Lời giải:
Để hàm đồng biến trên $R$ thì:

$m+1>0$

$\Leftrightarrow m>-1$

Mà $m$ nguyên và $m\in [-3;3]$ nên $m\in\left\{0;1;2;3\right\}$

Vậy có 4 giá trị thỏa mãn.

Đáp án là A

Đáp án A

 Với mọi m phương trình luôn có nghiệm x = 0

Ta có: \(f'\left(x\right)=3x^2+2\ge2;\forall x\)

Đặt \(g\left(x\right)=f\left(f\left(x\right)\right)-x\Rightarrow g'\left(x\right)=f'\left(x\right).f'\left(f\left(x\right)\right)-1\ge2.2-1>0;\forall x\)

\(\Rightarrow g\left(x\right)\) đồng biến trên R

\(\Rightarrow\min\limits_{\left[2;6\right]}g\left(x\right)=g\left(2\right)=f\left(f\left(2\right)\right)-2\)

Ta cần tìm m để \(f\left(f\left(2\right)\right)-2\ge0\)

Đặt \(5^m=t\Rightarrow f\left(2\right)=12-t\)

\(\left(1\right)\Leftrightarrow\left(12-t\right)^3+2\left(12-t\right)-t-2\ge0\) 

\(\Leftrightarrow\left(10-t\right)\left(t^2-26t+175\right)\ge0\)

\(\Rightarrow t\le10\)

\(\Rightarrow5^m\le10\Rightarrow m\le log_510\)

\(g\left(x\right)=x^4-4x^3+4x^2+a\)

\(g'\left(x\right)=4x^3-12x^2+8x=0\Leftrightarrow4x\left(x^2-3x+2\right)\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

\(f\left(0\right)=f\left(2\right)=\left|a\right|\) ; \(f\left(1\right)=\left|a+1\right|\)

TH1: \(\left\{{}\begin{matrix}M=\left|a\right|\\m=\left|a+1\right|\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|a\right|\ge\left|a+1\right|\\\left|a\right|\le2\left|a+1\right|\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\dfrac{2}{3}\le a\le-\dfrac{1}{2}\\a\le-2\end{matrix}\right.\) \(\Rightarrow a=\left\{-3;-2\right\}\)

TH2: \(\left\{{}\begin{matrix}M=\left|a+1\right|\\m=\left|a\right|\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|a+1\right|\ge\left|a\right|\\\left|a+1\right|\le2\left|a\right|\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}\le a\le-\dfrac{1}{3}\\a\ge1\end{matrix}\right.\) \(\Rightarrow a=\left\{1;2;3\right\}\)