Question

A function $y=f\left(x\right)$ satisfies the differential equation $\frac{dy}{dx}\sin x-y\cos x+\frac{{\sin }^{2}x}{x^2}=0$ such that ${lim}_{x\to \infty }f\left(x\right)=0$ then which of the following is not true $\to$

• A. $1<\stackrel{\pi /2}{\underset{0}{\int }}f\left(x\right)dx<\frac{\pi }{2}$
• B. $f\left(x\right)$ is an odd function
• C. ${lim}_{x\to 0}f\left(x\right)=2$
• D. both (B) & (C)

Answer 1

Answer: (A)

$1<\stackrel{\pi /2}{\underset{0}{\int }}f\left(x\right)dx<\frac{\pi }{2}$

Given equation $\frac{\left(dy\right)\sin x-y\left(\cos x\right)dx}{{\sin }^{2}x}+\frac{dx}{x^2}=0$
$\int d\left(y/\sin x\right)+\int \frac{1}{x^2}dx=0$
hence $y/\sin x=\frac{1}{x}+C$
$\Rightarrow y=\frac{\sin x}{x}+C\sin x$
Now $\because {lim}_{x\to \infty }\left(f\right(x\left)\right)=0\Rightarrow c=0$
$\Rightarrow y=\left[\frac{\sin x}{x}\right],{lim}_{x\to 0}\left[\frac{\sin x}{x}\right]=1$
In $(0,\pi /2),y_{min}=\frac{2}{\pi }and{y}_{max}=1$
$\Rightarrow 1<\stackrel{\pi /2}{\underset{0}{\int }}\frac{\sin x}{x}dx<\frac{\pi }{2}$