Question

设函数f(x)=-2sin^2-2acosx-2a+1的最小值为g(a)。写出g(a)的函数解析式，求g(a)=1/2时a的值，并对此时的a

（接上：）求f(x)的最大值。 求答案！

Answer 1

1.是f(x)=-2sin^2 x-2acosx-2a+1吧f(x)=-2(1-cos^2 x)-2acosx-2a+1=2cos^2x-2acosx-2a-1=2(cosx-a/2)^2-a^2/2-2a-1令t=cosx-1<=t<=1h(t)=2(t-a/2)^2-a^2/2-2a-1a/2>1,a>2g(a)=h min=h(1)=-4a+1-1<=a/2<=1, -2<=a<=2g(a)=h min=h(a/2)= -a^2/2-2a-1a/2<-1,a<-2g(a)=h min=h(-1)=1g(a)的函数解析式分段函数g(a)= ①-4a+1，（a>2）②-a^2/2-2a-1，（-2<=a<=2）③1, (a<-2)2.g(a)=1/2①-4a+1=1/2，（a>2）a=1/8无解②-a^2/2-2a-1＝1/2，（-2<=a<=2）a=-1f(x)= 2(cosx+1/2)^2+1/2f max=1/2 懂了么亲，求采纳