y = √(1-x)+√(x-0.5)
1-x≥0,x-0.5≥0
定義域0.5≤x≤1
y' = -1/[2√(1-x)] + 1/[2√(x-0.5)] = [ √(1-x) - √(x-0.5) ] / { 2 √ [(1-x)(x-0.5)] }
當(dāng)x∈(0.5,0.75)時,√(1-x) - √(x-0.5)＞0,y‘＞0,y單調(diào)增；
當(dāng)x∈(0.75,1)時,√(1-x) - √(x-0.5)＜0,y‘＜0,y單調(diào)減.
x=0.75時有最大值：ymax = y = √(1-0.75)+√(0.75-0.5) = 1
x=0.5時,f(0.5) = √(1-0.5)+√(0.5-0.5) = √0.5 = √2/2
x=1時,f(1) = √(1-1)+√(1-0.5) = √0.5 = √2/2
∴最小值√2/2
a=1,b=√2/2
a^2+b^2 = 1+1/2 = 3/2