2022-2023学年高二上学期期末考试数学评分标准

2022-2023学年度第一学期期末学业水平检测高二数学评分标准
一、单项选择题：本题共8小题，每小题5分，共40分。

1--8：B D D A C C A B
二、多项选择题：本题共4小题，每小题5分，共20分。

9．BC 10．CD 11．AD 12．BCD
三、填空题：本题共4个小题，每小题5分，共20分。

13．26；14
．5；15．2；16．（1）221n n -；（2）1(1)242n n --++．四、解答题：本题共6小题，共70分。

解答应写出文字说明，证明过程或演算步骤。

17.（10分）
解：（1）因为22a =
，所以a =·································································1分
因为渐近线方程为y x =±，所以1b a =
，b =···············································2分所以C 的方程为222x y -=···········································································4分
（2）由（1）知，C 的右焦点坐标为(2,0)························································5分若直线l 斜率不存在，则直线l 的方程2x =，
此时(2,A B ，1246x x =<，不合题意··········································6分若直线l 斜率存在，则设直线l 的方程为(2)y k x =-将(2)y k x =-代入222x y -=得：2222
(1)4420k x k x k -+--=······················7分所以2
122
2461k x x k +=-=-···············································································8分即24k =，解得2k =±················································································9分所以，直线l 的方程为2(2)y x =±-······························································10分18.（12分）
解：（1）若选择①；
由题知，若数列{}n a 的公比1q =，则41214,2S a S a ==，
与4215,3S S ==矛盾···················································································1分
数列{}n a 的公比1q ≠，则421142(1)(1),11a q a q S S q q
--==--···································2分所以424221151S q q S q
-==+=-，解得2q =（2q =-舍）······································4分所以，212(12)312a S -==-，解得11a =····························································5分所以12n n a -=······························································································6分
若选择②；
由题知：数列{}n a 是各项均为正数的等比数列
又因为123a a +=，34a =，所以2
11(1)3,4a q a q +==······································1分
所以
121(1)34
a q a q +=，所以23440q q --=·························································3分解得2q =或23q =-（舍）···········································································4分所以2314a a q ==，所以11a =·······································································5分所以12n n a -=······························································································6分
（2）由（1）知：12
n n n n a -=···········································································7分所以012211231...22222n n n n n T ---=+++++12311231...222222n n n T n n --=+++++···························································8分两式相减得：2321111111...2222222n n n n T n --=++++++-·································9分1121212n n n -=--222n n +=-所以1242n n n T -+=-······················································································12分19．（12分）解：（1）因为
(0,1)在直线l 上，所以直线l 的方程为：1y x =-+··························1分因为CP l ⊥，所以直线CP 的方程为：1y x =+·················································2分所以C 点的坐标为(1,0)-···············································································3分设圆C 的半径为r ，又因为||AB =·········································4分解得3=r ···································································································5分所以，圆C 的标准方程为22(1)9x y ++=························································6分
（2）若该直线斜率不存在，则其方程为2x =，显然符合题意······························8分若该直线斜率存在，设其方程为(2)4y k x =-+，设点C 到该直线的距离为d 因为该直线与圆C 相切，所以3d =
=·················································10分解得：724k =····························································································11分综上，过点(2,4)Q 与圆C 相切的直线的方程为：2x =或724820x y --=···········12分
20.（12分）解：（1）该校男生支持方案一的概率为
2001200+4003=········································2分该校女生支持方案一的概率为3003300+1004=······················································4分全科试题免费下载公众号《高中僧课堂》
（2）3人中恰有2人支持方案一分两种情况，
①仅有两个男生支持方案一，记为事件A ；
②仅有一个男生支持方案一、一个女生支持方案一，记为事件B ；
显然事件,A B 互斥························································································5分所以，2131()()(134
36P A =-=
······································································7分1131()2()(1)3343P B =⨯-=································································9分所以3人中恰有2人支持方案一概率为：13()()()36
P P A B P A P B =+=+=·········10分（3）21p p >·····························································································11分显然112p =，该校学生支持方案二的概率估计值也为12，因为高一年级学生对方案二的支持率估计为35015050050011491600400()500500224242⨯+⨯=+<+所以，2112p p >=·····················································································12分21.（12分）解：（1）将1=
1n n n a a a ++两边同取倒数得：111=1n n a a ++，即1111n n a a +-=·················2分又11a =，111a =，所以数列1{}n a 以1为首项，公差是1的等差数列·····················3分所以11(1)1n n n a =+-⨯=·············································································4分所以1n a n
=·································································································5分（2）（ⅰ）因为
2111n n i i i i b a -===∑∑，所以1211112321n n b b b +++=++++- 所以121111112321n n b b b --+++=++++- （2n ≥）···································6分两式减得11111112212221n n n n n b ---=++++++- ············································8分111111112222
n n n n ----<++++ 111112[21(21)]122n n n n n ----=---==······10分当1n =时，11b =所以1n b ≤·································································································11分（ⅱ）所以11112321
n ++++- 121111n b b b n =+++≤++++= ···············12分
22.（12分）
解：（1）由题知：2|||WC WC =······································································1分
所以121112||||||||||4||2WC WC WC WC C C C C +=+==>=···························2分所以，点W 的轨迹为以12,C C 为焦点的椭圆······················································3分
所以曲线D 的方程为2
214
x y +=····································································4分（2）设1122(,),(,)A x y B x y ，
将y kx m =+代入2
214
x y +=得：222(14)8440k x kmx m +++-=·····················5分所以222121222844,,16(41)01414--+==∆=-+>++km m x x x x k m k k ···························6分
因为11AC BC k k ==
2k =，11=+y kx m ，22=+y kx m
整理得，12()(0m x x -++=····························································8分
若0m =，则直线l 经过1C 点，不合题意
所以122814km x x k -+=
=-+
，即1(44m k k =+··········································9分因为2216(41)0∆=-+>k m ，所以2223114(4)16+>=+k m k k ，解得234
>k ·······························································································10分
则211|2|2|44d k k k ==+=+
，||PQ =
令(1,)3=t ，则23737((444=+=+t d t t t ，53||8
PQ =，
所以73176||(424
t d PQ t +=+≥（当且仅当t =，1k =±时等号成立）所以||d PQ +的最小值为764
·····································································12分。