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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 0 \right)=0$, $f\left( 2 \right)=2$ và $\left| {f}'\left( x \right) \right|\le 2$, $\forall x\in \mathbb{R}$. Biết rằng tập hợp tất cả các giá trị của tích phân $\int\limits_{0}^{2}{f\left( x \right) \text{d}x}$ là đoạn $\left[ a;b \right]$. Tính $b-a$.
A. $4$.
B. $2$.
C. $1$.
D. ${{x}^{3}}$
A. $4$.
B. $2$.
C. $1$.
D. ${{x}^{3}}$
Với mọi $x\in \left[ 0;2 \right]$, ta có $\left| f\left( x \right)-f\left( 2 \right) \right|=\left| \int\limits_{x}^{2}{{f}'\left( x \right)dx} \right|\le \int\limits_{x}^{2}{\left| {f}'\left( x \right) \right|dx}\le \int\limits_{x}^{2}{2dx}=2\left( 2-x \right)$
và $\left| f\left( x \right)-f\left( 0 \right) \right|=\left| \int\limits_{0}^{x}{{f}'\left( x \right)dx} \right|\le \int\limits_{0}^{x}{\left| {f}'\left( x \right) \right|dx}\le \int\limits_{0}^{x}{2dx}=2x$.
Suy ra, với mọi $x\in \left[ 0;2 \right]$, ta có: $\left\{ \begin{aligned}
& 2x-4\le f\left( x \right)-f\left( 2 \right)\le 4-2x \\
& -2x\le f\left( x \right)-f\left( 0 \right)\le 2x \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2x-2\le f\left( x \right)\le 6-2x \\
& -2x\le f\left( x \right)\le 2x \\
\end{aligned} \right.$..
Do đó, ta có $\max \left\{ 2x-2;-2x \right\}\le f\left( x \right)\le \min \left\{ 6-2x;2x \right\}$, $\forall x\in \left[ 0;2 \right]$
$\Rightarrow $ $\int\limits_{0}^{2}{\max \left\{ 2x-2;-2x \right\}dx}\le \int\limits_{0}^{2}{f\left( x \right)dx}\le \int\limits_{0}^{2}{\min \left\{ 6-2x;2x \right\}dx}$.
Xét $- 2x\ge 2x-2\Leftrightarrow 4x\le 2\Leftrightarrow x\le \dfrac{1}{2}$. Suy ra
$\int\limits_{0}^{2}{\max \left\{ - 2x; 2x-2 \right\} \text{d}x}=\int\limits_{0}^{\dfrac{1}{2}}{\left( - 2x \right) \text{d}x}+\int\limits_{\dfrac{1}{2}}^{2}{\left( 2x-2 \right) \text{d}x}=-\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{1}{2}$
Tương tự, ta có $\int\limits_{0}^{2}{\min \left\{ 2x; 2x+6 \right\} \text{d}x}=\dfrac{7}{2}$.
Vậy $\dfrac{1}{2}\le \int\limits_{0}^{2}{f\left( x \right) \text{d}x}\le \dfrac{7}{2}$ nên $a=\dfrac{1}{2}$, $b=\dfrac{7}{2}$ ; suy ra $b-a=3$.
và $\left| f\left( x \right)-f\left( 0 \right) \right|=\left| \int\limits_{0}^{x}{{f}'\left( x \right)dx} \right|\le \int\limits_{0}^{x}{\left| {f}'\left( x \right) \right|dx}\le \int\limits_{0}^{x}{2dx}=2x$.
Suy ra, với mọi $x\in \left[ 0;2 \right]$, ta có: $\left\{ \begin{aligned}
& 2x-4\le f\left( x \right)-f\left( 2 \right)\le 4-2x \\
& -2x\le f\left( x \right)-f\left( 0 \right)\le 2x \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2x-2\le f\left( x \right)\le 6-2x \\
& -2x\le f\left( x \right)\le 2x \\
\end{aligned} \right.$..
Do đó, ta có $\max \left\{ 2x-2;-2x \right\}\le f\left( x \right)\le \min \left\{ 6-2x;2x \right\}$, $\forall x\in \left[ 0;2 \right]$
$\Rightarrow $ $\int\limits_{0}^{2}{\max \left\{ 2x-2;-2x \right\}dx}\le \int\limits_{0}^{2}{f\left( x \right)dx}\le \int\limits_{0}^{2}{\min \left\{ 6-2x;2x \right\}dx}$.
Xét $- 2x\ge 2x-2\Leftrightarrow 4x\le 2\Leftrightarrow x\le \dfrac{1}{2}$. Suy ra
$\int\limits_{0}^{2}{\max \left\{ - 2x; 2x-2 \right\} \text{d}x}=\int\limits_{0}^{\dfrac{1}{2}}{\left( - 2x \right) \text{d}x}+\int\limits_{\dfrac{1}{2}}^{2}{\left( 2x-2 \right) \text{d}x}=-\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{1}{2}$
Tương tự, ta có $\int\limits_{0}^{2}{\min \left\{ 2x; 2x+6 \right\} \text{d}x}=\dfrac{7}{2}$.
Vậy $\dfrac{1}{2}\le \int\limits_{0}^{2}{f\left( x \right) \text{d}x}\le \dfrac{7}{2}$ nên $a=\dfrac{1}{2}$, $b=\dfrac{7}{2}$ ; suy ra $b-a=3$.
Đáp án D.