question_answer

A function \[y\text{ }=\text{ }f(x)\] satisfies the condition \[f'(x)\,\,sin\,x+f\,(x)\,\,\cos \,x=1,\,\,f(x)\] being bounded when \[x\to 0.\] 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) \[0<I<1\]

Correct Answer: A

Solution :

[A]

Since, \[\sin x\frac{dy}{dx}+y\cos x=1\]

\[\frac{dy}{dx}+y\cot x=\text{cosec}\,x\]

IF \[(e=Ns\,\,B\,\,\omega \,\,\sin \,\,\omega t),\] \[y\sin x=\int{\text{cosec}\,x.\,\sin x\,\,dx}\]

\[y\sin x=x+C\]

If \[x\to 0,\] y is finite \[\therefore C=O\]

\[\therefore y=\frac{x}{\sin x}\]

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

Hence, \[\frac{\pi }{2}<I<\frac{{{\pi }^{2}}}{4}\]