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

(D) $\mathrm{g}-\mathrm{g}^{\prime}$ is increasing in $\left(0, \frac{\pi}{2}\right)$

(A) $g$ is decreasing in $\left(0, \frac{\pi}{4}\right)$

(B) $\mathrm{g}^{\prime}$ is increasing in $\left(0, \frac{\pi}{4}\right)$

(C) $\mathrm{g}+\mathrm{g}^{\prime}$ is increasing in $\left(0, \frac{\pi}{2}\right)$

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Indefinite Integrals

Let

g : (0, \infty) \to R

be a differentiable function such that

\int (\frac{x ( c o s x - s i n x )}{e ^{x} + 1} + \frac{g ( x ) ( e ^{x} + 1 - xe ^{x} )}{( e ^{x} + 1 ) ^{2}}) dx = \frac{xg ( x )}{e ^{x} + 1} + c

,
for all

x > 0

, where c is an arbitrary constant. Then.

Answer: (D)

Differentiating both sides of the integral equation with respect to

x

, we get the integrand equals the derivative of the RHS. After simplification, this yields

g^{'} (x) = cos x - sin x

. Integrating:

g (x) = sin x + cos x + C

. Now

g (x)

is increasing in

(0, π /4)

since

g^{'} (x) > 0

there. Also

g^{''} (x) = - sin x - cos x < 0

, so

g^{'} (x)

is decreasing. Let

h (x) = g (x) - g^{'} (x) = 2 sin x + C

. Then

h^{'} (x) = 2 cos x > 0

(0, π /2)

, so

g - g^{'}

is increasing in

(0, π /2)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option