Question

A function $y=f\left(x\right)$ satisfies the condition $f^{\prime }\left(x\right)sinx+f\left(x\right)cosx=1,f\left(x\right)$ being bounded when $x\to 0$. if $I={\int }_0^{\frac{\pi }{2}}f\left(x\right)dx,$ 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. $0<I<1$

Answer 1

The correct option is A

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

$sinx\frac{dy}{dx}+ycosx=1$
$\frac{dy}{dx}+ycotx=cosecx$

$IF=e^{\int cotxdx}=e^{ln\left(sinx\right)}=sinxysinx=\int cosecxsinxdx=x+cIfx=0,yisfinite\therefore c=0$
$y=x\left(cosecx\right)=\frac{x}{sinx}$
Now $I<\frac{{\pi }^2}{4}andI>\frac{\pi }{2}$
Hence $\frac{\pi }{2}<I<\frac{{\pi }^2}{4}$