Question

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

• A. $e^\pi<\pi^c$
• B. $e^{2\pi}<(2\pi)^c$
• C. $e^\pi>\pi^c$
• D. $(2e)^\pi>\pi^{(2e)}$

Answer 1

Answer: (C)

$e^\pi>\pi^c$

Answer 2

Step 1 : Let
$y = \left( \frac{1}{x} \right)^{2x}$
Taking the natural logarithm on both sides:
$\ln y = 2x \ln \left( \frac{1}{x} \right)$
Simplify:
$\ln y = -2x \ln x$
Step 2 : Differentiating with respect to $x$
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = -2(1 + \ln x)$
Multiply through by $y$:
$\frac{dy}{dx} = y \cdot (-2)(1 + \ln x)$
Step 3 : Behavior of the function
For $x > \frac{1}{e}$, the function $f^n$ is decreasing.
Thus, we can establish the following inequalities:
$e < \pi$
$\left( \frac{1}{e} \right)^{2e} > \left( \frac{1}{\pi} \right)^{2\pi}$
$e^\pi > \pi^e$