Câu hỏi 25

Ta có: \(f'\left( x \right) = \dfrac{2}{{2x - 1}}\)

\( \Rightarrow f\left( x \right) = \int {\dfrac{2}{{2x - 1}}dx}  = \ln \left| {2x - 1} \right| + C\)\( = \left[ \begin{array}{l}\ln \left( {2x - 1} \right) + {C_1}\,\,\,khi\,\,x \ge \dfrac{1}{2}\\\ln \left( {1 - 2x} \right) + {C_2}\,\,\,khi\,\,x < \dfrac{1}{2}\end{array} \right.\).

Với \(x = 0\) ta có \(f\left( 0 \right) = \ln 1 + {C_2} = 1\)\( \Rightarrow {C_2} = 1\).

Với \(x = 1\) ta có \(f\left( 1 \right) = \ln 1 + {C_1} = 2 \Rightarrow {C_1} = 2\).

\(\begin{array}{l} \Rightarrow f\left( x \right) = \left[ \begin{array}{l}\ln \left( {2x - 1} \right) + 2\,\,\,khi\,\,x \ge \dfrac{1}{2}\\\ln \left( {1 - 2x} \right) + 1\,\,\,khi\,\,x < \dfrac{1}{2}\end{array} \right.\\ \Rightarrow f\left( { - 1} \right) = \ln 3 + 1;\,\,f\left( 3 \right) = \ln 5 + 2\\ \Rightarrow f\left( { - 1} \right) + f\left( 3 \right) = \ln 3 + \ln 5 + 3 = 3 + ln15\end{array}\) 

Chọn C.