Câu hỏi 10

Với mọi \(x \ne 0\), ta có: \(f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x\,\, \Rightarrow f\left( {\frac{1}{x}} \right) + 2f\left( x \right) = \frac{3}{x}\,\)\(\begin{array}{l} \Rightarrow \left( {f\left( x \right) + 2f\left( {\frac{1}{x}} \right)} \right) - 2\left( {f\left( {\frac{1}{x}} \right) + 2f\left( x \right)} \right) = 3x - \frac{6}{x} \Leftrightarrow  - 3f\left( x \right) = 3x - \frac{6}{x}\\ \Leftrightarrow f\left( x \right) = \frac{2}{x} - x\end{array}\)

Khi đó: \(I = \int\limits_1^2 {f\left( x \right)dx}  = \int\limits_1^2 {\left( {\frac{2}{x} - x} \right)dx}  = \left. {\left( {2\ln \left| x \right| - \frac{1}{2}{x^2}} \right)} \right|_1^2 = 2\ln 2 - \frac{3}{2}\).

Chọn: C