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Question Number 44369 by Necxx last updated on 27/Sep/18
$${The}\:{domain}\:{of}\: \\ $$$${f}\left({x}\right)=\sqrt{{cos}\left({sinx}\right)}\:+\left(\mathrm{1}−{x}\overset{−\mathrm{1}} {\right)}+\:\mathrm{sin}^{−\mathrm{1}} \frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}} \\ $$$${is}.........=. \\ $$
Commented by Necxx last updated on 28/Sep/18
$${please}\:{help} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18
$${sin}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}}\right)\:{is}\:{defined}\:{when}\:\mid\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}}\mid\leqslant\mathrm{1} \\ $$$$\bigstar{x}^{\mathrm{2}} +\mathrm{1}\leqslant\mathrm{2}\mid{x}\mid \\ $$$$\mid{x}\mid^{\mathrm{2}} −\mathrm{2}\mid{x}\mid+\mathrm{1}\leqslant\mathrm{0} \\ $$$$\left(\mid{x}\mid−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{0}\:{but}\:\left(\mid{x}\mid−\mathrm{1}\right)^{\mathrm{2}} \:{can}\:{not}\:{be}\:{less}\:{than} \\ $$$${zero}\:\:{so}\:\left(\mid{x}\mid−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mid{x}\mid=\mathrm{1}\:\:\:\:{x}=\pm\mathrm{1} \\ $$$${so}\:{domain}\:{of}\:{sin}^{−\mathrm{1}} \frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}}\:\:\:\:{x}\epsilon\left\{\pm\mathrm{1}\right\} \\ $$$$\bigstar\bigstar{domain}\:{for}\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:\:\:{x}\epsilon{R}−\left\{\mathrm{1}\right\} \\ $$$$\bigstar\bigstar\bigstar{domain}\:{for}\:\sqrt{{cos}\left({sinx}\right)}\:\:\:\: \\ $$$${from}\:{attachex}\:{graph}\:{of}\:\sqrt{{cos}\left({sinx}\right)}\: \\ $$$${the}\:{domain}\:{x}\epsilon\left(−\infty,\infty\right) \\ $$$${so}\:{required}\:{ans}\:{is}\:\:{x}\epsilon\:\left[−\mathrm{1},\mathrm{1}\right) \\ $$$${the}\:{value}\:{of}\:{x}\:{from}\:−\mathrm{1}\:{to}\:+\mathrm{1}\:{excluding}\:\mathrm{1} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$$\:\: \\ $$$$ \\ $$$$\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18
