Area Bounded by Function Composition Curve and Lines

To find the area of the region bounded by the given curves, we first determine the explicit form of $g(x)$.

Step 1: Determine the composite function $f^{26}(x)$

The function $f:(1, \infty) \rightarrow \mathbb{R}$ is defined as: $f(x) = \frac{x-1}{x+1}$

Let us compute the successive compositions $f^n(x) = f(f^{n-1}(x))$:

1. $f^1(x) = \frac{x-1}{x+1}$

2. $f^2(x) = f(f(x)) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{(x-1) - (x+1)}{(x-1) + (x+1)} = \frac{-2}{2x} = -\frac{1}{x}$

3. $f^3(x) = f(f^2(x)) = f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{-1-x}{-1+x} = \frac{x+1}{1-x}$

4. $f^4(x) = f(f^3(x)) = f\left(\frac{x+1}{1-x}\right) = \frac{\frac{x+1}{1-x} - 1}{\frac{x+1}{1-x} + 1} = \frac{(x+1) - (1-x)}{(x+1) + (1-x)} = \frac{2x}{2} = x$

Since $f^4(x) = x$, the sequence of functions $f^n(x)$ is periodic with a period of $4$.

Thus, for $n = 26$: $26 \equiv 2 \pmod 4 \implies f^{26}(x) = f^2(x) = -\frac{1}{x}$

Step 2: Find $g(x)$

We are given $g(x) + f^{26}(x) = 0$ for $x \in (1, \infty)$, so: $g(x) = -f^{26}(x) = -\left(-\frac{1}{x}\right) = \frac{1}{x}$

Step 3: Identify the region and boundaries

The region is bounded by the following curves and lines:

1. $y = g(x) = \frac{1}{x}$
2. Line $2y = 2x - 3 \implies y = x - \frac{3}{2}$
3. $y = 0$ (the x-axis)
4. Line $x = 4$

Now, let us find the points of intersection:

• Intersection of $y = x - \frac{3}{2}$ and $y = 0$: $x - \frac{3}{2} = 0 \implies x = \frac{3}{2}$

• Intersection of $y = x - \frac{3}{2}$ and $y = \frac{1}{x}$ for $x \in (1, \infty)$: $x - \frac{3}{2} = \frac{1}{x} \implies 2x^2 - 3x - 2 = 0 \implies (2x + 1)(x - 2) = 0$ Since $x > 1$, we take $x = 2$. The point of intersection is $\left(2, \frac{1}{2}\right)$.

• Intersection of $y = \frac{1}{x}$ and $x = 4$: $y = \frac{1}{4}$

Thus, the region starts at $x = \frac{3}{2}$ and ends at $x = 4$.

• For $x \in \left[\frac{3}{2}, 2\right]$, the region is bounded between the line $y = x - \frac{3}{2}$ and $y = 0$.
• For $x \in [2, 4]$, the region is bounded between the curve $y = \frac{1}{x}$ and $y = 0$.

Step 4: Compute the area

The total area $A$ is given by: $A = \int_{3/2}^{2} \left(x - \frac{3}{2}\right) \, dx + \int_{2}^{4} \frac{1}{x} \, dx$

Calculating each integral separately: $\int_{3/2}^{2} \left(x - \frac{3}{2}\right) \, dx = \left[ \frac{\left(x - \frac{3}{2}\right)^2}{2} \right]_{3/2}^{2} = \frac{\left(2 - \frac{3}{2}\right)^2}{2} - 0 = \frac{\left(\frac{1}{2}\right)^2}{2} = \frac{1}{8}$

$\int_{2}^{4} \frac{1}{x} \, dx = \left[ \log_e x \right]_{2}^{4} = \log_e 4 - \log_e 2 = \log_e \left(\frac{4}{2}\right) = \log_e 2$

Adding the two parts together: $\text{Area} = \frac{1}{8} + \log_e 2$

Thus, the correct option is A.