• # question_answer If $\int\limits_{\sin x}^{l}{{{t}^{2}}f(t)dt]=1-\sin x,x\in \left( 0,\frac{\pi }{2} \right)}$then $f\left( \frac{1}{\sqrt{2}} \right)$ is equal to A)  1                     B)  2 C)  3                                 D)  4

On differentiating both sides with respect to x, we get $0-{{\sin }^{2}}\,x.f(\sin x)\,\cos x=-\cos x\,$ $\Rightarrow \,\,f(\sin x)=\frac{1}{{{\sin }^{2}}x}$ $\therefore \,\,f\left( \frac{1}{\sqrt{2}} \right)\,={{(\sqrt{2})}^{2}}=2$