已知函数f(x)=x²(x∈[-2,2]),g(x)=a²sin(2x+π/6)+3a(x∈[0,π/2]),
问题描述：
已知函数f(x)=x²(x∈[-2,2]),g(x)=a²sin(2x+π/6)+3a(x∈[0,π/2]),
∃x1∈[-2,2],∀x2∈[0,π/2],使得f(x1)=g(x2)成立,则实数a的取值范围是?
答
x∈[0,π/2]sin(2x+π/6)∈[-1/2,1]g(x)=a²sin(2x+π/6)+3a∈[-a²/2+3a,a²+3a]f(x)=x²(x∈[-2,2])f(x)∈[0,4]∃x1∈[-2,2],∀x2∈[0,π/2],使得f(x1)=g(x2)成立即交集不为空交集为空...