Question

If $\underset{\sin x}{\overset{1}{\int }}{t}^{2}f(t)dt=1-\sin x,x\in \left(0,\frac{\pi }{2}\right)$ then $f\left(\frac{1}{\sqrt{2}}\right)$ is equal to

Answer 1

Answer: 2

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