Question

Let $g:(0,\infty )\to R$ be a differentiable function such that
$\int \left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x{e}^x\right)}{{\left(e^x+1\right)}^2}\right)dx=\frac{xg(x)}{e^x+1}+c$
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)

$\int \left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x{e}^x\right)}{{\left(e^x+1\right)}^2}\right)dx=\frac{xg(x)}{e^x+1}+c$
On differentiating both sides w.r.t. $x$, we get
$\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)(e^x+1-x{e}^x}{{\left(e^x+1\right)}^2}\right)$
$=\frac{\left(e^x+1\right)\left(g(x)+x{g}^{\prime }(x)\right)-e^x\cdot x\cdot g(x)}{{\left(e^x+1\right)}^2}$
$\left(e^x+1\right)x(\cos x-\sin x)+g(x)\left(e^x+1-x{e}^x\right)$
$=\left(e^x+1\right)\left(g(x)+x{g}^{\prime }(x)\right)-e^x\cdot x\cdot g(x)$
$\Rightarrow g{}^{\prime }(x)=\cos x-\sin x$
$\Rightarrow g(x)=\sin x+\cos x+C$
$g(x)$ 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 $\varphi (x)=g(x)-g^{\prime }(x)=2\sin x+C$
$\Rightarrow {\varphi }^{\prime }(x)=g^{\prime }(x)-g^{\prime \prime }(x)=2\cos x>0$
$\Rightarrow \varphi$ is increasing