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(A) Given $f(x) = \frac{1}{e^x + 2e^{-x}}$.Using the $AM$-$GM$ inequality for $e^x$ and $2e^{-x}$:$\frac{e^x + 2e^{-x}}{2} \geq \sqrt{e^x \cdot 2e^{-x

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$(A)$ and $(R)$ are true. $(R)$ is the correct explanation of $(A)$

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(A) Given $f(x) = \frac{1}{e^x + 2e^{-x}}$.Using the $AM$-$GM$ inequality for $e^x$ and $2e^{-x}$:$\frac{e^x + 2e^{-x}}{2} \geq \sqrt{e^x \cdot 2e^{-x}} = \sqrt{2}$.So, $e^x + 2e^{-x} \geq 2\sqrt{2}$.Therefore, $f(x) = \frac{1}{e^x + 2e^{-x}} \leq \frac{1}{2\sqrt{2}}$.Since $e^x + 2e^{-x} > 0$, we have $0 Thus, Reason $(R)$ is true.Since $\frac{1}{2\sqrt{2}} \approx \frac{1}{2.828} \approx 0.3535$, and $\frac{1}{3} \approx 0.3333$, we see that $\frac{1}{3} By the Intermediate Value Theorem, since $f(x)$ is continuous, $f(c) = \frac{1}{3}$ must hold for some $c \in R$.Thus, Assertion $(A)$ is true and $(R)$ is the correct explanation.

$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

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