JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
The local maximum value of the function $f(x)=\left(\frac{2}{x}\right)^{x^{2}}$, $x>0$, is :
Differentiation and its Application
Solution
Concept. Maximise $\ln f=x^2\ln\frac2x$ instead of $f$. Formulas used. (1) $\dfrac{d}{dx}\left(x^2\ln\dfrac2x\right)=2x\ln\dfrac2x-x$Step 1 - Critical point. By (1), $x\left(2\ln\frac2x-1\right)=0$ gives $\ln\frac2x=\frac12$, so $x=\frac{2}{\sqrt e}$. Step 2 - Max. For smaller $x$, $\ln\frac2x>\frac12$ and the derivative is positive; for larger $x$ it is negative. So it is a maximum. Step 3 - Value. $\ln f=\frac4e\cdot\frac12=\frac2e$, so $f=e^{2/e}$. Answer: (B)