Question

If ${\int }_{\sin x}^1{t}^{2}f\left(t\right)=1,x\in (0,\frac{\pi }{2})$, then $f\left(\frac{1}{\sqrt{3}}\right)$ is equal to

• A. $3$
• B. $\frac{1}{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

The correct option is A $3$

Given ${\int }_{\sin x}^1{t}^{2}f\left(t\right)=1-\sin x$
On differentiating both sides, we get
$-{\sin }^{2}xf\left(\sin x\right)-\cos x=-\cos x$
$\Rightarrow f\left(\sin x\right)=\frac{1}{{\sin }^{2}x}$
$\Rightarrow f\left(x\right)=\frac{1}{x^2}$
Therefore, $f\left(\frac{1}{\sqrt{3}}\right)=3$