Question

Let $f:R\to R$ be a continuous function defined by $f\left(x\right)=\frac{1}{e^x+2e^{-x}}$

Statement I : $f\left(c\right)=\frac{1}{3}$, for some $c\in R$.

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

• A. Statement I is correct, Statement II is also correct; Statement II is the correct explanation of Statement I
• B. Statement I is correct, Statement II is also correct; Statement II is not the correct equation of Statement I
• C. Statement I is correct, statement II is correct
• D. Statement I is incorrect, Statement II is correct

Answer 1

The correct option is B

Statement I is correct, Statement II is also correct; Statement II is not the correct equation of Statement I

Explanation for correct answer Option(s)

Step 1: Analysing Statement II

We have the given Equation as:

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

Finding the maximum value of $f\left(x\right)$

$$\begin{gathered} \frac{dy}{dx}=\frac{-1}{{\left(e^x+2e^{-x}\right)}^2}\times e^x-2e^{-x} \\ \frac{dy}{dx}=\frac{-e^x-2e^{-x}}{{\left(e^x+2e^{-x}\right)}^2} \end{gathered}$$

for the maximum value of $x$, $\frac{dy}{dx}=0$

$$\begin{gathered} \Rightarrow \frac{-e^x-2e^{-x}}{{\left(e^x+2e^{-x}\right)}^2}=0 \\ \Rightarrow e^x-2e^{-x}=0 \\ \Rightarrow e^x=2e^{-x} \\ \Rightarrow e^{2x}=2 \\ \Rightarrow e^x=\sqrt{2} \\ \Rightarrow x=\log \sqrt{2} \end{gathered}$$

Then,

$$\begin{gathered} f{\left(x\right)}_{max}=\frac{1}{e^x+2e^{-x}} \\ \Rightarrow f{\left(x\right)}_{max}=\frac{1}{\sqrt{2}+{\displaystyle \frac{2}{\sqrt{2}}}} \\ \Rightarrow f{\left(x\right)}_{max}=\frac{\sqrt{2}}{4}\times \frac{1}{2\sqrt{2}} \\ \Rightarrow f{\left(x\right)}_{max}=\frac{1}{2\sqrt{2}} \end{gathered}$$

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

Thus, statement II is true.

Step 2: Analysing Statement I.

We know that the arithmetic mean is greater than equal to the geometric mean

$$\begin{gathered} \Rightarrow \frac{-e^x-{\displaystyle \frac{2}{e^x}}}{2}\ge {\left(e^x\times \frac{2}{e^x}\right)}^{\frac{1}{2}} \\ \Rightarrow \frac{-e^x-{\displaystyle \frac{2}{e^x}}}{2}\ge \sqrt{2} \\ \Rightarrow -e^x-\frac{2}{e^x}\ge 2\sqrt{2} \\ \Rightarrow 0<\frac{1}{-e^x-\frac{2}{e^x}}\le \frac{1}{2\sqrt{2}} \\ \Rightarrow \frac{1}{3}\in \frac{1}{-e^x-\frac{2}{e^x}} \\ \Rightarrow \frac{1}{3}\in f\left(x\right) \end{gathered}$$

Hence, the statement I is true.

Therefore, the correct answer is option B.