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Câu hỏi: Cho hàm số $y=\left| \dfrac{1}{4}{{x}^{4}}-{{x}^{3}}+{{x}^{2}}+m \right|$. Tính tổng tất cả các số nguyên m để $\underset{\left[ -1;2 \right]}{\mathop{\max }} y\le 11$.
A. –19.
B. –37.
C. –30.
D. –11.
A. –19.
B. –37.
C. –30.
D. –11.
Xét hàm số $y=\dfrac{1}{4}{{x}^{4}}-{{x}^{3}}+{{x}^{2}}+m$ liên tục trên đoạn $\left[ -1;2 \right]$.
Ta có ${f}'\left( x \right)={{x}^{3}}-3{{x}^{2}}+2x=0\Leftrightarrow \left[ \begin{aligned}
& x=0\in \left[ -1;2 \right] \\
& x=1\in \left[ -1;2 \right] \\
& x=2\in \left[ -1;2 \right] \\
\end{aligned} \right.$
Ta lại có: $f\left( -1 \right)=\dfrac{9}{4}+m$ ; $f\left( 0 \right)=m$ ; $f\left( 1 \right)=\dfrac{1}{4}+m$ ; $f\left( 2 \right)=m$.
Khi đó: $\left\{ \begin{aligned}
& \underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=\max \left\{ f\left( -1 \right);f\left( 0 \right);f\left( 1 \right);f\left( 2 \right) \right\}=f\left( -1 \right)=m+\dfrac{9}{4} \\
& \underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=\min \left\{ f\left( -1 \right);f\left( 0 \right);f\left( 1 \right);f\left( 2 \right) \right\}=f\left( 0 \right)=f\left( 2 \right)=m \\
\end{aligned} \right.$.
Suy ra: $\underset{\left[ -1;2 \right]}{\mathop{\max }} y=\max \left\{ \left| m \right|;\left| m+\dfrac{9}{4} \right| \right\}$.
Theo yêu cầu bài toán $\underset{\left[ -1;2 \right]}{\mathop{\max }} y\le 11\Leftrightarrow \left[ \begin{aligned}
& \left| m+\dfrac{9}{4} \right|\le 11 \\
& \left| m \right|\le \left| m+\dfrac{9}{4} \right| \\
& \left| m \right|\le 11 \\
& \left| m+\dfrac{9}{4} \right|\le \left| m \right| \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& -\dfrac{53}{4}\le m\le \dfrac{35}{4} \\
& m\ge -\dfrac{9}{8} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -11\le m\le 11 \\
& m\le -\dfrac{9}{8} \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& -\dfrac{9}{8}\le m\le \dfrac{35}{4} \\
& -11\le m\le -\dfrac{9}{8} \\
\end{aligned} \right.\Leftrightarrow -11\le m\le \dfrac{35}{4}$.
Vì m nguyên nên $m=\left\{ -11;-10;...;8 \right\}$.
Kết luận: tổng các số nguyên m thỏa mãn yêu cầu bài toán là: $-11-10-9-...+8=-30$.
Tìm tham số để $\underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} \left| f\left( x \right) \right|\le a$ (với $a>0$ ).
Phương pháp:
Tìm $\left\{ \begin{aligned}
& \underset{\left[ \alpha ;\beta \right]}{\mathop{\min }} f\left( x \right)=m \\
& \underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} f\left( x \right)=M \\
\end{aligned} \right.\left( M>m \right)$.
Suy ra: $\underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} f\left( x \right)=\max \left\{ \left| m \right|,\left| M \right| \right\}$.
Theo bài ra: $\underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} \left| f\left( x \right) \right|\le a$, nên ta có hai trường hợp:
TH1: $\left\{ \begin{aligned}
& \left| M \right|\le a \\
& \left| m \right|\le \left| M \right| \\
\end{aligned} \right.$.
TH2: $\left\{ \begin{aligned}
& m\le a \\
& \left| M \right|\le \left| m \right| \\
\end{aligned} \right.$.
Ta có ${f}'\left( x \right)={{x}^{3}}-3{{x}^{2}}+2x=0\Leftrightarrow \left[ \begin{aligned}
& x=0\in \left[ -1;2 \right] \\
& x=1\in \left[ -1;2 \right] \\
& x=2\in \left[ -1;2 \right] \\
\end{aligned} \right.$
Ta lại có: $f\left( -1 \right)=\dfrac{9}{4}+m$ ; $f\left( 0 \right)=m$ ; $f\left( 1 \right)=\dfrac{1}{4}+m$ ; $f\left( 2 \right)=m$.
Khi đó: $\left\{ \begin{aligned}
& \underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=\max \left\{ f\left( -1 \right);f\left( 0 \right);f\left( 1 \right);f\left( 2 \right) \right\}=f\left( -1 \right)=m+\dfrac{9}{4} \\
& \underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=\min \left\{ f\left( -1 \right);f\left( 0 \right);f\left( 1 \right);f\left( 2 \right) \right\}=f\left( 0 \right)=f\left( 2 \right)=m \\
\end{aligned} \right.$.
Suy ra: $\underset{\left[ -1;2 \right]}{\mathop{\max }} y=\max \left\{ \left| m \right|;\left| m+\dfrac{9}{4} \right| \right\}$.
Theo yêu cầu bài toán $\underset{\left[ -1;2 \right]}{\mathop{\max }} y\le 11\Leftrightarrow \left[ \begin{aligned}
& \left| m+\dfrac{9}{4} \right|\le 11 \\
& \left| m \right|\le \left| m+\dfrac{9}{4} \right| \\
& \left| m \right|\le 11 \\
& \left| m+\dfrac{9}{4} \right|\le \left| m \right| \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& -\dfrac{53}{4}\le m\le \dfrac{35}{4} \\
& m\ge -\dfrac{9}{8} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& -11\le m\le 11 \\
& m\le -\dfrac{9}{8} \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& -\dfrac{9}{8}\le m\le \dfrac{35}{4} \\
& -11\le m\le -\dfrac{9}{8} \\
\end{aligned} \right.\Leftrightarrow -11\le m\le \dfrac{35}{4}$.
Vì m nguyên nên $m=\left\{ -11;-10;...;8 \right\}$.
Kết luận: tổng các số nguyên m thỏa mãn yêu cầu bài toán là: $-11-10-9-...+8=-30$.
Tìm tham số để $\underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} \left| f\left( x \right) \right|\le a$ (với $a>0$ ).
Phương pháp:
Tìm $\left\{ \begin{aligned}
& \underset{\left[ \alpha ;\beta \right]}{\mathop{\min }} f\left( x \right)=m \\
& \underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} f\left( x \right)=M \\
\end{aligned} \right.\left( M>m \right)$.
Suy ra: $\underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} f\left( x \right)=\max \left\{ \left| m \right|,\left| M \right| \right\}$.
Theo bài ra: $\underset{\left[ \alpha ;\beta \right]}{\mathop{\max }} \left| f\left( x \right) \right|\le a$, nên ta có hai trường hợp:
TH1: $\left\{ \begin{aligned}
& \left| M \right|\le a \\
& \left| m \right|\le \left| M \right| \\
\end{aligned} \right.$.
TH2: $\left\{ \begin{aligned}
& m\le a \\
& \left| M \right|\le \left| m \right| \\
\end{aligned} \right.$.
Đáp án C.