问题
填空题
x,y∈(0,2],且xy=2,且6-2x-y≥a(2-x)(4-y)恒成立,则实数a取值范围是______.
答案
令2x+y=t,
∵x,y∈(0,2],且xy=2,
∴2x+y=2x+
2 |
x |
2x•
|
∴t∈[4,5]
∵6-2x-y≥a(2-x)(4-y)=a(8-4x-2y+xy)=a(10-4x-2y)
∴6-t≥a(10-2t),
a≤
6-t |
10-2t |
∴当t=4时,a≤(
6-t |
10-2t |
故答案为(-∞,1].