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(B) Given $f(x) = \frac{1}{e^x + 2e^{-x}} = \frac{e^x}{e^{2x} + 2}$.To find the range,we find the maximum value of $f(x)$.$f'(x) = \frac{e^x(e^{2x} +

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Statement-$1$ is true,Statement-$2$ is false.

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(B) Given $f(x) = \frac{1}{e^x + 2e^{-x}} = \frac{e^x}{e^{2x} + 2}$.To find the range,we find the maximum value of $f(x)$.$f'(x) = \frac{e^x(e^{2x} + 2) - e^x(2e^{2x})}{(e^{2x} + 2)^2} = \frac{e^x(2 - e^{2x})}{(e^{2x} + 2)^2}$.Setting $f'(x) = 0$,we get $e^{2x} = 2$,so $e^x = \sqrt{2}$.The maximum value is $f(\ln \sqrt{2}) = \frac{\sqrt{2}}{2 + 2} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$.Since $e^x + 2e^{-x} > 0$,$f(x) > 0$. Thus,$0 Statement-$2$ claims $0 For Statement-$1$,we check if $\frac{1}{3}$ is in the range $(0, \frac{1}{2\sqrt{2}}]$.Since $\sqrt{2} \approx 1.414$,$2\sqrt{2} \approx 2.828$. Thus $\frac{1}{2\sqrt{2}} \approx \frac{1}{2.828} \approx 0.353$.Since $\frac{1}{3} \approx 0.333$,we have $0 By the Intermediate Value Theorem,there exists $c$ such that $f(c) = \frac{1}{3}$.Thus,Statement-$1$ is true and Statement-$2$ is false.

Let $f:R \to R$ be a continuous function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$.
Statement-$1$: $f(c) = \frac{1}{3}$ for some $c \in R$.
Statement-$2$: $0 < f(x) < \frac{1}{2\sqrt{2}}$ for all $x \in R$.

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Let $f:R \to R$ be a continuous function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$.Statement-$1$: $f(c) = \frac{1}{3}$ for some $c \in R$.Statement-$2$: $0 < f(x) < \frac{1}{2\sqrt{2}}$ for all $x \in R$.