已知函数y=asin(2x+π/6)+b在x属于[0,π/2]上的值域为[-5,1],求a,b.

题目

已知函数y=asin(2x+π/6)+b在x属于[0,π/2]上的值域为[-5,1],求a,b.

答案

0<=x<=π/2
0<=2x<=π
π/6<=2x+π/6<=7π/6
π/6<=2x+π/6<=π,则0<=sin(2x+π/6)<=1
π<2x+π/6<=7π/6,则-1/2<=sin(2x+π/6)<0
所以-1/2<=sin(2x+π/6)<=1
若a>0
则y最大=a+b,最小=-1/2*a+b
所以a+b=1
-a/2+b=-5
a=4,b=-3
若a<0
则y最大=-1/2*a+b,最小=a+b
所以a+b=-5
-a/2+b=1
a=-4,b=-1
所以
a=4,b=3或a=-4,b=-1