题目内容
已知命题A“?x∈R,x2+(a-1)x+1<0”.(1)写出命题A的否定;
(2)若命题A是假命题,求出实数a的取值范围.
【答案】分析:(1)命题A的否定:?x∈R,x2+(a-1)x+1≥0;
(2由题设知?x∈R,x2+(a-1)x+1≥0,即△=(a-1)2-4≤0,由此能求出实数a的取值范围.
解答:解:(1)命题A的否定:?x∈R,x2+(a-1)x+1≥0;
(2)∵?x∈R,x2+(a-1)x+1<0为假命题,
∴?x∈R,x2+(a-1)x+1≥0,
即△=(a-1)2-4≤0,
解得-1≤a≤3.
点评:本题考查命题的真假判断和应用,解题时要注意不等式性质的合理运用.
(2由题设知?x∈R,x2+(a-1)x+1≥0,即△=(a-1)2-4≤0,由此能求出实数a的取值范围.
解答:解:(1)命题A的否定:?x∈R,x2+(a-1)x+1≥0;
(2)∵?x∈R,x2+(a-1)x+1<0为假命题,
∴?x∈R,x2+(a-1)x+1≥0,
即△=(a-1)2-4≤0,
解得-1≤a≤3.
点评:本题考查命题的真假判断和应用,解题时要注意不等式性质的合理运用.
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