函数y=log1/2 (cosx+sinx)(0
问题描述:
函数y=log1/2 (cosx+sinx)(0
答
y为减函数.求y最小 也就是求cosx+sinx最大 令t=cosx+sinx=根号2*sin(x+π/4)
x+π/4=π/2 x=π/4 t最大为根号2
y=log1/2(根号2)=-log2(2^1/2)=-1/2