Question

If $\int_{sin x}^1 \, t^2 f(t) dt =1-sin x, \forall x \in (0,\pi/2), \, then \, f\bigg(\frac{1}{\sqrt 3}\bigg)$ is

• A. 3
• B. $\sqrt 3$
• C. 44564
• D. None of these

Answer 1

Answer: (A)

3

Answer 2

Since $\int_{sin x}^1 \, t^2 f(t) dt =1-sin x$ thus to find f(x)
On differentiating both sides using Newton Leibnitz
formula
i.e. \hspace20mm \frac{d}{dx} \int_{sin x}^1 t^2 f(t) dt =\frac{d}{dx}(1-sin x)
$\Rightarrow \, \, \, \\{ 1^2f(-1)\\}.(0)-(sin^2x).f(sin x).cos x=-cos x$
\Rightarrow \, \, \, \, \, \hspace20mm f(sin x)=\frac{1}{sin^2 x}
For f $\bigg(\frac{1}{\sqrt 3}\bigg)$ is obtained when sin x = $1\sqrt 3$
i.e \hspace20mm f\bigg(\frac{1}{\sqrt 3}\bigg) =(\sqrt 3)^2 =3