Question

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

Answer: (A)

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

$\text{ I.F. }=e^{\int \cot xdx}=e^{\ln (\sin x)}=\sin x$
$y\sin x=\int \mathrm{cosec}x\cdot \sin xdx$
$y\sin x=x+C$
$\text{ if }x=0,y\text{ is finite }$
$\therefore CC=0$
$y=x(\mathrm{cosec}x)=\frac{x}{\sin x}$
$\text{ Now }I<\frac{{\pi }^2}{4}\text{ and }I>\frac{\pi }{2}$
$\text{ Hence }\frac{\pi }{2}<I<\frac{{\pi }^2}{4}\Rightarrow (A)$