Question

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

• A. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
• B. Statement-1 is true, Statement-2 is false
• C. Statement-1 is false, Statement-2 is true
• D. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Answer 1

Answer: (D)

$f\left(x\right)=\frac{1}{e^x+2e^{-x}}=\frac{e^x}{e^{2x}+2}$
$f^{\prime }\left(x\right)=\frac{\left(e^{2x}+2\right)e^x-2e^{2x}\cdot e^x}{{\left(e^{2x+2}\right)}^2}$
$f^{\prime }\left(x\right)=0\Rightarrow e^{2x}+2=2e^{2x}$
$e^{2x}=2\Rightarrow e^x=\sqrt{2}$
maximum $f\left(x\right)=\frac{\sqrt{2}}{4}=\frac{1}{2\sqrt{2}}$
$0<f\left(x\right)\le \frac{1}{2\sqrt{2}}\forall x\in R$
Since $0<\frac{1}{3}<\frac{1}{2\sqrt{2}}$ for some $c\in R$
$f\left(c\right)=\frac{1}{3}$