The domain of the real-valued function $f (x) = {log}_{5 + x} (x^{2} - 2) + {[{log}_{10} (\frac{x^{2} - x}{2})]}_{10}^{1 / 2}$ is

(- \infty, - 4) \cup (- 4, - 1] \cup [2, \infty)

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(- 5, - 4) \cup (- 4, - \sqrt{2}) \cup [2, \infty)

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(- 5, - 4) \cup (- 4, - \sqrt{2}) \cup (\sqrt{2}, \infty)

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(- 5, - 4) \cup (- 4, - 1] \cup [2, \infty)

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Solution

The correct option is B $(- 5, - 4) \cup (- 4, - \sqrt{2}) \cup [2, \infty)$
For ${log}_{5 + x} (x^{2} - 2)$ ,
$5 + x > 0 and 5 + x \neq 1 \Rightarrow x \in (- 5, - 4) \cup (- 4, \infty)$
Also, $(x^{2} - 2) > 0$
$\Rightarrow x^{2} > 2 \Rightarrow x \in (- 5, - 4) \cup (- 4, - \sqrt{2}) \cup (\sqrt{2}, \infty) \dots (1)$

For ${[{log}_{10} (\frac{x^{2} - x}{2})]}_{10}^{1 / 2}$ ,
${log}_{10} (\frac{x^{2} - x}{2}) \geq 0$
$\Rightarrow \frac{x^{2} - x}{2} \geq 1$
$\Rightarrow x^{2} - x \geq 2$
$\Rightarrow x^{2} - x - 2 \geq 0$
$\Rightarrow (x + 1) (x - 2) \geq 0 \Rightarrow x \in (- \infty, - 1] \cup [2, \infty) \dots (2)$

Hence, from $(1)$ and $(2)$ ,
$x \in (- 5, - 4) \cup (- 4, - \sqrt{2}) \cup [2, \infty)$

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