问题补充:
填空题若命题“?x∈R,使x2+(a-1)x+1<0”是假命题,则实数a的取值范围为________.
答案:
-1≤a≤3解析分析:先求出命题的否定,再用恒成立来求解解答:命题“?x∈R,使x2+(a-1)x+1<0”的否定是:““?x∈R,使x2+(a-1)x+1≥0”即:△=(a-1)2-4≤0,∴-1≤a≤3故
时间:2019-10-13 22:26:31
填空题若命题“?x∈R,使x2+(a-1)x+1<0”是假命题,则实数a的取值范围为________.
-1≤a≤3解析分析:先求出命题的否定,再用恒成立来求解解答:命题“?x∈R,使x2+(a-1)x+1<0”的否定是:““?x∈R,使x2+(a-1)x+1≥0”即:△=(a-1)2-4≤0,∴-1≤a≤3故
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