若当P(m,n)为圆x2+(y-1)2=1上任意一点,不等式m+n+c>=0恒成立,求实数c的取值范围.

题目

若当P(m,n)为圆x2+(y-1)2=1上任意一点,不等式m+n+c>=0恒成立,求实数c的取值范围.

答案

设圆上点为x=cosα,y=sinα+1
所以m=cosα,n=sinα+1
m+n+c=cosα+sinα+1+c
=√2[(√2/2)cosα+(√2/2)sinα]+1+c
=√2[sin(π/4)cosα+cos(π/4)sinα]+1+c
=√2sin(α+π/4)+1+c
所以√2sin(α+π/4)+1+c≥0
c≥-1-√2sin(α+π/4)
-1≤sin(α+π/4)≤1
-√2≤√2sin(α+π/4)≤√2
-√2≤-√2sin(α+π/4)≤√2
-1-√2≤-1-√2sin(α+π/4)≤-1+√2
c取最大值-1+√2
c≥-1+√2