Let g:(0, ) R be a differentiable function such that (x( x- x) e^… - Vidyadip

MCQ

Let $g:(0, \infty) \rightarrow R$ be a differentiable function such that $\int\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right)}{\left(e^{x}+1\right)^{2}}\right) d x=\frac{x g(x)}{e^{x}+1}+c$ for all $x >0$, where $c$ is an arbitrary constant. Then.

A
$g$ is decreasing in $\left(0, \frac{\pi}{4}\right)$
B
$g^{\prime}$ is increasing in $\left(0, \frac{\pi}{4}\right)$
C
$g+g^{\prime}$ is increasing in $\left(0, \frac{\pi}{2}\right)$
✓
$g - g ^{\prime}$ is increasing in $\left(0, \frac{\pi}{2}\right)$

✓

Answer

Correct option: D.

$g - g ^{\prime}$ is increasing in $\left(0, \frac{\pi}{2}\right)$

d
$\int\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right)}{\left(e^{x}+1\right)^{2}}\right) d x=\frac{x g(x)}{e^{x}+1}+c$

On differentiating both sides w.r.t. $x$, we get

$\left(\frac{x(\cos x-\sin x)}{e^{x}+1}+\frac{g(x)\left(e^{x}+1-x e^{x}\right.}{\left(e^{x}+1\right)^{2}}\right)$

$=\frac{\left(e^{x}+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^{x} \cdot x \cdot g(x)}{\left(e^{x}+1\right)^{2}}$

$\left(e^{x}+1\right) x(\cos x-\sin x)+g(x)\left(e^{x}+1-x e^{x}\right)$

$=\left(e^{x}+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^{x} \cdot x \cdot g(x)$

$\Rightarrow g^{\prime}(x)=\cos x-\sin x$

$\Rightarrow g(x)=\sin x+\cos x+C$

$g ( x )$ is increasing in $(0, \pi / 4)$

$g ^{\prime \prime}( x )=-\sin x -\cos x <0$

$\Rightarrow g ^{\prime}( x )$ is decreasing function

let $h ( x )= g ( x )+ g ^{\prime}( x )=2 \cos x + C$

$\Rightarrow h ^{\prime}( x )= g ^{\prime}( x )+ g ^{\prime \prime}( x )=-2 \sin x <0$

$\Rightarrow h$ is decreasing

let $\phi( x )= g ( x )- g ^{\prime}( x )=2 \sin x + C$

$\Rightarrow \phi^{\prime}( x )= g ^{\prime}( x )- g { }^{\prime \prime}( x )=2 \cos x >0$

$\Rightarrow \phi$ is increasing

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