Question

8．设命题p：实数x满足x²-4x+3＜0，命题q：满足$\left\{\begin{array}{l}{{x}^{2}-x-6≤0}\\{{x}^{2}+2x-8＞0}\end{array}\right.$，p∧q为假，p∨q为真，求实数x的取值范围．

Answer 1

分析 由p∧q为假，p∨q为真可得：p，q一真一假，分别求解不等式（组），进而可得数x的取值范围．

解答 解：∵p∧q为假，p∨q为真
∴p，q一真一假
p真：1＜x＜3
q真：$\left\{\begin{array}{l}{x^2}-x-6≤0\\{x^2}+2x-8＞0\end{array}\right.⇒\left\{\begin{array}{l}-2≤x≤3\\ x＜-4或x＞2\end{array}\right.⇒2＜x≤3$
p假：x≤1或x≥3
q假：x≤2或x＞3
当p真q假时：$\left\{\begin{array}{l}1＜x＜3\\ x≤2或x＞3\end{array}\right.⇒1＜x≤2$
当p假q真时：$\left\{\begin{array}{l}x≤1或x≥3\\ 2＜x≤3\end{array}\right.⇒x=3$
综上所述：x∈{x|1＜x≤2或x=3}

点评 本题以命题的真假判断与应用为载体，考查了不等式组的解法，复合命题，难度不大，属于基础题．