（全国高考甲卷9科）2023年普通高等学校招生全国统一考试高考语数英+物化生+政史地真题试卷

2023年普通高等学校招生全国统一考试（全国甲卷）语文一、现代文阅读（36分）（一）论述类文本阅读（3小题，9分 （二）实用类文本阅读（3小题，12分220 （三）文学类文本阅读（3小题，15分巴金 二、古代诗文阅读（34分（一）文言文阅读（5小题，20分母之言。其于昆弟，尤笃有爱。执母丧，倚庐三年，．．ABCDEFGH （二）古代诗歌阅读（2小题，9分 （三）名篇名句默写（1小题，6分《邹忌讽齐王纳谏》中记载，齐王接受邹忌的意见，广开言路。一开始有很多人进谏，以至 三、语言文字运用（20分（5小题，20分 们能古相并。于代局，人尤是建治级人做的的今看来多成题，定心好是当，果决坏，就应了。”二教是一讲，说“釜是，‘’是。破和沉是词。釜’‘釜’意，就把砸；舟’‘舟’意，就把凿。样法动叫‘动’同做的具古叫釜现叫‘’同水运工，代‘’现叫船这是今汇演。古叫冠现叫‘子古叫‘’现叫鞋都这情。曹步有‘在中’句 里的釜和舟跟釜舟’的釜和舟思同。三教讲比简，说比少他样：“羽河攻朝军，河后把饭锅碎把凿，了己退，示进退决，于秦打了后大就‘釜舟个语示定大决，顾何牲意。” B.又吃鱼又嫌C.前怕狼后怕 D.首尾不能兼解要点。要求阐释准确，语言流畅，不超过60个字。 四、写作（60分8002023年普通高等学校招生全国统一考试（全国甲卷）语文一、现代文阅读（36分）（一）论述类文本阅读（3小题，9分从重构上古史体系角度看，2070—80年代，夏鼐、苏秉琦等系统地构建了新石器 【答案】1. 2. 3.【1【2【3（二）实用类文本阅读（3小题，12分220220赫 【答案】4. 5.【4220赫兹”，据此看出，谷物幼苗根部发出的轻响是主动发出的声音。【5B.“下面的发现可以作为证明第一段中心观点的材料”错。原文第一段“那么树木会如何.D.“都使用了打比方和举例子的说明方法”错。原文第二段“树木为了确保信息能够快速220赫兹。……这些幼苗的顶端总会往声源方向生长，这意【6（三）文学类文本阅读（3小题，15分巴金 【答案】7.C 【7C.“依据是这些材料本身缺乏生命力”错误，由文中“诗应该给人以创造的喜悦，诗应该【8【9二、古代诗文阅读（34分（一）文言文阅读（5小题，20分ABCDEFGH 【答案】10. 11. 12.【10之”主谓宾齐全，B【12【13（二）古代诗歌阅读（2小题，9分 【答案】14.A 15.①用动词“垂”“谢”赋予“柳”“梅”动态的美感，运用虚写的手【14【15（三）名篇名句默写（1小题，6分 两句诗“ “ ①.门庭若②.时时而间进 ③.停杯投箸不能食 ④.拔剑四顾心茫然 ⑤.忽如一夜春 ⑥.千树万树梨花开（砌下落梅如雪乱，拂了一身还满）三、语言文字运用（20分（5小题，20分 ‘ B.又吃鱼又嫌C.前怕狼后怕 D.首尾不能兼解方法，拟出讲解要点。要求阐释准确，语言流畅，不超过60个字。 【答案】17. 18.语句：④；修改为：古人不能跟我们相提并论【17【18【19【20【21四、写作（60分800劳动的生活模式，进入了节时省力、高能高速的全新状态。随着智能搜索引擎、OSO图文人像识别、ChatGPT遗憾的是，在技术发展使时间变得充裕、人们对时间的使用更加自由的同时，现代人5GGSAppeVisonPro2023年普通高等学校招生全国统一考试（全国甲卷） {x|x3k,k {∣x{∣x D.若复数ai1ai2,aR，则a A.- B. C. D. A. B. C. D. ab1

