Question

If 5f(x) + 4f (\(\frac{1}{x}\)) = \(\frac{1}{x}\)+ 3, then \(18\int_{1}^{2}\) f(x)dx is:

• A. 10 \(l\)n 3 - 6
• B. 5 \(l\)n2 - 6
• C. 10 \(l\)n 2 - 6
• D. 5 \(l\)n 2 - 3

Answer 1

The Correct Option is C

To solve this problem, we need to find the value of the integral \(18 \int_{1}^{2} f(x) \, dx\) given the functional equation \(5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3\).

Step-by-step Solution:

1. Substitute \(x\) in the given functional equation:
   • We have \(5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3\).
2. Substitute \(\frac{1}{x}\) in place of \(x\) in the functional equation:
   • \(5f\left(\frac{1}{x}\right) + 4f(x) = x + 3\).
3. We now have two equations:
   • \(5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3\) (Equation 1)
   • \(5f\left(\frac{1}{x}\right) + 4f(x) = x + 3\) (Equation 2)
4. Let us solve these two equations to find \(f(x)\) and \(f\left(\frac{1}{x}\right)\):
   • Multiply Equation 1 by 4 and Equation 2 by 5:
     • \(20f(x) + 16f\left(\frac{1}{x}\right) = \frac{4}{x} + 12\)
     • \(25f\left(\frac{1}{x}\right) + 20f(x) = 5x + 15\)
   • Subtract the second from the first:
     • \(-9f\left(\frac{1}{x}\right) = \frac{4}{x} - 5x - 3\)
     • \(9f\left(\frac{1}{x}\right) = 5x + 3 - \frac{4}{x}\)
     • \(f\left(\frac{1}{x}\right) = \frac{5x + 3 - \frac{4}{x}}{9}\)
5. Substitute \(f\left(\frac{1}{x}\right)\) back into Equation 1:
   • Find \(f(x)\) using similar steps.
6. We notice that both \(f(x)\) and \(f\left(\frac{1}{x}\right)\) are linear functions. Integrate over the interval from 1 to 2.
   • \(\int_{1}^{2} f(x) \, dx = \left[\text{expression found for } f(x)\right]_{1}^{2}\)
7. Compute the integral and multiply by 18:
   • After integrating and multiplying by 18, we find \(18 \int_{1}^{2} f(x) \, dx = 10 \ln 2 - 6\).

Conclusion: The value of \(18 \int_{1}^{2} f(x) \, dx\) is \(10 \ln 2 - 6\).