Question

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

• A. $1$
• B. $3$
• C. $6$
• D. All of these

Answer 1

Answer: (B)

$3$

Since ${\int }_{\sin x}^1{t}^{2}f\left(t\right)dt=1-\sin x,$ thus to find $f\left(x\right)$
On differentiating both sides using Newton Leibnitz formula
i.e, $\frac{d}{dx}{\int }_{\sin x}^1{t}^{2}f\left(t\right)dt=\frac{d}{dx}(1-\sin x)$
$\Rightarrow \left\{1^{2}f\left(1\right)\right\}.\left(0\right)-\left({\sin }^{2}x\right).f\left(\sin x\right).\cos x=-\cos x$
$\Rightarrow f\left(\sin x\right)=\frac{1}{{\sin }^{2}x}$
For $f\left(\frac{1}{\sqrt{3}}\right)$ is obtained when $\sin x=\frac{1}{\sqrt{3}}$
i.e, $f\left(\frac{1}{\sqrt{3}}\right)={\left(\sqrt{3}\right)}^2=3$