已知y=acos(2x-π/3)+b的定义域为[0,π/2]值域是[1/2 ,5]
1求a,b的值,
0≤x≤π/2
0≤2x≤π
-π/3≤2x-π/3≤2π/3
2x-π/3=0 cos(2x-π/3)=1 y有最大值
y=a+b=5
2x-π/3=-π/3 cos(-π/3)=1/2
2x-π/3=2π/3 cos(2π/3)=-1/2 y有最小值
y=-a/2+b=1/2
解a+b=5 -a/2+b=1/2得
a=3 b=2
2若f(x)在[0,π/6]上递增,设g(x)=asin(bx-π/4)……??
