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Câu hỏi: Cho hai hàm số $f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+3x$ và $g\left( x \right)=m{{x}^{3}}+n{{x}^{2}}-x$, với $a, b, c, m, n \in \mathbb{R}$. Biết hàm số $y=f\left( x \right)-g\left( x \right)$ có ba điểm cực trị là $-3$, $1$ và $4$. Diện tích hình phẳng giới hạn bởi hai đồ thị $y=f'\left( x \right)$ và $y=g'\left( x \right)$ bằng
A. $\dfrac{935}{36}.$
B. $\dfrac{941}{36}.$
C. $\dfrac{937}{36}.$
D. $\dfrac{939}{36}.$
A. $\dfrac{935}{36}.$
B. $\dfrac{941}{36}.$
C. $\dfrac{937}{36}.$
D. $\dfrac{939}{36}.$
Ta có: $f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+3x\Rightarrow f'\left( x \right)=4a{{x}^{3}}+3b{{x}^{2}}+2cx+3$.
Ta có: $g\left( x \right)=m{{x}^{3}}+n{{x}^{2}}-x\Rightarrow g'\left( x \right)=3m{{x}^{2}}+2nx-1$.
$y=f\left( x \right)-g\left( x \right)\Rightarrow y'=f'\left( x \right)-g'\left( x \right)=4a{{x}^{3}}+3\left( b-m \right){{x}^{2}}+2\left( c-n \right)x+4$.
$y'=0\Leftrightarrow \left[ \begin{aligned}
& x=-3 \\
& x=1 \\
& x=4 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 4a{{\left( -3 \right)}^{3}}+3\left( b-m \right){{\left( -3 \right)}^{2}}+2\left( c-n \right)\left( -3 \right)+4=0 \\
& 4a{{.1}^{3}}+3\left( b-m \right){{.1}^{2}}+2\left( c-n \right).1+4=0 \\
& 4a{{.4}^{3}}+3\left( b-m \right){{.4}^{2}}+2\left( c-n \right).4+4=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{12} \\
& b-m=\dfrac{-2}{9} \\
& c-n=\dfrac{-11}{6} \\
\end{aligned} \right.$
$S=\int\limits_{-3}^{4}{\left| f'\left( x \right)-g'\left( x \right) \right|}\text{d}x=\int\limits_{-3}^{4}{\left| \dfrac{1}{3}{{x}^{3}}-\dfrac{2}{3}{{x}^{2}}-\dfrac{11}{3}x+4 \right|}\text{d}x=\dfrac{937}{36}$.
Ta có: $g\left( x \right)=m{{x}^{3}}+n{{x}^{2}}-x\Rightarrow g'\left( x \right)=3m{{x}^{2}}+2nx-1$.
$y=f\left( x \right)-g\left( x \right)\Rightarrow y'=f'\left( x \right)-g'\left( x \right)=4a{{x}^{3}}+3\left( b-m \right){{x}^{2}}+2\left( c-n \right)x+4$.
$y'=0\Leftrightarrow \left[ \begin{aligned}
& x=-3 \\
& x=1 \\
& x=4 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 4a{{\left( -3 \right)}^{3}}+3\left( b-m \right){{\left( -3 \right)}^{2}}+2\left( c-n \right)\left( -3 \right)+4=0 \\
& 4a{{.1}^{3}}+3\left( b-m \right){{.1}^{2}}+2\left( c-n \right).1+4=0 \\
& 4a{{.4}^{3}}+3\left( b-m \right){{.4}^{2}}+2\left( c-n \right).4+4=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{12} \\
& b-m=\dfrac{-2}{9} \\
& c-n=\dfrac{-11}{6} \\
\end{aligned} \right.$
$S=\int\limits_{-3}^{4}{\left| f'\left( x \right)-g'\left( x \right) \right|}\text{d}x=\int\limits_{-3}^{4}{\left| \dfrac{1}{3}{{x}^{3}}-\dfrac{2}{3}{{x}^{2}}-\dfrac{11}{3}x+4 \right|}\text{d}x=\dfrac{937}{36}$.
Đáp án C.