已知f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最小值为-4,求a
问题描述:
已知f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最小值为-4,求a
答
f(x)=2(cosx)^2+√3sin2x+a
=cos2x+1+√3sin2x+a
=2sin(2x+π/6)+a+1
x∈[0,π/2]
则π/6≤2x+π/6≤7π/6
所以f(x)的最小值是2*(-1/2)+a+1=a
所以a=-4