江西宜春市丰城市东煌学校2025-2026学年高二下学期7月期末数学试题

标签:
特供解析文字版答案
切换试卷
| 9页
| 65人阅读
| 0人下载

2025-2026学年度高二数学期末考试卷

第I卷(选择题)

 

一、单选题

1.函数eqId055f35f80a1514d7bcbecef1f83c6f12 (e为自然对数的底数)的图象大致是(  )

A.@@@26eac04b-02ee-4bb0-ba1d-493721319c42 B.@@@6896a4bc-5f89-4f69-a36a-8e1e2f5fe72f C.@@@ba60be0c-fc9b-4adc-916f-2e0b7a6d883e D.@@@8d5235eb-ef62-495b-a74d-76358139c655

2.已知数列eqId83cf38189d5cbf627d2b82ac0eb76006满足eqIdab694854363d1ae3e44c74904843be32,eqIdf1fb73c3135e576d762bbae078ae2684,则eqId83cf38189d5cbf627d2b82ac0eb76006的前10项和等于(    )

A.eqIde35bc4d9b5d0ccf0b4c015e2ef872f9d   B.eqIdd7f31e6df3bdd9658381fea248d16d0d 

C.eqId6cd7ff42a146a026598d75f7c33906c4 D.eqId0cddeec723a2329333db9e06f23b9a66

3.函数eqIdfbbf465c215c08f44502c7ce4b8bb601的导数为(    )

A.eqId9e3ef005ab89c01e4d85373e38c3d671 B.eqId62b4c332ba40fbda9aeafe2de4e472c7

C.eqId096fd2bb6d22f7af3d0cbdbad5e7f3f7 D.eqIdff1267a4efefee9c234e428d07a36ac9

4.已知函数eqId09f86f37ec8e15846bd731ab4fcdbacd的导函数为eqId724340d69477c0ec2418c392b22b1cab,且满足eqId3e85b6d5108d11d0c263507582f5c775,则eqId0d73bb75b197203cd0bd5a6e12aaf4a3(    )

A.eqId4e8b4bbd2f9912adfc9864c0e1e76a9d B.-1 C.eqIdf89eef3148f2d4d09379767b4af69132 D.0

5.已知等比数列eqIdb4be2164a2c67d6163faee87a10942bb前eqIdb6a24198bd04c29321ae5dc5a28fe421项的和为eqId08eb71ecf8d733b6932f4680874dbbf3,若eqId66df06d2928f3bb775dbf7c78c3d67a1,则eqId294f5ba74cdf695fc9a8a8e52f421328值为(    )

A.1 B.eqIdacbc6a613224461ade69362d46550474 C.2 D.eqId274a9dc37509f01c2606fb3086a46f4f

6.曲线eqId5b7186a0a143460905e117dd9dc17c45在eqIdf23d29646155e27b172ecdf263e2d702处切线的斜率为(    )

A.4 B.2 C.eqId56d266a04f3dc7483eddbc26c5e487db D.eqIdf89eef3148f2d4d09379767b4af69132

7.函数eqId89f066c412963644721afb7da99708fe的单调递减区间为(    )

A.eqId455ba3d3e46977fcbe5b71f8bb9df4be B.eqId7160d93f92089ef36f3dab809d3114b8 C.eqIded6d804ef44bfc64f824b0ccef71765e D.eqId5ed2f490aac02631c2ed9e6b76354a49

8.若函数eqIda00f01d4816c6e6a6daed2d8be0c738a有两个极值点,则eqId0a6936d370d6a238a608ca56f87198de的取值范围是(    )

A.eqId5e5e7569ea6cf2b0d62cb3cac98311dc B.eqIdbbf61c1e96802cfff4541f641641c320

C.eqIdb6864594f5b19905511b99ac4c762f04 D.eqIde2ebcb26273dcef72e9fa55e4bd183b8

 

