Question

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

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

Answer 1

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

$x^2\frac{dy}{dx}=y^2{e}^{\frac{1}{x}}\Rightarrow \frac{dy}{y^2}=\frac{e^{\frac{1}{x}}}{x^2}dx$ integrating both sides,
$\frac{-1}{y}=\int \frac{e^{\frac{1}{x}}}{x^2}dx+C\Rightarrow \text{ putting }\frac{1}{x}=t\Rightarrow \frac{-1}{x^2}dx=dt$
$\frac{-1}{y}=-\int e^{t}dt+C\Rightarrow \frac{-1}{y}=e^{-t}+C\Rightarrow \frac{-1}{y}=-e^{\frac{1}{x}}+C$
$\because \underset{x\to 0^{-}}{\mathrm{Lim}}f(x)=1\Rightarrow -1=0+C\Rightarrow C=-1$
$\frac{-1}{y}=-e^{\frac{1}{x}}-1\Rightarrow y=\frac{1}{1+e^{\frac{1}{x}}}\Rightarrow \frac{dy}{dx}=\frac{1}{{\left(1+e^{\frac{1}{x}}\right)}^2}{e}^{\frac{1}{x}}\left(\frac{-1}{x^2}\right)=\frac{e^{\frac{1}{x}}}{x^2{\left(1+e^{\frac{1}{x}}\right)}^2}$

$\therefore \frac{dy}{dx}>0\forall x\in R-\{0\}$
$\underset{x\to \pm \infty }{\mathrm{Lim}}\frac{1}{1+e^{\frac{1}{x}}}=\frac{1}{2}$
$\text{ and }\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{1}{1+e^{\frac{1}{x}}}=0$
$\therefore$ Graph of the function is
From the graph options (A), (B) and (D) are correct.