Question

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

• 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)

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

$f\left(x\right)=\frac{1}{e^x+2e^{-x}}=\frac{e^x}{e^{2x}+2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{(e^{2x}+2)e^x-2e^{2x}\cdot e^x}{(e^{2x+}+2{)}^2}$

For critical point $f^{\prime }\left(x\right)=0$

$\Rightarrow e^x=\frac{\sqrt{2}}{4}$

Maximum value $f\left(x\right)=\frac{\sqrt{2}}{4}=\frac{1}{2\sqrt{2}}$

$0<f\left(x\right)\le \frac{1}{2\sqrt{2}}\forall x\in R$

Since $0<\frac{1}{3}<\frac{1}{2\sqrt{2}}=$ for some $c\in R$

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

Hence, option 'D' is correct.