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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục, có đạo hàm trên đoạn $\left[ a;b \right]$ và đồ thị của hàm số $f'\left( x \right)$ trên $\left[ a;b \right]$ là đường cong như hình vẽ bên. Khi đó, mệnh đề nào
A. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( b \right)$
B. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( {{x}_{1}} \right)$
C. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( a \right)$
D. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( {{x}_{2}} \right)$
A. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( b \right)$
B. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( {{x}_{1}} \right)$
C. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( a \right)$
D. $\underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( {{x}_{2}} \right)$
Ta có ${{S}_{1}}=\int\limits_{a}^{{{x}_{1}}}{-f'\left( x \right)dx}=f\left( a \right)-f\left( {{x}_{1}} \right)>0\Rightarrow f\left( a \right)>f\left( {{x}_{1}} \right)$
${{S}_{2}}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{-f'\left( x \right)dx}=f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right)>0\Rightarrow f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right)$
${{S}_{3}}=\int\limits_{{{x}_{2}}}^{b}{-f'\left( x \right)dx}=f\left( b \right)-f\left( {{x}_{2}} \right)>0\Rightarrow f\left( b \right)>f\left( {{x}_{2}} \right)$
Do đó ta có $\left\{ \begin{aligned}
& f\left( a \right)>f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right) \\
& f\left( b \right)>f\left( {{x}_{2}} \right) \\
\end{aligned} \right.\Rightarrow $ $ \underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( {{x}_{2}} \right)$
${{S}_{2}}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{-f'\left( x \right)dx}=f\left( {{x}_{1}} \right)-f\left( {{x}_{2}} \right)>0\Rightarrow f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right)$
${{S}_{3}}=\int\limits_{{{x}_{2}}}^{b}{-f'\left( x \right)dx}=f\left( b \right)-f\left( {{x}_{2}} \right)>0\Rightarrow f\left( b \right)>f\left( {{x}_{2}} \right)$
Do đó ta có $\left\{ \begin{aligned}
& f\left( a \right)>f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right) \\
& f\left( b \right)>f\left( {{x}_{2}} \right) \\
\end{aligned} \right.\Rightarrow $ $ \underset{x\in \left[ a;b \right]}{\mathop{\min }} f\left( x \right)=f\left( {{x}_{2}} \right)$
Đáp án D.