Question

Let \( f(x) = \dfrac{x-1}{x+1} \), \( f^{(1)}(x) = f(x) \), \( f^{(2)}(x) = f(f(x)) \), and \( g(x) + f^{(2)}(x) = 0 \). The area of the region enclosed by the curves \( y = g(x) \), \( y = 0 \), \( x = 4 \), and \( 2y = 2x - 3 \) is:

• A. \(\frac{1}{4} + \ln 2\)
• B. \(\frac{1}{8} + 2\ln 2\)
• C. \(\frac{1}{4} + 2\ln 2\)
• D. \(\frac{1}{8} + \ln 2\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We first find the function \(f^{(2)}(x)\) by composition. Then we determine \(g(x)\). Finally, we calculate the area using integration between the specified vertical lines and the curves.

Step 2: Key Formula or Approach:
1. \( f^{(2)}(x) = f\left(\frac{x-1}{x+1}\right) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{x-1 - (x+1)}{x-1 + x+1} = \frac{-2}{2x} = -\frac{1}{x} \). 2. \( g(x) = -f^{(2)}(x) = \frac{1}{x} \).

Step 3: Detailed Explanation:
1. The curves are \(y = 1/x\) and \(y = x - 3/2\). 2. Intersection: \(1/x = x - 1.5 \implies 2 = 2x^2 - 3x \implies 2x^2 - 3x - 2 = 0\). Roots are \((2x+1)(x-2) = 0 \implies x=2\) (ignoring negative). 3. The area required is between \(x=2\) and \(x=4\) under the curves as bounded. 4. Area = \(\int_2^4 \frac{1}{x} dx - \int_2^4 (x - 1.5) dx\). (Or similar depending on the specific geometry). Actually, the region is between \(y=g(x)\), \(y=0\), \(x=4\), and the line \(y = x - 1.5\). Area = \(\int_2^4 \frac{1}{x} dx + \text{Triangle Area}\). Upon integration: \(\ln(4) - \ln(2) = \ln 2\). The linear part provides the constant fraction.

Step 4: Final Answer:
The area is \(\frac{1}{8} + \ln 2\).