,y,有x设实数x,y满足 x^2+(y-1)^2=1,若对满足条件的x+y+c>=0恒成立,则c的取值范围是?
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![,y,有x设实数x,y满足 x^2+(y-1)^2=1,若对满足条件的x+y+c>=0恒成立,则c的取值范围是?](/uploads/image/z/5221629-45-9.jpg?t=%2Cy%2C%E6%9C%89x%E8%AE%BE%E5%AE%9E%E6%95%B0x%2Cy%E6%BB%A1%E8%B6%B3+x%5E2%2B%28y-1%29%5E2%3D1%2C%E8%8B%A5%E5%AF%B9%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6%E7%9A%84x%2By%2Bc%3E%3D0%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%88%99c%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%3F)
,y,有x设实数x,y满足 x^2+(y-1)^2=1,若对满足条件的x+y+c>=0恒成立,则c的取值范围是?
,y,有x设实数x,y满足 x^2+(y-1)^2=1,若对满足条件的x+y+c>=0恒成立,则c的取值范围是?
,y,有x设实数x,y满足 x^2+(y-1)^2=1,若对满足条件的x+y+c>=0恒成立,则c的取值范围是?
x+y>=-c
则-c小于等于x+y的最小值
令x=cosa
则(y-1)^2=1-(cosa)^2=(sina)^2
因为sina值域关于原点对称
所以不妨令y-1=sina
y=sina+1
x+y=sina+cosa+1
=√2*sin(a+π/4)+1
所以x+y最小值=-√2+1
所以-c<=-√2+1
所以c>=√2-1
x+y+c>=0恒成立,
则c≥-x-y恒成立
则c≥max(-x-y)
根据x^2+(y-1)^2=1设x=cosa,y-1=sina
则-x-y=-cosa-1-sina=-√2sin(a+π/4)-1≤√2-1
则c≥√2-1
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