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Câu hỏi: Biết rằng hàm số $f\left( x \right)=a{{x}^{2}}+bx+c$ thỏa mãn $\int\limits_{0}^{1}{f\left( x \right)dx}=-\dfrac{7}{2}$, $\int\limits_{0}^{2}{f\left( x \right)dx}=-2$ và $\int\limits_{0}^{3}{f\left( x \right)dx}=\dfrac{13}{2}$ (với a, b, $c\in \mathbb{R}$ ). Giá trị của biểu thức $P=a+b+c$ là
A. $P=-\dfrac{3}{4}$.
B. $P=-\dfrac{4}{3}$.
C. $P=\dfrac{4}{3}$.
D. $P=\dfrac{3}{4}$.
A. $P=-\dfrac{3}{4}$.
B. $P=-\dfrac{4}{3}$.
C. $P=\dfrac{4}{3}$.
D. $P=\dfrac{3}{4}$.
Ta có $\int\limits_{0}^{d}{f\left( x \right)dx}=\left( \dfrac{a}{3}{{x}^{3}}+\dfrac{b}{2}{{x}^{2}}+cx \right)\left| \begin{aligned}
& d \\
& 0 \\
\end{aligned} \right.=\dfrac{a}{3}{{d}^{3}}+\dfrac{b}{2}{{d}^{2}}+cd$
Do đó $\left\{ \begin{aligned}
& \int\limits_{0}^{1}{f\left( x \right)dx}=-\dfrac{7}{2}\Leftrightarrow \dfrac{a}{3}+\dfrac{b}{2}+c=-\dfrac{7}{2} \\
& \int\limits_{0}^{2}{f\left( x \right)dx}=-2\Leftrightarrow \dfrac{8}{3}a+2b+2c=-2 \\
& \int\limits_{0}^{3}{f\left( x \right)dx}=\dfrac{13}{2}\Leftrightarrow 9a+\dfrac{9}{2}b+3c=\dfrac{13}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=3 \\
& c=-\dfrac{16}{3} \\
\end{aligned} \right.$
Vậy $P=a+b+c=-\dfrac{4}{3}$.
& d \\
& 0 \\
\end{aligned} \right.=\dfrac{a}{3}{{d}^{3}}+\dfrac{b}{2}{{d}^{2}}+cd$
Do đó $\left\{ \begin{aligned}
& \int\limits_{0}^{1}{f\left( x \right)dx}=-\dfrac{7}{2}\Leftrightarrow \dfrac{a}{3}+\dfrac{b}{2}+c=-\dfrac{7}{2} \\
& \int\limits_{0}^{2}{f\left( x \right)dx}=-2\Leftrightarrow \dfrac{8}{3}a+2b+2c=-2 \\
& \int\limits_{0}^{3}{f\left( x \right)dx}=\dfrac{13}{2}\Leftrightarrow 9a+\dfrac{9}{2}b+3c=\dfrac{13}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=3 \\
& c=-\dfrac{16}{3} \\
\end{aligned} \right.$
Vậy $P=a+b+c=-\dfrac{4}{3}$.
Đáp án B.