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已知命题p:存在x属于R,使得x^2-2ax+2a^2-5a+4=0;命题q:曲线x^2/3+y^2/a-3=1是双曲线.若"p或q"为真,"p且q"为假,求实数a的取值范围
人气:354 ℃ 时间:2019-10-17 07:24:44
解答
x^2-2ax+2a^2-5a+4=0判别式:
(-2a)^2-4(2a^2-5a+4)
=-4a^2+20a-16
=-4(a^2-5a+4)
=-4(a-1)(a-4)
P为真,判别式≥0-4(a-1)(a-4)≥0,(a-1)(a-4)≤01≤a≤4
P为假,a4
Q为真,a-3
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