Let g:(0, \infty) \to R be a differentiable function such that \int(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)(e^x+1-x e^x)}{(e^x+1)^2}) d x=\frac{x g(x)}{e^x+1}+c, for all {\x}>0, where {\c} is an arbitrary constant. Then : [25-Jun-2022-Shift-1]

Correct Answer is : g−g ' is increasing in (0,π2)

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

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

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

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

Differential Equations Part 2

Section: Mathematics

Question : 95 of 100

Marks: +1, -0

Let g:(0,∞)→R be a differentiable function such that

∫(‌

x(cos‌x−sin‌x)

ex+1

+‌

g(x)(ex+1−xex)

(ex+1)2

)dx=‌

xg(x)

ex+1

, for all x>0, where c is an arbitrary constant. Then :

[25-Jun-2022-Shift-1]

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