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Câu hỏi: Cho hàm số $f\left( x \right)={{x}^{3}}-4x\int\limits_{0}^{1}{\left| f\left( x \right) \right|}dx$ và $f\left( 1 \right)>0$. Khi đó $f\left( 4 \right)$ bằng?
A. $64$.
B. $60$.
C. $62$.
D. $63$.
A. $64$.
B. $60$.
C. $62$.
D. $63$.
Ta có: $f\left( x \right)={{x}^{3}}-4mx$ với $m\ge 0$. Khi $f\left( x \right)=0\Leftrightarrow {{x}^{3}}-4mx=0\Leftrightarrow \left[ \begin{matrix}
x=0 \\
x=2\sqrt{m} \\
x=-2\sqrt{m} \\
\end{matrix} \right.$
Do $f\left( 1 \right)>0$ $\Rightarrow {{1}^{3}}-4m.1>0\Leftrightarrow m<\dfrac{1}{4}\Leftrightarrow \sqrt{m}<\dfrac{1}{2}\Leftrightarrow 2\sqrt{m}<1$
Suy ra: $m=\int\limits_{0}^{1}{\left| f(x) \right|}dx=\int\limits_{0}^{1}{\left| {{x}^{3}}-4mx \right|}dx=\int\limits_{0}^{2\sqrt{m}}{\left| {{x}^{3}}-4mx \right|}dx+\int\limits_{2\sqrt{m}}^{1}{\left| {{x}^{3}}-4mx \right|}dx$
$=\int\limits_{0}^{2\sqrt{m}}{\left( -{{x}^{3}}+4mx \right)}dx+\int\limits_{2\sqrt{m}}^{1}{\left( {{x}^{3}}-4mx \right)}dx=\left( \dfrac{-{{x}^{4}}}{4}+2m{{x}^{2}} \right)\left| \begin{matrix}
2\sqrt{m} \\
0 \\
\end{matrix} \right.+\left( \dfrac{{{x}^{4}}}{4}-2m{{x}^{2}} \right)\left| \begin{matrix}
1 \\
2\sqrt{m} \\
\end{matrix} \right.$
$=\left[ \dfrac{-{{\left( 2\sqrt{m} \right)}^{4}}}{4}+2m{{\left( 2\sqrt{m} \right)}^{2}} \right]-0+\left( \dfrac{{{1}^{4}}}{4}-2m{{.1}^{2}} \right)-\left[ \dfrac{{{\left( 2\sqrt{m} \right)}^{4}}}{4}-2m{{\left( 2\sqrt{m} \right)}^{2}} \right]$
$=\left( -4{{m}^{2}}+8{{m}^{2}} \right)+\dfrac{1}{4}-2m-\left( 4{{m}^{2}}-8{{m}^{2}} \right)=8{{m}^{2}}-2m+\dfrac{1}{4}$
$\Rightarrow m=8{{m}^{2}}-2m+\dfrac{1}{4}\Leftrightarrow 8{{m}^{2}}-3m+\dfrac{1}{4}=0\Leftrightarrow 32{{m}^{2}}-12m+1=0\Leftrightarrow \left[ \begin{matrix}
m=\dfrac{1}{4} \\
m=\dfrac{1}{8} \\
\end{matrix} \right.$.
Do $m<\dfrac{1}{4}$ nên $m=\dfrac{1}{8}$ thỏa mãn. Vậy $f\left( x \right)={{x}^{3}}-\dfrac{x}{2}\Rightarrow f\left( 4 \right)={{4}^{3}}-\dfrac{4}{2}=64-2=62$
x=0 \\
x=2\sqrt{m} \\
x=-2\sqrt{m} \\
\end{matrix} \right.$
Do $f\left( 1 \right)>0$ $\Rightarrow {{1}^{3}}-4m.1>0\Leftrightarrow m<\dfrac{1}{4}\Leftrightarrow \sqrt{m}<\dfrac{1}{2}\Leftrightarrow 2\sqrt{m}<1$
Suy ra: $m=\int\limits_{0}^{1}{\left| f(x) \right|}dx=\int\limits_{0}^{1}{\left| {{x}^{3}}-4mx \right|}dx=\int\limits_{0}^{2\sqrt{m}}{\left| {{x}^{3}}-4mx \right|}dx+\int\limits_{2\sqrt{m}}^{1}{\left| {{x}^{3}}-4mx \right|}dx$
$=\int\limits_{0}^{2\sqrt{m}}{\left( -{{x}^{3}}+4mx \right)}dx+\int\limits_{2\sqrt{m}}^{1}{\left( {{x}^{3}}-4mx \right)}dx=\left( \dfrac{-{{x}^{4}}}{4}+2m{{x}^{2}} \right)\left| \begin{matrix}
2\sqrt{m} \\
0 \\
\end{matrix} \right.+\left( \dfrac{{{x}^{4}}}{4}-2m{{x}^{2}} \right)\left| \begin{matrix}
1 \\
2\sqrt{m} \\
\end{matrix} \right.$
$=\left[ \dfrac{-{{\left( 2\sqrt{m} \right)}^{4}}}{4}+2m{{\left( 2\sqrt{m} \right)}^{2}} \right]-0+\left( \dfrac{{{1}^{4}}}{4}-2m{{.1}^{2}} \right)-\left[ \dfrac{{{\left( 2\sqrt{m} \right)}^{4}}}{4}-2m{{\left( 2\sqrt{m} \right)}^{2}} \right]$
$=\left( -4{{m}^{2}}+8{{m}^{2}} \right)+\dfrac{1}{4}-2m-\left( 4{{m}^{2}}-8{{m}^{2}} \right)=8{{m}^{2}}-2m+\dfrac{1}{4}$
$\Rightarrow m=8{{m}^{2}}-2m+\dfrac{1}{4}\Leftrightarrow 8{{m}^{2}}-3m+\dfrac{1}{4}=0\Leftrightarrow 32{{m}^{2}}-12m+1=0\Leftrightarrow \left[ \begin{matrix}
m=\dfrac{1}{4} \\
m=\dfrac{1}{8} \\
\end{matrix} \right.$.
Do $m<\dfrac{1}{4}$ nên $m=\dfrac{1}{8}$ thỏa mãn. Vậy $f\left( x \right)={{x}^{3}}-\dfrac{x}{2}\Rightarrow f\left( 4 \right)={{4}^{3}}-\dfrac{4}{2}=64-2=62$
Đáp án C.