Question bank/JEE Main/Mathematics

The complete solution set of
$\frac{x^{2} - |x| - 12}{x - 3} \geq 0$ is

A$\ \lbrack - 4,3)$B$\lbrack 4,\infty)$C$\lbrack - 4,3) \cup \lbrack 4,\infty)$D$\lbrack - 4,4) \cup \lbrack 4,\infty)$

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$\frac{x^{2} - |x| - 12}{x - 3} \geq 0$
Case 1: $x < 0$
$\frac{x^{2} + ( - x) - 12}{x - 3} \geq 0$
$\frac{x^{2} + x - 12}{x - 3} \geq 0$
$\frac{(x - 3)\ (x + 4)}{x - 3} \geq 0$
but $x < 0$
$\Rightarrow \ \ x \in \lbrack - 4,0)$
Case 2: $x \geq 0$
$\frac{x^{2} - x - 12}{x - 3} \geq 0$
$\frac{n^{2} - 4n + 3n - 12}{n - 3} \geq 0$
$\left( \frac{n - 4)(n + 3)}{x - 3} \geq 0 \right.\$
but $x \geq 0$
$\Rightarrow$ $x \in \lbrack 0,3) \cup \lbrack 4,\infty)$
Taking union of both the cases.
$x \in \lbrack - 4,3) \cup \lbrack 4,\infty)$

Source: Grass-G / M-05 Mathematics Paper 10 October.docx