已知f(x)=√2asin(x-π/4)+a+b当a＜0时f(x)在[0,π]上的值域为[2,3]求a,b

Question

已知f(x)=√2 a sin(x-π/4)+a+b
当a＜0时f(x)在[0,π]上的值域为[2,3]
求a,b

Answer 1

当x∈[0,π]时 ,则有(x-π/4)∈[-π/4 ,3π/4]
∴sin(x-π/4)∈[-√2/2 ,1]
则f(x)=√2a×(-√2/2) + a + b =b
f(x)=√2a×1 + a + b =(√2+1)a + b
∵a＜0
∴f(x)∈[(√2+1)a + b ,b]
∵f(x)在[0,π]上的值域为[2,3]
∴b=3
(√2+1)a + b=2
解得：a=1-√2 ,b=3