辽宁省大连市第11中学2019高三第二次模拟考试(文科)数学2019.05

发布时间:2019-05-21 22:42:13   来源:文档文库   
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大连第11中学2019高三第二次模拟考试

(文科)数学

命题人:

I

一.选择题:(本大题共12小题,每小题5分,共60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)

1. 复数289beca47072023f35fe95cce9762ba8.png,则z的共轭复数是

Aff283b6f546f3dc760d43413af28eb72.png B08c94306b73d1e24268dc827c259e080.png Cd3c3a3297f0a69567159a1b0c0a1f944.png D7a2f67bce71d51452a856db37a15cd4e.png

2.已知集合a4752ed93283497b7ab37de92c24f023.png,则9bc02d5f5bd148ba3a4e2517066ae085.png=( )

Ae4efbaa6c2952da86fa96dac18a839a4.png B4bcb1a7cc74ab60e228a77115b5c5b39.png Cb21188cc702dc9f6a0ab619f6ce05900.png D35d34bd77cedc4ce50b6f59331b60092.png

3.已知等差数列02731f49cef7ec135770140b199cf7cb.png满足4883a8bd1833f58ddbf50d51f79baedf.png9e5d5ee7f6d2d5c900b3f7b953b0db15.png,则其前n项和的最大值是

A32 B30 C28 D24

4. 已知命题ed5c7bc73125280da2340e93d0b62a70.png,则该命题的否定为

A794eef1c8038f0844ba704e8e65c815d.png B732a695f626aec1b9151309a84afdb80.png

Ccab1134ce5ef69e004dc394c832f83b1.png Dcfa715decee5558193c2bdf6cfd8143c.png

5 抛物线C49059373893ad4eddbe00b7ade4d2065.png焦点F,过C上一点D作直线DE垂直准线于E,△DEF恰好为等腰直角三角形,其面积为4,则抛物线方程为

Af2384216107333a6e55784e492b6f0df.png B796b62ea095a9eac7059dd518c0567b6.png

C7cdd10dd857d8d8d5071ff43625ebbad.png Da57c140a655068c4ec3842aa33c31c78.png

6 执行右图所示程序框图,若输入n=3,则输出c=

Abd8eacd6ef8c460fea72f998c06d4e7e.png B6bbf6a0b4badb095886da44504a88a64.png

C1ab424e34339a41547e06f7217733dc8.png D2d51988e69678e707c1134ff12f357f0.png

word/media/image31.emf

7一棱长为2的正方体被一圆锥面截去一部分后,剩余部分几何体三视图如图所示,则该几何体体积为(

A73a25e3a41189322fbb740c7299978cc.png B1e57f8a421350aa535701d31f792ca64.png

C2968eaad01bbef535cd41e0caa21b0c6.png D18ad01f5160acaf6d1490bb5713fae48.png

8. 在面积为S的△ABC内任取一点P,则△PBC的面积大于40a86aad5b34e0fdeac786bf3c48bad0.png且小于001d2c1ed59db21d3f6c997bf08855aa.png的概率为(

A6c2e3e2e98abd1fd9a66519db9da8d90.png B22b3abac47092d2f8d865ad1bf6e79a9.png Cbb0695289f737b14ad09e2ee77c5942f.png D9df743fb4a026d67e85ab08111c4aedd.png

9 将函数d1ec09cc19c9928b4d0c1201c5494df8.png的图象向右平移t个单位长度t>0得到函数e84fec1e074026d6fa8e3155482c35c3.png,若e84fec1e074026d6fa8e3155482c35c3.png62d8a90931ff1ab35845f2d7a5654da5.png处取得最大值,则t的最小值为

A035bc15da451c1dc88c666a1aa37e358.png B6d1a6127d3610e7b68659478ed0c2ae2.png C 4aec10557b2285f585c704ef8cfb8848.png D15a638c09d3e9bb4597ce8bf69141c36.png

10. 已知d3ebeac861a49f098e26dd22e603ea9a.png,若848e3c556b35c6332b3cde398352d743.png的最大值为4,则实数a的取值是(

