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The domain of the definition of the function $f\left( x \right) = \frac{1}{{4 - {x^2}}} + \log \,\left( {{x^3} - x} \right)$ is
$\left( {1,2} \right) \cup \left( {2,\infty } \right)$
$\left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {3,\infty } \right)$
$\left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right)$
$\left( { - 2, - 1} \right) \cup \left( { - 1,0} \right) \cup \left( {2,\infty } \right)$
Solution
$f\left( x \right) = \frac{1}{{4 – {x^2}}} + {\log _{10}}\left( {{x^3} – x} \right)$
Let ${f_1} = \frac{1}{{4 – {x^2}}}$ and ${f_2} = {\log _{10}}\left( {{x^3} – x} \right)$
$ \Rightarrow 4 – {x^2} \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} – x > 0$
$ \Rightarrow x \ne \pm 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x\left( {x + 1} \right)\left( {x + 1} \right) > 0$
$x \in \left( { – 1,0} \right) \cup \left( {1,\infty } \right) – \left\{ 2 \right\}$
$x \in \left( { – 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right)$
