Cho hàm số y=f(x)có đạo hàm, liên tục trên R

Đáp án C

Ta có: \(y = {x^2}f\left( {2{\rm{x}} - 1} \right) \Rightarrow y' = 2{\rm{x}}f\left( {2{\rm{x}} - 1} \right) + 2f'\left( {2{\rm{x}} - 1} \right){x^2}\)

Thay \(x = 1 \Rightarrow {k_2} = 2f\left( 1 \right) + 2f'\left( 1 \right)\), mặt khác \({k_1} = f'\left( 1 \right)\)

Do \({d_1} \bot {{\rm{d}}_2}\) nên \({k_1}.{k_2} = - 1 \Leftrightarrow 2f\left( 1 \right).f'\left( 1 \right) + 2{f'^2}\left( 1 \right) = - 1\)

\( \Leftrightarrow {f'^2}\left( 1 \right) + f\left( 1 \right).f'\left( 1 \right) = - \frac{1}{2}\)

Suy ra \({f'^2}\left( 1 \right) + f\left( 1 \right).f'\left( 1 \right) + \frac{{{f^2}\left( 1 \right)}}{4} = \frac{{{f^2}\left( 1 \right)}}{4} - \frac{1}{2}\)

\( \Leftrightarrow {\left[ {f'\left( 1 \right) + \frac{{f\left( 1 \right)}}{2}} \right]^2} = \frac{{{f^2}\left( 1 \right)}}{4} - \frac{1}{2} \ge 0 \Rightarrow {f^2}\left( 1 \right) \ge 2\)

\( \Leftrightarrow \left| {f\left( 1 \right)} \right| \ge \sqrt 2 \).