二、多选题

9.若函数eqId7e8df43936ac37134adc09bf9a259e74恰好有三个单调区间,则实数a的取值可以是(   )

A.eqId81fb134b2b48acc99213fff6ccfee65f B.eqIdacbc6a613224461ade69362d46550474 C.0 D.2

10.若eqId08eb71ecf8d733b6932f4680874dbbf3为数列eqId83cf38189d5cbf627d2b82ac0eb76006的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和,且eqId1864a821776d4a596493f1b9ee176856,则下列说法正确的是(    )page number 0

试卷第1页,共1页

学科网 zxxk.com学科网 zxxk.com学科网(北京)股份有限公司


A.eqIdc0863cf59114f905e9ad3debc5572792 B.eqIdbbe7bdaaf8b0adf10bf2ef6c1255b1dc

C.数列eqId83cf38189d5cbf627d2b82ac0eb76006是等比数列 D.数列eqId34f86e5aaea193d51fa06c58abb3898b是等比数列

11.已知数列eqId83cf38189d5cbf627d2b82ac0eb76006满足eqId780a1b00ea3a4fec3069509041c84511,eqId4b9cff1a9ceaafab92feca53e701b150,则下列各数是eqId83cf38189d5cbf627d2b82ac0eb76006的项的有(    )

A.eqIdbf31876698721a199c7c53c6b320aa86 B.eqIda4b8503f4706b8321e4e79a87eadea84 C.2 D.3

第II卷(非选择题)

请点击修改第II卷的文字说明

 

三、填空题

12.等比数列eqIdb4be2164a2c67d6163faee87a10942bb中,eqIda316124e688e76d6f330ffbea49d427d,eqIdb5b38eccdc733944bfe49b24a35b5a4d,则eqIdb4be2164a2c67d6163faee87a10942bb的前eqIdd91e07104b699c4012be2d26160976a2项的和是________.

13.已知a1=2,an+1=2nan,则数列{an}的通项公式an=________.

14.已知eqId93ab7e09e52b1b48e1ee96c4b9333dff对任意的eqId69303621c56f67b4ec4e0ac575deb554恒成立,则实数eqId0a6936d370d6a238a608ca56f87198de的最大值为_____.

 

四、解答题

15.已知等差数列eqId83cf38189d5cbf627d2b82ac0eb76006满足eqIda8a0eecb5b800fce9ae10aed86ffee62,eqIdc47d65fdf9e1f5cca7325fb4324a6fb7.

(1)求eqId83cf38189d5cbf627d2b82ac0eb76006的通项公式;

(2)若对一切eqIda37a59558292ad6b3d0978bfd7484990,eqIdfa72d5c20ad97322156f1f73668c0941恒成立,求eqIddf64046e91b047037f19e4032e3b6de3的取值范围.

16.已知数列eqId83cf38189d5cbf627d2b82ac0eb76006是等差数列,其前eqIdb6a24198bd04c29321ae5dc5a28fe421项和为eqId08eb71ecf8d733b6932f4680874dbbf3,且eqId73f376cf4ca3eb2167fe5c9cb2b36fe9,eqId5731f65834e58bb01c8d21a695e395ce.

(1)求eqId83cf38189d5cbf627d2b82ac0eb76006的通项公式;

(2)设eqIde44e8d0669e5bb98993cb10e0e7899b7,求数列eqId034ba25825c13725931c483aa47c9363的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和eqIdf1ae9a3b0b7aeb1545b65d91aa371b3c.

17.求函数eqIdb5efc44f035595cb2c12b7d99f32ba29在区间eqId2a5c4c4e50a0a67ce5fa0e422d2eb4ef上的最大值和最小值.

18.已知数列eqId83cf38189d5cbf627d2b82ac0eb76006的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和为eqId57c599e7cec6d192fb73218e7882ceca.

(1)求数列eqId83cf38189d5cbf627d2b82ac0eb76006的通项公式;

(2)求证:数列eqId83cf38189d5cbf627d2b82ac0eb76006是等差数列.

