Q46 / 90 · JEE Main 2010 · Mathematics

Let $f: R \rightarrow R$ be a continuous function defined by $f(x)=\frac{1}{e^x+2 e^{-x}}$. Statement-1: $f(c)=\frac{1}{3}$, for some $c \in R$. Statement-2: $0 < f(x) \leq \frac{1}{2 \sqrt{2}}$, 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
  • Correct answer
    D
    Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Solution

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

Hence, the correct answer is option: 'D'.

Relevant Questions