Question

设a≥0,若y=cos^2x-asinx+b的最大值为0,最小值为-4,试求a,b的值

Answer 1

y = (cosx)^2 - asinx + b
= 1- (sinx)^2 - asinx +b
= -(sinx + a/2)^2 + (1+b+a^2/4 )
case 1: 0≤a ≤2
max y =0
1+b+a^2/4 = 0 (1)
min y = -4
-(1 + a/2)^2 + (1+b+a^2/4 ) =-4 (2)
from (1) and (2)
-(1 + a/2)^2 = -4
1+a/2 = 2
a = 2
from (1)
b = -2

case 2: a>2
max y =0
-(-1+ a/2)^2+1+b+a^2/4 = 0 (1)
min y = -4
-(1 + a/2)^2 + (1+b+a^2/4 ) =-4 (2)
(1) -(2)
2a= 4
a= 2
rejected case 2

ie
a = 2 ,b=-2