Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

${\sin }^{-1}x=2{\sin }^{-1}a$ $-\frac{\pi }{2}\le {\sin }^{-1}\mathrm{x}\le \frac{\pi }{2};\therefore -\frac{\pi }{2}\le 2{\sin }^{-1}\mathrm{a}\le \frac{\pi }{2}-\frac{\pi }{4}\le {\sin }^{-1}\mathrm{a}\le \frac{\pi }{4}$ or $\frac{-1}{\sqrt{2}}\le \mathrm{a}\le \frac{1}{\sqrt{2}}$ $\therefore |\mathrm{a}|\le \frac{1}{\sqrt{2}}($ As $\frac{1}{\sqrt{2}}>\frac{1}{2}$ ). Out of given four option no one is absolutely correct but (c) could be taken into consideration. $\to |\mathrm{a}|\le \frac{1}{\sqrt{2}}$ is correct, if $\mathrm{a}<\frac{1}{\sqrt{2}}$ is taken as correct then it domain satisfy for $\mathrm{a}=\frac{1}{\sqrt{3}}$ but equation is satisfied. $\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>\frac{1}{2}$