问题 选择题
已知在平面直角坐标系中,O(0,0),A(1,-2),B(1,1),C(2,-1),动点M(x,y)满足条件
-2≤
OM
OA
≤2
1≤
OM
OB
≤2
,则z=
OM
OC
的最大值为(  )
A.-1B.0C.3D.4
答案

设M(x,y)则

OM
=(x,y),
OA
=(1,-2),
OB
=(1,1)
OC
=(2,-1)

-2≤
OM
OA
≤2
1≤
OM
OB
≤2
,∴
-2≤x-2y≤2
1≤x+y≤2

OM
OC
=2x-y=(x-2y)+(x+y),-1≤
OM
OC
≤4

故选D.

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