广西钦州市第四中学2025-2026学年高一上学期第三周考试数学试卷

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广西钦州市第四中学 2025-2026 学年高一上学期第三周考试数学试卷

注意事项:

1.答题前,考生务必将自己的姓名、准考证号填写在答题卡上,

2.四答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑。如需改动,用橡皮擦干净后,再选涂其他答案标号。四答非选择题时,将答案写在签题卡上。写在本试卷上无效。

3.考试结来后,.将本试卷和答题卡一并交回

一、单选题(共8小题,每小题5分,共40分)

1.下列命题为真命题的是(   )

A.eqId9308709cc7102d2c2895a9853a2aa3e2 B.eqId5f1a071309764f321fbe82861f04c6e0

C.eqIdbb4c670f1d88b9ce03d60ab997586b65 D.eqId890e1b0b67e6f99692d248eaa6a809cd

2.下列命题中错误的是(    )

A.若eqId432d77fe5ad3032d59a237dd94c8a638,eqIddcf7adcc976209d4b686156120bea276,则eqId125415eb6733489cdc96ee48eb88aee7 B.若eqIdbab5cc3c1f4073f786f31ecef107fb3f,则eqId432d77fe5ad3032d59a237dd94c8a638

C.若eqId432d77fe5ad3032d59a237dd94c8a638,则eqId0cb54b1b3617ebc502cb44194cbcd1dc D.若eqId2d100c22435a23e017cfe6f535379d3c,eqId58b140e221ddf537b8964fff8557cca0,则eqId58900a2822b9ff972d90440fced21584

3.若正数x,y满足eqId21c7501096f4be07c98e97e29db21a21,则eqId1cf99adccc80f28343fedd8d0aad7429的最小值是(    )

A.6 B.eqIdf16a68cf5a2fd75300fca439d39bdbce C.eqId5acc735c6274673a049a11f25ac86d3b D.eqIdb6cc12b8602c976e9aadac51f2d6ea2a

4.已知eqId3c11f7110b7dd464c1be6980f072c359,则下列大小关系不正确的是(    )

A.eqId8a534643d79364509b3e14b17f45750f B.eqId27f50ce6b511e6b928796e048fc7fa5c

C.eqId3c22b9d08cf536bbb76bce1b0f135772 D.eqIdb977abc63e6de32827fc8f49694abafa

5.若a>b,c>d,则(   )

A.eqId3c22b9d08cf536bbb76bce1b0f135772 B.a-c>b-d

C.a-d>b-c D.ac>bd

6.设eqId08115d6d9f876dea921a4d32260ff1fb,eqId061813f1ec633c5c4c393c4de7938322,且不等式eqId090d584536ad2890bf5cac8d1d3eeb0a有解,则正实数eqId0a6936d370d6a238a608ca56f87198de的取值范围是(    )

A.eqIdf184ef9e0d57554e95f369c9d4bbfea1 B.eqIdb1eaf48f1ad368af0b0961322e50d74e C.eqIdd26a8f1a01b4c0ea95cf543bcdb49ba4 D.eqId655b06387179d53c1e474fcfcb408b1e

7.已知eqIdb2a9344f4fca7b9779ca7720e5277ea6且eqId5558c083d34cbb0a58d3ce1dc6f5778e,则eqIde9d68de1c5ddf822f57b56b191bd582d的最小值为(   )

A.4eqIdcf298f00799cbf34b4db26f5f63af92f B.8 C.2eqIdcf298f00799cbf34b4db26f5f63af92f D.4

8.已知a,eqId67ca5fd57c2c2fcc3c7a574fdd1467d9,且eqId2b6320fbdc13c4ab88cf8f577cce4001,则eqIdac42025e439a68768819900999631ed3的最小值是(    )

A.6 B.9 C.13 D.eqId5c62b79bd14d9f73fd0a3b6ef154ea92

二、多选题(共3小题,每小题6分,共18分)

9.已知x,y为正实数,eqIdcbe0890043f34b4575bf7bb3a773e32b,则(    )

A.xy的最大值为4 B.eqId1295333db25226b7e4d5af5caeeed28d的最小值为eqId95bacae35b6e16a0a33c2bdc6bc07df7page number 0


