Question

If $\underset{\sin x}{\overset{1}{\int }}{t}^{2}f\left(t\right)dt=1-\sin x,$ where $x\in (0,\frac{\pi }{2}),$ then the value of $f\left(\frac{1}{\sqrt{3}}\right)$ is

• A. $2$
• B. $0$
• C. $4$
• D. $3$

Answer 1

The correct option is D $3$

$\underset{\sin x}{\overset{1}{\int }}{t}^{2}f\left(t\right)dt=1-\sin x$
Differentiating both sides using Leibnitz theorem, we get:
$1^2\times f\left(1\right)0-{\sin }^{2}xf\left(\sin x\right)\cos x=-\cos x$
$\Rightarrow f\left(\sin x\right)=\frac{1}{{\sin }^{2}x}\therefore f\left(\frac{1}{\sqrt{3}}\right)=3$