充殿器
花朵
共回答了10个问题采纳率:90% 举报
(1)当a=2时,f(x)=sin2x+2cosx=-cos2x+2cosx+1=-(cosx-1)2+2,
∵0≤x≤[π/3],∴[1/2]≤cosx≤1,
则当cosx=[1/2],即x=[π/3]时,[f(x)]min=f([π/3])=[7/4];当cosx=1,即x=0时,[f(x)]max=2;
(2)f(x)=sin2x+acosx=-cos2x+acosx+1=-(cosx-[a/2])2+1+
a2
4,
∵[1/2]≤cosx≤1,
∴当a≥0,cosx=-1时,[f(x)]min=-(1+[a/2])2+1+
a2
4=a;当a<0,cosx=1时,[f(x)]min=-(1-[a/2])2+1+
a2
4=-a.
1年前
9