Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 2\left| x \right| + 1,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,2,\,\,x = 0\end{array} \right.\).  Chọn phát biểu đúng?

Xét hàm số: \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 2\left| x \right| + 1,\,\,\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\end{array} \right.\)

TXĐ : \(D = \mathbb{R} \Rightarrow \) đáp án C sai.

+) \(f\left( { - x} \right) = {\left( { - x} \right)^2} - 2\left| { - x} \right| + 1\) \( = {x^2} - 2\left| x \right| + 1\)\( = f\left( x \right)\)\( \Rightarrow f\left( x \right)\) là hàm số chẵn \( \Rightarrow \) đáp án A sai.

+) \(f\left( { - 3} \right) + f\left( 3 \right)\)\( = {\left( { - 3} \right)^2} - 2\left| { - 3} \right| + 1 + {3^2} - 2\left| 3 \right| + 1 = 8\) \( \Rightarrow \)Đáp án B sai.

+) Lấy \({x_1};{x_2} \in \left( {1; + \infty } \right)\), có

\(\begin{array}{l}\dfrac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}}\\ = \dfrac{{x_2^2 - 2\left| {{x_2}} \right| + 1 - \left( {x_1^2 - 2\left| {{x_1}} \right| + 1} \right)}}{{{x_2} - {x_1}}}\\ = \dfrac{{x_2^2 - x_1^2 - 2\left| {{x_2}} \right| + 2\left| {{x_1}} \right| + 1 - 1}}{{{x_2} - {x_1}}}\\ = \dfrac{{x_2^2 - x_1^2 - 2{x_2} + 2{x_1}}}{{{x_2} - {x_1}}}\,\,\,\,\left( {{x_1};{x_2} \in \left( {1; + \infty } \right)} \right)\\ = \dfrac{{\left( {x_2^{} - x_1^{}} \right)\left( {x_2^{} + x_1^{}} \right) - 2\left( {{x_2} - {x_1}} \right)}}{{{x_2} - {x_1}}}\\  = {x_2} + {x_1} - 2\end{array}\)

Mà \({x_1};{x_2} \in \left( {1; + \infty } \right)\)\( \Rightarrow {x_2} + {x_1} - 2 > 0 \)

\(\Rightarrow \) Hàm số đồng biến trên \(\left( {1; + \infty } \right)\) \( \Rightarrow \) Đáp án D đúng.