Monotonicity – JEE Main Previous Year Question with Solutions
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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]

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Sol. (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]

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Sol. (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 $\mathrm{h}^{\prime}(\mathrm{c})=0=\mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{g}^{\prime}(\mathrm{c})$

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Comments
  • LOL
    May 25, 2021 at 11:09 pm

    iski koi aukat nahi hai kya , itne kam questions.

    1
  • god
    September 4, 2020 at 11:01 pm

    bekar

    0
  • August 15, 2020 at 8:17 pm

    Incomplete qution

    0
  • August 15, 2020 at 8:16 pm

    Incomplete qution

    0