If \( 2f(x)+f\left(\displaystyle\frac{1}{x}\right)=\frac{1}{x}-5,\ x\ne 0 \), then \( \int_{1}^{2}\displaystyle f(x)\,dx \) is equal to:
(consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{6}[\log 16-13] \)
Replacing x by \( \frac{1}{x} \) in the given relation gives a second equation:
\( \displaystyle 2f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}-5\quad\ldots(1) \)
\( \displaystyle 2f\left(\frac{1}{x}\right)+f(x)=x-5\quad\ldots(2) \)
Multiplying (1) by 2 and subtracting (2) eliminates \( f\left(\frac{1}{x}\right) \):
\( \displaystyle 4f(x)-f(x)=\frac{2}{x}-10-x+5 \)
\( \displaystyle 3f(x)=\frac{2}{x}-x-5 \)
\( \displaystyle f(x)=\frac{1}{3}\left(\frac{2}{x}-x-5\right) \)
Integrating from 1 to 2,
\( \displaystyle \int_1^2 f(x)\,dx=\frac{1}{3}\left[2\log x-\frac{x^2}{2}-5x\right]_1^2 \)
\( \displaystyle =\frac{1}{3}\left[(2\log 2-2-10)-\left(0-\frac{1}{2}-5\right)\right] \)
\( \displaystyle =\frac{1}{3}\left[2\log 2-12+\frac{11}{2}\right] \)
\( \displaystyle =\frac{1}{3}\left[2\log 2-\frac{13}{2}\right] \)
\( \displaystyle =\frac{1}{6}\left[4\log 2-13\right] \)
\( \displaystyle =\frac{1}{6}[\log 16-13] \)