Let g:(0,∞) → R be a differentiable function such that ∫ ((x(cos x - sin x))/(eˣ + 1) + g(x)(eˣ + 1 - x eˣ) (eˣ + 1)²)dx = (x g(x))/(eˣ + 1) + c, for all x > 0, where c is an arbitrary constant. Then:

(D) g - g' is increasing in (0,(π)/2)

(A) g is decreasing in (0,(π)/4)

(B) g' is increasing in (0,(π)/4)

(C) g + g' is increasing in (0,(π)/2)

Differential Equations: JEE Main 2022 Mathematics question

JEE Main 2022 · 25 Jun, Morning shift · Q3625

JEE Main 202225 Jun, Morning shiftMathematicsHardDifferential Equations163× since 2002

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 :

JEE Main 2022 · 25 Jun, Morning shift · Q3625

JEE Main 202225 Jun, Morning shiftMathematicsHardDifferential Equations163× since 2002

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 :