Question

A function $y=f\left(x\right)$ satisfies the differential equation $\frac{dy}{dx}sinx-y/cosx+\frac{si{n}^{2}x}{x^2}=0$ and is such that $y\to 0$ as $x\to \infty$, then

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

Answer 1

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

$li{m}_{x\to 0}f\left(x\right)=1$

From the giiven differential equation,

$\frac{sinx\frac{dy}{dx}-ycosx}{si{n}^{2}x}=-\frac{1}{x^2}$

$\Rightarrow \int d\left(\frac{y}{sinx}\right)=\int \frac{dx}{x^2}$

$\Rightarrow \frac{y}{sinx}=\frac{1}{x}+c$...(i)

But given $y\to 0$ as $x\to \infty$ and $-1sinx\le 1$
Then from (i)
$0=\frac{1}{\infty }+c\Rightarrow c=0$

$\therefore$ from (i) $\frac{y}{sinx}=\frac{1}{x}+0$

$\Rightarrow y=\frac{sinx}{x}$

$\therefore f\left(x\right)=/dfracsinxx$

$\Rightarrow li{m}_{x\to 0}f\left(x\right)=li{m}_{x\to 0\left(\frac{sinx}{x}\right)}=1$

Also, $f\left(x\right)=\frac{sinx}{x}$

$\Rightarrow f(-x)=\frac{sin(-x)}{(-x)}=\frac{-sinx}{-x}=\frac{sinx}{x}=f\left(x\right)$

$\Rightarrow f\left(x\right)$ is an even function but from $x-\frac{x^3}{3!}+\frac{x^5}{5!}...to\infty$

$\Rightarrow$$\sin x<x$

$\Rightarrow sinx<x$

$\Rightarrow dfracsinxx<1$

$\Rightarrow \frac{sinx}{x}<1$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{sinxdx}{x}<{\int }_0^{\frac{\pi }{2}}dx$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{sinxdx}{x}<\frac{\pi }{2}$

and so, $\frac{\pi }{2}<{\int }_0^{\frac{\pi }{2}}\frac{sinxdx}{x}<\pi$