Lời giải

•  \( 12 =  \int_{-1}^{0} f(-x) \, dx = -\int_{-1}^{0} f(x) \, dx = \int_{0}^{-1} f(x) \, dx\)  

•  Đặt \( t = -2x \Rightarrow dt = -2 \, dx \)

\( \begin{cases}
x = \frac{1}{2} \Rightarrow t = -1 \\
x = 1 \Rightarrow t = -2
\end{cases} \)    

\( \int_{\frac{1}{2}}^1 f(-2x) \, dx = \int_{-1}^{-2} f(t) \left(-\frac{1}{2} \, dt \right) = -\frac{1}{2} \int_{-1}^{-2} f(t) \, dt \)

                            \(= \frac{1}{2} \int_{-2}^{-1} f(t) \, dt = 2 \)              

                            \( \Rightarrow \int_{-2}^{-1} f(t) \, dt = 4 \)

\( 12 + 4 = \int_{0}^{-1} f(x) \, dx + \int_{-1}^{-2} f(x) \, dx = \int_{0}^{-2} f(x) \, dx = -\int_{-2}^0 f(x) \, dx\)

              \(= \int_{0}^2 f(x) \, dx = 16\)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)