First slide

Introduction to linear inequalities

Question

Solution set of $\log x^{2} (\frac{x}{| x |} - x) \geq 0,$ is

Moderate

A

$(- \infty, 0) \cup (1, 2)$
B

$(- \infty, 1) \cup (2, \infty)$
C

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

$(- \infty, - 2] \cup (0, 1)$

Solution

The given expression is $\log x^{2} (\frac{x}{| x |} - x) \geq 0$
Case 1: If $x^{2} > 1,$ then $\frac{x}{| x |} - x \geq 1$

$\Rightarrow$ $1 - x \geq 1, (x > 1) \Rightarrow x \leq 0, x > 1$ (not valid)

or $x < - 1, - 1 - x \geq 1$ $\Rightarrow x < - 1, x \leq - 2$

$∴$     $x \in (- \infty, - 2]$

Case II: If $0 < x^{2} < 1,$ then $0 < \frac{x}{| x |} \leq 1$

$∴$     for $- 1 < x < 0, 0 < - 1 - x \leq 1$

i.e.,       $- 1 < x < 0, - 2 \leq x < - 1$

Hence from ( I ) and ( II ),

we get the solution set $(- \infty, - 2] \cup (0, 1)$

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