化成普通方程为y?x?2.

?x2?y2?2y?2xx?0x??2由?解得{或{

y?2y?0?y?x?2所以直线l与曲线C交点的极坐标分别为?2,?????,?2,?? 2?23.1.当a?2?时,因为f?x??x?2?x?1??x?2???x?1??1,

所以f?x??1的解集为R,

由f?x??0,得x?2?x?1,则x?2?x?1,即x2?4x?x2?2x?1,解得x?故不等式0?f?x??1的解集为???,223, 2??3??; 2?2.当a?0,x??0,???时, f?x??x?a?x?1???1?a,x?1,

2x?a?1,0?x?1?1?17. 2则f?x?max?f?1??1?a?a?3,又a?0,所以a?22当0?a?1,x??1,???时, f?x??1?a?0?a?3,故0?a?1不合题意, 当a?1,x??0,???时, f?x??x?a?x?1??x?a???x?1??a?1?a?1 当且仅当0?x?1时等号成立,则a2?3?a?1,又a?1,所以a?2? 综上: a的取值范围为???,?

???1?17??U?2,???. 2?