C.eqId191ece4c91081e4cc219ca36f9c80648的最小值为3 D.eqId5753e7dc1f08e9c6d52d65ff98829556的最小值为16

10.下列说法正确的有(   )

A.若eqId432d77fe5ad3032d59a237dd94c8a638,则eqIdd2128a00f52af4427721f0ebba591daa B.若eqId08115d6d9f876dea921a4d32260ff1fb,则eqId7f17607be154c2e0d31c44f0af4e2202有最小值2

C.若eqIdca24341509c05e672999202f2df0ebaf,则eqIdec8201ff29a2091d40eee10db6bbc1f2 D.若eqId24a57996290794e082b21d8f1dfc322a,则eqIdf6d4ceb780d147ef10d3eaf29dda5327有最大值1

11.已知eqId94440d3e4c073f94f2b266ff99d50e74,eqId67ca5fd57c2c2fcc3c7a574fdd1467d9,eqId5be97cd1c7111b654d87d8fbb63b6a84,则下列结论成立的是(    )

A.eqId7f7266b2ef457b8ddeee3fa2cc24022e的最小值为eqIdb8860d9787671b53b1ab68b3d526f5ca B.eqIdcfde095e760c84e51c2793f0d3f34fef的最小值为eqId5ca7d1107389675d32b56ec097464c14

C.eqIdf1d6742f574916317608c73ad9e7d083的最小值为eqId61128ab996360a038e6e64d82fcba004 D.eqId2582892eba557b99dbe879efb9754d3d的最小值为eqIdbdaa19de263700a15fcf213d64a8cd57

第II卷(非选择题)

三、填空题 (共3小题,每小题5分,共18分)

12.已知eqId0fb7b4b17841b422c111cef772d642f4,eqId646d4e6fdd638752ee3cd977c51fc097,则eqId909a4f858f645bfb056687e7c9ba4c76的取值范围为      .

13.若直角三角形斜边长等于4eqIdcf298f00799cbf34b4db26f5f63af92f cm,则直角三角形面积的最大值为          .

14.已知eqId39d5f0d374837655cc286d326305da36,eqId5be97cd1c7111b654d87d8fbb63b6a84,则eqIdcfde095e760c84e51c2793f0d3f34fef的取值范围为          .

四、解答题  (共5小题,共77分)

15.某学校引入种植类劳动教育课程,打算围成如图所示的四块全等的长方形田地种植不同种类的蔬菜,其中一面可以利用原有的墙(足够长),其他各面需要用篱笆围成,设其中一块田地为矩形eqId411b38a18046fea8e9fab1f9f9b80a5f.

@@@3ec599dc-1a03-4cf4-a026-5855c704d0b4

(1)若每块田地的面积为eqId7ecd4c649e43aab78653a1adc890a508,要使围成四块田地的篱笆总长最小,应该设计田地的长eqIdf52a58fbaf4fea03567e88a9f0f6e37e和宽eqId03902478df1a55bc99703210bccab910各为多少?

(2)现有40m长的篱笆,要使每块田地的面积最大,应该设计田地的长eqIdf52a58fbaf4fea03567e88a9f0f6e37e和宽eqId03902478df1a55bc99703210bccab910各为多少?

16.高一的珍珍阅读课外书籍时,发现笛卡尔积是代数和图论中一个很重要的课题.对于非空数集eqId01c74a907dda6bb7d9d56d009d9df253,定义eqId887670c49c897d11f06d668eb282cb27,将eqId8df69c6a544f62a016490b012c353ae7称为“eqId5963abe8f421bd99a2aaa94831a951e9与eqId7f9e8449aad35c5d840a3395ea86df6d的笛卡尔积”.

(1)若eqIdbfc03ec3d78487844b44cd273efc9188,eqIdf808f81b6ea9da53d51c549be04f4267,求eqId8df69c6a544f62a016490b012c353ae7和eqId38146b730fc539c6cf22dc0e8430429e.

