已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值

问题描述:

已知a>0,函数f(x)=-2acos(2x+π/6)+2a+b,当x∈(-π/4,π/4)时,-5≤f(x)≤1.(1)求常数a,b的值
(2)设g(x)=f(x+π/4),且满足lg【g(x)】>0,求g(x)的单调区间.

(1)因为x∈(-π/4,π/4),所以2X+π/6∈(-2π/3,2π/3)
所以得出2acos(2x+π/6)∈(-1/2,1],
a>0,当2acos(2x+π/6)=-1/2时,f(x)max=1=-2a*(-1/2)+2a+b,得出:a+b=1
当2acos(2x+π/6)=1时,f(x)min=-5=-2a*1+2a+b,得出:b=-5
a+b=1,b=-5 ,所以a=6
(2)g(x)=f(x+π/4)=-2*6cos(2(x+π/4)+π/6)+2*6-5=-12cos(2X+2π/3)+7
因为x∈(-π/4,π/4),所以2X+2π/3∈(π/6,7π/6)
g(x)在 (π/6 ,π]上单调递减,在[π,7π/6)上单调递增.
因为考虑要满足lg【g(x)】>0=lg1,根据lgX在定义域范围内是单调递增的,
所以得出:g(x)>1,-12cos(2X+2π/3)+7>1,cos(2X+2π/3)