Dựa vào đồ thị hàm số \(y={f}’\left( 1+x \right)\) ta có \({f}’\left( 1+x \right)=0\).
\(\Leftrightarrow \left[ \begin{align}
& x=0 \\
& x=1 \\
& x=2 \\
\end{align} \right.\)
Đặt \(t=1+x\Rightarrow {f}’\left( t \right)=0.\)
\(\Leftrightarrow \left[ \begin{align}
& t=1 \\
& t=2 \\
& t=3 \\
\end{align} \right.\)
Vậy
\({f}’\left( t \right)\ge 0\Leftrightarrow \left[ \begin{align}
& t\le 1 \\
& 2\le t\le 3 \\
\end{align} \right.\)\(\,,\,{f}’\left( t \right)\le 0\) \(\Leftrightarrow \left[ \begin{align}
& 1\le t\le 2 \\
& t\ge 3 \\
\end{align} \right.\).
\(g\left( x \right)=f\left( -{{x}^{2}}+2x-2022+m \right)\)\( \Rightarrow {g}’\left( x \right)=\left( 2-2x \right){f}’\left( -{{x}^{2}}+2x-2022+m \right)\).
Hàm số \(g\left( x \right)=f\left( -{{x}^{2}}+2x-2022+m \right)\) đồng biến trên \(\left( 0\,;\,1 \right)\)\( \Leftrightarrow \) \(\left( 2-2x \right){f}’\left( -{{x}^{2}}+2x-2022+m \right)\ge 0\,,\,\forall x\in \left( 0;1 \right)\)\( \Leftrightarrow {f}’\left( -{{x}^{2}}+2x-2022+m \right)\ge 0\,,\,\forall x\in \left( 0;1 \right)\).
\( \Leftrightarrow \left[ \begin{array}{l}
– {x^2} + 2x – 2022 + m \le 1\\
\left\{ \begin{array}{l}
– {x^2} + 2x – 2022 + m \ge 2\\
– {x^2} + 2x – 2022 + m \le 3
\end{array} \right.
\end{array} \right.{\kern 1pt} {\kern 1pt} \forall x \in \left( {0;1} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}
m \le {x^2} – 2x + 2023\\
\left\{ \begin{array}{l}
m \ge {x^2} – 2x + 2024\\
m \le {x^2} – 2x + 2025
\end{array} \right.
\end{array} \right.{\kern 1pt} {\kern 1pt} \forall x \in \left( {0;1} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}
m \le 2022\\
2024 \le m \le 2024
\end{array} \right.\)
Vậy có \(2023\) số.