Question

Let the function $y=f\left(x\right)$ satisfies the differential equation $x^2\frac{dy}{dx}=y^2{e}^{1/x}(x\ne 0)$ and $\underset{x\to 0^{-}}{lim}f\left(x\right)=1.$ Identify the CORRECT statement(s) ?

• A. Range of $f$ is $(0,1)-\left\{\frac{1}{2}\right\}$
• B. $f\left(x\right)$ is bounded
• C. $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
• D. $\underset{0}{\overset{e}{\int }}f\left(x\right)dx>\underset{0}{\overset{1}{\int }}f\left(x\right)dx$

Answer 1

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

The correct options are
A Range of $f$ is $(0,1)-\left\{\frac{1}{2}\right\}$
B $f\left(x\right)$ is bounded
D $\underset{0}{\overset{e}{\int }}f\left(x\right)dx>\underset{0}{\overset{1}{\int }}f\left(x\right)dx$

$x^2\frac{dy}{dx}=y^2{e}^{\frac{1}{x}}$
$\Rightarrow \frac{dy}{y^2}=\frac{e^{\frac{1}{x}}}{x^2}dx$
Integrating both sides,
$\int \frac{dy}{y^2}=\int \frac{e^{\frac{1}{x}}}{x^2}dx$
$\Rightarrow \frac{-1}{y}=-e^{\frac{1}{x}}+C$
Given, $\underset{x\to 0^{-}}{lim}f\left(x\right)=1$
$\Rightarrow -1=0+C\Rightarrow C=-1$
$\therefore \frac{-1}{y}=-e^{\frac{1}{x}}-1$
$\Rightarrow y=\frac{1}{1+e^{\frac{1}{x}}}$

$\frac{dy}{dx}=\frac{e^{\frac{1}{x}}}{x^2{(1+e^{\frac{1}{x}})}^2}$
$\therefore \frac{dy}{dx}>0\forall x\in R-\left\{0\right\}$
$\underset{x\to \pm \infty }{lim}\frac{1}{1+e^{\frac{1}{x}}}=\frac{1}{2}$
and $\underset{x\to 0^{+}}{lim}\frac{1}{1+e^{\frac{1}{x}}}=0$
$\underset{x\to 0^{-}}{lim}\frac{1}{1+e^{\frac{1}{x}}}=1$

Graph of the function is

Clearly, $f\left(x\right)>0$ for all $x\in D_f$
$\therefore \underset{0}{\overset{e}{\int }}f\left(x\right)dx>\underset{0}{\overset{1}{\int }}f\left(x\right)dx$