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Let $f: R \rightarrow R$ be a continuous function defined by $f(x)=\frac{1}{e^x+2 e^{-x}}$.
Statement-1: $f(c)=\frac{1}{3}$, for some $c \in R$.
Statement-2: $0 < f(x) \leq \frac{1}{2 \sqrt{2}}$, for all $x \in R$
Options:
Statement-1: $f(c)=\frac{1}{3}$, for some $c \in R$.
Statement-2: $0 < f(x) \leq \frac{1}{2 \sqrt{2}}$, for all $x \in R$
Solution:
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Verified Answer
The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1
$f(x)=\frac{1}{e^x+2 e^{-x}}=\frac{e^x}{e^{2 x}+2}$
$f^{\prime}(x)=\frac{\left(e^{2 x}+2\right) e^x-2 e^{2 x} \cdot e^x}{\left(e^{2 x+2}\right)^2}$
$f^{\prime}(x)=0 \quad \Rightarrow e^{2 x}+2=2 e^{2 x}$
$\mathrm{e}^{2 \mathrm{x}}=2 \quad \Rightarrow \mathrm{e}^{\mathrm{x}}=\sqrt{2}$
maximum $f(x)=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}}$
$0 < f(x) \leq \frac{1}{2 \sqrt{2}} \quad \forall x \in R$
Since $0 < \frac{1}{3} < \frac{1}{2 \sqrt{2}} \quad \Rightarrow$ for some $c \in R$
$f(c)=\frac{1}{3}$
$f^{\prime}(x)=\frac{\left(e^{2 x}+2\right) e^x-2 e^{2 x} \cdot e^x}{\left(e^{2 x+2}\right)^2}$
$f^{\prime}(x)=0 \quad \Rightarrow e^{2 x}+2=2 e^{2 x}$
$\mathrm{e}^{2 \mathrm{x}}=2 \quad \Rightarrow \mathrm{e}^{\mathrm{x}}=\sqrt{2}$
maximum $f(x)=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}}$
$0 < f(x) \leq \frac{1}{2 \sqrt{2}} \quad \forall x \in R$
Since $0 < \frac{1}{3} < \frac{1}{2 \sqrt{2}} \quad \Rightarrow$ for some $c \in R$
$f(c)=\frac{1}{3}$
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