Question

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

• A. $3$
• B. $\sqrt{3}$
• C. $1/3$
• D. None of these

Answer 1

The correct option is A $3$

${\int }_{\sin x}^1{t}^{2}f\left(t\right)dt=1-\sin x\forall xϵ(0,\pi /2)$
Differentiating both sides we get
$\frac{d}{dx}\left(1\right)[1\cdot f(1\left)\right]-\cos x\left({\sin }^{2}x\right)f\left(\sin x\right)=-\cos x$
$\Rightarrow f\left(\sin x\right)=\frac{1}{{\sin }^{2}x}$
$\therefore f\left(\frac{1}{\sqrt{3}}\right)=f\left(\sin \left({\sin }^{-1}\frac{1}{\sqrt{3}}\right)\right)$
$={\left[\frac{1}{\sin \left({\sin }^{-1}\frac{1}{\sqrt{3}}\right)}\right]}^2=3$