已知函数f(x)=x2(x∈-2,2),g(x)=a2sin(2x+π/6)+3a,x∈0,π/2,对于任意的x1∈-2,2,总存在x0∈0,π/2,使得g(x0)=f(x1),求a范围

Question

已知函数f(x)=x2(x∈【-2,2】),g(x)=a2sin(2x+π/6)+3a,x∈【0,π/2】,对于任意的x1∈【-2,2】,
总存在x0∈【0,π/2】,使得g(x0)=f(x1),求a范围

Answer 1

f(x)=x^2在x∈[-2,2]的值域是[-4,4].面g(x)的值域是D=[-a^2+3a,a^2+3a].
由题意可知,[-4,4]是D的子集,即-a^2+3a<=-4且a^2+3a>=4,则a<=-4或a>=4