A2 B3 C-2 D-3

11. 双曲线4f147f1f0a2cb408151f8935f052e41d.png的左右顶点分别是A1A2PC上任意一点,若直线P A1P A2的斜率之积1d85ae76a6227a8e6678fb6300116aed.png,则双曲线C的离心率为

A5 Baa4e3cfb024c7ff30a8846913966dfb1.png C2 D545d7812bbf630d7929e6c909cdbdae5.png

word/media/image55.wmf12. word/media/image56.wmf已知函数38c6787bd0d415aaddd3d37304bce385.png,若|word/media/image58.wmf|word/media/image59.wmf,则的取值范围是

A. word/media/image60.wmf B C83ceeab417cfd138b3e1f5ff1ce27aa7.png D.5ad8d442c8c1807c04a1335c5939cb66.png

II

.填空题:(本大题共4,每小题5,20,把答案填在答卷纸的相应位置上13设向量212b9bb30d9f82be72c08e290b582cdb.pnge3df65a6b88d462076995877406a591e.png,且166d019db3dfb9eb0cc7ab35de10b375.png,则afa7007067049f2ddde95b14deabef23.png

14.曲线9483e75665e16df2172895243b9a6cbd.png在点(0,1)处的切线方程为

15数列02731f49cef7ec135770140b199cf7cb.pngn项和44d853a7808a331d95220fcb38095649.png,已知3a17e215a2744c5a88eee49e63d50b71.png,则其通项公式6b264e52453f89797441f58c9a874006.png

16.已知三棱锥word/media/image72.wmfABC,外接球O,且03d1eddf115f49889b426b371b0e4776.png

ecb68477794b0da56d150a5883a5f42e.png三棱锥外接球的表面积________.

.解答题:(本大题共6小题,70,解答应写出文字说明、证明过程或演算步骤)

17(本小题满分12)

533c003242f935720a3ff6d1bc2c631e.png的内角ABC的对边分别为abc已知05a0337f1cf8054543698ef4bc410c8e.png

1)求角C

2)若533c003242f935720a3ff6d1bc2c631e.png外接圆半径为2,求533c003242f935720a3ff6d1bc2c631e.png周长的最大值

word/media/image78.emf18(本小题满分12)

某高校统计大一学生每周的自习时间(单位:小时),随机抽取400学生,制成了如图所示的频率分布直方图,其中自习时间的范围是fc84a9cb794cae483d040ca0e05dac3e.png,样本数据分组为7d1c3c26b19fc14247b65a68d3d8734a.png8139444a5825d56102db8a40871fac80.png0d0dc188a13f065825d1618cc30c8976.png8ffe9e46407f54fae6433605ff093a41.pngfdebe08bd60193fa279d3363d645d58a.png. 期末结束后学校为学期综合成绩的前30%发放奖学金,400学生样本中每周自习时间超过25小时的学生有60人获得奖学金

1根据上述数据完成下列249543bbeb4e6a832f228d60a680ff60.png列联表,

word/media/image86.wmfword/media/image87.wmf根据此数据你认为是否有99%以上的把握认为综合成绩排名与自习时间长短有关?

参考公式:,

2)学校为鼓励学生多去自习,通过分层抽样的方式在每周自习时间超过25小时的学生中抽取6名学生代表,在这6名代表中随机抽取2人为全校作报告,求这两人中一人每周自习时间fdebe08bd60193fa279d3363d645d58a.png,另一人自习时间在8ffe9e46407f54fae6433605ff093a41.png中的概率

word/media/image88.emf

19(本小题满分12)

如图,在四棱锥81a76cc5dcec10bda7298f04289fb2c7.png中,ABCD4dc5c7084e4226cba28ec6f22ccf0e92.png PBAD.

