f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/20 05:53:48
![f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值](/uploads/image/z/1994502-30-2.jpg?t=f%28x%29%3Dcos%28asinx-cosx%29%2Bcos%5E2%28%CF%80%2F2-x%29%E6%BB%A1%E8%B6%B3f%28-%CF%80%2F3%29%3Df%280%29%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B%CF%80%2F4%2C11%CF%80%2F24%5D%E4%B8%8A%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值
f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值
f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值
解 原题应为f(x)=cosx(asinx-cosx)+cos^2(π/2-x)=acosxsinx-cos^2(x)+cos^2(π/2-x)
=(a/2)sin2x-(1+cos2x)/2+(1+cos2(π/2-x))/2=(a/2)sin2x-cos2x
所以f(-π/3)=(a/2)sin2(-π/3)-cos2(-π/3)=(a/2)*(-√3/2)+1/2
f(0)==(a/2)sin2*0-cos2*0=1
即(a/2)*(-√3/2)+1/2=1 a=-2/√3
周期函数啊
答案是 2 和 根号2 这是重庆的高考题