Câu hỏi 17

Ta có \(\left[ {2f\left( x \right) + 1 - {x^2}} \right]f'\left( x \right) = 2x\left[ {1 + f\left( x \right)} \right]\)

\(\begin{array}{l} \Leftrightarrow 2f\left( x \right).f'\left( x \right) + f'\left( x \right)\left( {1 - {x^2}} \right) = 2x.\left( {1 + f\left( x \right)} \right)\\ \Leftrightarrow 2f\left( x \right).f'\left( x \right) = \left( {{x^2} - 1} \right)f'\left( x \right) + 2x\left( {1 + f\left( x \right)} \right)\\ \Leftrightarrow {\left[ {{f^2}\left( x \right)} \right]^\prime } = {\left[ {\left( {{x^2} - 1} \right)\left( {f\left( x \right) + 1} \right)} \right]^\prime }\end{array}\)

Lấy nguyên hàm hai vế ta được \({f^2}\left( x \right) = \left( {{x^2} - 1} \right)\left( {f\left( x \right) + 1} \right) + C\)

Lại có \(f\left( 1 \right) = 1 \Rightarrow 1 = \left( {1 - 1} \right).2 + C \Rightarrow C = 1\)

Nên \({f^2}\left( x \right) = \left( {{x^2} - 1} \right)\left( {f\left( x \right) + 1} \right) + 1\)

\(\begin{array}{l} \Leftrightarrow {f^2}\left( x \right) = {x^2}f\left( x \right) + {x^2} - f\left( x \right)\\ \Leftrightarrow f\left( x \right)\left( {{x^2} - f\left( x \right)} \right) + {x^2} - f\left( x \right) = 0\\ \Leftrightarrow \left( {{x^2} - f\left( x \right)} \right)\left( {f\left( x \right) + 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}f\left( x \right) =  - 1\left( {ktm} \right)\\f\left( x \right) = {x^2}\left( {tm} \right)\end{array} \right.\end{array}\)

Suy ra \(\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {{x^2}dx}  = \dfrac{1}{3}.\)

Chọn C