河北省承德市双滦区实验中学2026-2027学年高三上学期数学第一 周周末自主学习检测试卷(1)

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河北省承德市双滦区实验中学

2026-2027学年上学期高三数学第一周末自主学习检测试卷(1)

一、单选题(本大题共8小题,共40分。在每小题列出的选项中,选出符合题目的一项)

1.命题“eqIdf5c105c658cc0c36c8faca5cac7a60eb” 的否定是(   )

A.eqIda6882ecc37ea16132236561763e58efe B.eqIdb3d7e8ed55a592ab622948cdc70a943d

C.eqId63f572e4450ade83b814a9ded2676cc5 D.eqId44625e02187d6ef27ffd68b2c2d42cdc

2.若集合eqId53c9e6ca86296df87bbd7edd6d2c20fe,则eqId5963abe8f421bd99a2aaa94831a951e9的真子集个数为(    )

A.eqId5ca7d1107389675d32b56ec097464c14 B.eqIdb8860d9787671b53b1ab68b3d526f5ca C.eqId8b06e95b57b7a81cd81d05557a11fa92 D.eqIdd8d4cb09f73abcf5c0d93c6644d362af

3.若集合eqId65916a9cd31967a28cbaa25b44a402b8内的元素个数为eqIda0123b5e8d6a6430fb38a7bd75c94e51,则eqId380bbacf854e30e2e747fc286d2b9997(    )

A.eqId11efbc8f14ba04f358d714d468bdbbfd B.eqIdda8c4f6bcfc959569a7e2bc8bc1cd600 C.eqId09d6843d2c2585cd7b2b9c4df4a0b5cf D.eqId275b1eea3706e7c37977c694da712992

4.设集合eqId18a44e76ca16012f4a4fab7dde62a361,集合eqIdef21161892b68f1903fa1bd0891b84ee,则(     )

A.eqIdac047e91852b91af639feec23a9598b2⫋eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3    B.eqIdad96b50521f4bbf3d436d05dc258083d    C.eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3⫋eqIdac047e91852b91af639feec23a9598b2 D.eqId84c4227b2ccc3c89877628a5b1a1a9ad

5.设集合eqId875ea21ac299a385b2b69f1dec3f32f4,集合eqId2aaf4fb6429d4e06485e9c7276216d36,若eqId96f1ba0a1129741502600e47bf058c98,则实数eqId0a6936d370d6a238a608ca56f87198de的取值范围为(    )

A.eqId3e10140ab3cdc13d710a65b2287c892b B.eqIde5d6243e93c41978871cb23d8e66148d C.eqId9d188ec2580e273ce87e51653a2177ee D.eqIda8e8432a3bda0db5b552568a673c0ffb

6.设集合eqIdc697e6f67d02b8a1a77cafc821e2f885,则eqIdb4b9b470218359a4a47be9244980489e(   )

A.eqId6170ed4a10e57487d525e71610dff82f B.eqId6b51d6256271c2d202c776dca91455db C.eqId29343388ca8b33dc98325e65382b38a0 D.eqId9a837165ca03f9e4ea8964979c95e3bb

7.“eqId523a21fee5ba9bf63986d5c9c748e0ec,eqId3037bb58e170b7c279ef4401f9f36ba8”的一个充分不必要条件是(   )

A.eqIdaf170c939d2e18faf44ce09761d29a3c B.eqId93a078cca891304dfb3887278b5fbc70 C.eqId4526d19896bdff6cb66b4aea9a6ef24d D.eqId23660aaa7c5547ee166dc985812853db

8.已知不等式eqId2c762f97c305215402280050fe1c4556成立的充分不必要条件是eqIdd47f8d234c1df11e957b9bd7d3f2da47,则实数eqId294f5ba74cdf695fc9a8a8e52f421328的取值范围是(    )

A.eqIdfeab5f8ba4c43734887c48a3918d3e08或eqIdea02853e8f9d1a925846ffd18c26ab90  B.eqIdfeab5f8ba4c43734887c48a3918d3e08或eqIdba3c82ed3506c03bea5bd55c84543810  C.eqIdfe0111555f3e0f62b215aefa8eef770b D.eqId50db592033df9f51f5c3a3509ce79bf4

二、多选题(本大题共3小题,共18分。在每小题有多项符合题目要求)

@@@06b6c5fa-47d2-486e-9c84-b527c159ef309.已知集合U是全集,集合M,N的关系如图所示,则(    )

