题目

设函数f(x)=2mcos^2x-2√3msinx*cosx+n(m>0)的定义域为[0,pai/2]值域[1,4]
（1）求m,n的值.（2）若f（x）=2,求x

提问时间：2020-08-11

答案

1.
f(x)=2m(cosx)^2-m-3^0.5×2msinxconx+m+n
=mcos2x-3^0.5×msin2x+m+n
=2m(1/2×cos2x-3^0.5/2×sin2x)+m+n
=2mcos(π/3+2x)+m+n
x∈[0,π/2]
2x∈[0,π]
2x+π/3∈[π/3,4π/3]
cos(2x+π/3)∈[-1,1/2]
2mcos(π/3+2x)+m+n∈[-m+n,2m+n]
-m+n=1 2m+n=4
m=1 n=2
2.
f(x)=2cos(2x+π/3)+3=2
cos(2x+π/3)=-1/2
2x+π/3=2π/3 或 2x+π/3=4π/3
x=π/6 或 x=π/2