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Câu hỏi: Cho hàm số $y=\dfrac{2x+m}{x+1}$. Biết $\underset{\left[ 0;2 \right]}{\mathop{\min }} y+3\underset{\left[ 0;2 \right]}{\mathop{\max }} y=10$. Chọn khẳng định đúng
A. $m\in \left( 1;3 \right)$.
B. $m\in \left[ 3;5 \right)$.
C. $m\in \left( 5;7 \right)$.
D. $m\in \left[ 7;9 \right)$.
A. $m\in \left( 1;3 \right)$.
B. $m\in \left[ 3;5 \right)$.
C. $m\in \left( 5;7 \right)$.
D. $m\in \left[ 7;9 \right)$.
Ta có ${y}'=\dfrac{2-m}{{{\left( x+1 \right)}^{2}}}$
Trường hợp 1: Nếu $2-m>0\Leftrightarrow m<2$ thì $\underset{\left[ 0; 2 \right]}{\mathop{\min y}} =f\left( 0 \right)=m; \underset{\left[ 0; 2 \right]}{\mathop{\max y}} =f\left( 2 \right)=\dfrac{m+4}{3}$
Khi đó $\underset{\left[ 0;2 \right]}{\mathop{\min }} y+3\underset{\left[ 0;2 \right]}{\mathop{\max }} y=10$ $\Leftrightarrow m+m+4=10\Leftrightarrow m=3$ ( loại)
Trường hợp 2: Nếu $2-m<0\Leftrightarrow m>2$ thì $\underset{\left[ 0; 2 \right]}{\mathop{\max y}} =f\left( 0 \right)=m; \underset{\left[ 0; 2 \right]}{\mathop{\min y}} =f\left( 2 \right)=\dfrac{m+4}{3}$
Khi đó $\underset{\left[ 0;2 \right]}{\mathop{\min }} y+3\underset{\left[ 0;2 \right]}{\mathop{\max }} y=10$ $\Leftrightarrow 3m+\dfrac{m+4}{3}=10\Leftrightarrow m=2,6$ ( tm). Vậy $m=2,6\in \left( 1; 3 \right)$.
Trường hợp 1: Nếu $2-m>0\Leftrightarrow m<2$ thì $\underset{\left[ 0; 2 \right]}{\mathop{\min y}} =f\left( 0 \right)=m; \underset{\left[ 0; 2 \right]}{\mathop{\max y}} =f\left( 2 \right)=\dfrac{m+4}{3}$
Khi đó $\underset{\left[ 0;2 \right]}{\mathop{\min }} y+3\underset{\left[ 0;2 \right]}{\mathop{\max }} y=10$ $\Leftrightarrow m+m+4=10\Leftrightarrow m=3$ ( loại)
Trường hợp 2: Nếu $2-m<0\Leftrightarrow m>2$ thì $\underset{\left[ 0; 2 \right]}{\mathop{\max y}} =f\left( 0 \right)=m; \underset{\left[ 0; 2 \right]}{\mathop{\min y}} =f\left( 2 \right)=\dfrac{m+4}{3}$
Khi đó $\underset{\left[ 0;2 \right]}{\mathop{\min }} y+3\underset{\left[ 0;2 \right]}{\mathop{\max }} y=10$ $\Leftrightarrow 3m+\dfrac{m+4}{3}=10\Leftrightarrow m=2,6$ ( tm). Vậy $m=2,6\in \left( 1; 3 \right)$.
Đáp án A.