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Question Number 121002 by Ar Brandon last updated on 04/Nov/20

Answered by MJS_new last updated on 04/Nov/20

Commented by Ar Brandon last updated on 04/Nov/20

Alright Sir. Thanks for your attention

Commented by MJS_new last updated on 04/Nov/20

$$\mathrm{I}\mathrm{simply}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{it}$$

Answered by 675480065 last updated on 04/Nov/20

$$\mid {\log }_2\left({\mathrm{x}}^2/2\right)\mid ⩽1$$$$\Rightarrow -1⩽{\log }_2\left({\mathrm{x}}^2/2\right)⩽1$$$$\mathrm{Domain}:\mathrm{x}\ne 0$$$$\Rightarrow \frac{1}{2}⩽{\mathrm{x}}^2/2⩽2$$$$\Rightarrow 1⩽{\mathrm{x}}^2⩽4$$$$\mathrm{Case}1:{\mathrm{x}}^2⩽4\Rightarrow -2⩽\mathrm{x}⩽2$$$$\mathrm{Case}2:{\mathrm{x}}^2⩾1\Rightarrow \mathrm{x}⩽-1\cup \mathrm{x}⩾1$$$$\mathrm{Hence}$$$$-2⩽\mathrm{x}⩽-1\cup 1⩽\mathrm{x}⩽2$$$$\mathrm{Ans}:\mathrm{B}$$

Commented by MJS_new last updated on 04/Nov/20

$$\mathrm{good},\mathrm{I}\mathrm{just}\mathrm{was}\mathrm{about}\mathrm{to}\mathrm{post}\mathrm{the}\mathrm{same}...$$

Commented by Ar Brandon last updated on 04/Nov/20

Thanks

Answered by mathmax by abdo last updated on 04/Nov/20

$$\mid {\log }_2\left(\frac{{\mathrm{x}}^2}{2}\right)\mid ⩽1\iff -1⩽{\log }_2\left(\frac{{\mathrm{x}}^2}{2}\right)⩽1\Rightarrow -1⩽\frac{\ln \left(\frac{{\mathrm{x}}^2}{2}\right)}{\ln 2}⩽1\Rightarrow$$$$-\ln 2⩽\ln \left(\frac{{\mathrm{x}}^2}{2}\right)⩽\ln 2\Rightarrow \frac{1}{2}⩽\frac{{\mathrm{x}}^2}{2}⩽2\Rightarrow 1⩽{\mathrm{x}}^2⩽4\Rightarrow 1⩽\sqrt{{\mathrm{x}}^2}⩽2\Rightarrow$$$$1⩽\mid \mathrm{x}\mid ⩽2\Rightarrow \mathrm{x}\in [-2,-1]\cup [1,2]$$

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