JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q12

← JEE Main 2022 · 25 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 12 of 90

Let $g:(0,\infty)\to\mathbb R$ be a differentiable function such that
$\int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-xe^x\right)}{\left(e^x+1\right)^2}\right)dx=\frac{x\,g(x)}{e^x+1}+c,$
for all $x>0$ , where $c$ is an arbitrary constant. Then :

g

is decreasing in

\left(0,\dfrac\pi4\right)

g'

is increasing in

\left(0,\dfrac\pi4\right)

g+g'

is increasing in

\left(0,\dfrac\pi2\right)

g-g'

is increasing in

\left(0,\dfrac\pi2\right)

Answer and worked solution

Official answer: D

Differentiate the right side. $\dfrac d{dx}\dfrac{xg}{e^x+1}=\dfrac{(g+xg')(e^x+1)-xge^x}{(e^x+1)^2}$ .

Compare with the integrand: the $g$ terms match, leaving $g'(x)=\cos x-\sin x$ , so $g=\sin x+\cos x+C$ .

Check. $g-g'=2\sin x+C$ , increasing on $\left(0,\frac\pi2\right)$ . (Here $g'>0$ on $(0,\frac\pi4)$ , $g''<0$ , and $g+g'=2\cos x+C$ decreases, so A, B, C fail.)

(D)

Topics:Integration by parts Monotonicity

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JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 12