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

The Correct Option is D

\(∫(\frac {x(cos⁡x−sin⁡x)}{e^x+1} + \frac {g(x)(e^x+1−xe^x)}{(e^x+1)2})dx = \frac {xg(x)}{e^x+1}+c,\)
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 \((0, \frac \pi2)\)