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Câu hỏi: Cho hàm số $y={{x}^{3}}-3x+m$. Gọi S là tập hợp tất cả các giá trị của tham số thực m sao cho $\underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6$. Số phần tử của S là:
A. 0.
B. 6.
C. 1.
D. 2.
A. 0.
B. 6.
C. 1.
D. 2.
Xét hàm số $y={{x}^{3}}-3x+m,x\in \left[ 0;2 \right]$
$y=3{{x}^{2}}-3=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=-1\left( l \right) \\
\end{aligned} \right.$
Ta có $y\left( 0 \right)=m;y\left( 1 \right)=m-2;y\left( 2 \right)=m+2$.
Suy ra: $\underset{\left[ 0;2 \right]}{\mathop{\min }} y=m-2;\underset{\left[ 0;2 \right]}{\mathop{\max }} y=m+2$.
TH1: $\left( m+2 \right)\left( m-2 \right)\le 0\Rightarrow -2\le m\le 2$.
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|=0;\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=\left\{ \left| m-2 \right|;\left| m+2 \right| \right\}$.
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6\Leftrightarrow \left[ \begin{aligned}
& 0+2-m=6 \\
& m+2=6 \\
\end{aligned} \right.\Leftrightarrow m=\pm 4$, không thỏa mãn.
TH2: $m-2>0\Leftrightarrow m>2\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|=\left| m-2 \right|=m-2;\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=\left| 2+m \right|=m+2$
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6\Leftrightarrow m-2+m+2=6\Leftrightarrow m=3$ (thỏa mãn).
TH3: $2+m<0\Leftrightarrow m<-2\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|=\left| 2+m \right|=-2-m$ ;
$\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=\left| -2+m \right|=-\left( -2+m \right)=2-m$
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6\Leftrightarrow -2-m+2-m=6\Leftrightarrow m=-3$ (thỏa mãn).
Vậy có 2 số nguyên thỏa mãn.
$y=3{{x}^{2}}-3=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=-1\left( l \right) \\
\end{aligned} \right.$
Ta có $y\left( 0 \right)=m;y\left( 1 \right)=m-2;y\left( 2 \right)=m+2$.
Suy ra: $\underset{\left[ 0;2 \right]}{\mathop{\min }} y=m-2;\underset{\left[ 0;2 \right]}{\mathop{\max }} y=m+2$.
TH1: $\left( m+2 \right)\left( m-2 \right)\le 0\Rightarrow -2\le m\le 2$.
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|=0;\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=\left\{ \left| m-2 \right|;\left| m+2 \right| \right\}$.
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6\Leftrightarrow \left[ \begin{aligned}
& 0+2-m=6 \\
& m+2=6 \\
\end{aligned} \right.\Leftrightarrow m=\pm 4$, không thỏa mãn.
TH2: $m-2>0\Leftrightarrow m>2\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|=\left| m-2 \right|=m-2;\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=\left| 2+m \right|=m+2$
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6\Leftrightarrow m-2+m+2=6\Leftrightarrow m=3$ (thỏa mãn).
TH3: $2+m<0\Leftrightarrow m<-2\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|=\left| 2+m \right|=-2-m$ ;
$\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=\left| -2+m \right|=-\left( -2+m \right)=2-m$
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{\min }} \left| y \right|+\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| y \right|=6\Leftrightarrow -2-m+2-m=6\Leftrightarrow m=-3$ (thỏa mãn).
Vậy có 2 số nguyên thỏa mãn.
Đáp án D.