设函数f(x)=mcos^2x+根号3msinxcosx+n(m>0)的定义域为[0,派/6]值域为[3,4].求m.n的值

题目

设函数f(x)=mcos^2x+根号3msinxcosx+n(m>0)的定义域为[0,派/6]值域为[3,4].求m.n的值

答案

(x)=mcos²x+√3*m*sinxcosx+n
=m*[cos2x/2+√3/2*sin2x]+n+m/2
=m*sin(2x+π/6)+n+m/2
∴ 值域为[n-m/2,n+3m/2]=[3,4].
n-m/2=3
n+3m/2=4
m=1/2,n=13/4