1)证明:平面PBD平面ABCD

2)若AB=2,△PBD为等边三角形,Q为底面对角线交点,求三棱锥P-BCQ的体积

20.(本小题满分12分)已知函数bb2ca679242d8ad2b134040585a5c775.png887fb68a10cbd4369b27c90bee0334d8.png

1)讨论50bbd36e1fd2333108437a2ca378be62.png的单调性;

2887fb68a10cbd4369b27c90bee0334d8.png时,若关系式71febe95bdec76b8ea852249fdf8bead.png恒成立,求AB的最值

21.8b73bdbf452fd7cf6840411396bfb043.png分别是椭圆9c698ab85cbdf615f3b418c2049365d3.png085ff2aeb6555e7447244313f1db6891.png的左右焦点,离心率e29392220eb76ad7b91325d5102216a7.png44c29edb103a2872f519ad0c9a0fdaaa.png0d61f8370cad1d412f80b84d143e1257.png上一点b196a6c1c80244eac787de5be00027ea.pngdd5b99e0c74e856d63373e4de507811c.png

1)求C的标准方程;

2直线l过点281735ccbc3bbe0aaea3c64e04bc47c8.png,且与C有两个不同交点ABAB不是左右顶点),在x轴上是否存在定点N,使得直线ANBN的斜率互为相反数?若存在,求出点N坐标(用m表示);若不存在,说明理由

请考生在2223题中任选一题作答,如果多做,则按所做的第一题记分.做答时,用2B铅笔在答题卡上把所选题目对应的标号涂黑.

22.(本小题满分10分)选修44:坐标系与参数方程

已知曲线9824b26a51714309aa4afd370035ce53.png的参数方程是4473afd895696795bc48b942d2c46738.png9824b26a51714309aa4afd370035ce53.png上每一点的纵坐标保持不变,横坐标变为原来的93b05c90d14a117ba52da1d743a43ab1.png,得曲线932d0ec79260e01afd1dd960c7bc69bb.png,以坐标原点为极点,word/media/image108.wmf轴的正半轴为极轴建立坐标系,直线l的极坐标方程是ee2aa0bd03408f673123ed180637f680.png

1)求曲线932d0ec79260e01afd1dd960c7bc69bb.png的极坐标方程及直线l的直角坐标方程;

2直线l与曲线932d0ec79260e01afd1dd960c7bc69bb.png交于AB两点,P为曲线9824b26a51714309aa4afd370035ce53.png上任意一点,求5e9720b2b258552156e6013d3723cbb5.png的取值范围.

23.(本小题满分10分)选修45:不等式选讲

已知函数9a96640beff42e461c5684577192c5d6.png,不等式5da31a14fd918f3bcb2e0d980ea37610.png的解集为ec8073dfa724c2c930b96f45b6a41668.png

1)求m的值;

2)若323c5f97105643bc61e288fe596194ca.png539fa66a54d60fdbd6278ccebed13ddd.png96df96dd95bbf61a8224853c7a06c48c.pngdfc10c2ff8dfe85ac0e680a7cc41fb57.png,证明:7b55b5082a6be0dc190c4253f3f3e19c.png

参考答案

一、选择题(文理相同)

C A B B D C D B A A B C

二、填空题

13.62833b72398bda7036a87285f9d29e9e.png 14.、(文)45046de61b285009b23590450000d87f.png

15. 200 16. 7232b1cb6f9fcd8e09561fc0939b8b90.png

三、解答题

17.解:(1533c003242f935720a3ff6d1bc2c631e.png中,9a71ab271478ee04d1d91b2b6978e7d4.png

由正弦定理88c0252b1c9e5540836c91f51f152ccc.png可得9d79247a771f263b257b8763a4e1c0fa.png

由余弦定理可得f4b12167eb258ec565560c527122e928.png 69b401088ec9df0fba13d527f90b9a07.png

(理)(2)由正弦定理ba95bde14a58a443492dd352cdcb9025.png

147ee89f4bd1482d5ca7c27d394553b2.png

df75aa57506c06822cf2dd37964319e3.png时,533c003242f935720a3ff6d1bc2c631e.png周长最大值为77bd132ba061a527c73dc8a7620e1f6a.png

