若函数f(x)=2asinxcosx-a*(根号2)*(sinx+cosx)+a+b的定义域为[0,派/2],值域为[-5,1],求a,b的值
来源:百度知道 编辑:UC知道 时间:2024/09/24 22:17:16
若函数f(x)=2asinxcosx-a*(根号2)*(sinx+cosx)+a+b的定义域为[0,派/2],值域为[-5,1],求a,b的值
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f'(x)=2a[cos2x-cos(x+派/4)]
令f'(x)=0 =〉
x=派/4
又 f(0)=(1-根号2)a+b
=f(派/2)=(1-根号2)a+b
所以f(派/4)=1 f(0)=-5
或f(派/4)=-5 f(0)=1
解之a=6根号2+6 b=1
或a=--6根号2-6 b=-5
函数f(x)=2asinxcosx-a*(根号2)*(sinx+cosx)+a+b的定义域为[0,派/2],值域为[-5,1],求a,b的值
f(x)=[(sinx+cosx)²-1]a-a√2(sinx+cosx)+a+b
因x∈[0,π/2]
则x+π/4∈[π/4,3π/4],
sinx+cosx=√2sin(x+π/4)
则1≤sinx+cosx≤√2
设sinx+cosx=y
则f(x)=ay²-a√2y+b=a(y-√2/2)²+b-a/2
则当a>0时
f(x)max=2a-2a+b=1,f(x)min=a-√2a+b=-5
=>b=1,a=6(√2+1)
若a<0
f(x)min=2a-2a+b=-5,f(x)max=a-√2a+b=1
=>b=-5,a=-6(√2+1)
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