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Câu hỏi: Cho hàm số $f\left( x \right)={{x}^{3}}-{{x}^{2}}+x-m-2$ (m là tham số thực). Gọi S là tập hợp tất cả các giá trị của m sao cho $\underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|+\underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=16$. Tổng các phần tử của S là:
A. 3.
B. 17.
C. 34.
D. 31.
A. 3.
B. 17.
C. 34.
D. 31.
Xét hàm số $f\left( x \right)={{x}^{3}}-{{x}^{2}}+x-m-2$ trên đoạn $\left[ 0;3 \right]$
Ta có: ${f}'\left( x \right)=3{{x}^{2}}-2x+1>0,\forall x\in \mathbb{R}$
Ta lại có: $f\left( 0 \right)=-m-2;f\left( 3 \right)=-m+19$.
TH1: $\left( m+2 \right)\left( m-19 \right)\le 0\Leftrightarrow -2\le m\le 19\Rightarrow \left\{ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=0 \\
& \underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=\max \left\{ \left| m+2 \right|,\left| m-19 \right| \right\} \\
\end{aligned} \right.$
$\Rightarrow \left[ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=m+2,\text{ khi }\dfrac{17}{2}\le m\le 19 \\
& \underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=19-m,\text{ khi }-2\le m\le \dfrac{17}{2} \\
\end{aligned} \right.$
Vậy $\underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|+\underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=16\Rightarrow \left[ \begin{aligned}
& m+2=16,\text{ khi }\dfrac{17}{2}\le m\le 19 \\
& 19-m=16,\text{ khi }0\le m\le \dfrac{17}{2} \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& m=14 \\
& m=3 \\
\end{aligned} \right.$
TH2: $\left( m+2 \right)\left( m-19 \right)>0\Leftrightarrow \left[ \begin{aligned}
& m>19 \\
& m<-2 \\
\end{aligned} \right.$
Suy ra $\underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|+\underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=\left| m+2 \right|+\left| m-19 \right|=\left| 2m-17 \right|=16\Leftrightarrow \left[ \begin{aligned}
& m=\dfrac{1}{2}\left( \text{loai} \right) \\
& m=\dfrac{33}{2}\left( \text{loai} \right) \\
\end{aligned} \right.$.
Vậy $S=\left\{ 3;14 \right\}$.
Ta có: ${f}'\left( x \right)=3{{x}^{2}}-2x+1>0,\forall x\in \mathbb{R}$
Ta lại có: $f\left( 0 \right)=-m-2;f\left( 3 \right)=-m+19$.
TH1: $\left( m+2 \right)\left( m-19 \right)\le 0\Leftrightarrow -2\le m\le 19\Rightarrow \left\{ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=0 \\
& \underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=\max \left\{ \left| m+2 \right|,\left| m-19 \right| \right\} \\
\end{aligned} \right.$
$\Rightarrow \left[ \begin{aligned}
& \underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=m+2,\text{ khi }\dfrac{17}{2}\le m\le 19 \\
& \underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=19-m,\text{ khi }-2\le m\le \dfrac{17}{2} \\
\end{aligned} \right.$
Vậy $\underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|+\underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|=16\Rightarrow \left[ \begin{aligned}
& m+2=16,\text{ khi }\dfrac{17}{2}\le m\le 19 \\
& 19-m=16,\text{ khi }0\le m\le \dfrac{17}{2} \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& m=14 \\
& m=3 \\
\end{aligned} \right.$
TH2: $\left( m+2 \right)\left( m-19 \right)>0\Leftrightarrow \left[ \begin{aligned}
& m>19 \\
& m<-2 \\
\end{aligned} \right.$
Suy ra $\underset{\left[ 0;3 \right]}{\mathop{\min }} \left| f\left( x \right) \right|+\underset{\left[ 0;3 \right]}{\mathop{\max }} \left| f\left( x \right) \right|=\left| m+2 \right|+\left| m-19 \right|=\left| 2m-17 \right|=16\Leftrightarrow \left[ \begin{aligned}
& m=\dfrac{1}{2}\left( \text{loai} \right) \\
& m=\dfrac{33}{2}\left( \text{loai} \right) \\
\end{aligned} \right.$.
Vậy $S=\left\{ 3;14 \right\}$.
Đáp án B.