贵州省凯里市第一中学2018-2019学年高一下学期期末考试数学试题 下载本文

所以二面角P?CD?A的平面角是?PEA. 设PA?“1”,由?BAD??BPA?60?, 可得BA?AD?3,

进而可计算AE?33AD?, 2213, 2PE?PA2?AE2?∴cos?PEA?AE313?. PE1320.解:(Ⅰ)由S?22abc1?absinC得c?2sinC ① 422于是a?b?2csinC?ab?c?ab, 即a?b?c?ab

222a2?b2?ab1? ∴cosC?2ab2又C??0,??,所以c?

?3

(Ⅱ)c?2sinC?3 由S?11absinC?ah得b?2, 22将b?2,c?3,C?解得a?1.

?3代入a?b?2csinC?ab中,

2221.(Ⅰ)解:设等差数列?an?的公差为d(d?0),

??a2?5?a1?d?5因为?,所以?2 2??a1a11?a3?a1?a1?10d???a1?2d?解得??a1?2,

?d?3所以an?3n?1,n?N*. (Ⅱ)bn?33? 22an?5an?4?3n?1??5?3n?1??4?1111?(?),

3n?n?1?3nn?11?111111?Sn??(?)?(?)?L?(?)?

3?1223nn?1?11?(1?). 3n?1因为

11?0,所以Sn?, n?1311?0,所以数列?Sn?是递增数列,于是Sn?S1=.

3n?n?1?6又因为bn?综上,

11?Sn?. 6322.解:(Ⅰ)当a?2,圆心C为??1,4?, 圆C的方程为?x?1???y?4??4, 设圆心C到直线l的距离为d,则d?224?(MN2)?1. 2①若直线l的斜率存在,设直线l的方程为y?kx?1,即kx?y?1?0,

d??k?34?1,解得k??,

3k2?14x?1,即4x?3y?3?0. 3此时l的方程为y??②若直线l的斜率不存在,直线l的方程为x?0,验证满足d?1,符合题意.

综合①②可知,直线l的方程为4x?3y?3?0或x?0.

uuuruuuruuur(Ⅱ)设P(x,y),则AP??x,y?1?,OA??0,1?,OP??x,y?,

uuuruuur于是OA?OP??x,y?1?

uuuruuuruuur2222由AP?OA?OP?0得x?y?1?0,即x?y?1,

????所以点P在圆O:x?y?1上,又点P在圆C上, 故圆C与圆O有公共点,即1?OC?3,

22于是1??a?3?2?4a2?3,解得0?a?6, 5因此实数a的取值范围是?0,?.

5(答案、评分标准仅供参考;如有其它解法,酌情给分)

?6???