Đáp án đúng: C
Giải chi tiết:Đặt \(\int\limits_0^1 {\left| {f\left( x \right)} \right|dx}  = a\), hàm số đã cho trở thành \(f\left( x \right) = {x^3} - 4ax\).Vì \(f\left( 1 \right) > 0 \Leftrightarrow 1 - 4a > 0 \Leftrightarrow a < \dfrac{1}{4}\).Xét \(f\left( x \right) = {x^3} - 4xa = 0 \Leftrightarrow x\left( {{x^2} - 4a} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\{x^2} = 4a\end{array} \right.\).+) Với \(a \le 0\) thì phương trình \(f\left( x \right) = 0\) có nghiệm duy nhất\(x = 0\), khi đó ta có \(f\left( x \right) > 0\,\,\forall x \in \left[ {0;1} \right]\) nên \(\begin{array}{l}a = \int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {\left( {{x^3} - 4ax} \right)dx} \\\,\,\,\, = \left. {\left( {\dfrac{{{x^4}}}{4} - 2a{x^2}} \right)} \right|_0^1 = \dfrac{1}{4} - 2a\\ \Leftrightarrow 4a = 1 - 8a \Leftrightarrow 12a = 1 \Leftrightarrow a = \dfrac{1}{{12}}\,\,\left( {ktm} \right)\end{array}\)+) Với \(a > 0\) ta có \(f\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x =  \pm 2\sqrt a \end{array} \right.\).Vì \(a < \dfrac{1}{4} \Leftrightarrow \sqrt a  < \dfrac{1}{2} \Leftrightarrow 2\sqrt a  < 1\).Ta có bảng xét dấu:Khi đó ta có:\(\begin{array}{l}a = \int\limits_0^1 {\left| {f\left( x \right)} \right|dx}  = \int\limits_0^{2\sqrt a } {\left| {f\left( x \right)} \right|dx}  + \int\limits_{2\sqrt a }^1 {\left| {f\left( x \right)} \right|dx} \\\,\,\,\, =  - \int\limits_0^{2\sqrt a } {f\left( x \right)dx}  + \int\limits_{2\sqrt a }^1 {f\left( x \right)dx} \\\,\,\,\, =  - \int\limits_0^{2\sqrt a } {\left( {{x^3} - 4ax} \right)dx}  + \int\limits_{2\sqrt a }^1 {\left( {{x^3} - 4ax} \right)dx} \\\,\,\,\, =  - \left. {\left( {\dfrac{{{x^4}}}{4} - 2a{x^2}} \right)} \right|_0^{2\sqrt a } + \left. {\left( {\dfrac{{{x^4}}}{4} - 2a{x^2}} \right)} \right|_{2\sqrt a }^1\\ \Leftrightarrow a =  - \dfrac{{16{m^2}}}{4} + 8{a^2} + \left( {\dfrac{1}{4} - 2a} \right) - \dfrac{{16{a^2}}}{4} + 8{a^2}\\ \Leftrightarrow 8{a^2} - 3a + \dfrac{1}{4} = 0 \Leftrightarrow \left[ \begin{array}{l}a = \dfrac{1}{4}\,\,\,\left( {ktm} \right)\\a = \dfrac{1}{8}\,\,\left( {tm} \right)\end{array} \right.\end{array}\)\( \Rightarrow a = \dfrac{1}{8} \Rightarrow f\left( x \right) = {x^3} - \dfrac{1}{2}x\).Vậy \(f\left( 4 \right) = 62\).Chọn C