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Let g:(0,∞ )→ R be a differentiable function such that ∫ ( x( cosx- sinx) e^x+1+ g(x)(e^x+1-xe^x) (e^x+1)^2)dx= xg(x) e^x+1+C, for all x>0, where C…

JEE Main 2022 — Mathematics Calculus

2022mcqhard

Let g:(0,∞)→Rg:(0,\infty )\rightarrow Rg:(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+C\displaystyle \int (\frac{x(\mathrm{cos}x-\mathrm{sin}x)}{{e}^{x}+1}+\frac{g(x)({e}^{x}+1-x{e}^{x})}{{({e}^{x}+1)}^{2}})dx=\frac{xg(x)}{{e}^{x}+1}+C∫(ex+1x(cosx−sinx)​+(ex+1)2g(x)(ex+1−xex)​)dx=ex+1xg(x)​+C, for all x>0x>0x>0, where CCC is an arbitrary constant. Then

Official previous-year question

Held on 25 Jun 2022 · Verified 6 Jul 2026.

Options

  1. A

    ggg is decreasing in (0,π4)(0,\frac{\pi }{4})(0,4π​)

  2. B

    g−g′g-{g}^{'}g−g′ is increasing in (0,π2)(0,\frac{\pi }{2})(0,2π​)

  3. C

    g′{g}^{'}g′ is increasing in (0,π4)(0,\frac{\pi }{4})(0,4π​)

  4. D

    g+g′g+{g}^{'}g+g′ is increasing in (0,π2)(0,\frac{\pi }{2})(0,2π​)

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