18. [解析](1)时，点在圆上.又

，圆心在直线直线上，故.　　　　　　　　　 ………………………..2分

(2)设.

联立方程组，

，

．………………………………………………………. 　　 4分

即

又，………………. 　 6分

当时，此式不成立，

从而．…………………………. 　 9分

又令令函数当时，从而，……………………………… 11分

解此不等式，可得或．……………………　 13分

17.[解析]

(I) 共有种结果····································································· 4分

(II) 若用来表示两枚骰子向上的点数所构成的点的坐标，满足的结果有：

，(3，1)，(4，1)(5，1)，(6，1)(3，2)，(4，2)(5，2)，(6，2)(4，3)，

(5，3)(6，3)，(5，4)(6，4)，(6，5)共15种．·································· 8分

(III)满足的概率是：P＝． 　··········································· 13分

16.[解析](Ⅰ)由题意得

············································································ 2分

因为cosA≠0, ······················································································ 3分

所以tanA=1. ······················································································· 4分

(Ⅱ)由(Ⅰ)知tanA=1得

······································································ 5分

=········································································· 6分

=······························································· 7分

因为xR,所以.···································································· 8分

当时，f(x)有最大值，···························································· 10分

当sinx=-1时，f(x)有最小值-2，····························································· 11分

所以所求函数f(x)的值域是······················································ 　 13分

(二)选做题(14-15题，考生只能从中选做一题)

14．(坐标系与参数方程选做题)[解析].将极坐标方程和分别化为普通方程，，然后就可解得答案．

15．(几何证明选讲选做题)[解析]15．如图，由相交弦定理可知，

由切割线定理可知

(一)必做题(11-13题)

11.[解析], ,从而数列的前项和为.

12.[解析]20.由题意可知

13.[解析]. 由，又由双曲线 的对称性可知又双曲线的渐近线方程为.又因为.因为

10.[解析]C.由，考虑到斜率以及由x,y满足约束条件所确定的可行域，数形结合，易得答案为C．

9.[解析]C.由题意“函数f(x)、f(x+2)均为偶函数”可知，的周期为.从而从而选C.

8．[解析]B. 因为函数在区间是增函数，又因，所以时，

从而从而选B．

7．[解析]A． ．

6．[解析]A.随机取出2个小球得到的结果数有种(提倡列举).取出的小球标注的数字之和为3或6的结果为共3种,故所求答案为(A).