已知函数f(x)=-sin²x-sinx+a(1)当f(x)=0有实数解时,求a的取值范围(2)若x∈[6/π,3/2π],
1≤f(x)≤4/17,求a的取值范围
人气:379 ℃ 时间:2019-08-22 14:38:40
解答
(1)令t=sinx,则f(x)=f(t)=-t^2-t+a=-(t+1/2)^2+a+1/4当f(x)=0时-(t+1/2)^2+a+1/4=0-(t+1/2)^2=-(a+1/4)(t+1/2)^2=a+1/4a+1/4≥0a≥-1/4解法2:当-t^2-t+a=0时有实数解时,依据△=b^2-4ac=1+4a≥0,得4a≥-1a≥-1...
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