，且abc0，则cosac,bc

C. D. A. B. C. D.有50人报名足球俱乐部，60人报名乒乓球俱乐部，70人报名足球或乒乓球俱乐部，若已知某人报足球俱乐部，则其报乒乓球俱乐部的概率为（ A. B. C. D.“sin2sin21”是“sincos0”的 B.必要条件但不是充分条C.充要条 D.既不是充分条件也不是必要条 0)的离心率

，其中一条渐近线与圆(x

2

2

B两点，则|AB|

2

4有五名志愿者参加社区服务，共服务星期六、星期天两天，每天从中任选两人参加服务，则恰有1人连续参加两天服务的选择种数为（ A. B. C. D.fxycos2xππyfxy1x1 6 B. C. D.在四棱锥PABCD中，底面ABCD为正方形，AB4,PCPD3,PCA45，则PBC的面 A.

B.

C.

D.x2y2FF为两个焦点，O为原点，PcosFPF3，则|PO

A. C.

若y(x1)2axsinxπ为偶函数，则a 2 2x3yx，y满足约束条件3x2yxy

，设z3x2y，则z的最大值 ，D为BC上一点，AD为BAC的平分线，AD 已知数列ana21Sn为ann2Snnan求an求数列an1n项和T2n XX单位：g（PPk2k已知直线x2y10与抛物线C:y22px(p0)交于A,B两点，且|AB| （1）pf(xaxsinxx0,πcos3

2 若a8f(xf(xsin2xax2t|PA||PB|4f(x2xaaa0yfx2，求a2023年普通高等学校招生全国统一考试（全国甲卷）

A.{x|x3k,k{∣x

{∣xD.若复数ai1ai2,aR，则a A.- B. C. D.【详解】因为ai1aiaa2iia2a1a2i22a

0a1 A. B. C. D.n1A123B325n112n2时，判断框条件满足，第二次执行循环体，A358B8513n213n3A81321B211334n314n4B34． ab1

，且abc0，则cosac,bc

C. D. 【详解】因为 ,所以 r,

abc →2,

a+b=- ,所以 a

2ab

112ab

ab如图,设OA b,OC→由题知OAOB1OC

AB边上的高OD

2,AD 2 所以CDCOOD

232 tanACD

AD1,cosACD c, c, 3 2

1 A B. C. D.【分析】根据题意列出关于q的方程,计算出q,S4q3q44q4q2,q3q24q40,即(q2)(q1)(q2)0q0,q2S4124815有50人报名足球俱乐部，60人报名乒乓球俱乐部，70人报名足球或乒乓球俱乐部，若 A. B. C. D.【详解】报名两个俱乐部的人数为50607040记“某人报足球俱乐部”为事件ABPA505PAB404 P(∣AP(

70.8“sin2sin21”是“sincos0”的 B.必要条件但不是充分条 D.既不是充分条件也不是必要条【详解】当sin2sin21时，例如π,0但sincos0即sin2sin21推不出sincos0当sincos0sin2sin2(cos)2sin21，即sincos0能推出sin2sin21.sin2sin21是sincos0成立的必要不充分条件.x2y2

0)

(x2)2(y3)21交于A，B两点，则|AB|

2

4【详解】由e ，则a2b2

a2

5y2x22则圆心(2,3)到渐近线的距离d22所以弦长

4511则恰有1人连续参加两天服务的选择种数为（ A. B. C. D.假设a42人各参加星期六与星期天的社区服务，共有A212种方法，1人连续参加了两天社区服务的选择种数有51260种.fx为函数ycos2xπ向左平移π个单位所得函数，则yfx 6 y1x1的交点个数为 B. C. D.fxsin2xfxy1x1 fxy1x1的大小关系，从而精确图像，由此得解 ycos2xππ 6 y

πcos2x66cos2x2sin2xfxsin2x y1x1显然过01与10 fxy1x1 2x3π2x3π2x7πx3πx3πx7πfxy1x1

x3πf3πsin3π1y13π13π41 4 2

4 x3πf3πsin3π1y13π13π41 4

x7πf7πsin7π1y17π17π4 4

fxy1x1的交点个数为3 在四棱锥PABCD中，底面ABCD为正方形，AB4,PCPD3,PCA45，则PBC的面积为（ A.

