Question

Let $g : (0, ∞) →R$ be a differentiable function such that $∫(\frac {x(cos⁡x−sin⁡x)}{e^x+1} + \frac {g(x)(e^x+1−xe^x)}{(e^x+1)2})dx = \frac {xg(x)}{e^x+1}+c,$ for all $x>0$, where c is an arbitrary constant. Then:

• A. g is decreasing in $(0, \frac \pi4)$
• B. g′ is increasing in $(0, \frac \pi4)$
• C. g+g′ is increasing in $(0, \frac \pi2)$
• D. g-g' is increasing in $(0, \frac \pi2)$

Answer 1

Answer: (D)

g-g' is increasing in $(0, \frac \pi2)$

Answer 2

$∫(\frac {x(cos⁡x−sin⁡x)}{e^x+1} + \frac {g(x)(e^x+1−xe^x)}{(e^x+1)2})dx = \frac {xg(x)}{e^x+1}+c,$
Differentiating on both sides
$\frac {x(cos⁡x−sin⁡x)}{e^x+1} + \frac {g(x)(e^x+1−xe^x)}{(e^x+1)2}$

$=\frac {(e^x+1)(g(x)+xg^{\frac 1x})−xg(x)e^x}{(e^x+1)^2}$
$=\frac {g(x)[e^x+1−xe^x]}{(e^x+1)^2} +\frac {xg'(x)(e^x+1)}{(e^x+1)^2}$

$=\frac {x(cos⁡x−sin⁡x)}{e^x+1}$
$= \frac {xg'(x)}{e^x+1}$

⇒ g‘(x)=cos⁡x−sin⁡x>0 in $(0, \frac \pi4)$
⇒ g(x) is increasing in $(0, \frac \pi4)$
⇒ Option (A) is wrong.

Now,
g”(x)=−sin⁡x−cos⁡x<0 in $(0, \frac \pi4)$
⇒ g(x) is increasing in $(0, \frac \pi4)$
⇒ Option (B) is wrong.

Let h(x) = g(x) + g′(x)
⇒ ℎ‘(x)=g‘(x)+g”(x)=−2sin⁡x<0 in x∈$(0, \frac \pi2)$
⇒ g + g' is decreasing in $(0, \frac \pi2)$
⇒ Option (C) is wrong.

Let J(x) = g(x) – g′(x)
J‘(x)=g‘(x)−g”(x)=2cos⁡x>0 in $(0, \frac \pi2)$
⇒ g – g′ is increasing in $(0, \frac \pi2)$
⇒ Option (D) is correct.

So, the correct option is (D): g-g' is increasing in $(0, \frac \pi2)$