Question

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

Accepted Answer

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

Answer

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

Answer

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

Answer

$g + g'$ is increasing in $\left( {0,\frac{\pi }{2}} \right)$

JEE MATHEMATICS MCQ: INCREASING & DECREASING

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