(2)若集合eqId73465a1f9aa03481295bf6bd3c6903ac是有限集,将集合eqId73465a1f9aa03481295bf6bd3c6903ac的元素个数记为eqId2ec8996421ea2bdb85b9f29c714d6a8b.已知eqIdb12eb88345234db73636d4bf3c55e7ca,且存在实数eqId0a6936d370d6a238a608ca56f87198de满足eqId6390d1cf4d00d975f4f6ba19a5aac290对任意eqIdc7905fd422e78a1d22ff6f11950bc5cb恒成立,求eqId0a6936d370d6a238a608ca56f87198de的取值范围,并指明当eqId0a6936d370d6a238a608ca56f87198de取到最值时eqId223df28e586d0f67cdb8b675cec0a59a和eqIdb7ac94bced60536f5595d1ffecf875ff满足的关系式及eqId294f5ba74cdf695fc9a8a8e52f421328应满足的条件.

17.已知eqIdb1010846eeec6c9da29640f5aa3f8738克糖水中含有eqId9aa8a716a31b0f51b70fdf9bdb257909克糖eqId715a04391bf471b55c301cf278a9510b,再添加eqId294f5ba74cdf695fc9a8a8e52f421328克糖eqId344cc24b575f4fd1ea7fe8ce5612fa9a(假设糖全部溶解),糖水变甜了.请将这一事实表示为一个不等式,不必证明.并利用此结论证明:若eqId76f0649064a085fb74c997fb507a9b6d为三角形的三边长,则eqId01721633154e61aa2650bf0b8b10e666.page number 1


18.关于实数大小关系的基本事实是解决等式或不等式问题的逻辑基础.两个正数的大小关系是完全确定的,但通过运算就会产生非常奇妙的变化基本不等式就是其中之一.通过运算(代数变形)可以解决很多关于基本不等式的问题.例如此题:已知eqId632244ea6931507f8656e1cc3437d392为正实数,且eqId5be97cd1c7111b654d87d8fbb63b6a84,则eqIdfdc176569ec8d428788cda28f6cac07a的最小值为_____.

其解法如下:eqId6a7ec255caf8eebf1e3399daf5801750,当且仅当eqIdf2b650b5bc58b918b70ba663283f7125,即eqIdf072c3a564e36a26d722e585379161ff时,等号成立,因此eqIdfdc176569ec8d428788cda28f6cac07a的最小值为3.

根据上述材料解决以下问题.

(1)已知eqId76f0649064a085fb74c997fb507a9b6d为正实数,且eqId751e274e9107d780c39ba9c49d6daefb,求证:eqId227d4e6ac62442cfdca656d07ed0b122;

(2)已知eqId2684b72f9f38f5046c8ecd4280b7b14b,且eqId5be97cd1c7111b654d87d8fbb63b6a84,则eqIdfe7a493d9f3460532a58ff6fa0a76e0a的最小值是多少?

(3)某同学在解决题目“已知eqId81dea63b8ce3e51adf66cf7b9982a248为正实数,eqIdd053b14c8588eee2acbbe44fc37a6886为非负实数,且eqId2436951e9a6f5ca0fc0ae65b9d787119,则eqId7af5a38d61d02f28f4a2792a06f2a65b的最小值是多少?”时,给出如下解法:

令eqId25849e741f1634153e80e70c465ec01e,则eqId2436951e9a6f5ca0fc0ae65b9d787119化为eqIdb095ce1274ec0b6d3c971485532b9a9d.

原式eqId270dc746ec2ebbb75060b97b68553961eqId9d555ebb74bef5e06085545a2d2053bb

eqIde1b8c4641d34371960e86457c49cf392当且仅当eqIda970675c8d897cd497355745ec53d273,即eqId9fbdafc4b61c2fd0cf6f4e3da0bfd51b,即eqId5b028286277ca44bf85a50ef447dbfd6,eqId1b6a27a9d5d345f1e04020bfea2770ac时,等号成立.

利用上述解题思路和数学逻辑思维,解决如下问题:已知eqIdea5191b46feeb8b4e60a400bd02d99b0,则eqId404c3be5eb6a7c2d3c6011c58cd2056f的最大值是多少?

19.已知eqId2684b72f9f38f5046c8ecd4280b7b14b,且eqId50893c29f6224aece1b113eb575f1740.

(1)求eqId18f0281e6bbdbe08beeccb55adf84536的最大值;

(2)求eqId1d57f3696dd2be39aedc35242e8c518d的最大值;

(3)求eqIdb0cf6ce5637d34ebdc5118e908bd7dba的最小值.

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