The set of all real x satisfying the inequality 3 - | x | 4 - | x | 0 is

Question

The set of all real $x$ satisfying the inequality $\frac{3 - | x |}{4 - | x |} \geq 0$ is

Smart Achievers · Accepted Answer

(1)

$Case I: When x \geq 0$

$∴$ $| x |$ $= x$

$\Rightarrow$ $\frac{3 - | x |}{4 - | x |} = \frac{3 - x}{4 - x} \geq 0 \Rightarrow \frac{(3 - x) (4 - x)}{(4 - x)^{2}} \geq 0$

$\Rightarrow$ $(x - 3) (x - 4) \geq 0 and x \neq 4 \Rightarrow x \leq 3 or x > 4$

$But x \geq 0, so x \in [0, 3] \cup (4, \infty)$ ...(i)

$Case II: When x < 0$

$∴$ $| x |$ $= - x$

$\Rightarrow$ $\frac{3 - | x |}{4 - | x |} = \frac{3 + x}{4 + x} \geq 0 \Rightarrow \frac{(x + 3) (x + 4)}{{(x + 4)}^{2}} \geq 0$

$\Rightarrow$ $(x + 3) (x + 4) \geq 0 and x \neq - 4$

$\Rightarrow$ $x < - 4 or x \geq - 3$

$But x < 0, so x \in (- \infty, - 4) \cup [- 3, 0)$ ...(ii)

$From (i) and (ii), x lies in the interval$

$(- \infty, - 4) \cup [- 3, 3] \cup (4, \infty)$

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