已知y=-acos2x-根号3asin2x+2a+b,x属于[0,π/4】,函数的值域为[-5,1] 求常数a b
问题描述:
已知y=-acos2x-根号3asin2x+2a+b,x属于[0,π/4】,函数的值域为[-5,1] 求常数a b
答
y=-2a(1/2·cos2x+√3/2·sin2x)+2a+b
=-2asin(2x+π/6)+2a+b
由x∈[0,π/4],可知sin(2x+π/6)∈[1/2,1]
将sin(2x+π/6)=1/2与sin(2x+π/6)=1分别代入
则原函数的两个极值分别为y=a+b和y=b
①若a+b=-5,b=1
则a=-6,b=1
②若b=5,a+b=-5
则a=-10,b=5
答
y=-acos2x-√3asin2x+2a+b
= -asin(2x+π/6)+2a+b
x属于[0,π/4],
2x+π/6属于[π/6,2π/3],
-a/2+2a+b=-5
-a+2a+b=1
解得
a=-12,b=13