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 $xϵR$

Answer 1

$f\left(x\right)=\frac{1}{e^x+2e^{-x}}=\frac{e^x}{e^{2x}+2}{f}^1\left(x\right)=\frac{\left(e^{2x+2}\right)e^x-2e^{2x}.e^x}{(e^{2x+2}{)}^2}{f}^1\left(x\right)=0e^{2x}=2\Rightarrow e^{2x}+2=2e^{2x}\Rightarrow e^x=\sqrt{2}Maximumf\left(x\right)=\frac{\sqrt{2}}{4}=\frac{1}{2\sqrt{2}}0<f\left(x\right)\le \frac{1}{2\sqrt{2}}\forall xϵR$

Since $0<f\left(x\right)\le \frac{1}{2\sqrt{2}}or,0<\frac{1}{3}<\frac{1}{2\sqrt{2}}$

For sum $CϵR$

$f\left(c\right)=\frac{1}{3}$

Statement $1$ is true

Statement $2$ is true

Statement $2$ is correct explanation of statement $1$