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
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