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已知函数f(x)=sin2x+根号3cos2x-1,x属于R

求函数f(x)的最小正周期

若存在x0属于[0,7π/12],使不等式f(x0)<a成立,求实数a的取值范围
人气:315 ℃ 时间:2019-09-23 04:15:04
解答
答:
f(x)=sin2x+√3cos2x-1
=2*[(1/2)sin2x+(√3/2)cos2x]-1
=2sin(2x+π/3)-1
最小正周期T=2π/2=π
0<=x<=7π/12,0<=2x<=7π/6
π/3<=2x+π/3<=9π/6=3π/2
所以:-1<=f(x)<=1
因为:f(x)所以:a>1
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