Verified Solution • 2025–26 Session moderate

QJEEM22Ma7885: JEE Main 2022 Mathematics - Class 12 JEEM22Ma7885 Solution: Let be a differentiable function such that , for all x > 0,…

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

∫(x(cos⁡x−sin⁡x)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} , for all x > 0, where c is an arbitrary constant. Then :

Multiple Choice Options

  • A g is decreasing in (0,π4)\left( {0,{\pi \over 4}} \right)
  • B g' is increasing in (0,π4)\left( {0,{\pi \over 4}} \right)
  • C g + g' is increasing in (0,π2)\left( {0,{\pi \over 2}} \right)
  • D g −- g' is increasing in (0,π2)\left( {0,{\pi \over 2}} \right)

Step-by-Step Verified Solution

∫(x(cos⁡x−sin⁡x)ex+1+g(x)(ex+1−xex)(ex+1)2)dx=xg(x)ex+1+c\int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right)}{\left(e^x+1\right)^2}\right) d x=\frac{x g(x)}{e^x+1}+c

On differentiating both sides w.r.t. x\mathrm{x} , we get

amp;(x(cos⁡x−sin⁡x)ex+1+g(x)(ex+1−xex(ex+1)2)amp;=(ex+1)(g(x)+xg′(x))−ex⋅x⋅g(x)(ex+1)2amp;(ex+1)x(cos⁡x−sin⁡x)+g(x)(ex+1−xex)amp;=(ex+1)(g(x)+xg′(x))−ex⋅x⋅g(x)amp;⇒gg′(x)=cos⁡x−sin⁡xamp;⇒g(x)=sin⁡x+cos⁡x+C\begin{aligned}&\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\\\&=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\\\&\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\\\&=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\\\&\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\\\&\Rightarrow g(x)=\sin x+\cos x+C\end{aligned}

g(x)\mathrm{g}(\mathrm{x}) is increasing in (0,π/4)(0, \pi / 4)

g "(x)=-\sin x-\cos x<0

⇒g′(x)\Rightarrow \mathrm{g}^{\prime}(\mathrm{x}) is decreasing function

let h(x)=g(x)+g′(x)=2cos⁡x+Ch(x)=g(x)+g^{\prime}(x)=2 \cos x+C

\Rightarrow \mathrm{h}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime \prime}(\mathrm{x})=-2 \sin \mathrm{x}<0

⇒h\Rightarrow \mathrm{h} is decreasing

let ϕ(x)=g(x)−g′(x)=2sin⁡x+C\phi(\mathrm{x})=\mathrm{g}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=2 \sin \mathrm{x}+\mathrm{C}

\Rightarrow \phi^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})-\mathrm{g}{ }^{\prime \prime}(\mathrm{x})=2 \cos \mathrm{x}>0

⇒ϕ\Rightarrow \phi is increasing

Hence option D is correct.

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