Question

If ${\int }_{sinx}^1{t}^2.f\left(t\right)dt=1-sinx,\forall xϵ(0,\frac{\pi }{2})$ then the value of $f\left(\frac{1}{\sqrt{3}}\right)$ is

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

Answer 1

The correct option is D $3$

${\int }_{sinx}^1{t}^2.f\left(t\right)dt=1-sinx$
Differentiating both sides with respect to ‘x’
$0-si{n}^{2}x.f\left(sinx\right).cosx=-cosx\Rightarrow cosx[1-si{n}^{2}x.f(sinx\left)\right]=0Butcosx\ne 0$
so, $f\left(sinx\right)=\frac{1}{si{n}^{2}x}f\left(\frac{1}{\sqrt{3}}\right)=3$