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If $f(x)=\sqrt{\log _{10} x^{2}}$. The set of all values of $x$ for which $f(x)$ is real, is
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Verified Answer
The correct answer is:
$(-\infty,-1] \cup[1, \infty)$
$f(x)=\sqrt{\log _{10} x^{2}}$ is real, if
$$
\begin{array}{l}
\log _{10} x^{2} \geq 0 \\
\Rightarrow x^{2} \geq 1 \\
\Rightarrow x < -1 \text { and } x>1 \\
\Rightarrow x \in(-\infty,-1] \cup[1, \infty)
\end{array}
$$
$$
\begin{array}{l}
\log _{10} x^{2} \geq 0 \\
\Rightarrow x^{2} \geq 1 \\
\Rightarrow x < -1 \text { and } x>1 \\
\Rightarrow x \in(-\infty,-1] \cup[1, \infty)
\end{array}
$$
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