已知函数f(x)=2x^2-3x+1,g(x)=Asin(x-π/6)(A>0),当o

题目

已知函数f(x)=2x^2-3x+1,g(x)=Asin(x-π/6)(A>0),当o<=x<=π/2时,求y=f(sinx)的最大值
若对任意的x1∈[0,3],总存在x2∈[0,π],使f(x1)=g(x2)成立,求实数A的取值范围
若方程f(sinx)=a-sinx在[0,2π)上有两解,求实数a的取值范围

答案

已知函数f(x)=2x^2-3x+1,g(x)=Asin(x-π/6)(A>0),(1)当ox3=-π/2,x4=π/2F’’(x)=4cos2x+3sinx=4-8(sinx)^2+3sinx==> F’’(x1)= F’’(x2)=4-9/2+9/4=7/4>0∴F(x)在x1,x2处取极小值F’’(x3)=4-8-3=-7...