Question

设α∈R,f(x)=cosx(asinx-cosx)+cos2( π 2 -x)满足f(- π 3 )=f(0),求函数f(x)在[ π 4 , 11π 24 ]

、
上的最大值和最小值

Answer 1

f(x)=cosx(asinx-cosx)+cos2(π/2-x)
=a/2*sin2x-cos²x+cos(π-2x)
f(0)=-2
f(-π/3)=-√3a/4-1/4+1/2
f(0)=f(-π/3) 则a=3√3
f(x)=3√3/2sin2x-1/2 cos2x+1/2+cos2x
f(x)=3sin(2x+π/6)+1/2
x∈[ π/4 ,11π/24]
2x+π/6∈[2π/3，13π/12]
sin(2x+π/6)∈[-0.26，√3/2]
f(x)的最大值：(3√3+1)/2
f(x)的最小值：-0.28

Answer 2

π 3\π 4\11π 24 是什么意思

追问

-3/π,4/π,11π/24