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Câu hỏi: Cho hàm số bậc nhất $f\left( x \right)$ thỏa mãn $\int\limits_{0}^{1}{f\left( x \right)}dx=4;$ $\int\limits_{2}^{3}{f\left( x \right)}dx=2.$ Tính $I=\int\limits_{0}^{1}{f\left( f\left( 2x-5 \right) \right)}dx$
A. $6$.
B. $\dfrac{7}{2}$.
C. $-4$.
D. $\dfrac{3}{2}$.
A. $6$.
B. $\dfrac{7}{2}$.
C. $-4$.
D. $\dfrac{3}{2}$.
Hàm số bậc nhất $f\left( x \right)=ax+b$
$4=\int\limits_{0}^{1}{f\left( x \right)}dx=\int\limits_{0}^{1}{\left( ax+b \right)}dx=\left. \left( \dfrac{a{{x}^{2}}}{2}+bx \right) \right|_{0}^{1}=\dfrac{a}{2}+b$ $\Rightarrow \dfrac{a}{2}+b=4\text{ }\left( 1 \right)$
$2=\int\limits_{2}^{3}{\left( ax+b \right)}dx=\left. \left( \dfrac{a{{x}^{2}}}{2}+bx \right) \right|_{2}^{3}=\dfrac{9a}{2}+3b-\dfrac{4a}{2}-2b=\dfrac{5a}{2}+b$ $\Rightarrow \dfrac{5a}{2}+b=2\text{ }\left( 2 \right)$
Từ $\left( 1 \right)$ và $\left( 2 \right)$ ta có: $\begin{aligned}
& \left\{ \begin{aligned}
& \dfrac{a}{2}+b=4 \\
& \dfrac{5a}{2}+b=2\text{ } \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b=\dfrac{9}{2} \\
\end{aligned} \right.\text{ } \\
& \\
\end{aligned} $ $ \Rightarrow f\left( x \right)=-x+\dfrac{9}{2}.$
$f\left( 2x-5 \right)=-2x+5+\dfrac{9}{2}.$
$f\left( f\left( 2x-5 \right) \right)=2x-5-\dfrac{9}{2}+\dfrac{9}{2}=2x-5$
$I=\int\limits_{0}^{1}{f\left( f\left( 2x-5 \right) \right)}dx=\int\limits_{0}^{1}{\left( 2x-5 \right)}dx=-4.$.
$4=\int\limits_{0}^{1}{f\left( x \right)}dx=\int\limits_{0}^{1}{\left( ax+b \right)}dx=\left. \left( \dfrac{a{{x}^{2}}}{2}+bx \right) \right|_{0}^{1}=\dfrac{a}{2}+b$ $\Rightarrow \dfrac{a}{2}+b=4\text{ }\left( 1 \right)$
$2=\int\limits_{2}^{3}{\left( ax+b \right)}dx=\left. \left( \dfrac{a{{x}^{2}}}{2}+bx \right) \right|_{2}^{3}=\dfrac{9a}{2}+3b-\dfrac{4a}{2}-2b=\dfrac{5a}{2}+b$ $\Rightarrow \dfrac{5a}{2}+b=2\text{ }\left( 2 \right)$
Từ $\left( 1 \right)$ và $\left( 2 \right)$ ta có: $\begin{aligned}
& \left\{ \begin{aligned}
& \dfrac{a}{2}+b=4 \\
& \dfrac{5a}{2}+b=2\text{ } \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b=\dfrac{9}{2} \\
\end{aligned} \right.\text{ } \\
& \\
\end{aligned} $ $ \Rightarrow f\left( x \right)=-x+\dfrac{9}{2}.$
$f\left( 2x-5 \right)=-2x+5+\dfrac{9}{2}.$
$f\left( f\left( 2x-5 \right) \right)=2x-5-\dfrac{9}{2}+\dfrac{9}{2}=2x-5$
$I=\int\limits_{0}^{1}{f\left( f\left( 2x-5 \right) \right)}dx=\int\limits_{0}^{1}{\left( 2x-5 \right)}dx=-4.$.
Đáp án C.