{x + 2} \leq 1$
$\frac{x | x | - x - 2}{x + 2} \leq 0$
$\begin{array}{l} Case I: x \in [0, \infty) \\ \begin{array}{c} \frac{x^{2} - x - 2}{x + 2} \leq 0 \\ \Rightarrow \frac{(x - 2) (x + 1)}{x + 2} \leq 0 \end{array} \end{array} \Rightarrow 0 \leq x \leq 2 (∵ x \in [0, \infty))$
i.e., integral values of x are 0, 1 , 2.
$\begin{aligned} Case II: x \in (- \infty, 0) \\ \frac{- x^{2} - x - 2}{x + 2} \leq 0 \end{aligned}$
$\begin{array}{l} \Rightarrow - 2 < x < 0 \\ \Rightarrow x = - 1 \end{array} (∵ x \in (- \infty, 0))$

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