黑龙江齐齐哈尔市2021-2022学年高二上学期期末考试数学答案(2)

高二数学试卷参考答案及评分标准第1 页（共4 页） 高二数学试卷参考答案及评分标准 一. 单选题：本大题共8 小题，每小题5 分，共40 分.在每小题给出的四个选项中，只有一项是符合题目要求的. 题号 1 2 3 4 5 6 7 8 答案 B C D C A A C D 二. 多选题：本大题共4 小题，每小题5 分，共20 分.在每小题给出的选项中，有多项符合题目要求，全部选 对的得5 分，有选错的得0 分，部分选对得2 分. 题号 9 10 11 12 答案 AB AB ACD ABD 三. 填空题：本大题共4 小题，每小题5 分，共20 分. 把正确答案写在答题卡相应题的横线上. 13、2 3 14、  4 8 15、1 16、 2 x 1 2  n n （注：第一空2 分；第二空3 分） 四. 解答题：共70 分，解答应写出文字说明、解答过程或演算步骤. 17.（10 分） 解：（1）【解法一】 直线l 的斜率为 3 2   l k ··················································································································································1分 m l //  直线m 的斜率为 3 2   m k ·························································································································2分 直线m 的方程为 ) 2 ( 3 2 3     x y ······················································································································3分 即 0 13 3 2   y x ····························································································································································4分 【解法二】 m l //  直线m 的方程为 0 3 2    C y x ············································································································1分 直线m 过点 13 0 3 3 2 2 ) 3 , 2 (          C C ························································································3分 直线m 的方程为 0 13 3 2   y x ··························································································································4分 （2）易知圆心 ) 2 , 1 ( C ，半径 1  r ······························································································································5分 圆心C 到直线l 的距离 13 13 2 13 2 3 2 | 6 2 3 1 2 | 2 2         d ······································································7分 13 13 6 13 4 1 2 2 | | 2 2       d r AB ··················································································································9分 线段AB 的长为13 13 6 ···············································································································································10分 高二数学试卷参考答案及评分标准第2 页（共4 页） 18.（12 分） 解：（1）设等差数列 } { n a 的公差为d                10 9 12 9 3 10 2 3 12 2 1 1 1 4 5 2 d a d a a a a a        1 1 1 d a ·····································3 分 n n an       1 ) 1 ( 1 等差数列 } { n a 的通项公式为 n an  ···········································································5 分 （2） n n n b 2    ··································································································································································6 分 n n n n n S 2 2 ) 1 ( 2 3 2 2 2 1 1 3 2 1                  ①·····················································································7 分 1 4 3 2 2 2 ) 1 ( 2 3 2 2 2 1 2                  n n n n n S ②·····················································································8 分 ①－②得： 1 3 2 2 2 2 2 2             n n n n S ·······································································································9 分  1 2 2 1 ) 2 1 ( 2        n n n n S ············································································································································10 分 数列 } { n b 的前n 项和 2 2 ) 1 ( 1      n n n S ············································································································12 分 19.（12 分） 解：（1） ) 3 cos( sin 3    C b B c  ，由正弦定理可得： ) 3 cos( sin sin sin 3    C B B C ····················1分 C C C B B sin 2 3 cos 2 1 sin 3 0 sin 0          ·····················································································3分 3 3 tan   C ··········································································································································································5分 6 0       C C  角C 的大小为6 ·····················································································································6分 （2） C ab b a c cos 2 2 2 2     b b c 6 12 2 2     ·······························································································7分 2 2 2 2 2 8 2 3 3 b c a b c       0 2 3 6 12 8 2 2 2          b b b b b 1  b ，或 2  b ···················9分 ①当 1  b 时， 7  c 2 3 sin 2 1     C ab S ABC ·····························································································10分 ②当 2  b 时， 2  c 3 sin 2 1     C ab S ABC ································································································11分  ABC  的面积为 3 或2 3 ·········································································································································12 分 高二数学试卷参考答案及评分标准第3 页（共4 页） 20.（12 分） 证明：（1）四边形ABCD 是等腰梯形， BC AD // 2 , 1    OB OD ··················1 分 连接EO ， PD EO DB DO PB PE // 3 1    ··································································2 分  EO  平面AEC ，  PD 平面AEC ·····································································3 分 // PD  平面AEC ·············································································································4 分 （2）  PO 平面ABCD ，  AC 平面ABCD AC PO    2 tan  PAC 2  OA OP 2  OP ·························································································································5 分  OC OB  以O 为坐标原点， 分别以 OP OC OB ， ， 所在直线为 z y x ， ， 轴， 建立空间直角坐标系 Oxyz ) 0 , 0 , 0 ( O  ， ) 2 , 0 , 0 ( P ， ) 0 , 1 , 0 (  A ， ) 0 , 2 , 0 ( C ， ) 0 , 0 , 2 ( B ······························································7 分 ) 2 , 1 , 0 (    PA ， ) 2 , 0 , 2 ( ) 2 , 2 , 0 (     PB PC ， ，···························································································8 分 设平面PBC 的法向量为 ) , , ( z y x n                0 2 2 0 2 2 z x PB n z y PC n 令  1 x 1  y ， 1  z ) 1 , 1 , 1 (  n ···························································································································10 分 设PA 与平面PBC 所成角为        | || | | | | , cos | sin n PA n PA n PA  5 15 3 5 3   ··························11 分 PA  与平面PBC 所成角的正弦值为 5 15 ···············································································································12 分 21.（12 分） 解：（1）由题，可知        3 2 2 2 3 b a b ········································································································································2 分 解得      3 2 b a ·······································································································································································3 分 双曲线C 的方程为 1 3 4 2 2  y x .·······························································································································4 分 （2） , 1 1 3 4 2 2          kx y y x 消y 得: 0 16 8 ) 4 3 ( 2 2     kx x k ·······················································································6 分 高二数学试卷参考答案及评分标准第4 页（共4 页）               0 ) 4 3 ( 64 ) 8 ( 0 4 3 2 2 2 k k k 2 3 1 1       k k ，且 ··············································································7分 设 ) , ( ) , ( 2 2 1 1 y x B y x A ， 2 2 1 2 2 1 4 3 16 4 3 8 k x x k k x x        ， ··············································································8分 1 1 ) ( ) 1 ( ) 1 )( 1 ( 2 1 2 1 2 2 1 2 1 2 1 2 1               x x k x x k kx kx x x y y x x OB OA  ·······························9分 0 2 4 3 8 4 3 ) 1 ( 16 2 2 2 2         k k k k 0 4 3 16 10 2 2      k k 2 3 2 3 0 4 3 2        k k k ，或 ··················11分 k 的取值范围为 1 2 3 2 3 1       k k ，或 ································································································12 分 22.（12 分） 解：由题可知  n