Mathematics · 2022
JEE Main · 25 June 2022, Shift 1 · Q12
Let g:(0, ∞) → R be a differentiable function such that ∫((x( cos x- sin x))/(e^x+1)+(g(x)( e^x+1-x e^x))/(( e^x+1)^2)) d x=(x g(x))/(e^x+1)+ c, for…
Let $\displaystyle g:(0, \infty) \rightarrow \mathbf{R}$ be a differentiable function such that
$\displaystyle \int\left(\frac{x(\cos x-\sin x)}{\mathrm{e}^x+1}+\frac{g(x)\left(\mathrm{e}^x+1-x \mathrm{e}^x\right)}{\left(\mathrm{e}^x+1\right)^2}\right) \mathrm{d} x=\frac{x g(x)}{\mathrm{e}^x+1}+\mathrm{c}$, for all $\displaystyle x>0$, where c is an arbitrary constant. Then :
Official answer
(4)
$\displaystyle g-g^{\prime}$ is increasing in $\displaystyle \left(0, \frac{\pi}{2}\right)$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.