Solution set of inequality {log _{10}}({x^2} | vedclass

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Question

(c) ${\log _{10}}({x^2} - 2x - 2) \le 0$&hellip;..$(i)$

For logarithm to be defined,

${x^2} - 2x - 2 > 0$

$ \Rightarrow $ ${(x - 1)^2} &gt

Vedclass · Accepted Answer

$[ - 1,\,1 - \sqrt 3 ) \cup (1 + \sqrt 3 \,,\,3]$

Vedclass · Answer

(c) ${\log _{10}}({x^2} - 2x - 2) \le 0$…..$(i)$ For logarithm to be defined, ${x^2} - 2x - 2 > 0$ $ \Rightarrow $ ${(x - 1)^2} > 3$ ==> $x - 1 < - \sqrt 3 $ or$x - 1 > \sqrt 3 $ ==> $x < 1 - \sqrt 3 $ or $x > 1 + \sqrt 3 $ i.e., $x < - (\sqrt 3 - 1)$ or $x > (\sqrt 3 + 1)$ Now from $(i),$ ${x^2} - 2x - 2 \le 1$ ==> ${x^2} - 2x - 3 \le 0$ ==> $(x - 3)\,(x + 1) \le 0$ $ \Rightarrow $ $ - 1 \le x \le 3$ $ herefore $ $x \in [ - 1,\, - (\sqrt 3 - 1)\,[\, \cup \,]\,\,\sqrt 3 + 1,\,3]$. i.e., $x \in [ - 1,\,1 - \sqrt 3 )\,\, \cup (1 + \sqrt 3 ,\,3)$.

Solution set of inequality ${\log _{10}}({x^2} - 2x - 2) \le 0$ is

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