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

 To solve the given problem, we need to evaluate the properties of the function \( g(x) \) based on the information given in the integral equation:

\(\int\left(\frac {x(\cos x- \sin x)}{e^x+1} + \frac {g(x)(e^x+1-xe^x)}{(e^x+1)^2}\right)dx = \frac {xg(x)}{e^x+1}+c,\)for all \( x > 0 \).

We start by differentiating both sides with respect to \( x \). For the right side, we apply the product rule to the term \( \frac{xg(x)}{e^x+1} \).

Step 1: Differentiate the right hand side

The right-hand side becomes:

\(\frac{d}{dx}\left(\frac{xg(x)}{e^x+1}\right) = \frac{g(x) + xg'(x)}{e^x+1} - \frac{xg(x)e^x}{(e^x+1)^2}\\)

Step 2: Differentiate the left hand side

As per the fundamental theorem of calculus, differentiating the left-hand side under the integral gives:

\(\frac{x(\cos x - \sin x)}{e^x+1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x+1)^2}\\)

Step 3: Equate both sides and simplify

By equating the differentiated left side and the differentiated right side, we get:

\(\frac{x(\cos x - \sin x)}{e^x+1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x+1)^2} = \frac{g(x) + xg'(x)}{e^x+1} - \frac{xg(x)e^x}{(e^x+1)^2}\\)

Simplifying shows:

\(\frac{x(\cos x - \sin x)}{e^x+1} + \frac{g(x)}{e^x+1} - \frac{xg(x)}{e^x+1} = \frac{g(x) + xg'(x)}{e^x+1} - \frac{xg(x)e^x}{(e^x+1)^2}\\)

Further simplification yields:

\(x(\cos x - \sin x) = xg'(x).\)

This implies:

\(g'(x) = \cos x - \sin x.\)

Step 4: Analyze the behavior of \( g(x) \) and \( g'(x) \)

The derivative \( g'(x) = \cos x - \sin x \) suggests that \( g(x) \) is decreasing for \( x \in \left(0, \frac{\pi}{4}\right) \) as \( \cos x - \sin x \) is positive there.

Also, to check the option for \( g - g' \) increasing, calculate:

\(g(x) = \int (\cos x - \sin x) \, dx + C \).\)

The expression \( g(x) - g'(x) = g(x) - (\cos x - \sin x) \) is constant when evaluated over the range as differentiation leads to zero, indicating a constant or increasing function.

Conclusion: The correct statement is "\(g - g'\)is increasing in \( \left(0, \frac{\pi}{2}\right) \)".