(文)(2e5d6cbf2fcad0c9fdfc98d00ac95ef0a.png e3c477be410d2e3986616978578e5c9e.png

4be00b21971ba3cfc7bbd99de3907fa7.png 527d6e3f05bdfa0a546e8ce809e8ba0d.png

18.解:(1)由直方图可知,样本中自习时间在fdebe08bd60193fa279d3363d645d58a.png中的频率为7857f593b0c4c15f0d0e238cd439574e.png自习时间在8ffe9e46407f54fae6433605ff093a41.png中的频率为174c1cc5ed1a2d19f3630876a13620ab.png

f5ca2eb7aa99fcdf7d12540a7286897b.png

∴有99%以上的把握认为综合成绩排名与自习时间长短有关

(理)(2)设抽出一名学生每周自习时间在8ffe9e46407f54fae6433605ff093a41.png中为事件A1,在fdebe08bd60193fa279d3363d645d58a.png中为事件A2,则f3b56ee16eaee4d4d3072e9efbf70232.pnge33dad2f0d66070993f1638d56c2d7fc.png抽出的两名学生中有一名学生每周自习时间在fdebe08bd60193fa279d3363d645d58a.png中为事件A,则4ad70d5fdbfc7c5f904384c2c6a76deb.png,两个人一人每周自习时间在8ffe9e46407f54fae6433605ff093a41.png中、一人在fdebe08bd60193fa279d3363d645d58a.png中概率7701cacc35a5cc9d8cdc67a66274c4d7.png

∴另一人每周自习时间8ffe9e46407f54fae6433605ff093a41.png中的概率30c0111c6fa6c92484898dc0cb66e648.png

33人中每周自习时间超过27.5小时的人数X服从二项分布:5cb40ba61f3eae3e4e23699047b91fce.png

X的数学期望8d4e1060419c6f5fea13b51ce2f45e75.png

(文)(2)根据直方图自习时间在8ffe9e46407f54fae6433605ff093a41.png中与在fdebe08bd60193fa279d3363d645d58a.png中的人数比例为21

∴抽取的6人中自习时间在8ffe9e46407f54fae6433605ff093a41.png中的有4人,用A1A2A3A4代表;在fdebe08bd60193fa279d3363d645d58a.png中的有2人,用B1B2代表.

抽取2人基本事件空间为da974a73278e90e99dbac87a25526d6e.png{A1A2),(A1A3),(A1A4),(A1B1,A1B2,A2A3),(A2A4),(A2B1,A2B2,A3A4),(A3B1,A3B2,A4B1,A4B2, B1B2}

这两人中一人每周自习时间fdebe08bd60193fa279d3363d645d58a.png中,另一人自习时间在8ffe9e46407f54fae6433605ff093a41.png中为事件A={A1B1,A1B2,A2B1,A2B2,A3B1,A3B2,A4B1,A4B2}

word/media/image146.emf147689b5ab159663c03c3994e3882864.png

19. 1)证明:四边形ABCD中,

ABCD4dc5c7084e4226cba28ec6f22ccf0e92.png∴四边形ABCD是等腰梯形,取AB中点E,连接CE,则四边形AECD为菱形,∴△EBC为等边三角形,∴∠BAD=ABC=60°,△ABD中,由余弦定理得368cfab9d4b5e4086e5146bf7e2c77c2.png,∴∠ADC=90°,∴ADBD,又∵PBADAD⊥平面PBD3fe16da4a5185dc07dcc88704db9382d.png

