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: Find pattern of composition Given \[ f(x)=\frac{x-1}{x+1} \] \[ f^2(x)=f(f(x))=-\frac1x \] \[ f^3(x)=f\left(-\frac1x\right)=\frac{x+1}{1-x} \] \[ f^4(x)=x \] Hence cycle length is \(4\). \[ f^{26}(x)=f^{24+2}(x)=f^2(x)=-\frac1x \] Given \[ g(x)+f^{26}(x)=0 \] \[ g(x)=\frac1x \] Step 2: Find bounded area Line: \[ 2y=2x-3 \] \[ y=x-\frac32 \] Intersection with hyperbola: \[ \frac1x=x-\frac32 \] \[ 2=2x^2-3x \] \[ 2x^2-3x-2=0 \] \[ (x-2)(2x+1)=0 \] \[ x=2 \] Area: \[ A=\int_{3/2}^{2}\left(x-\frac32\right)dx+\int_2^4\frac1x dx \] First part: \[ =\frac18 \] Second part: \[ =\ln4-\ln2=\ln2 \] Hence \[ \boxed{A=\frac18+\ln2} \]