Câu hỏi 36

Ta  có: \(VT = \left[ {f\left( x \right).f'\left( x \right)} \right]' = f'\left( x \right).f'\left( x \right) + f\left( x \right).f''\left( x \right) = {\left[ {f'\left( x \right)} \right]^2} + f\left( x \right).f''\left( x \right)\)

\( \Rightarrow \left[ {f'\left( x \right).f\left( x \right)} \right]' = {x^3} - 2x\;\;\;\left( * \right)\)

Nguyên hàm hai vế của \(\left( * \right)\) ta được: \(f'\left( x \right).f\left( x \right) = \dfrac{{{x^4}}}{4} - {x^2} + C.\;\;\left( 1 \right)\)

Lại có: \(f'\left( 0 \right) = f\left( 0 \right) = 2 \Rightarrow C = 2.2 = 4.\) 

\(\begin{array}{l} \Rightarrow \left( 1 \right) \Leftrightarrow f\left( x \right).f'\left( x \right) = \dfrac{{{x^4}}}{4} - {x^2} + 4\\ \Rightarrow \int {f\left( x \right)f'\left( x \right)dx = \int {\left( {\dfrac{{{x^4}}}{4} - {x^2} + 4} \right)} } dx \Leftrightarrow \int {f\left( x \right)df\left( x \right) = \dfrac{{{x^5}}}{{20}} - \dfrac{{{x^3}}}{3} + 4x + A} \\ \Leftrightarrow \dfrac{{{f^2}\left( x \right)}}{2} = \dfrac{{{x^5}}}{{20}} - \dfrac{{{x^3}}}{3} + 4x + A \Leftrightarrow {f^2}\left( x \right) = \dfrac{{{x^5}}}{{10}} - \dfrac{{2{x^3}}}{3} + 8x + 2A.\end{array}\)

Có \(f\left( 0 \right) = 2 \Rightarrow 4 = 2A \Leftrightarrow A = 2\)

\(\begin{array}{l} \Rightarrow {f^2}\left( x \right) = \dfrac{{{x^5}}}{{10}} - \dfrac{{2{x^3}}}{3} + 8x + 4\\ \Rightarrow {f^2}\left( 2 \right) = \dfrac{{{2^5}}}{{10}} - \dfrac{{{{2.2}^3}}}{3} + 8.2 + 4 = \dfrac{{268}}{{15}}.\end{array}\)

Chọn A.