Question

A function y=f(x) satisfying the differential equation $\frac{dy}{dx}.sinx-ycosx+\frac{si{n}^{2}x}{x^2}=0$ is such that, $y\to 0$ as $x\to \infty$ then the statement which is correct is

• A. $li{m}_{x\to 0},f\left(x\right)=1$
• B. ${\int }_0^{\frac{\pi }{2}}$ f(x) dx is less than $\frac{\pi }{2}$
• C. ${\int }_0^{\frac{\pi }{2}}$ f(x)dx is greater than unity
• D. f(x) is an odd function

Answer 1

Answer: (A), (B), (C)

The correct options are
A $li{m}_{x\to 0},f\left(x\right)=1$
B ${\int }_0^{\frac{\pi }{2}}$ f(x) dx is less than $\frac{\pi }{2}$
C ${\int }_0^{\frac{\pi }{2}}$ f(x)dx is greater than unity

$\frac{dy}{dx}\sin x-y\cos x=-\frac{{\sin }^{2}x}{x^2}$

$\Rightarrow \frac{\frac{dy}{dx}\sin x-y\cos x}{{\sin }^{2}x}=\frac{d}{dx}\left(\frac{y}{\sin x}\right)=-\frac{1}{x^2}$

On integrating both sides,

$\frac{y}{\sin x}=\frac{1}{x}+C\Rightarrow y=\frac{\sin x}{x}+C\sin x$ where C is the constant of integration.

So, ${lim}_{x\to \infty }(1+Cx)\frac{\sin x}{x}=0$

But, it is only possible iff $C=0$

ie, $y=f\left(x\right)=\frac{\sin x}{x}$

and ${lim}_{x\to 0}\frac{\sin x}{x}=1$

We know that $cosx<\frac{\sin x}{x}<cos\frac{x}{2}$ when $x\in (0,\pi )$

So, ${\int }_0^{\pi /2}\cos xdx<{\int }_0^{\pi /2}\frac{\sin x}{x}dx<{\int }_0^{\pi /2}\cos \frac{x}{2}$

$\Rightarrow 1<{\int }_0^{\pi /2}\frac{\sin x}{x}<\sqrt{2}<\frac{\pi }{2}$

And as $f\left(x\right)$ is an even function,

Options A, B and C are correct.