word/media/image150.emf(理)(2)在平面PBD中,过点PPOBD=O

061aad5beaed674bc5263e05e5ab8553.png04108e35c56a2b0e5af9bf06d4a75508.png

∴∠PBO直线PB与平面ABCD成角,

PBO =60°,

AD⊥平面PBD,∴ADPDADBD

∴∠PDB =60°,∴△PBD为等边三角形,OBD中点,∴OCE上,且为CE中点,∵CEBD

O为原点,OEOBOP为坐标轴建立空间直角坐标系,设AB=4,则各点坐标分别为8f8e362351bfcd4c13d11d5e026f7c73.png8e7a6a868031944205433a1c2254fb52.pngf25fc3dd43165ea7a4099500332fd00c.pngf49b2a95346e951efd58db2b68cb5570.png,平面PAC中,524213d53c087444e5fcf009e660b1e6.png72525a2ef0c60d4ee1893a7a5a0d7051.png,设平面PAC法向量db08cced38f3965437b5ec79e1cf48ce.pngf6e6e557a9f9497e9fafa5c2e7046262.pngf8ef7b906795e4b1a0ff75ea7e30dc9b.png,不妨设9360d2c79de73e141e391d96ae0770ba.png,则f1d5b9db5a7054e9a0b5d4f28fbc8829.png9a4ef40d1cc546f801e31f5890936bd1.png

word/media/image165.emf平面PBC中,9ce556f8e23b813a864f30fd4b2f862d.png72525a2ef0c60d4ee1893a7a5a0d7051.png,设平面PBC法向量2a6223a1beb9c992654bc5eabe58edb3.pnge3c2c6a6a56568bf034f79307d5d2324.pngb6d5a42272114e4fe09fb902fae368eb.png,不妨设9360d2c79de73e141e391d96ae0770ba.png,则3ea855b517fbb952a2456c4d7217d47a.pngbc5c0e09876d5012f01514ef25c89702.png,设二面角A-PC-Bα,则7dbce71bc6474133e12937ef2a4df3b8.png

(文)(2)取BD中点O,连接PO,∵PBD为等边三角形,POBD061aad5beaed674bc5263e05e5ab8553.png04108e35c56a2b0e5af9bf06d4a75508.png

727dac765927377526f5089ab16feede.png

∵四边形ABCD是等腰梯形,由第(1)问过程可得, ACBC,∠CBD=30°,

AB=2BC=1RtBCQ中,be5c30698276014f6a0df6d11ce92d28.png,∴268e2e7ce080778d46b6d4cc31471314.png

b1d5df4f50389b27a335297582577a78.png

word/media/image177.emf20.:1)如图,由已知得9c00bdea34174deccd28fabf1c37dd9f.png67d243c74a6813524eee9b3d1ff16f5a.png6dc750c89fc60362bb7cd951eeb7c304.png86546065714541caa0b8c293acb25103.png,

0114ca724d2acbc00e6e1b4cea3e8e40.png

e4b6a4478a34b88381fe13419237a3e1.png,ff001ca93a711a067c67e6a7a2836f54.png4f875a12768677105befe43a6b4cc424.png

word/media/image186.emfC的标准方程为714b14fa99b57735db4fa0c5277a02f4.png

2)存在定点N

AB不是左右顶点,837b42df81c2810b441714ce8a72ab66.png

l的斜率不存在时,点MC的内部,此时78517aa0148fe0553a43d810c20e93d0.png,由于AB关于x轴对称,故x轴上除M外的任意一点结论均成立;现讨论l斜率存在时,设l的方程为3cd23ed62f654bfe71c0ca6b7bc51721.png,点ABN坐标分别为a3f156f63021c1015fe414a6cacce46f.png

故有75b15d8ca565b4a2e8f3d7188284a062.png

lC联立得726ba77537d2b44b1580df2400f8081b.png,消元得:8602631ba902482eb7542b3be225bbf8.png

f598f66f92eed3cccd6e68bf0eabc5ab.png a720822b511a24ac283c11ca7a6bec97.png

整理得4bdaf3490a9cfc71a55addc80e5a7714.png

∴定点242dded231775369a1bb55366fde7be0.png7de1d32298e92abfc1113be441eaeb34.png

(理)21.解:(1bf096815c856e4e53bf57a6513b39259.png

cae9743b2aa30af47283cd8d49c0b452.png时,634e0231a21a388254e137f1d5518da8.png,解得cad2db31b343e1739847bd772b939b68.png0f840bc0209c8857ac5d2a2a1a3bd596.png,解得5186c2de83001c971834318555e7623f.png

f(x)73b4481cc08c0aa2e7f9f3331662c756.png上是增函数,在fcedc2d9e7e7a046bc49a18acde3545f.png上是减函数;

