Question

Let function $y=f(x)$ satisfies the differential equation $x^2 \frac{d y}{d x}=y^2 e^{\frac{1}{x}}(x \neq 0)$ and $\underset{x \rightarrow 0^{-}}{\text{Lim}} f(x)=1$. Identify the correct statement(s)?

• A. Range of $f(x)$ is $(0,1)-\left\{\frac{1}{2}\right\} $
• B. $ f(x)$ is bounded
• C. $\underset{x \rightarrow 0^{-}}{\text{Lim}} f(x)=1$
• D. $\int\limits_0^e f(x) d x >\int\limits_0^1 f(x) d x$

Answer 1

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

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

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