 Most Affordable JEE | NEET | 8,9,10 Preparation by Kota's Top IITian Doctor Faculties

# Monotonicity - JEE Main Previous Year Question with Solutions `
JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a continuous function defined by $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{\mathrm{x}}+2 \mathrm{e}^{-\mathrm{x}}}$ Statement-1 : $\mathrm{f}(\mathrm{c})=\frac{1}{3},$ for some $\mathrm{c} \in \mathrm{R}$ Statement-2 : $0<\mathrm{f}(\mathrm{x}) \leq \frac{1}{2 \sqrt{2}},$ for all $\mathrm{x} \in \mathrm{R}$ (1) Statement–1 is true, Statement–2 is true ; Statement–2 is a correct explanation for Statement–1. (2) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for statement–1. (3) Statement–1 is true, Statement–2 is false. (4) Statement–1 is false, Statement–2 is true. [AIEEE-2010]
Ans. (1) $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{\mathrm{x}}+2 \mathrm{e}^{-\mathrm{x}}}$ $\mathrm{y}=\frac{1}{\mathrm{e}^{\mathrm{x}}+2 \mathrm{e}^{-\mathrm{x}}}$ Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t} \in(0, \infty)$ $\mathrm{y}=\frac{1}{\mathrm{t}+\frac{2}{\mathrm{t}}} \Rightarrow \mathrm{y}=\frac{\mathrm{t}}{\mathrm{t}^{2}+2}$ $\Rightarrow \mathrm{t}^{2} \mathrm{y}-\mathrm{t}+2 \mathrm{y}=0$ $\mathrm{D} \geq 0$ $1-8 \mathrm{y}^{2} \geq 0$ $\Rightarrow 8 \mathrm{y}^{2}-1 \leq 0$ $\Rightarrow \mathrm{y} \in\left[\frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}\right]$ but $\mathrm{y}>0$ $\therefore \mathrm{y} \in\left(0, \frac{1}{2 \sqrt{2}}\right]$ $\therefore \mathrm{f}(0)=\frac{1}{3}$ $\therefore \mathrm{f}(\mathrm{c})=\frac{1}{3}(\mathrm{c} \in \mathrm{R})$ So Statement- 1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement- $1 .$
Q. If $\mathrm{f}$ and $\mathrm{g}$ are differentiable functions in $[0,1]$ satisfying $\mathrm{f}(0)=2=\mathrm{g}(1), \mathrm{g}(0)=0$ and $\mathrm{f}(1)=6, \text { then for some } \mathrm{c} \in] 0,1[:$ (1) 2f'(c) = g'(c) (2) 2f'(c) = 3g'(c) (3) f'(c) = g'(c) (4) f'(c) = 2g'(c) [JEE-MAIN 2014]
Ans. (4) Consider a function $\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{x})-2 \mathrm{g}(\mathrm{x})$ $\therefore \mathrm{h}(0)=2, \mathrm{h}(1)=2$ $\because \mathrm{h}(\mathrm{x})$ is continuous and diffrentiable in $[0,1]$ $\therefore$ by Rolle's theorem, there exist at least one 'such that $\mathrm{h}^{\prime}(\mathrm{c})=0=\mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{g}^{\prime}(\mathrm{c})$