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: Idea.
We will integrate the given relation directly, without finding $f(x)$. Let $I=\int_{1}^{2}f(x)dx$ and $J=\int_{1}^{2}f\left(\frac{1}{x}\right)dx$.

Step 2: Integrate the given equation.
Integrate $2f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}-5$ from 1 to 2.
\[ 2I+J=\int_{1}^{2}\left(\frac{1}{x}-5\right)dx=\log 2-5 \]

Step 3: Second relation.
Now use the second form $2f\left(\frac{1}{x}\right)+f(x)=x-5$. Integrating it from 1 to 2 gives
\[ 2J+I=\int_{1}^{2}(x-5)dx=\frac{3}{2}-5=-\frac{7}{2} \]

Step 4: Solve the pair.
We have $2I+J=\log 2-5$ and $I+2J=-\frac{7}{2}$.
Multiply the first by 2: $4I+2J=2\log 2-10$.
Subtract the second: $3I=2\log 2-10+\frac{7}{2}=2\log 2-\frac{13}{2}$.
\[ I=\frac{4\log 2-13}{6}=\frac{\log 16-13}{6} \]

Step 5: Compare.
This equals option 1. The other options carry a different constant or coefficient, so they cannot match.

Final Answer:
Option 1 is correct. \[ \boxed{\frac{1}{6}\left[\log 16-13\right]} \]