若函数f(x)=sinx/2cosx/2+cos2x/2-1/2(1)若f(α)=√2/4,α属于(0,2π),求α的值(2)求f(x)在-四分之派,π上的最大值和最小值及相应的x值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 21:54:34
![若函数f(x)=sinx/2cosx/2+cos2x/2-1/2(1)若f(α)=√2/4,α属于(0,2π),求α的值(2)求f(x)在-四分之派,π上的最大值和最小值及相应的x值](/uploads/image/z/1042152-24-2.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsinx%2F2cosx%2F2%2Bcos2x%2F2-1%2F2%281%29%E8%8B%A5f%28%CE%B1%29%3D%E2%88%9A2%2F4%2C%CE%B1%E5%B1%9E%E4%BA%8E%EF%BC%880%2C2%CF%80%EF%BC%89%2C%E6%B1%82%CE%B1%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E6%B1%82f%28x%29%E5%9C%A8%EF%BC%8D%E5%9B%9B%E5%88%86%E4%B9%8B%E6%B4%BE%2C%CF%80%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%E5%8F%8A%E7%9B%B8%E5%BA%94%E7%9A%84x%E5%80%BC)
若函数f(x)=sinx/2cosx/2+cos2x/2-1/2(1)若f(α)=√2/4,α属于(0,2π),求α的值(2)求f(x)在-四分之派,π上的最大值和最小值及相应的x值
若函数f(x)=sinx/2cosx/2+cos2x/2-1/2(1)若f(α)=√2/4,α属于(0,2π),求α的值
(2)求f(x)在-四分之派,π上的最大值和最小值及相应的x值
若函数f(x)=sinx/2cosx/2+cos2x/2-1/2(1)若f(α)=√2/4,α属于(0,2π),求α的值(2)求f(x)在-四分之派,π上的最大值和最小值及相应的x值
(1) 若函数f(x)=sinx/2cosx/2+cos2x/2-1/2 => 化简得f(x)=1/2*sinx+1/2*cos2x-1/2
=1/2sinx+1/2*[1-2sin^2(x)]-1/2
=-sin^2(x)+1/2sinx
当f(α)=√2/4时,有 -sin^2(x)+1/2sinx=√2/4 => -(sinx-1/4)^2=√2/4-1/16>0
但是左边 √2sin(x+π/4)-2=√2 => √2 * sin(x+π/4)=√2+2
sin(x+π/4)=1+√2>1 因为|sin(x+π/4)|《1,所以在(0,2π)上,无α满足上面条件.
(2)当x 在[-π/4,π]上时 由 -π/4《x《π => 0《x《5/4π
所以 -1《√2 * sin(x+π/4)《√2 => -1/4《√2 /4 sin(x+π/4)《√2 /4
=> -1/4-1/2《√2/4 sin(x+π/4)-1/2《√2/4-1/2
=> f(x)最大值√2/4-1/2 ;最小值-3/4
f(x)化简=sin(x+pi/4)-1/2 计算就是了啊