9b30fdf9a09acbb5d41e727cbd07569c.png时,634e0231a21a388254e137f1d5518da8.png,解得5a328951584ec54650a63efc013df35a.png0f840bc0209c8857ac5d2a2a1a3bd596.png,解得e4f63075bfe799063fe0e1e6a0c34003.png

f(x)39957d3c14298094c34af9cb1e90527a.png上是增函数,在eda3529d050635f8fe88ce5f4c7b0102.png上是减函数;

3872c9ae3f427af0be0ead09d07ae2cf.png时,cf513776df4dea5963ed6b4afad9593e.png恒成立,且只在e28b7964a01e56385d1d9fa4da54388c.png4a3fc52d7e6d57bcf9bad4287f7c9ad0.pngf(x)R上是增函数.

2d85e3634d8dc269f39031f837410de87.png时,5930338fee74d45f36bf96c1cf8e04a1.png

若要7e4dcf16bb924dec942d44fce8c0fe01.png使得14186a311faa7ebed6e8a658b3982c80.png成立,

只需21699f97913f0b950eabcd6ee7af651b.png时,a75443124726f8f8368c07a658f64db2.png成立,

由(1)知当43be9163c38281625cf713051581195b.png时,f(x)10397fd67370c312d6931e0f4f8dedc7.png上是增函数,715dd8fc8328b0e998b31d31df92df6b.png

f207aade4ecf6e4421f694d6bec63f34.png时,f(x) e2321b139a153e923f4f6cd0eeeb9bdc.png上是减函数,在2b5b7c1b2a31f40a362041992aa25198.png上是增函数,

e26837a3a0dd3d12b4619078d93afb7f.png

f94ee3db8e9262705f005581ecda890f.png时,f(x)10397fd67370c312d6931e0f4f8dedc7.png上是减函数,cf61207e8e8001902f0870b431369bb5.png

3266025e7296bc32c4cfc4115af3fda3.png,对称轴50a20ce04c291ac897290a4f2e3bc9e2.png

43be9163c38281625cf713051581195b.png时,g(x)10397fd67370c312d6931e0f4f8dedc7.png上是增函数,d839aec0da3736a3ff4444d0e6e69709.png

99d2980c12bbc8b001c43c2a3087cc05.png,解得ad162210189e6f72fe9710ce9a441908.png43be9163c38281625cf713051581195b.png

f207aade4ecf6e4421f694d6bec63f34.png时,g(x) 9384b4461d6fc233b08eadbbe1eb73c0.png上是增函数,在4b01978b77be38bfa26be75f89eb5322.png上是减函数,

2ed5726e09bda55a7635ac7045c0dee7.pngd45926dcc80e00f59a8a2218f36c229a.png

整理得dbd043ede67dcf9baf982ecf2c00d18a.pngf207aade4ecf6e4421f694d6bec63f34.png∴只需282a4df27b6e0b7ad8674426a014c54e.png

537c157109d15b1bfa2d89fbbb6b50fe.png2f251b4a73e60e5ec4d530ba1d3c1b10.png,当f440dc0236526042f52a3d40435a2d22.png时,fb896f2cd237a2ff94052b38bef41621.pngca8e608169b20a94570ac837e8ba0833.png92a558fbe599d60e7a43c16264e1423d.png上是增函数,又352383e4e798148b05ec354cff23c231.png77b4599594def71ab1fd041bbd4e3bc7.png时,2ccb86ce824e8cf6f083bb35df4d9e1c.pngd0fbdd69e908c07e927c1fb9940a7440.png

f94ee3db8e9262705f005581ecda890f.png时,g(x)10397fd67370c312d6931e0f4f8dedc7.png上是减函数,809b9dc9bddcf3a5f5f9a1593e834937.png

ea82ef168cf45561d3463b4977683de6.png,解得f94ee3db8e9262705f005581ecda890f.png

综上所述,77b4599594def71ab1fd041bbd4e3bc7.pngf94ee3db8e9262705f005581ecda890f.png

(文)20.解:(1f(x),定义域为b921db311612fd3665c51872c7a83455.png5cf7716d8f426d07c1913d3cf4aa79b2.png

