Question

已知函数f(x)=a+根号下2bsin(x+π/4)(a,b属于Z,当x属于闭区间0到π/2时,f(x)的最大值为2根号

Answer 1

过（0,1）

1=a+b

0<=x<=π/4

π/4<=x+π/4<=π/2

所以√2/2<=sin(x+π/4)<=1

若b>0

则最大=a+√2b=2√2-1

a+b=1

则a=-1,b=2

若b<0

则最大=a+b=2√2-1

和a+b=1矛盾

所以f(x)=-1+2√2sin(x+π/4)