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"$5z1fi" '已知圆C:X^2+y^2-2x-2y+1=0,直线l:y=kx,且直线l与圆C相交于P、Q两个不同点。有一点M（0，b)，满足MP⊥MQ。
当b∈（1，2/3）时，求k的取值范围。
把y=kx代入x^2+y^2-2x-2y+1=0,整理得
(1+k^2)x^2-(2+k)x+1=0,
设P(x1,y1),Q(x2,y2),则
x1+x2=(2+k)/(1+k^2),x1x2=1/(1+k^2).
向量MP=(x1,y1-b),MQ=(x2,y2-b),
由MP⊥MQ得
0=x1x2+(y1-b)(y2-b)
=x1x2+(kx1-b)(kx2-b)
=(1+k^2)x1x2-bk(x1+x2)+b^2
=1-bk(2+k)/(1+k^2)+b^2,
∴（1-b+b^2)k^2-2bk+1+b^2=0,
△=4b^2-4(1-b+b^2)(1+b^2)
=-4(1-b+b^2-b^3+b^4)<0(b∈（2/3，1）），
∴k的取值范围是空集。
其他答案（共1个回答）
f(1)=a+2b+c奇数，f(0)=c奇数，
所以a+2b是偶数,
所以a是偶数。
反设，如果存在整数零点k，
则ak^2+2bk+c=0
因为ak^2,2b...
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