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Câu hỏi: Cho hai hàm số ${f\left(x \right)=ax^3+bx^2+cx+d}$, $g\left( x \right)=a{{x}^{2}}+bx+e\left( a,b,c,d,e\in \mathbb{R},a\ne 0 \right)$ có đồ thị lần lượt là hai đường cong ${\left(C_1 \right)}$, ${\left(C_2 \right)}$ ở hình vẽ bên.
Diện tích hình phẳng giới hạn bởi hai đồ thị ${\left(C_1 \right)}$, ${\left(C_2 \right)}$ bằng ${\dfrac{8}{3}}$. Tính $f\left( 2 \right)-g\left( -1 \right)$.
A. $f\left(2 \right)-g\left(-1 \right)=-26$.
B. $f\left(2 \right)- g\left(-1 \right)=-24$.
C. $f\left(2 \right)-g\left(-1 \right)=-28$.
D. $f\left(2 \right)-g\left(-1 \right)=-30$.
Dựa vào đồ thị, ta có $f\left( x \right)-g\left( x \right)=a\left( x-1 \right){{\left( x-3 \right)}^{2}}$ và $a>0$
Ta có: $S=\int\limits_{1}^{3}{\left| f\left( x \right)-g\left( x \right) \right|\text{d}x}=\dfrac{8}{3}\Leftrightarrow \int\limits_{1}^{3}{\left| a\left( x-1 \right){{\left( x-3 \right)}^{2}} \right|\text{d}x}=\dfrac{8}{3}\Leftrightarrow \int\limits_{1}^{3}{a\left( x-1 \right){{\left( x-3 \right)}^{2}}\text{d}x}=\dfrac{8}{3}$
$\Leftrightarrow \int\limits_{1}^{3}{a\left( {{x}^{3}}-7{{x}^{2}}+15x-9 \right)\text{d}x}=\dfrac{8}{3}\Leftrightarrow \left. a\left( \dfrac{1}{4}{{x}^{4}}-\dfrac{7}{3}{{x}^{3}}+\dfrac{15}{2}{{x}^{2}}-9x \right) \right|_{1}^{3}=\dfrac{8}{3}\Leftrightarrow \dfrac{4}{3}a=\dfrac{8}{3}\Leftrightarrow a=2$.
Do đó $f\left( x \right)-g\left( x \right)=2\left( x-1 \right){{\left( x-3 \right)}^{2}}$ $\Leftrightarrow \left( a{{x}^{3}}+b{{x}^{2}}+cx+d \right)-\left( a{{x}^{2}}+bx+e \right)=2\left( x-1 \right){{\left( x-3 \right)}^{2}}$
$\Leftrightarrow a{{x}^{3}}+\left( b-a \right){{x}^{2}}+\left( c-b \right)x+d-e=2\left( {{x}^{3}}-7{{x}^{2}}+15x-9 \right)$
Đồng nhất hệ số ta có
$\Rightarrow \left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& a=2 \\
& b-a=-14 \\
& c-b=30 \\
& d-e=-18 \\
\end{aligned} \\
\end{array}\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& a=2 \\
& b=-12 \\
& c=18 \\
& d=e-18 \\
\end{aligned} \\
\end{array} \right. \right.$
$\begin{array}{*{35}{l}}
\Rightarrow f\left( x \right)=2{{x}^{3}}-12{{x}^{2}}+18x+e-18;g\left( x \right)=2{{x}^{2}}-12x+e \\
\end{array} $ $ \Rightarrow f\left( 2 \right)-g\left( -1 \right)=-28$
Vậy ${f(2)-g(-1)=-28}$.
A. $f\left(2 \right)-g\left(-1 \right)=-26$.
B. $f\left(2 \right)- g\left(-1 \right)=-24$.
C. $f\left(2 \right)-g\left(-1 \right)=-28$.
D. $f\left(2 \right)-g\left(-1 \right)=-30$.
Dựa vào đồ thị, ta có $f\left( x \right)-g\left( x \right)=a\left( x-1 \right){{\left( x-3 \right)}^{2}}$ và $a>0$
Ta có: $S=\int\limits_{1}^{3}{\left| f\left( x \right)-g\left( x \right) \right|\text{d}x}=\dfrac{8}{3}\Leftrightarrow \int\limits_{1}^{3}{\left| a\left( x-1 \right){{\left( x-3 \right)}^{2}} \right|\text{d}x}=\dfrac{8}{3}\Leftrightarrow \int\limits_{1}^{3}{a\left( x-1 \right){{\left( x-3 \right)}^{2}}\text{d}x}=\dfrac{8}{3}$
$\Leftrightarrow \int\limits_{1}^{3}{a\left( {{x}^{3}}-7{{x}^{2}}+15x-9 \right)\text{d}x}=\dfrac{8}{3}\Leftrightarrow \left. a\left( \dfrac{1}{4}{{x}^{4}}-\dfrac{7}{3}{{x}^{3}}+\dfrac{15}{2}{{x}^{2}}-9x \right) \right|_{1}^{3}=\dfrac{8}{3}\Leftrightarrow \dfrac{4}{3}a=\dfrac{8}{3}\Leftrightarrow a=2$.
Do đó $f\left( x \right)-g\left( x \right)=2\left( x-1 \right){{\left( x-3 \right)}^{2}}$ $\Leftrightarrow \left( a{{x}^{3}}+b{{x}^{2}}+cx+d \right)-\left( a{{x}^{2}}+bx+e \right)=2\left( x-1 \right){{\left( x-3 \right)}^{2}}$
$\Leftrightarrow a{{x}^{3}}+\left( b-a \right){{x}^{2}}+\left( c-b \right)x+d-e=2\left( {{x}^{3}}-7{{x}^{2}}+15x-9 \right)$
Đồng nhất hệ số ta có
$\Rightarrow \left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& a=2 \\
& b-a=-14 \\
& c-b=30 \\
& d-e=-18 \\
\end{aligned} \\
\end{array}\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& a=2 \\
& b=-12 \\
& c=18 \\
& d=e-18 \\
\end{aligned} \\
\end{array} \right. \right.$
$\begin{array}{*{35}{l}}
\Rightarrow f\left( x \right)=2{{x}^{3}}-12{{x}^{2}}+18x+e-18;g\left( x \right)=2{{x}^{2}}-12x+e \\
\end{array} $ $ \Rightarrow f\left( 2 \right)-g\left( -1 \right)=-28$
Vậy ${f(2)-g(-1)=-28}$.
Đáp án C.