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Câu hỏi: Tìm $m$ để giá trị lớn nhất của hàm số $y=\left| {{x}^{3}}-3x+2m-1 \right|$ trên đoạn $\left[ 0 ; 2 \right]$ là nhỏ nhất. Giá trị của $m$ thuộc khoảng nào?
A. $\left( -\dfrac{3}{2} ; -1 \right)$.
B. $\left( \dfrac{2}{3} ; 2 \right)$.
C. $\left[ -1 ; 0 \right]$.
D. $\left( 0 ; 1 \right)$.
A. $\left( -\dfrac{3}{2} ; -1 \right)$.
B. $\left( \dfrac{2}{3} ; 2 \right)$.
C. $\left[ -1 ; 0 \right]$.
D. $\left( 0 ; 1 \right)$.
Đặt $f\left( x \right)={{x}^{3}}-3x+2m-1\Rightarrow {f}'\left( x \right)=3{{x}^{2}}-3$ .
Xét ${f}'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-3=0\Leftrightarrow \left[ \begin{aligned}
& x=-1\notin \left[ 0 ; 2 \right] \\
& x=1\in \left[ 0 ; 2 \right] \\
\end{aligned} \right.$.
Ta có : $\left\{ \begin{aligned}
& f\left( 0 \right)=2m-1 \\
& f\left( 1 \right)=2m-3 \\
& f\left( 2 \right)=2m+1 \\
\end{aligned} \right. \Rightarrow \left\{ \begin{aligned}
& \underset{\left[ 0 ; 2 \right]}{\mathop{\max }} f\left( x \right)=2m+1, \\
& \underset{\left[ 0 ; 2 \right]}{\mathop{\min }} f\left( x \right)=2m-3. \\
\end{aligned} \right.$
Gọi $M=\underset{\left[ 0 ; 2 \right]}{\mathop{\max }} \left| {{x}^{3}}-3x+2m-1 \right|$, khi đó $M=\max \left\{ \left| 2m-3 \right|,\left| 2m+1 \right| \right\}$ .
Ta có: $2M\ge \left| 2m-3 \right|+\left| 2m+1 \right|=\left| 3-2m \right|+\left| 2m+1 \right|\ge \left| \left( 3-2m \right)+\left( 2m+1 \right) \right|=4$ .
Suy ra $M\ge 2\Rightarrow \min M=2$ .
Dấu “ $=$ ” xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& \left| 3-2m \right|=\left| 2m+1 \right| \\
& \left( 3-2m \right)\left( 2m+1 \right)>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& 3-2m=2m+1 \\
& 3-2m=-\left( 2m+1 \right) \\
\end{aligned} \right. \\
& 4{{m}^{2}}-4m-3>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m=\dfrac{1}{2} \\
& -\dfrac{1}{2}<m<\dfrac{3}{2} \\
\end{aligned} \right.\Leftrightarrow m=\dfrac{1}{2}$.
Vậy $m\in \left( 0 ; 1 \right)$ .
Xét ${f}'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-3=0\Leftrightarrow \left[ \begin{aligned}
& x=-1\notin \left[ 0 ; 2 \right] \\
& x=1\in \left[ 0 ; 2 \right] \\
\end{aligned} \right.$.
Ta có : $\left\{ \begin{aligned}
& f\left( 0 \right)=2m-1 \\
& f\left( 1 \right)=2m-3 \\
& f\left( 2 \right)=2m+1 \\
\end{aligned} \right. \Rightarrow \left\{ \begin{aligned}
& \underset{\left[ 0 ; 2 \right]}{\mathop{\max }} f\left( x \right)=2m+1, \\
& \underset{\left[ 0 ; 2 \right]}{\mathop{\min }} f\left( x \right)=2m-3. \\
\end{aligned} \right.$
Gọi $M=\underset{\left[ 0 ; 2 \right]}{\mathop{\max }} \left| {{x}^{3}}-3x+2m-1 \right|$, khi đó $M=\max \left\{ \left| 2m-3 \right|,\left| 2m+1 \right| \right\}$ .
Ta có: $2M\ge \left| 2m-3 \right|+\left| 2m+1 \right|=\left| 3-2m \right|+\left| 2m+1 \right|\ge \left| \left( 3-2m \right)+\left( 2m+1 \right) \right|=4$ .
Suy ra $M\ge 2\Rightarrow \min M=2$ .
Dấu “ $=$ ” xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& \left| 3-2m \right|=\left| 2m+1 \right| \\
& \left( 3-2m \right)\left( 2m+1 \right)>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& 3-2m=2m+1 \\
& 3-2m=-\left( 2m+1 \right) \\
\end{aligned} \right. \\
& 4{{m}^{2}}-4m-3>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m=\dfrac{1}{2} \\
& -\dfrac{1}{2}<m<\dfrac{3}{2} \\
\end{aligned} \right.\Leftrightarrow m=\dfrac{1}{2}$.
Vậy $m\in \left( 0 ; 1 \right)$ .
Đáp án D.