19.已知eqId809002602966005cc46049fcf7b83214.

(1)求曲线eqId942c2141d01bde6b48210c56a17fc75e在eqId9b384412acba251d87902ab928902f16处的切线方程;

(2)若eqId0fde64f4d3c38e43fbdee24eadc4b0dd,解不等式eqIdcedf3794f76c7d668e3de82bfc5c9513.

page number 1

试卷第2页,共1页

学科网 zxxk.com学科网 zxxk.com学科网(北京)股份有限公司


 

学科网 zxxk.com学科网 zxxk.com

《2025-2026学年度高二数学期末考试卷》参考答案

题号

1

2

3

4

5

6

7

8

9

10

答案

D

C

B

A

D

D

B

C

BD

BC

题号

11

 

 

 

 

 

 

 

 

 

答案

AD

 

 

 

 

 

 

 

 

 

 

1.D

【分析】利用导数可得函数的单调性,结合条件即得.

【详解】由题意,知eqId01bea8bf593f594c51fc7cc547482bee,且eqIde0ee58be9f4937eb14ab900866ebc7db,

由eqIdb0c72d250a079379c5175693c165248c,可得eqIda71aa7636848cac20b2bb9752d8ccf56,由eqId7667e3a020f9d2883ffbaaed15e271b1,可得eqIde33817cce2b910818ef6d2ea1205ad74,

所以函数eqId09f86f37ec8e15846bd731ab4fcdbacd在eqId71ca214955cfbea2423456f7765f7cee上单调递减,在eqId1b94adb6c1d5ba48e929b53f581ce586上单调递增,而eqId7d365ce7e1224e03a40ff1491b7b8b14,

结合图象可得只有选项D符合题意.

故选:D.

2.C

【分析】由已知可知,数列eqId63d471926f7b27322d90c82b9ce21d3d是以eqIdf30d314a642667fef559032264647366为公比的等比数列,结合已知eqIdf1fb73c3135e576d762bbae078ae2684,可求eqIde72adb45c60c2f63b46e65ff787302bf,然后代入等比数列的求和公式即可得解.

【详解】解:eqId18ccac57032015a47bd9c290f11f59f4,

eqId2de0d10ef8b748d4531250c37c5d3f9eeqId04c67015059e1b6c05ab020f9a7e74e7,

eqId2de0d10ef8b748d4531250c37c5d3f9e数列eqId63d471926f7b27322d90c82b9ce21d3d是以eqIdf30d314a642667fef559032264647366为公比的等比数列,

eqId5d19b1cf608adff76c488c65cd6e5129,

eqId9e72f5e3057bdaf0f68db74c42f2ef74,

由等比数列的求和公式可得eqId2dbdd01efcbe38cd285e26152d75a79c.

故选:C.

3.B

【分析】根据复合函数的求导法则,两函数乘积求导的运算法则求解.

【详解】若eqId325f9e2f7784ebdd64292e805884dfa9,根据复合函数的求导法则,eqId2da83b5eda1801934521aff264e003bc,

根据两函数乘积的求导公式,eqIdfbbf465c215c08f44502c7ce4b8bb601的导数为eqIdd94c6a80f2bcb0369010eade34b5e0ce.

故选:B

4.A

【分析】对函数进行求导,然后运用赋值法进行求解即可.

【详解】eqId827f3e441b4997a599dfa1db591465e4,因此有eqId200b1c0b21f94d3f14884d515099787eeqId9d6b4c7c3878ef8178aa970b1f17dfc3,page number 2

答案第1页,共1页

学科网 zxxk.com学科网 zxxk.com学科网(北京)股份有限公司


由于学科网是一个信息分享及获取的平台,不确保部分用户上传资料的 来源及知识产权归属。如您发现相关资料侵犯您的合法权益,请联系学科网,我们核实后将及时进行处理。