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

Answer 2

To find the value of \(c\) and verify the correct answer, we need to find the critical point of the function \(f(x) = \left(\frac{1}{x}\right)^{2x}\). The function is defined for \(x > 0\). Our goal is to determine the condition at which this function attains its maximum value at \(x = \frac{1}{c}\).

First, rewrite the function using exponent and logarithms for easier manipulation:

\(f(x) = \left(\frac{1}{x}\right)^{2x} = x^{-2x} = e^{\ln(x^{-2x})} = e^{-2x \ln x}\)

To find the critical points, we need to differentiate the exponent function \(-2x \ln x\) with respect to \(x\) and set the derivative to zero.

Let \(g(x) = -2x \ln x\). Then:

\(\frac{d}{dx}[-2x \ln x] = -2 \left(\ln x + 1\right)\)

The critical points occur where:

\((2 (\ln x + 1) = 0\)

Simplifying:

\(\ln x + 1 = 0\) \(\ln x = -1\) \(x = e^{-1} = \frac{1}{e}\)

Therefore, the function attains its maximum value at \(x = \frac{1}{e}\). Given in the question that this happens at \(x = \frac{1}{c}\), it follows that:

\(c = e\)

Now let's evaluate the conditions in the options given:

• Option 1: \(e^{\pi} < \pi^c\) where \(c = e\) implies \(e^{\pi} < \pi^e\). This option needs verification.
• Option 2: \(e^{2\pi} < (2\pi)^c\) with \(c = e\) implying \(e^{2\pi} < (2\pi)^e\). We need to verify this.
• Option 3: \(e^{\pi} > \pi^c\) with \(c = e\) resulting in \(e^{\pi} > \pi^e\), which is our given correct answer.
• Option 4: \((2e)^{\pi} > \pi^{(2e)}\), which needs verification as well.

To verify the given correct answer (Option 3), use a comparison or numerical approximation to confirm if \(e^{\pi} > \pi^e\). Numerical approximations give \(e \approx 2.718\) and \(\pi \approx 3.141\). Calculating approximately these values:

• \(e^{\pi} \approx 2.718^{3.141} \approx 23.1407\)
• \(\pi^e \approx 3.141^{2.718} \approx 22.4591\)

Therefore, \(e^{\pi} > \pi^e\), confirming Option 3 as the correct answer. Option 3: \(e^\pi > \pi^c\) is the correct choice.