Question

If \(2f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}-5,\ x\neq 0\), then \(\displaystyle\int_{1}^{2}f(x)dx\) is equal to:
(consider \(\log_e x = \log x\))

• A. \(\frac{1}{6}\left[\log 16-13\right]\)
• B. \(\frac{1}{3}\left[\log 8-13\right]\)
• C. \(\frac{1}{3}\left[\log 8-14\right]\)
• D. \(\left[\log 16-14\right]\)

Hint:

Replace \(x\) by \(1/x\) to get a second equation, solve for \(f(x)=\frac{1}{3}\left(\frac{2}{x}-x-5\right)\), then integrate.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given one equation linking \(f(x)\) and \(f\left(\frac{1}{x}\right)\). We need \(f(x)\) itself before we can integrate it from 1 to 2.

Step 2: Key Formula or Approach:
Replace \(x\) by \(\frac{1}{x}\) to get a second equation. Then solve the two equations for \(f(x)\) like a pair of simultaneous equations.

Step 3: Form the two equations.
Equation (1): \(2f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}-5\).
Replace \(x\) by \(\frac{1}{x}\): Equation (2): \(2f\left(\frac{1}{x}\right)+f(x)=x-5\).

Step 4: Solve for \(f(x)\).
Multiply equation (1) by 2: \(4f(x)+2f\left(\frac{1}{x}\right)=\frac{2}{x}-10\).
Subtract equation (2): \(3f(x)=\frac{2}{x}-10-x+5=\frac{2}{x}-x-5\).
\[ f(x)=\frac{1}{3}\left(\frac{2}{x}-x-5\right) \]

Step 5: Integrate from 1 to 2.
\[ \int_{1}^{2}f(x)dx=\frac{1}{3}\left[2\log x-\frac{x^{2}}{2}-5x\right]_{1}^{2} \]
At \(x=2\): \(2\log 2-2-10=2\log 2-12\).
At \(x=1\): \(0-\frac{1}{2}-5=-\frac{11}{2}\).
Difference: \(2\log 2-12+\frac{11}{2}=2\log 2-\frac{13}{2}\).
So the integral is \(\frac{1}{3}\left(2\log 2-\frac{13}{2}\right)=\frac{4\log 2-13}{6}=\frac{\log 16-13}{6}\).

Step 6: Check the other options.
Options 2 and 3 have the factor \(\frac{1}{3}\) with \(\log 8\), which would come from a wrong coefficient of \(\log 2\). Option 4 has no factor \(\frac{1}{6}\). Only option 1 matches our value.

Final Answer:
The value is \(\frac{1}{6}\left[\log 16-13\right]\), option 1. \[ \boxed{\frac{1}{6}\left[\log 16-13\right]} \]