Question

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

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

Answer 1

Answer: (C)

$\left(2ny+xy{\log }_{e}x\right)dx=x{\log }_{e}xdy$
$\Rightarrow \frac{dy}{y}=\left(\frac{2n}{x{\log }_{e}x}+1\right)dx$
$\Rightarrow \log (y)=2n\log |\log x|+x+c$ and $c=0$
$(\because$ curve passes through $\left(e,e^e\right))$
$y=e^{x+\log (\log x{)}^{2n}}=e^x(\log x{)}^{2n}$
$\Rightarrow f(x)=e^x(\log x{)}^{2n}$
Now, $g(x)=\underset{n\to \infty }{\lim }f(x)=\{\begin{array}{llc}\to \infty & \text{ if }& x<\frac{1}{e}\\ 0& \text{ if }& \frac{1}{e}<x<e\\ \to \infty & \text{ if }& x>e\end{array}$
$\therefore {\int }_{1/e}^{e}g(x)dx=0$