Aspire Faculty ID #14429 · Topic: JEE Main 2024 (6 April Evening Shift) · Just now
JEE Main 2024 (6 April Evening Shift)

If the function $f(x)=\left(\dfrac1x\right)^{2x}, x>0$ attains its maximum at $x=\dfrac1e$, then:

Solution

Take log: $y = \left(\frac{1}{x}\right)^{2x}$
$\ln y = 2x \ln\left(\frac{1}{x}\right) = -2x \ln x$
Differentiate: $\frac{1}{y}\frac{dy}{dx} = -2(\ln x + 1)$
For extrema: $\ln x + 1 = 0 \Rightarrow x = \frac{1}{e}$
Now value at maximum: $f\left(\frac{1}{e}\right) = \left(e\right)^{2/e} = e^{2/e}$
Hence: Maximum value $= e^{2/e}$


Now compare options using logs: Take $\ln$ on both sides 

Option (1): $e^\pi < \pi^e$
$\Rightarrow \frac{\pi}{e} < \ln \pi$ (false, since $\frac{\pi}{e} > \ln \pi$)

Option (2): $e^{2\pi} < (2\pi)^e$
$\Rightarrow \frac{2\pi}{e} < \ln(2\pi)$ (false)

Option (3): $(2e)^\pi > \pi^{2e}$
$\Rightarrow \frac{\pi}{2e} > \frac{\ln \pi}{\ln(2e)}$ (true comparison holds)

Option (4): $e^\pi > \pi^e$
$\Rightarrow \frac{\pi}{e} > \ln \pi$ (true)

Final Answer: $ \boxed{\text{Option (4)}} $
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