Question

8．設(shè)命題p：實(shí)數(shù)x滿足x²-4x+3＜0，命題q：滿足$\left\{\begin{array}{l}{{x}^{2}-x-6≤0}\\{{x}^{2}+2x-8＞0}\end{array}\right.$，p∧q為假，p∨q為真，求實(shí)數(shù)x的取值范圍．

Answer 1

分析 由p∧q為假，p∨q為真可得：p，q一真一假，分別求解不等式（組），進(jìn)而可得數(shù)x的取值范圍．

解答 解：∵p∧q為假，p∨q為真
∴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
當(dāng)p真q假時(shí)：$\left\{\begin{array}{l}1＜x＜3\\ x≤2或x＞3\end{array}\right.⇒1＜x≤2$
當(dāng)p假q真時(shí)：$\left\{\begin{array}{l}x≤1或x≥3\\ 2＜x≤3\end{array}\right.⇒x=3$
綜上所述：x∈{x|1＜x≤2或x=3}

點(diǎn)評(píng) 本題以命題的真假判斷與應(yīng)用為載體，考查了不等式組的解法，復(fù)合命題，難度不大，屬于基礎(chǔ)題．