Question

If ${\int }_{\sin x}^1{t}^2\cdot f\left(t\right)dt=1-\sin x$

then $f\left(\frac{1}{\sqrt{3}}\right)$ is equal to?

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

Answer 1

Answer: (C)

$3$

${\int }_{sinx}^1{t}^2\cdot f\left(t\right)dt=1-sinx$
Differentiating both sides w.r.t $x$, we get
$-\cos x\times {\sin }^{2}xf\left(\sin x\right)=-\cos x$
$\Rightarrow f\left(\sin x\right)=\frac{1}{{\sin }^{2}x}\Rightarrow f\left(x\right)=\frac{1}{x^2}$
$\therefore f\left(\frac{1}{\sqrt{3}}\right)=3$