The trigonometric equation sin ^-1 x=2 sin ^-1 a has a solution for

$\sin ^{-1} x=2 \sin ^{-1} a$sin−1x=2sin−1a $-\frac{\pi}{2} \leq \sin ^{-1} \mathrm{x} \leq \frac{\pi}{2} ; \therefore-\frac{\pi}{2} \leq 2 \sin ^{-1} \mathrm{a} \leq \frac{\pi}{2} \quad-\frac{\pi}{4} \leq \sin ^{-1} \mathrm{a} \leq \frac{\pi}{4}$−2π​≤sin−1x≤2π​;∴−2π​≤2sin−1a≤2π​−4π​≤sin−1a≤4π​ or $\frac{-1}{\sqrt{2}} \leq \mathrm{a} \leq \frac{1}{\sqrt{2}}$2​−1​≤a≤2​1​ $\therefore|\mathrm{a}| \leq \frac{1}{\sqrt{2}}\left(\right.$∴∣a∣≤2​1​( As $\frac{1}{\sqrt{2}}>\frac{1}{2}$2​1​>21​ ). Out of given four option no one is absolutely correct but (c) could be taken into consideration. $\rightarrow|\mathrm{a}| \leq \frac{1}{\sqrt{2}}$→∣a∣≤2​1​ is correct, if $\mathrm{a} < \frac{1}{\sqrt{2}}$a<2​1​ is taken as correct then it domain satisfy for $\mathrm{a}=\frac{1}{\sqrt{3}}$a=3​1​ but equation is satisfied. $\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>\frac{1}{2}$2​1​>3​1​>21​