⑵ C2：??4cos?

两边同乘?得?2?4?cos?2?2?x2?y2，?cos??x

?x2?y2?4x 即?x?2??y2?4 ②

C3：化为普通方程为y?2x

由题意：C1和C2的公共方程所在直线即为C3 ①—②得：4x?2y?1?a2?0，即为C3

∴1?a2?0 ∴a?1

24．⑴ 如图所示：

??x?4，x≤?1?3?⑵ f?x???3x?2，?1?x?

2?3?4?x，x≥??2f?x??1

当x≤?1，x?4?1，解得x?5或x?3

∴x≤?1

31，3x?2?1，解得x?1或x? 2313∴?1?x?或1?x?

323当x≥，4?x?1，解得x?5或x?3

23∴≤x?3或x?5 21综上，x?或1?x?3或x?5

31??3??5，??? ∴f?x??1，解集为???，??1，3??当?1?x?