设函数fx=sinx+cosx和gx=2sinxcosx 若a为实数,求Fx=af（x）+g（x）,x属于[0,π／2]

令sinx+cosx=2sin(x+π/4)=t ∵0≤x≤π/2,π/4≤x+π/4≤3π/4,∴-√2/2≤sin(x+π/4)≤1 即-√2≤t≤2 (sinx+cosx)^2=1+2sinxcosx=t^2 2sinxcosx=t^2-1 F(x)=t+a(t^2-1)=at^2+t-a,-√2≤t≤2 讨论a取最值 当0＜a＜√2/2时-√2＜-1/a＜0,最小值h（a）=-a 当√2/2≤a＜2时-√2≤-1/a＜-1/2,最小值h（a）=a-√2 当a≥2时,-1/2≤-1/a＜0,最小值为h(a)=3a+2