Q109 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics

$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ is a function defined by $f(x)=\frac{1}{e^x+2 e^{-x}}$ Assertion (A) : $\mathrm{f}(\mathrm{c})=\frac{1}{3}$ for some values of $\mathrm{c} \in \mathrm{R}$ Reason (R): $0 < \mathrm{f}(\mathrm{x}) \leq \frac{1}{2 \sqrt{2}}$ for all $\mathrm{x} \in \mathrm{R}$ Then which of the following options is correct?

  • Correct answer
    A
    (A) and (R) are true. (R) is the correct explanation of (A)
  • B
    (A) and (R) are true, but (R) is not the correct explanation for $(\mathrm{A})$
  • C
    (A) is true but $(R)$ is false
  • D
    (A) is false but (R) is true

Solution

$\begin{aligned} & \quad f(x)=\frac{1}{e^x+2 e^{-x}} \\ & \because \quad \frac{e^x+2 e^{-x}}{2} \geq \sqrt{e^x \times 2 e^{-x}} \\ & e^x+2 e^{-x} \geq 2 \sqrt{2} \\ & \therefore \quad \frac{1}{e^x+2 e^{-x}} \leq 2 \sqrt{2} \\ & \because \quad f(x)>0 \\ & \therefore \quad 0 < f(x) < \frac{1}{2 \sqrt{2}} \forall x \in R \\ & \because \quad \frac{1}{3} \in\left(0, \frac{1}{2 \sqrt{2}}\right) \end{aligned}$ $\therefore \quad(A)$ and $(R)$ both are true and $(R)$ is correct explanation of $(A)$.

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

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