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]

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]

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

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