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Let g:(0, \infty) \rightarrow \mathbf{R} be a differentiable function such that

\int\left(\frac{x(\cos x-\sin x)}{\mathrm{e}^{x}+1}+\frac{g(x)\left(\mathrm{e}^{x}+1-x \mathrm{e}^{x}\right)}{\left(\mathrm{e}^{x}+1\right)^{2}}\right) \mathrm{d} x=\frac{x g(x)}{\mathrm{e}^{x}+1}+\mathrm{c}, \text { for all } x>0 \text {, } where c is an arbitrary constant. Then:

Option: 1

g is decreasing in \left(0, \frac{\pi}{4}\right)


Option: 2

g' is increasing in \left(0, \frac{\pi}{4}\right)


Option: 3

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


Option: 4

g-g' is increasing in \left(0, \frac{\pi}{4}\right)


Answers (1)

best_answer

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

Integrating by parts in first integral
\mathrm{ =\frac{x}{e^{x}+1}(\sin x+\cos x)-\int \frac{e^{x}+1-x e^{x}}{\left(e^{x}+1\right)^{2}} ( \sin x+\cos x ) d x+\int gx \frac{e^{x}+1-x e^{x}}{\left(e^{x}+1\right)^{2}} \, dx}

By comparison, we get \mathrm{ g(x)=\sin x+\cos x}

g(x)-g'(x)= 2sinx

This is increasing in (0,\frac{\pi}{4})

Posted by

Divya Prakash Singh

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