Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$$\begin{array}{rl}& {\sin }^{-1}x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]\\ & \Rightarrow -\frac{3\pi }{4}\le \left({\sin }^{-1}x-\frac{\pi }{4}\right)\le \frac{\pi }{4}\\ & 0\le {\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2\le \frac{9}{16}{\pi }^2\end{array}$$ Statement II is true $$\begin{array}{rl}& {\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3=a{\pi }^3\\ \Rightarrow & \left({\sin }^{-1}x+{\cos }^{-1}x\right)[{\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^2-\\ & 3{\sin }^{-1}x{\cos }^{-1}x]=a{\pi }^3\\ \Rightarrow & \frac{{\pi }^2}{4}-3{\sin }^{-1}x{\cos }^{-1}x=2a{\pi }^2\\ \Rightarrow & {\sin }^{-1}x\left(\frac{\pi }{2}-{\sin }^{-1}x\right)=\frac{{\pi }^2}{12}(1-8a)\\ \Rightarrow & {\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2=\frac{{\pi }^2}{12}(8a-1)+\frac{{\pi }^2}{16}\\ \Rightarrow & {\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2=\frac{{\pi }^2}{48}(32a-1)\end{array}$$ Putting this value in equation (1) $$0\le \frac{{\pi }^2}{48}(32a-1)\le \frac{9}{16}{\pi }^2$$ $$\begin{array}{rl}& \Rightarrow 0\le 32a-1\le 27\\ & \frac{1}{32}\le a\le \frac{7}{8}\end{array}$$ Statement-I is also true.