Cho hàm số $f(x)=\dfrac{{{x}^{5}}}{5}-{{x}^{2}}+(m-1)x-4029$. Có...

Đặt $h\left( x \right)=f\left( x-1 \right)+2023$.
Ta có $y=\left| f\left( x-1 \right)+2023 \right|$ = $\left| h(x) \right|=\sqrt{h{{(x)}^{2}}}$
$\begin{aligned}
& y'=\dfrac{h(x).h'(x)}{\left| h(x) \right|}\le 0 \forall x<2 \\
& \leftrightarrow \left\{ \begin{matrix}
h(x)<0 \\
h'(x)\ge 0 \\
\end{matrix} \forall x<2 hoac \right.\left\{ \begin{matrix}
h(x)>0 \\
h'(x)\le 0 \\
\end{matrix} \forall x<2 \right. \\
\end{aligned}$
Trường hợp 1
$\left\{ \begin{aligned}
& h(2)\le 0 \\
& {h}'\left( x \right)\ge 0 \\
\end{aligned} \right.\forall x\in \left( -\infty ;2 \right) $ tương đương $ \Leftrightarrow \left\{ \begin{aligned}
& f(1)+2023\le 0 \left( 1 \right) \\
& f'\left( x-1 \right)\ge 0 \forall x\in \left( -\infty ;2 \right) (2) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& m\le \dfrac{10039}{5} \left( 1 \right) \\
& {{\left( x-1 \right)}^{4}}-2\left( x-1 \right)+m-1\ge 0 \forall x\in \left( -\infty ;2 \right) (2) \\
\end{aligned} \right.$
Đặt $t=x-1,$ $t\in \left( -\infty ;1 \right)$, khi đó ta có
$\left( 2 \right)\Leftrightarrow {{t}^{4}}-2t+m-1\ge 0 \forall t\in \left( -\infty ;1 \right)$
$\Leftrightarrow -{{t}^{4}}+2t+1\le m \forall t\in \left( -\infty ;1 \right)$
Đặt $g(t)=-{{t}^{4}}+2t+1$ $\Rightarrow {{g}^{'}}(t)=-4{{t}^{3}}+2$.
Xét ${{g}^{'}}(t)=0\Leftrightarrow -4{{t}^{3}}+2=0\Leftrightarrow t=\dfrac{1}{\sqrt[3]{2}}$.
Nên $\Leftrightarrow f\left( \dfrac{1}{\sqrt[3]{2}} \right)\le m \Leftrightarrow m\ge \dfrac{3}{2\sqrt[3]{2}}+1$
Từ và suy ra $\dfrac{3}{2\sqrt[3]{2}}+1\le m\le \dfrac{10039}{5}$
Trường hợp 2
$\left\{ \begin{aligned}
& h(2)\ge 0 \\
& {h}'\left( x \right)\le 0 \\
\end{aligned} \right.\forall x\in \left( -\infty ;2 \right) $ tương đương $ \Leftrightarrow \left\{ \begin{aligned}
& f(1)+2023\ge 0 \left( 1 \right) \\
& f'\left( x-1 \right)\le 0 \forall x\in \left( -\infty ;2 \right) (2) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& m\ge \dfrac{10039}{5} \left( 1 \right) \\
& {{\left( x-1 \right)}^{4}}-2\left( x-1 \right)+m-1\le 0 \forall x\in \left( -\infty ;2 \right) (2) \\
\end{aligned} \right.$
Đặt $t=x-1,$ $t\in \left( -\infty ;1 \right)$, khi đó ta có
$\left( 2 \right)\Leftrightarrow {{t}^{4}}-2t+m-1\le 0 \forall t\in \left( -\infty ;1 \right)$
$\Leftrightarrow -{{t}^{4}}+2t+1\ge m \forall t\in \left( -\infty ;1 \right)$
Đặt $g(t)=-{{t}^{4}}+2t+1$ $\Rightarrow {{g}^{'}}(t)=-4{{t}^{3}}+2$.
Xét ${{g}^{'}}(t)=0\Leftrightarrow -4{{t}^{3}}+2=0\Leftrightarrow t=\dfrac{1}{\sqrt[3]{2}}$.
Vô nghiệm
Vậy: $\dfrac{3}{2\sqrt[3]{2}}+1\le m\le \dfrac{10039}{5}$, mà $m\in \mathbb{Z}$ nên có 2005 giá trị $m$ thỏa mãn yêu cầu bài toán.