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Câu hỏi: Cho hàm số $f\left( x \right)=\left| {{x}^{4}}-2{{x}^{2}}+m+3 \right|$ (m là tham số thực ). Gọi S là tập hợp tất cả giá trị của m sao cho $2\underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)+\underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=2020$. Tổng giá trị tất cả các phần tử của S bằng
A. –718.
B. 650.
C. –68.
D. –132.
A. –718.
B. 650.
C. –68.
D. –132.
Xét $g\left( x \right)={{x}^{4}}-2{{x}^{2}}+m+3$ trên đoạn $\left[ 0;3 \right]$ $\Rightarrow {g}'\left( x \right)=4{{x}^{3}}-4x=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=1 \\
& x=-1\notin \left[ 1;3 \right] \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& g\left( 0 \right)=m+3 \\
& g\left( 1 \right)=m+2 \\
& g\left( 3 \right)=m+66 \\
\end{aligned} \right.$
Suy ra $\underset{\left[ 0;3 \right]}{\mathop{\min }} g\left( x \right)=m+2$, $\underset{\left[ 0;3 \right]}{\mathop{\max }} g\left( x \right)=m+66$
TH1: $\left( m+1 \right)\left( m+66 \right)\le 0\Leftrightarrow -66\le m\le -1$
$\left[ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=\underset{\left[ 0;3 \right]}{\mathop{\max }} g\left( x \right)=\dfrac{\left| m+66+m+2 \right|+\left| m+66-m-2 \right|}{2}=\left| m+34 \right|+32 \\
& \underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)=0 \\
\end{aligned} \right.$
Vậy $2\underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)+\underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=2020\Leftrightarrow \left| m+34 \right|+3=2020\Leftrightarrow \left| m+34 \right|=2017\Leftrightarrow \left[ \begin{aligned}
& m=1 \\
& m=- \\
\end{aligned} \right.$(loại)
TH2: $\left( m+1 \right)\left( m+66 \right)>0\Leftrightarrow \left[ \begin{aligned}
& m>-1 \\
& m<-66 \\
\end{aligned} \right.$
$\left[ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=\underset{\left[ 0;3 \right]}{\mathop{\max }} g\left( x \right)=\dfrac{\left| m+66+m+2 \right|+\left| m+66-m-2 \right|}{2}=\left| m+34 \right|+32 \\
& \underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)=\dfrac{\left| m+66+m+2 \right|-\left| m+66-m-2 \right|}{2}=\left| m+34 \right|-32 \\
\end{aligned} \right.$
$\Rightarrow 2\underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)+\underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=2020\Leftrightarrow \left| m+34 \right|+3=2020\Leftrightarrow 2\left( \left| m+34 \right|-32 \right)+\left| m+34 \right|+32=2020$
$\Leftrightarrow 3\left| m+34 \right|=2052\Leftrightarrow \left| m+34 \right|=684\Leftrightarrow \left[ \begin{aligned}
& m=650 \\
& m=-718 \\
\end{aligned} \right.\left( N \right)$
Suy ra ${{m}_{1}}+{{m}_{2}}=-718+650=-68$
& x=0 \\
& x=1 \\
& x=-1\notin \left[ 1;3 \right] \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& g\left( 0 \right)=m+3 \\
& g\left( 1 \right)=m+2 \\
& g\left( 3 \right)=m+66 \\
\end{aligned} \right.$
Suy ra $\underset{\left[ 0;3 \right]}{\mathop{\min }} g\left( x \right)=m+2$, $\underset{\left[ 0;3 \right]}{\mathop{\max }} g\left( x \right)=m+66$
TH1: $\left( m+1 \right)\left( m+66 \right)\le 0\Leftrightarrow -66\le m\le -1$
$\left[ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=\underset{\left[ 0;3 \right]}{\mathop{\max }} g\left( x \right)=\dfrac{\left| m+66+m+2 \right|+\left| m+66-m-2 \right|}{2}=\left| m+34 \right|+32 \\
& \underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)=0 \\
\end{aligned} \right.$
Vậy $2\underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)+\underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=2020\Leftrightarrow \left| m+34 \right|+3=2020\Leftrightarrow \left| m+34 \right|=2017\Leftrightarrow \left[ \begin{aligned}
& m=1 \\
& m=- \\
\end{aligned} \right.$(loại)
TH2: $\left( m+1 \right)\left( m+66 \right)>0\Leftrightarrow \left[ \begin{aligned}
& m>-1 \\
& m<-66 \\
\end{aligned} \right.$
$\left[ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=\underset{\left[ 0;3 \right]}{\mathop{\max }} g\left( x \right)=\dfrac{\left| m+66+m+2 \right|+\left| m+66-m-2 \right|}{2}=\left| m+34 \right|+32 \\
& \underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)=\dfrac{\left| m+66+m+2 \right|-\left| m+66-m-2 \right|}{2}=\left| m+34 \right|-32 \\
\end{aligned} \right.$
$\Rightarrow 2\underset{\left[ 0;3 \right]}{\mathop{\min }} f\left( x \right)+\underset{\left[ 0;3 \right]}{\mathop{\max }} f\left( x \right)=2020\Leftrightarrow \left| m+34 \right|+3=2020\Leftrightarrow 2\left( \left| m+34 \right|-32 \right)+\left| m+34 \right|+32=2020$
$\Leftrightarrow 3\left| m+34 \right|=2052\Leftrightarrow \left| m+34 \right|=684\Leftrightarrow \left[ \begin{aligned}
& m=650 \\
& m=-718 \\
\end{aligned} \right.\left( N \right)$
Suy ra ${{m}_{1}}+{{m}_{2}}=-718+650=-68$
Đáp án C.