A.eqId7eac19737d882225c903054eb4863501   B.eqId12ba583329b831d06312a1f20edb4a81  C.eqId16acc83acb8de823aaea2e6511adf06d D.eqId6c7a9322c4e8c1bcd7cf47a0349f2ffd

10.已知eqId81dea63b8ce3e51adf66cf7b9982a248,eqIdd053b14c8588eee2acbbe44fc37a6886均不为0,记eqId9f6b2146204fc19ea01e0abcb13e119b的所有可能取值组成的集合为eqIdac047e91852b91af639feec23a9598b2,则eqIdac047e91852b91af639feec23a9598b2中的元素为(    )page number 0

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A.eqId274a9dc37509f01c2606fb3086a46f4f B.0 C.2 D.4

11.若命题“eqId0584e91f2dc06af7946e8b97553931ed”是假命题,则m的值可以是(    )

A.0 B.eqIdacbc6a613224461ade69362d46550474 C.2 D.4

三、填空题(本大题共3小题,共15分)

12.已知集合eqId7aad6b0caf0694076bd1e7e3bb4fe457,eqId7c3fd89e088697930040012d047de79d,若eqIdc2ad78dc8b8aed907b4fe9640c997454,求实数eqId0a6936d370d6a238a608ca56f87198de的取值范围是______.

13.设集合eqId162630de7571032867fdbbf1e7c133d5,eqId510a9a436ebe9987b303621d6c3b01f3,若eqIdc81919684bc4047d376c7e57dc6c8f1c的真子集个数是1,则实数eqId0a6936d370d6a238a608ca56f87198de的取值范围是______.

14.已知eqId66a208792083df8d7b6b9f887c1afb7c,eqId7ac5c92fa7ed30fcb5cc66fa0232346c,若eqIdf8d6316aeb91aef9da57589f8253f03f,则实数eqIdb1010846eeec6c9da29640f5aa3f8738的取值范围为________.

四、解答题(本大题共5小题,共77分。解答应写出文字说明,证明过程或演算步骤)

15.(本小题13分)已知集合eqIda5903a2a1d029aef814ee556b0e25c00,全集eqIdfc0dc8098ad6f31bdd87771ca9cfa33a.

(1)求eqId8fdbfa7a63fdf5717d40c8c9a73ec160,eqId9fd1315ef75fa8897d12759b3aef5c45;

(2)若集合eqIdf26dbffcc654496d8f0fdb8d65b7ab40,且eqId2c6ca098b764e0366d043b69ae9d8f68,求实数eqId294f5ba74cdf695fc9a8a8e52f421328的取值范围.

16.(本小题15分)已知集合eqId074c4fae73b3d78a14d0ed96a8378249,eqId305b40710bd3ed129b0f05a41935174f

(1)当eqId711b21672fd907c5c92fee1d649e7003时,求eqIdf709e51fdadd6ce84a552e97ba943537;

(2)若eqId56552cb9c94f329a295d508567103f75,求实数eqId294f5ba74cdf695fc9a8a8e52f421328的取值范围.

17.(本小题15分)已知集合eqIdd1ed69a069bc34808dca8a20ca7f108f,eqIdb94a9f70f4d76b825e4d7b50a58f03fa,eqId02c59216bf1b56b5dff4f052fc07e177.

(1)若eqId07241606934d6c75cf04455979d0e1b0,求实数eqId294f5ba74cdf695fc9a8a8e52f421328的取值范围;

(2)命题eqIdb1010846eeec6c9da29640f5aa3f8738:“eqId728ca24844243ea9185f6f79c773145b,都有eqId6ed006b944ea64f970fee46e2f558467”,若命题eqIdb1010846eeec6c9da29640f5aa3f8738为真命题,求实数eqId0a6936d370d6a238a608ca56f87198de的值.

18.(本小题17分)已知集合eqIdcc6bd17cd87e0798b301f3d065331013,集合eqIda26e2c0fbd923486904a77a3cf320e47.

(1)若eqId0b05d2be27e8f53e4de3071846dffb41,求实数eqId294f5ba74cdf695fc9a8a8e52f421328的取值范围;

(2)若eqId96f1ba0a1129741502600e47bf058c98,求实数eqId294f5ba74cdf695fc9a8a8e52f421328的取值范围.

