若函数f(x)=2asinxcosx-√2a(sinx=cosx)+a+b的定义域为[0,π/2],值域为[-5,1],求a,b的值

Question

若函数f(x)=2asinxcosx-√2a(sinx=cosx)+a+b的定义域为[0,π/2],值域为[-5,1],求a,b的值

Answer 1

括号中是sinx+cosx吗?

f(x)=2asinxcosx-√2a(sinx=cosx)+a+b
=[(sinx+cosx)²-1]a-a√2(sinx+cosx)+a+b
因为x∈[0,π/2]
故x+π/4∈[π/4,3π/4],
sinx+cosx=√2sin(x+π/4)
得1≤sinx+cosx≤√2
设sinx+cosx=y
则f(x)=ay²-a√2y+b=a(y-√2/2)²+b-a/2
当a＞0时
f(x)max=2a-2a+b=1,f(x)min=a-√2a+b=-5
解得b=1,a=6(√2+1)
若a＜0
f(x)min=2a-2a+b=-5,f(x)max=a-√2a+b=1
解得b=-5,a=-6(√2+1)