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Câu hỏi: Cho $y=f\left( x \right)$ có đồ thị của $y={f}'\left( x \right)$ như hình vẽ dưới đây.
Đặt $M=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{max}}} f\left( x \right)$, $m=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{min}}} f\left( x \right)$. Giá trị của biểu thức $M+m$ bằng
A. $f\left( 0 \right)+f\left( 2 \right)$.
B. $f\left( 5 \right)+f\left( -2 \right)$.
C. $f\left( 5 \right)+f\left( 6 \right)$.
D. $f\left( 0 \right)-f\left( 2 \right)$.
Đặt $M=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{max}}} f\left( x \right)$, $m=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{min}}} f\left( x \right)$. Giá trị của biểu thức $M+m$ bằng
A. $f\left( 0 \right)+f\left( 2 \right)$.
B. $f\left( 5 \right)+f\left( -2 \right)$.
C. $f\left( 5 \right)+f\left( 6 \right)$.
D. $f\left( 0 \right)-f\left( 2 \right)$.
Từ đồ thị của $y={f}'\left( x \right)$, ta có bảng biến thiên của $y=f\left( x \right)$ trên $\left[ -2;6 \right]$ như sau
Từ bảng biến thiên ta có
$M=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{max}}} f\left( x \right)=\text{max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}$, $m=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{min}}} f\left( x \right)=\text{min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}$.
Từ đồ thị của $y={f}'\left( x \right)$ ta có $\int\limits_{0}^{2}{\left| {f}'\left( x \right) \right|\text{d}x}<\int\limits_{2}^{5}{\left| {f}'\left( x \right) \right|\text{d}x}$ $\Leftrightarrow \int\limits_{0}^{2}{-{f}'\left( x \right)\text{d}x<\int\limits_{2}^{5}{{f}'\left( x \right)\text{d}x}}$ $\Leftrightarrow f\left( 0 \right)-f\left( 2 \right)<f\left( 5 \right)-f\left( 2 \right)$ $\Leftrightarrow f\left( 0 \right)<f\left( 5 \right)$.
Suy ra $\text{max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}=f\left( 5 \right)$.
Mặt khác, cũng từ từ đồ thị của $y={f}'\left( x \right)$, ta có $\int\limits_{-2}^{0}{\left| {f}'\left( x \right) \right|\text{d}x}>\int\limits_{0}^{2}{\left| {f}'\left( x \right) \right|\text{d}x}$ $\Leftrightarrow \int\limits_{-2}^{0}{{f}'\left( x \right)\text{d}x>\int\limits_{0}^{2}{-{f}'\left( x \right)\text{d}x}}$ $\Leftrightarrow f\left( 0 \right)-f\left( -2 \right)>f\left( 0 \right)-f\left( 2 \right)$ $\Leftrightarrow f\left( -2 \right)<f\left( 2 \right)$.
Hơn nữa $\int\limits_{2}^{5}{\left| {f}'\left( x \right) \right|\text{d}x}>\int\limits_{5}^{6}{\left| {f}'\left( x \right) \right|\text{d}x}$ $\Leftrightarrow \int\limits_{2}^{5}{{f}'\left( x \right)\text{d}x>\int\limits_{5}^{6}{-{f}'\left( x \right)\text{d}x}}$ $\Leftrightarrow f\left( 5 \right)-f\left( 2 \right)>f\left( 5 \right)-f\left( 6 \right)$ $\Leftrightarrow f\left( 2 \right)<f\left( 6 \right)$.
Suy ra $\text{min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}=f\left( -2 \right)$.
Vậy $\text{M=max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}=f\left( 5 \right)$, $\text{m=min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}=f\left( -2 \right)$,
nên $M+m=f\left( 5 \right)+f\left( -2 \right)$.
Từ bảng biến thiên ta có
$M=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{max}}} f\left( x \right)=\text{max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}$, $m=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{min}}} f\left( x \right)=\text{min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}$.
Từ đồ thị của $y={f}'\left( x \right)$ ta có $\int\limits_{0}^{2}{\left| {f}'\left( x \right) \right|\text{d}x}<\int\limits_{2}^{5}{\left| {f}'\left( x \right) \right|\text{d}x}$ $\Leftrightarrow \int\limits_{0}^{2}{-{f}'\left( x \right)\text{d}x<\int\limits_{2}^{5}{{f}'\left( x \right)\text{d}x}}$ $\Leftrightarrow f\left( 0 \right)-f\left( 2 \right)<f\left( 5 \right)-f\left( 2 \right)$ $\Leftrightarrow f\left( 0 \right)<f\left( 5 \right)$.
Suy ra $\text{max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}=f\left( 5 \right)$.
Mặt khác, cũng từ từ đồ thị của $y={f}'\left( x \right)$, ta có $\int\limits_{-2}^{0}{\left| {f}'\left( x \right) \right|\text{d}x}>\int\limits_{0}^{2}{\left| {f}'\left( x \right) \right|\text{d}x}$ $\Leftrightarrow \int\limits_{-2}^{0}{{f}'\left( x \right)\text{d}x>\int\limits_{0}^{2}{-{f}'\left( x \right)\text{d}x}}$ $\Leftrightarrow f\left( 0 \right)-f\left( -2 \right)>f\left( 0 \right)-f\left( 2 \right)$ $\Leftrightarrow f\left( -2 \right)<f\left( 2 \right)$.
Hơn nữa $\int\limits_{2}^{5}{\left| {f}'\left( x \right) \right|\text{d}x}>\int\limits_{5}^{6}{\left| {f}'\left( x \right) \right|\text{d}x}$ $\Leftrightarrow \int\limits_{2}^{5}{{f}'\left( x \right)\text{d}x>\int\limits_{5}^{6}{-{f}'\left( x \right)\text{d}x}}$ $\Leftrightarrow f\left( 5 \right)-f\left( 2 \right)>f\left( 5 \right)-f\left( 6 \right)$ $\Leftrightarrow f\left( 2 \right)<f\left( 6 \right)$.
Suy ra $\text{min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}=f\left( -2 \right)$.
Vậy $\text{M=max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}=f\left( 5 \right)$, $\text{m=min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}=f\left( -2 \right)$,
nên $M+m=f\left( 5 \right)+f\left( -2 \right)$.
Đáp án B.