设实数XY满足x^2+(y-1)^2=1若对满足条件xy不等式y/x-3+c大于等于0恒成立,则c的取值范围
1个回答
x^2+(y-1)^2=1上点(X,Y)
Y/X就是直线y=kx斜率
y=kx带入圆
(1+k^2)x-3kx=0
(3k)^2>=0,k0
所以k没有最小
y/x-3+c大于等于0不可能恒成立
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