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

The Correct Option is C

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:
\[e < \pi\]
\[\left( \frac{1}{e} \right)^{2e} > \left( \frac{1}{\pi} \right)^{2\pi}\]
\[e^\pi > \pi^e\]