MathematicsHard38×since 2002Q3625Let g:(0,∞)→Rg:(0,\infty ) \to Rg:(0,∞)→R be a differentiable function such that ∫(x(cosx−sinx)ex+1+g(x)(ex+1−xex)(ex+1)2)dx=x g(x)ex+1+c\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}∫(ex+1x(cosx−sinx)+(ex+1)2g(x)(ex+1−xex))dx=ex+1xg(x)+c, for all x > 0, where c is an arbitrary constant. Then :Ag is decreasing in (0,π4)\left( {0,{\pi \over 4}} \right)(0,4π)Bg' is increasing in (0,π4)\left( {0,{\pi \over 4}} \right)(0,4π)Cg + g' is increasing in (0,π2)\left( {0,{\pi \over 2}} \right)(0,2π)Dg −-− g' is increasing in (0,π2)\left( {0,{\pi \over 2}} \right)(0,2π)Check answerSkip