Cho hàm số \(y=\frac{2x-m}{x+2}\) với m là tham số, \(m\ne -4.\) Biết \(\underset{x\in \left[ 0;2 \right]}{\mathop{\min }}\,f\left( x \right)+\underset{x\in \left[ 0;2 \right]}{\ma...

* Hướng dẫn giải

Ta có \(y'=\frac{4+m}{{{\left( x+2 \right)}^{2}}}.\)

TH1. Nếu \(4+m>0\Leftrightarrow m>-4\) thì \(y'>0,\forall x\in \mathbb{R}\backslash \left\{ -2 \right\}.\)

Khi đó \(\left\{ \begin{array}{l} \mathop {\min }\limits_{x \in \left[ {0;2} \right]} f\left( x \right) = f\left( 0 \right) = - \frac{m}{2}\\ \mathop {\max }\limits_{x \in \left[ {0;2} \right]} f\left( x \right) = f\left( 2 \right) = \frac{{4 - m}}{4} \end{array} \right.\)

Mà \(\underset{x\in \left[ 0;2 \right]}{\mathop{\min }}\,f\left( x \right)+\underset{x\in \left[ 0;2 \right]}{\mathop{\max }}\,f\left( x \right)=-8\Leftrightarrow -\frac{m}{2}+\frac{4-m}{4}=-8\Leftrightarrow m=12\) (nhận).

TH2. Nếu \(4+m<0\Leftrightarrow m<-4\) thì \(y'<0,\forall x\in \mathbb{R}\backslash \left\{ -2 \right\}.\)

Khi đó \(\left\{ \begin{array}{l} \mathop {\min }\limits_{x \in \left[ {0;2} \right]} f\left( x \right) = f\left( 0 \right) = - \frac{m}{2}\\ \mathop {\max }\limits_{x \in \left[ {0;2} \right]} f\left( x \right) = f\left( 2 \right) = \frac{{4 - m}}{4} \end{array} \right.\)

Mà \(\underset{x\in \left[ 0;2 \right]}{\mathop{\min }}\,f\left( x \right)+\underset{x\in \left[ 0;2 \right]}{\mathop{\max }}\,f\left( x \right)=-8\Leftrightarrow -\frac{m}{2}+\frac{4-m}{4}=-8\Leftrightarrow m=12\) (loại).

Vậy m=12 thỏa yêu cầu bài toán.