Question

7．已知p：x²+mx+1=0有兩個(gè)不相等的負(fù)實(shí)根，q：方程4x²+4（m-2）x+1=0無實(shí)根，若p∧q為假，p∨q為真求：m的取值范圍．

Answer 1

分析 p：x²+mx+1=0有兩個(gè)不相等的負(fù)實(shí)根，可得$\left\{\begin{array}{l}{△={m}^{2}-4＞0}\\{-m＜0}\end{array}\right.$，解得m范圍．q：方程4x²+4（m-2）x+1=0無實(shí)根，可得△＜0，解得m范圍．若p∧q為假，p∨q為真，則p與q必然一真一假．

解答 解：p：x²+mx+1=0有兩個(gè)不相等的負(fù)實(shí)根，∴$\left\{\begin{array}{l}{△={m}^{2}-4＞0}\\{-m＜0}\end{array}\right.$，解得m＞2．
q：方程4x²+4（m-2）x+1=0無實(shí)根，∴△=16（m-2）²-16＜0，解得1＜m＜3．
若p∧q為假，p∨q為真，則p與q必然一真一假．
∴$\left\{\begin{array}{l}{m＞2}\\{m≤1或m≥3}\end{array}\right.$，或$\left\{\begin{array}{l}{m≤2}\\{1＜m＜3}\end{array}\right.$，
解得m≥3或1＜m≤2．
∴m的取值范圍是m≥3或1＜m≤2．

點(diǎn)評(píng) 本題考查了不等式的性質(zhì)與解法、一元二次方程的實(shí)數(shù)根與判別式的關(guān)系、簡(jiǎn)易邏輯的判定方法，考查了推理能力與計(jì)算能力，屬于中檔題．