Question

A function $y=f(x)$ satisfies the condition $f^{\prime}(x) \sin x+f(x) \cos x=1, f(x)$ being bounded when $x \rightarrow 0$. If $I =\int\limits_0^{\pi / 2} f( x ) dx$ 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)

$\sin x \frac{d y}{d x}+y \cos x=1$
$\frac{d y}{d x}+y \cot x=\operatorname{cosec} x$

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