Question

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

• A. \(\frac{1}{8} + \log_e 2\)
• B. \(\frac{1}{4} + \log_e 2\)
• C. \(\frac{5}{6} + 3 \log_e 2\)
• D. \(\frac{5}{6} + \log_e 2\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
First, we find the cycle of the function iterations to determine \(f^{26}(x)\). Then we find \(g(x)\) and compute the area using definite integration over the bounded region.
Step 2: Key Formula or Approach:
1. Iterate $f(x)$ to see if it repeats.
2. Area = \(\int_a^b (y_{upper} - y_{lower}) dx\).
Step 3: Detailed Explanation:
\(f^1(x) = \frac{x-1}{x+1}\) \(f^2(x) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} = \frac{-2}{2x} = -1/x\) \(f^3(x) = \frac{-1/x-1}{-1/x+1} = \frac{-1-x}{-1+x} = \frac{x+1}{1-x}\) \(f^4(x) = f(\frac{x+1}{1-x}) = \frac{\frac{x+1}{1-x}-1}{\frac{x+1}{1-x}+1} = \frac{2x}{2} = x\). The function has a period of 4. \(f^{26}(x) = f^2(x) = -1/x\). Since \(g(x) + f^{26}(x) = 0\), \(g(x) = 1/x\). The region is bounded by \(y = 1/x\), \(y = x - 1.5\), \(y=0\), and \(x=4\). The integration is split at the intersection of $y=x-1.5$ and $y=0$ ($x=1.5$).
Step 4: Final Answer:
The area is \(\frac{1}{8} + \log_e 2\).