JEE MAIN - Mathematics (2022 - 25th June Morning Shift - No. 9)
Let $$g:(0,\infty ) \to R$$ be a differentiable function such that
$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $$, for all x > 0, where c is an arbitrary constant. Then :
$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $$, for all x > 0, where c is an arbitrary constant. Then :
g is decreasing in $$\left( {0,{\pi \over 4}} \right)$$
g' is increasing in $$\left( {0,{\pi \over 4}} \right)$$
g + g' is increasing in $$\left( {0,{\pi \over 2}} \right)$$
g $$-$$ g' is increasing in $$\left( {0,{\pi \over 2}} \right)$$
Explanation
$$
\int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right)}{\left(e^x+1\right)^2}\right) d x=\frac{x g(x)}{e^x+1}+c
$$
On differentiating both sides w.r.t. $\mathrm{x}$, we get
$$ \begin{aligned} &\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\\\ &=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\\\ &\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\\\ &=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\\\ &\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\\\ &\Rightarrow g(x)=\sin x+\cos x+C \end{aligned} $$
$\mathrm{g}(\mathrm{x})$ is increasing in $(0, \pi / 4)$
$g "(x)=-\sin x-\cos x<0$
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})$ is decreasing function
let $h(x)=g(x)+g^{\prime}(x)=2 \cos x+C$
$$ \Rightarrow \mathrm{h}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime \prime}(\mathrm{x})=-2 \sin \mathrm{x}<0 $$
$\Rightarrow \mathrm{h}$ is decreasing
let $\phi(\mathrm{x})=\mathrm{g}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=2 \sin \mathrm{x}+\mathrm{C}$
$$ \Rightarrow \phi^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})-\mathrm{g}{ }^{\prime \prime}(\mathrm{x})=2 \cos \mathrm{x}>0 $$
$\Rightarrow \phi$ is increasing
Hence option D is correct.
On differentiating both sides w.r.t. $\mathrm{x}$, we get
$$ \begin{aligned} &\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\\\ &=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\\\ &\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\\\ &=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\\\ &\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\\\ &\Rightarrow g(x)=\sin x+\cos x+C \end{aligned} $$
$\mathrm{g}(\mathrm{x})$ is increasing in $(0, \pi / 4)$
$g "(x)=-\sin x-\cos x<0$
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})$ is decreasing function
let $h(x)=g(x)+g^{\prime}(x)=2 \cos x+C$
$$ \Rightarrow \mathrm{h}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime \prime}(\mathrm{x})=-2 \sin \mathrm{x}<0 $$
$\Rightarrow \mathrm{h}$ is decreasing
let $\phi(\mathrm{x})=\mathrm{g}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=2 \sin \mathrm{x}+\mathrm{C}$
$$ \Rightarrow \phi^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})-\mathrm{g}{ }^{\prime \prime}(\mathrm{x})=2 \cos \mathrm{x}>0 $$
$\Rightarrow \phi$ is increasing
Hence option D is correct.
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