Question

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 $\left(0, \frac{\pi}{2}\right)$
• D. $g - g$ ' is increasing in $\left(0, \frac{\pi}{2}\right)$

Answer 1

Answer: (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