Question

Let $y=f(x)$ be a curve passing through $\left(e, e^{e}\right)$, which satisfy the differential equation $\left(2 n y+x y \log _{e} x\right) d x-x \log _{e} x d y=0$ $x > 0, y > 0$. If $g(x)=\lim _{n \rightarrow \infty} f(x)$, then $\int_{1 / e}^{e} g(x) d x$ equals to

• A. e
• B. 1
• C. 0
• D. None of these

Answer 1

Answer: (C)

$\left(2 n y+x y \log _{e} x\right) d x=x \log _{e} x d y$
$\Rightarrow \frac{d y}{y}=\left(\frac{2 n}{x \log _{e} x}+1\right) d x$
$\Rightarrow \log (y)=2 n \log |\log x|+x+c$ and $c=0$
$\left(\because\right.$ curve passes through $\left.\left(e, e^{ e }\right)\right)$
$y=e^{x+\log (\log x)^{2 n}}=e^{x}(\log x)^{2 n}$
$\Rightarrow f(x)=e^{x}(\log x)^{2 n}$
Now, $g(x)=\displaystyle\lim _{n \rightarrow \infty} f(x)=\begin{cases}\rightarrow \infty & \text { if } & x < \frac{1}{e} \\ 0 & \text { if } & \frac{1}{e} < x < e \\ \rightarrow \infty & \text { if } & x > e\end{cases}$
$\therefore \int_{1 / e}^{e} g(x) d x=0$