Let $$f : (1, \infty) \to \mathbf{R}$$ be a function defined as $$f(x) = \frac{x-1}{x+1}$$. Let $$f^{i+1}(x) = f(f^i(x))$$, $$i = 1, 2, ..., 25$$, where $$f^1(x) = f(x)$$. If $$g(x) + f^{26}(x) = 0$$, $$x \in (1, \infty)$$, then the area of the region bounded by the curves $$y = g(x)$$, $$2y = 2x - 3$$, $$y = 0$$ and $$x = 4$$ is :

To solve this problem, we first need to simplify the composition of the function $$f(x)$$ and then calculate the area using integration.

1. Simplify the Iterated Function $$f^{n}(x)$$

Let's find a pattern for $$f^{n}(x)$$ where $$f(x) = \frac{x-1}{x+1}$$.

• $$f^1(x)$$: $$\frac{x-1}{x+1}$$
• $$f^2(x)$$: $$f(f(x)) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} = \frac{(x-1)-(x+1)}{(x-1)+(x+1)} = \frac{-2}{2x} = -\frac{1}{x}$$
• $$f^3(x)$$: $$f(f^2(x)) = \frac{-1/x - 1}{-1/x + 1} = \frac{-1-x}{-1+x} = \frac{x+1}{1-x}$$
• $$f^4(x)$$: $$f(f^3(x)) = \frac{\frac{x+1}{1-x}-1}{\frac{x+1}{1-x}+1} = \frac{(x+1)-(1-x)}{(x+1)+(1-x)} = \frac{2x}{2} = x$$
• $$y = g(x) = \frac{1}{x}$$
• $$2y = 2x - 3 \implies y = x - 1.5$$
• $$y = 0$$ (x-axis)
• $$x = 4$$
• $$y = x - 1.5$$ and $$y = 0$$ intersect at $$x = 1.5$$.
• $$y = 1/x$$ and $$y = x - 1.5$$ intersect when $$\frac{1}{x} = x - \frac{3}{2} \implies 2 = 2x^2 - 3x \implies 2x^2 - 3x - 2 = 0$$.
• From $$x = 1.5$$ to $$x = 2$$: Area under the line $$y = x - 1.5$$.
• From $$x = 2$$ to $$x = 4$$: Area under the curve $$y = 1/x$$.

The function is periodic with a period of 4.

Since $$26 = 4 \times 6 + 2$$, we have:

$$f^{26}(x) = f^2(x) = -\frac{1}{x}$$

2. Determine $$g(x)$$

The problem states $$g(x) + f^{26}(x) = 0$$:

$$g(x) - \frac{1}{x} = 0 \implies g(x) = \frac{1}{x}$$

3. Find the Bounded Region

We need the area bounded by:

Finding Intersection Points:

Factoring: $$(2x+1)(x-2) = 0$$. Since $$x \in (1, \infty)$$, the intersection is at $$x = 2$$.

4. Calculate the Area

The region consists of two parts:

$$\text{Area} = \int_{1.5}^{2} (x - 1.5) \, dx + \int_{2}^{4} \frac{1}{x} \, dx$$

Part 1:

$$\left[ \frac{x^2}{2} - 1.5x \right]_{1.5}^{2} = \left( \frac{4}{2} - 3 \right) - \left( \frac{2.25}{2} - 2.25 \right) = -1 - (-1.125) = 0.125 = \frac{1}{8}$$

Part 2:

$$\left[ \ln x \right]_{2}^{4} = \ln 4 - \ln 2 = \ln\left(\frac{4}{2}\right) = \ln 2$$

Total Area:

$$\frac{1}{8} + \ln 2$$

Correct Option: A