Key Concepts and Formulas

• Fundamental Theorem of Calculus (FTC): If $F(x) = \int f(x) dx$, then $F'(x) = f(x)$.
• Product Rule: $(uv)' = u'v + uv'$
• Quotient Rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
• Monotonicity: A function $f(x)$ is increasing if $f'(x) > 0$ and decreasing if $f'(x) < 0$.

Step-by-Step Solution

Step 1: Differentiate both sides with respect to $x$

We are given the integral equation: $\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c}$ We differentiate both sides with respect to $x$ to eliminate the integral.

Applying the Fundamental Theorem of Calculus to the left-hand side (LHS): $\frac{d}{dx} \left[ \int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx} \right] = \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2}$

Differentiating the right-hand side (RHS) using the quotient rule: Let $u = xg(x)$ and $v = e^x + 1$. Then $u' = g(x) + xg'(x)$ (using the product rule) and $v' = e^x$. $\frac{d}{dx} \left( \frac{xg(x)}{e^x + 1} + c \right) = \frac{(g(x) + xg'(x))(e^x + 1) - xg(x)e^x}{(e^x + 1)^2}$

Equating the differentiated LHS and RHS: $\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2} = \frac{(g(x) + xg'(x))(e^x + 1) - xg(x)e^x}{(e^x + 1)^2}$

Step 2: Simplify the equation and solve for $g'(x)$

Multiply both sides by $(e^x + 1)^2$ to eliminate the denominators: $x(\cos x - \sin x)(e^x + 1) + g(x)(e^x + 1 - xe^x) = (g(x) + xg'(x))(e^x + 1) - xg(x)e^x$

Expand the terms: $x(\cos x - \sin x)(e^x + 1) + g(x)e^x + g(x) - xg(x)e^x = g(x)e^x + g(x) + xg'(x)e^x + xg'(x) - xg(x)e^x$

Notice that $g(x)e^x$, $g(x)$, and $-xg(x)e^x$ appear on both sides. Cancel these terms: $x(\cos x - \sin x)(e^x + 1) = xg'(x)(e^x + 1)$

Since $x > 0$ and $e^x + 1 > 0$, we can divide both sides by $x(e^x + 1)$: $g'(x) = \cos x - \sin x$

Step 3: Find $g(x)$ by integrating $g'(x)$

Integrate $g'(x)$ to find $g(x)$: $g(x) = \int (\cos x - \sin x) dx = \sin x + \cos x + C_1$ where $C_1$ is the constant of integration.

Step 4: Analyze the monotonicity of the options

We have:

• $g(x) = \sin x + \cos x + C_1$
• $g'(x) = \cos x - \sin x$
• $g''(x) = -\sin x - \cos x$

Now, analyze each option:

Option (A): $g$ is decreasing in $\left(0, {\pi \over 4}\right)$ Since $g'(x) = \cos x - \sin x > 0$ in $\left(0, {\pi \over 4}\right)$, $g(x)$ is increasing. This option is incorrect.

Option (B): $g'$ is increasing in $\left(0, {\pi \over 4}\right)$ Since $g''(x) = -(\sin x + \cos x) < 0$ in $\left(0, {\pi \over 4}\right)$, $g'(x)$ is decreasing. This option is incorrect.

Option (C): $g + g'$ is increasing in $\left(0, {\pi \over 2}\right)$ Let $h(x) = g(x) + g'(x) = (\sin x + \cos x + C_1) + (\cos x - \sin x) = 2\cos x + C_1$. Then $h'(x) = -2\sin x$. In $\left(0, {\pi \over 2}\right)$, $\sin x > 0$, so $h'(x) < 0$. Thus, $g + g'$ is decreasing. This option is incorrect.

Option (D): $g - g'$ is increasing in $\left(0, {\pi \over 2}\right)$ Let $k(x) = g(x) - g'(x) = (\sin x + \cos x + C_1) - (\cos x - \sin x) = 2\sin x + C_1$. Then $k'(x) = 2\cos x$. In $\left(0, {\pi \over 2}\right)$, $\cos x > 0$, so $k'(x) > 0$. Thus, $g - g'$ is increasing. This option is correct.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when differentiating and integrating.
• Quotient Rule: Double-check the application of the quotient rule, especially the order of terms in the numerator.
• Monotonicity: Remember that $f(x)$ is increasing when $f'(x) > 0$ and decreasing when $f'(x) < 0$.

Summary

We used the Fundamental Theorem of Calculus to differentiate both sides of the given integral equation. This allowed us to simplify the equation and solve for $g'(x)$. We then integrated $g'(x)$ to find $g(x)$. Finally, we analyzed the monotonicity of the functions in each option by examining the signs of their derivatives. We found that $g - g'$ is increasing in $\left(0, {\pi \over 2}\right)$, which corresponds to option (D).

Final Answer The final answer is \boxed{D}.