634e0231a21a388254e137f1d5518da8.png,解得1b1c3a4cf1e3b108a7314b85be7698ce.png0f840bc0209c8857ac5d2a2a1a3bd596.png,解得488155c4b1fd53acb1d43256c0ad83b0.png

f(x)1d5972818d73a2604170d0d4bd643547.png上是增函数,在fb4d1ba7dcc5fa9e2e9057f8dd20d482.png上是减函数;

2)显然,右边不等式等价于63d2c2272926300d377b9e34c9aae44d.png,由(1)得f(x)97e1663bd7f430420d0bb5f6c3a76716.png时取最大值b9ea4f445a2bb7ab6c151d132373c78e.png

42786e721cb3c84f92fd094d4a809e27.png,故B的最佳取值为042b3cb45c0ed22da8bf0b5b6289b63e.png

左边不等式等价于a1e88275aafd954bb833b6c7c7e2732c.png,令8d99d6a21e948963bb647bf85cdf94e5.png5bd608db1cfd4bf61a114af90eb0e1b6.png

fa766ee48ffb29d77fddad988c118cff.png,解得3d43cd43402bffe4e03927e4a6a6f932.png593b55fae41779a3941cf10822c20cb8.png,解得9602fcd17978ba4386d9a750616678ef.png

g(x)39153f5eda348a562fdc0cafac894132.png上是减函数,在46801dd9f4dac97cd51f1b573acde113.png上是增函数,

g(x)b83b9a850bbba82fae134b54367c124b.png时取最小值ae38dd9f368b79d40a83bbaeb51da71c.png9b20d7df88f171f156f21c6e2877a00a.png,故A的最佳取值为166485d3526d313c8bd076a6266f0b86.png

最佳的AB的值分别为6f552599b1869c007cd4799ae02e5a64.pnge77f10fbfd7f5bc87def9fcbe1f46219.png.

22.解:(1)经变换得曲线932d0ec79260e01afd1dd960c7bc69bb.png的参数方程是f7ea4f1fdbc104377de9a072d357520d.png,故其普通方程为0f8f39a23aa423a547f6432184b50b08.png∴极坐标方程为1708873598b84fc5e1cd9ee1a8f99cfc.png,直线l的直角坐标方程为26d1c9af87bca19230a20bbbfa18391a.png

2)联立l932d0ec79260e01afd1dd960c7bc69bb.png9d9db5aea0f4fa23c4886aff0d710f4f.png999010de1049c653abb2ea288d4e5006.png682e93a3b8f1fc4594241970c1e0bc13.png

P为曲线9824b26a51714309aa4afd370035ce53.png上任意一点,设P坐标为774c2d6dd1d7b56ea10546ec6d9da358.png

875fb7123f19708653cf25c159a10ed3.png

4b12bb49f2c31acc0b68a6a1c95cc293.png12d8e6ad558fc0b1c98362ce1e769c0b.png

23.解:(1)由不等式5da31a14fd918f3bcb2e0d980ea37610.png的解集为ec8073dfa724c2c930b96f45b6a41668.png可知fd05d8d90456c441c8f10641bd8576bc.png的解为-40

∴有0e711e0e8e7b5bae2669e293bd67ab5c.png,解得40aa2227f8ab9f9737e2ce467090bb9c.png

2323c5f97105643bc61e288fe596194ca.png539fa66a54d60fdbd6278ccebed13ddd.png96df96dd95bbf61a8224853c7a06c48c.png814715bb9bc7ac398e0bf7a4a88bff39.png

∴由柯西不等式可得6e7df355a3b468caf28f8c52a73ec762.png

7b55b5082a6be0dc190c4253f3f3e19c.png

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