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【题目】若命题“t∈R,t2﹣2t﹣a<0”是假命题,则实数a的取值范围是 .
【答案】(﹣∞,﹣1]
【解析】解:命题“t∈R,t2﹣2t﹣a<0”是假命题, 则t∈R,t2﹣2t﹣a≥0是真命题,
∴△=4+4a≤0,解得a≤﹣1.
∴实数a的取值范围是(﹣∞,﹣1].
故答案为:(﹣∞,﹣1].
命题“t∈R,t2﹣2t﹣a<0”是假命题,则t∈R,t2﹣2t﹣a≥0是真命题,可得△≤0.
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【题目】若命题“t∈R,t2﹣2t﹣a<0”是假命题,则实数a的取值范围是 .
【答案】(﹣∞,﹣1]
【解析】解:命题“t∈R,t2﹣2t﹣a<0”是假命题, 则t∈R,t2﹣2t﹣a≥0是真命题,
∴△=4+4a≤0,解得a≤﹣1.
∴实数a的取值范围是(﹣∞,﹣1].
故答案为:(﹣∞,﹣1].
命题“t∈R,t2﹣2t﹣a<0”是假命题,则t∈R,t2﹣2t﹣a≥0是真命题,可得△≤0.