Question

A function $y=f\left(x\right)$ satisfies the condition $f\left(x\right)\sin x+f\left(x\right)\cos x=1$, $f\left(x\right)$ being bounded when $x\to 0$. If $I={\int }_0^{\pi /2}f\left(x\right)dx$ then

• A. $\frac{\pi }{2}<I<\frac{{\pi }^2}{4}$
• B. $\frac{\pi }{4}<I<\frac{{\pi }^2}{2}$
• C. $1<I<\frac{\pi }{2}$
• D. $0<I<\frac{\pi }{2}$

Answer 1

The correct option is C $1<I<\frac{\pi }{2}$

$\sin x\frac{dy}{dx}+y\cos x=1$

$\frac{dy}{dx}+y\cot x=\csc x$

IF $=e^{\int \cot xdx}=e^{\ln \left(\sin x\right)}=\sin x$

$y\sin x=\int \csc x.\sin xdx$

$y\sin x=x+C$

If $x=0,y$ is finite $\therefore C=0$

$y=x\left(\csc x\right)=\frac{x}{\sin x}$

Now, $I<\frac{{\pi }^2}{4}$ and $I>\frac{\pi }{2}$

hence, $\frac{\pi }{2}<I<\frac{{\pi }^2}{4}$