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 $c \epsilon R.$
Statement-2: $0 < f\left(x\right) \le \frac{1}{2\sqrt{2}}$, for all $x \epsilon R$

• 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)

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