If the function f(x)= (1/x ) 2 x ; x 0 attains the maximum value at x=frac 1 e then :

Mathematics · Application of Derivatives

JEE Main 2024 — 6 April, Shift 2 — Question 10

If the function $f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$ attains the maximum value at $x=\frac{1}{\mathrm{e}}$ then :

Let $y=\left(\frac{1}{x}\right)^{2 x}$

$\ln y=2 x \ln \left(\frac{1}{x}\right)$

$\ell$ ny $=-2 x \ell n x$

$\frac{1}{y} \frac{d y}{d x}=-2(1+\ln x)$

for $\mathrm{x}>\frac{1}{\mathrm{e}} \mathrm{f}^{\mathrm{n}}$ is decreasing

so, $\mathrm{e}<\pi$

$\left(\frac{1}{\mathrm{e}}\right)^{2 \mathrm{e}}>\left(\frac{1}{\pi}\right)^{2 \pi}$

$\mathrm{e}^{\pi}>\pi^{\mathrm{e}}$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.