已知函数f(x)=2asin^2x-(2根号3)asinxcosx+a+b的定义域为[0,π⼀2],值域为[-5,1],求a b的值??

2025-05-14 13:38:26
推荐回答(2个)
回答(1):

2asin^x-2根号3asinxcosx+a+b

=2asin²x-√3a(2sinxcosx)+a+b

=-(a-2asin²x)+a-√3a(2sinxcosx)+a+b

=-acos2x-√3asin2x+2a+b

=-2a[1/2cos2x+√3/2sin2x]+2a+b

=-2asin(2x+π/6)+2a+b

又x∈[π/2,π],

则2x+π/6∈[7π/6,13π/6]

sin(2x+π/6)∈[-1,1/2]

①若a>0

则值域是[a+b,4a+b]

有a+b=2 4a+b=5

得a=b=1

②若a<0

则值域是[4a+b,a+b]

有4a+b=2 a+b=5

得a=-1 b=6

回答(2):

2asin^2x=a(1-cos2x),-(2根号3)asinxcosx=-(根号3)asin2x
f(x)=-(1/2)a((1/2)cos2x+(2分之根号3)sin2x)+2a+b
f(x)=-(1/2)a(sin(2x+π/6))+2a+b