2013海淀二模14 求详解
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2013海淀二模14 求详解
2013海淀二模14 求详解
2013海淀二模14 求详解
（1）
设P(x,y),根据题意：
W:|x|+|y|=√[(x-1)²+(y-1)²]
①将(x,y)换成(-x,-y)
|-x|+|-y|=√[(-x-1)²+(-y-1)²]
即x|+|y|=√[(x+1)²+(y+1)²]
方程改变,①不正确
②将(x,y)换成(y,x)
W方程可化为：
|y|+|x|=√[(y-1)²+(x-1)²]方程不变
∴W关于y=x对称
③
当x≥0且y≥0时,
W即 x+y=√[(x-1)²+(y-1)²]
∴x²+y²+2xy=x²+y²-2(x+y)+2
∴xy+x+y=1
∵xy>0
∴x+y<1
x=0时,解得 y=1
∴W与y轴正半轴交于点B(0,1)
根据对称性交x轴正半轴与A(1,0)
那么W在第一象限部分曲线在
线段AB的下方
∴W与x,y的非负半周围成的面积小于
SΔAOB=1/2
∴②③正确
(2)
若,x<0,y>0
W:y-x=√[(x-1)²+(y-1)²]
-2xy=-2（x+y)+2
xy-x-y+1=0
（x-1)(y-1)=0
∵x-1<0 ∴y-1=0
即 y=1(x<0),为射线
若x>0,y<0
W:x-y=√[(x-1)²+(y-1)²]
-2xy=-2（x+y)+2
xy-x-y+1=0
（x-1)(y-1)=0
∵y-1<0 ∴x-1=0
即 x=1(y<0),为射线
那么W上到原点距离的取最小值时的点在
第一象限,应该是曲线与y=x的交点(填空题可以这样做的)
∵xy+x+y=1
∴1=xy+x+y≤(x+y)²/4+(x+y)
∴(x+y)²+4(x+y)-4≥0
解得x+y≥(-4+4√2)/2=2(√2-1)
∵（x+y)/2≤√[(x²+y²)/2]
∴√[(x²+y²)/2]≥(√2-1)
∴√(x²+y²)≥√2(√2-1)=2-√2
即W上点到原点距离的最小值为2-√2
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