Cho hàm số \(f\left( x \right) = \frac{{{x^2} - mx - 2m}}{{x + 2}}\) (m là tham số thực). Gọi S là tập hợp tất cả các giá trị của m sao cho \(\mathop {\max }\limits_{\left[ {1;2} \...

* Hướng dẫn giải

Ta có : \({{f}^{/}}\left( x \right)=\frac{{{x}^{2}}+4x}{{{\left( x+2 \right)}^{2}}}>0,\forall x\in \left[ 1;2 \right]\Rightarrow \) Hàm số tăng trên \(\left[ 1;2 \right]\) và \(f\left( 1 \right)=\frac{1-3m}{3};\ f\left( 2 \right)=\frac{4-4m}{4}=1-m\)

+) \(f\left( 1 \right).f\left( 2 \right)>0\Leftrightarrow \frac{\left( 1-3m \right)\left( 1-m \right)}{3}>0\Leftrightarrow \left[ \begin{align} & m<\frac{1}{3} \\ & m>1 \\ \end{align} \right.\)

Ta có: \(\underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|+\underset{\left[ 1;2 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=\frac{4}{3}\Leftrightarrow \left| \frac{1-3m}{3} \right|+\left| 1-m \right|=\frac{4}{3}\Leftrightarrow \left| \frac{4-6m}{3} \right|=\frac{4}{3}\)

\(\Leftrightarrow \left[ \begin{align} & m=0\ (tmdk) \\ & m=\frac{4}{3}\left( tmdk \right) \\ \end{align} \right.\ \)

+) \(f\left( 1 \right).f\left( 2 \right)<0\Leftrightarrow \frac{\left( 1-3m \right)\left( 1-m \right)}{3}<0\Leftrightarrow \frac{1}{3}

Ta có : \(\underset{\left[ 1;2 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=0,\quad \underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left\{ \left| f\left( 1 \right) \right|,\left| f\left( 2 \right) \right| \right\}=\max \left\{ \left| \frac{1-3m}{3} \right|,\left| 1-m \right| \right\}\)

\(\mathop {\max }\limits_{\left[ {1;2} \right]} \left| {f\left( x \right)} \right| + \mathop {\min }\limits_{\left[ {1;2} \right]} \left| {f\left( x \right)} \right| = 1 \Leftrightarrow \left[ \begin{array}{l} \left| {\frac{{1 - 3m}}{3}} \right| = \frac{4}{3}\\ \left| {1 - m} \right| = \frac{4}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = \frac{5}{3}\;(l)\\ m = - 1\;(l)\\ m = - \frac{1}{3}\;(l)\\ m = \frac{7}{3}\;(l) \end{array} \right.\)

+) \(f\left( 1 \right)=0\Leftrightarrow m=\frac{1}{3}\)

Ta có: \(\underset{\left[ 1;2 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=0,\quad \underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\left| f\left( 2 \right) \right|=\left| 1-\frac{1}{3} \right|=\frac{2}{3}\ \Rightarrow \underset{\left[ 1;2 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|+\underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\frac{2}{3}\) (không thỏa)

+)\(f\left( 2 \right)=0\Leftrightarrow m=1\)

Ta có: \(\underset{\left[ 1;2 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=0,\quad \underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\left| f\left( 1 \right) \right|=\left| \frac{1-3}{3} \right|=\frac{2}{3}\ \Rightarrow \underset{\left[ 1;2 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|+\underset{\left[ 1;2 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\frac{2}{3}\) (không thỏa)

Vậy : \(S=\left\{ 0;\frac{4}{3} \right\}\)