Question

Let

\[ f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 \] 

Then

\[ \int_{1}^{2} 3f(x)\,dx \]

equals ______.

• A. \(2 + \ln 2\)
• B. \(2 - \ln 2\)
• C. \(2\)
• D. \(3 \ln 2\)
• E. \(\ln 2\)

Hint:

Always form a second equation using \(x \to 1/x\) when both \(f(x)\) and \(f(1/x)\) appear.

Answer 1

The Correct Option is B

Concept: When a functional equation contains \(f(x)\) and \(f(1/x)\), we form a second equation by replacing \(x\) with \(1/x\), then solve simultaneously.

Step 1: Write the given equation. \[ f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 5 ...(1) \]

Step 2: Replace \(x\) by \(1/x\). \[ f\left(\frac{1}{x}\right) + 2f(x) = x - 5 ...(2) \]

Step 3: Solve the two equations. Multiply equation (2) by 2: \[ 2f\left(\frac{1}{x}\right) + 4f(x) = 2x - 10 ...(3) \] Now subtract (1) from (3): \[ [2f(1/x) + 4f(x)] - [f(x) + 2f(1/x)] = (2x - 10) - \left(\frac{1}{x} - 5\right) \] Simplifying LHS: \[ 4f(x) - f(x) = 3f(x) \] Simplifying RHS: \[ 2x - 10 - \frac{1}{x} + 5 = 2x - \frac{1}{x} - 5 \] Thus: \[ 3f(x) = 2x - \frac{1}{x} - 5 \]

Step 4: Set up the integral. \[ \int_1^2 3f(x)\,dx = \int_1^2 \left(2x - \frac{1}{x} - 5\right) dx \]

Step 5: Integrate each term separately. \[ \int 2x\,dx = x^2 \] \[ \int \frac{1}{x}\,dx = \ln x \] \[ \int 5\,dx = 5x \] So, \[ \int_1^2 3f(x)\,dx = \left[x^2 - \ln x - 5x\right]_1^2 \]

Step 6: Evaluate at limits. At \(x=2\): \[ 4 - \ln 2 - 10 = -6 - \ln 2 \] At \(x=1\): \[ 1 - 0 - 5 = -4 \] Now subtract: \[ (-6 - \ln 2) - (-4) = -2 - \ln 2 \] \[ = 2 - \ln 2 \]