@@@bc96b52b-9914-433f-8705-c297abccda7e19.(本小题17分)已知全集eqId442be389c6471b36eff2652b75beb114,集合eqId829813e188ed9cc53c984661a41595a9,eqIdf4d7c71d086fe145682f8ecd49e38160.

(1)若集合eqIdc5db41a1f31d6baee7c69990811edb9f表示图中阴影部分,求集合eqIdc5db41a1f31d6baee7c69990811edb9f;

(2)若集合eqId773c9c217c2ce8639b4898f5b642e68d,eqId09a05a938526dc10a404767604317418,eqId91709f38ba81ad85cfb48d8e01e0291f,eqIda0ed1ec316bc54c37c4286c208f55667⊆eqId2a30f3a8b673cc28bd90c50cf1a35281,求实数eqId0a6936d370d6a238a608ca56f87198de的取值范围.

 

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参考答案:

题号

1

2

3

4

5

6

7

8

9

10

答案

D

C

C

B

D

D

A

D

ABD

ABC

题号

11

 

 

 

 

 

 

 

 

 

答案

AB

 

 

 

 

 

 

 

 

 

 

1.D

【详解】命题“eqIdf5c105c658cc0c36c8faca5cac7a60eb” 的否定是“eqId44625e02187d6ef27ffd68b2c2d42cdc”.

2.C

【详解】由题集合eqId53c9e6ca86296df87bbd7edd6d2c20fe可得集合eqId5e5ac52d57a15e5619fe867b8f7bbd73,

集合的真子集个数为eqId54eec8d8e965aa406a18de78a2addf19,故eqId5963abe8f421bd99a2aaa94831a951e9的真子集个数为eqId3ca820de7e47037bb05f6374905e5690.

3.C

【详解】若eqIdbb5f421939ee855f25927e7570d82c71,则eqIda707548d9d9c01a3e930189ad93346fb,不符合题意;若eqId94440d3e4c073f94f2b266ff99d50e74,则eqId57220617ee4a03b9bad1f313dc388985,所以集合eqId5963abe8f421bd99a2aaa94831a951e9中的元素为eqIdd360794828cac510eae1d40d88b64384、eqIda463aa057973b80a2964c646cf5cc97c、eqId4b41d43009a98c3c17f98931986e17c4、eqIddaa5e9bd516f6282483b92cfe6074623、eqId0a6936d370d6a238a608ca56f87198de,故集合eqId5963abe8f421bd99a2aaa94831a951e9中的元素个数为eqId72b6587ddffb762accf681f769f0d7e0,解得eqId24453a6db583f77daaa930a070a9138a,符合题意.综上所述,eqId24453a6db583f77daaa930a070a9138a.

4.B

【详解】因为eqId45a173784888adf2946382fa093ba53a,eqId1c11497da9dc1fee5a18f1e0de706c5a,所以eqId81dea63b8ce3e51adf66cf7b9982a248是任意奇数,所以集合eqIdac047e91852b91af639feec23a9598b2是所有奇数构成的集合.

同理,当eqId45a173784888adf2946382fa093ba53a,eqId3580e1194032fba47f397ea00f55b159时,eqId81dea63b8ce3e51adf66cf7b9982a248表示的是所有比4的整数倍多1的数,即eqId1cb0919b9a6e884c38f85c3a8c127f00①,

当eqIda91fee75fd9e8ff4f7b2971df7568afd时,eqId81dea63b8ce3e51adf66cf7b9982a248表示的是所有比4的整数倍少1的数,即eqId11fb949a7d6297a73d54342f82526dc8②,

所以①②合起来也是全部奇数,所以集合eqId54a5d7d3b6b63fe5c24c3907b7a8eaa3也是所有奇数构成的集合,

所以eqIdad96b50521f4bbf3d436d05dc258083d,故A,C错误,B正确;因为集合eqIdac047e91852b91af639feec23a9598b2是所有奇数构成的集合,所以eqId822672fe70ac445ec00a5684c72eb378是所有偶数构成的集合,所以eqId28621681893d53be2182115dc2cf919b,故D错误.

5.D

【详解】若eqId91719b9fd3cdc9d816949371e343b32e,则eqIdf3807adfc522c63cc1842f763c34baa8,解得eqId839013ee674de584aa7e0e0a1690deb8,因此当eqId40d5e131bfa717cd28891ac1df3c3319时,eqId0a6936d370d6a238a608ca56f87198de的取值范围为eqId6886c8666af017d2690e7c925281bb1dpage number 2

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