Key Concepts and Formulas

• Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For non-negative real numbers $a_1, a_2, \dots, a_n$, $\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \dots a_n}$. Equality holds when $a_1 = a_2 = \dots = a_n$.
• Intermediate Value Theorem (IVT): If $f$ is continuous on $[a, b]$, and $k$ is between $f(a)$ and $f(b)$, then there exists $c \in [a, b]$ such that $f(c) = k$.
• Range of a Continuous Function: If a function is continuous, its range is an interval.

Step-by-Step Solution

Step 1: Analyze the function and its continuity.

The given function is $f(x) = \frac{1}{e^x + 2e^{-x}}$. Since $e^x$ is continuous and $e^x > 0$ for all $x \in \mathbb{R}$, the denominator $e^x + 2e^{-x}$ is continuous and positive. Therefore, $f(x)$ is continuous on $\mathbb{R}$.

Step 2: Analyze Statement - 2: $0 < f(x) \le \frac{1}{2\sqrt{2}}$ for all $x \in \mathbb{R}$.

To verify this statement, we need to find the range of $f(x)$. We will use the AM-GM inequality to find the minimum value of the denominator.

Step 3: Apply AM-GM to the denominator.

Since $e^x > 0$ and $2e^{-x} > 0$, we can apply the AM-GM inequality to these two terms: $\frac{e^x + 2e^{-x}}{2} \ge \sqrt{e^x \cdot 2e^{-x}}$ $e^x + 2e^{-x} \ge 2\sqrt{2e^0} = 2\sqrt{2}$ This shows that the minimum value of the denominator is $2\sqrt{2}$.

Step 4: Find the maximum value of $f(x)$.

Since $f(x) = \frac{1}{e^x + 2e^{-x}}$, the maximum value of $f(x)$ occurs when the denominator is at its minimum: $f(x)_{\text{max}} = \frac{1}{2\sqrt{2}}$

Step 5: Determine the condition for equality in AM-GM.

Equality in AM-GM holds when $e^x = 2e^{-x}$. Multiplying both sides by $e^x$, we get $e^{2x} = 2$. Taking the natural logarithm of both sides, $2x = \ln 2$, so $x = \frac{1}{2} \ln 2 = \ln \sqrt{2}$.

Step 6: Find the lower bound of $f(x)$.

As $x \to \infty$, $e^x \to \infty$ and $2e^{-x} \to 0$. Therefore, $e^x + 2e^{-x} \to \infty$, and $f(x) \to 0$. As $x \to -\infty$, $e^x \to 0$ and $2e^{-x} \to \infty$. Therefore, $e^x + 2e^{-x} \to \infty$, and $f(x) \to 0$. Since $e^x + 2e^{-x} > 0$ for all $x$, we have $f(x) > 0$ for all $x$.

Step 7: Conclude about Statement - 2.

Thus, $0 < f(x) \le \frac{1}{2\sqrt{2}}$ for all $x \in \mathbb{R}$. Statement - 2 is true.

Step 8: Analyze Statement - 1: $f(c) = \frac{1}{3}$ for some $c \in \mathbb{R}$.

We know that the maximum value of $f(x)$ is $\frac{1}{2\sqrt{2}}$. Since $2\sqrt{2} \approx 2(1.414) = 2.828$, we have $\frac{1}{2\sqrt{2}} \approx \frac{1}{2.828}$. Also, $f(x) > 0$. So the range of $f(x)$ is $(0, \frac{1}{2\sqrt{2}}]$.

Since $f(x)$ is continuous, its range is an interval. We need to check if $\frac{1}{3}$ lies in the range $(0, \frac{1}{2\sqrt{2}}]$. We have $\frac{1}{3} \approx 0.333$ and $\frac{1}{2\sqrt{2}} \approx 0.354$. Thus, $0 < \frac{1}{3} < \frac{1}{2\sqrt{2}}$. Since $f(x)$ is continuous and $\frac{1}{3}$ is within its range, by the Intermediate Value Theorem, there exists a $c \in \mathbb{R}$ such that $f(c) = \frac{1}{3}$. Statement - 1 is true.

Step 9: Determine if Statement - 2 explains Statement - 1.

Statement - 2 tells us the maximum value of $f(x)$ and that $f(x)$ is always positive. Since $\frac{1}{3}$ is less than the maximum value and greater than 0, the Intermediate Value Theorem guarantees that there exists some $c$ such that $f(c) = \frac{1}{3}$. Therefore, Statement - 2 is true and Statement - 1 is true. However, Statement - 2 doesn't explain statement -1.

Common Mistakes & Tips

• Remember to check the condition for equality in AM-GM to ensure the minimum value is attainable.
• Make sure you understand the Intermediate Value Theorem and how it applies to continuous functions.
• Don't forget to check if the value in Statement - 1 lies within the range found in Statement - 2.

Summary

We analyzed the given function $f(x)$ and determined its range using the AM-GM inequality. We found that Statement - 2 is true, as $0 < f(x) \le \frac{1}{2\sqrt{2}}$. We also found that Statement - 1 is true, as $\frac{1}{3}$ lies within the range of $f(x)$, guaranteeing the existence of a $c$ such that $f(c) = \frac{1}{3}$ by the Intermediate Value Theorem. Statement - 2 does not explain statement - 1.

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).