Question

Let \( f(x) \) be a polynomial such that \( f(x) + f(1/x) = f(x)f(1/x) \), \( x > 0 \). If \( \int f(x)\,dx = g(x) + c \) and \( g(1) = \frac{4}{3} \), \( f(3) = 10 \), then \( g(3) \) is:

• A. 10
• B. 9
• C. 8
• D. 12

Hint:

For \(f(x) + f(1/x) = f(x)f(1/x)\), rewrite as \([f(x)-1][f(1/x)-1] = 1\). Then \(f(x)-1 = \pm x^n\).

Answer 1

The Correct Option is D

Concept: Given functional equation: \[ f(x) + f\left(\frac{1}{x}\right) = f(x) f\left(\frac{1}{x}\right) \]

Step 1: Rewrite the equation. \[ f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) = 0 \] \[ \Rightarrow f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) + 1 = 1 \] \[ \Rightarrow \left[f(x) - 1\right]\left[f\left(\frac{1}{x}\right) - 1\right] = 1 \] Let \(h(x) = f(x) - 1\). Then: \[ h(x) \cdot h\left(\frac{1}{x}\right) = 1 \]

Step 2: Determine the form of \(h(x)\). For a polynomial \(f(x)\), \(h(x)\) is also a polynomial. The condition \(h(x) \cdot h(1/x) = 1\) for all \(x>0\) implies that \(h(x)\) must be of the form: \[ h(x) = \pm x^n \] where \(n\) is an integer. Thus: \[ f(x) = 1 + x^n \quad \text{or} \quad f(x) = 1 - x^n \]

Step 3: Use \(f(3) = 10\). Try \(f(x) = 1 + x^n\): \[ 1 + 3^n = 10 \Rightarrow 3^n = 9 \Rightarrow n = 2 \] So \(f(x) = 1 + x^2\). Try \(f(x) = 1 - x^n\): \[ 1 - 3^n = 10 \Rightarrow -3^n = 9 \Rightarrow 3^n = -9 \quad \text{(no real solution)} \] Thus: \[ f(x) = x^2 + 1 \]

Step 4: Find \(g(x)\). \[ g(x) = \int f(x) \, dx = \int (x^2 + 1) \, dx = \frac{x^3}{3} + x + C \] Given \(g(1) = \frac{4}{3}\): \[ g(1) = \frac{1}{3} + 1 + C = \frac{4}{3} + C = \frac{4}{3} \Rightarrow C = 0 \] Thus: \[ g(x) = \frac{x^3}{3} + x \]

Step 5: Compute \(g(3)\). \[ g(3) = \frac{27}{3} + 3 = 9 + 3 = 12 \]