B.

C.

D.【分析】法一：利用全等三角形的证明方法依次证得OCO，PDBPCA，从而得到PAPB，再在C中利用余弦定理求得PA，从而求得PB，由此在C法二：先在△PAC中利用余弦定理求得PA ，cosPCB1，从而求PAPC3，再利用空间向量的数量积运算与余弦定理得到关于PB,BPD的方程组，从而求得PB ，由此在PBC中利用余弦定理与三角形面积公式即可得解.ACBD交于OPO，则OACBD因为底面ABCD为正方形，AB4，所以ACBD42，则DOCO2 又PCPD3，POOP，所以PDOPCO，则PDOPCO，PCPD3ACBD42，所以PDBPCAPAPB在△PACPC3AC42PCA45PA2AC2PC22ACPCcosPCA329242 ，则PB

217故在PBCPC3PB17BC4PC2BC2 916 所以cosPCB 2PC 23 1cos22又0PCBπ，所1cos22所以PBC的面积为S1PCBCsinPCB13422 ACBD交于OPO，则OACBDABCDAB4ACBD42，在△PACPC3PCA45，PA2AC2PC22ACPCcosPCA329242PA2PC2AC

217故PA 217所以cosAPC 2172PA

17PAPCPAPCcosAPC173173 PBmBPD

1 1

PO

PAPC

PBPD，所以PAPC

PBPD PAPC2PAPCPBPD2PBPD则17923m2923mcosm26mcos110又在△PBDBD2PB2PD22PBPDcosBPD，即32m296mcos，m26mcos230②，两式相加得2m2340，故PBm 故在PBCPC3PB17BC4PC2BC2 916 所以cosPCB 2PC 23 1cos22又0PCBπ，所1cos22所以PBC的面积为S1PCBCsinPCB13422 x2y2

F,

cosFPF

1， OP 5则|PO| A.

C.

【分析】方法一：根据焦点三角形面积公式求出△PF1F2P的坐标，从而得出OP的值；PFPF,PF2PF2 【详解】方法一：设FPF20π

b2tanF1PF2b2 cos2sin2

1tan2

由cosF1PF2cos2cos2sin21tan25tan2a29b26c2a2b23

1FF 123 61y23 1 x29139

x2yx2y 3PFPF2a6PF2PF22PFPFFPFFF2

1即PF2PF2 PFPF12②，联立 PFPF15,PF2PF221 PO 2PFPF

2PF1PF2

3 即PO

21 PFPF2a6PF2PF22PFPFFPFFF2

1即PF2PF2 PFPF12②，联立①②，解得：PF2PF221 由中线定理可知，2OP2FF22PF2PF242，易知FF ，解得OP

30

1 1若y(x1)2axsinxπ为偶函数，则a 2 fπfπ，从而求得a2，再检验即可得解 2 yfxx12axsinxπx12axcosx 2 义域为R π π

所以f2f2，即a sa ，

则πa21212π，故a2 fxx122xcosxx21cosx，fxx21cosxx21cosxf又定义域为R，故fx为偶函数，所以a2

x2x3yx，y满足约束条件3x2yxy

，设z3x2y，则z的最大值 y3xz过点Az 由2x3y3可得x3A(333x2y yzmax332315 2EF中点为OABBB1中点GMBB1C1C的NFGEGOMONMN，如图，即R ON2MN则球心OON2MN

FG2EGFG2EG22

1个交点，EF12.在ABC中，AB2，BAC60,BC ，D为BC上一点，AD为BAC的平分线，则AD ABcACbBCa方法一：由余弦定理可得，22b222bcos60∘6，因为b0，解得：b1 12bsin60∘12ADsin30∘1ADbsin30∘ AD

3b1

2313

2方法二：由余弦定理可得，22b222bcos60∘6，因为b0，解得：b1

sin

sinB

6 2，sinC 2 因为 ，所以C45∘，B180∘60∘45∘75∘又BAD30o，所以ADB75ADAB2．2．已知数列ana21Sn为ann2Snnan求an求数列an1n项和T2n （1）ann（2）n（2）n n

S1,n

,n1详解】2Snnan，n22Sn1n1an12SnSn1nann1an12an化简得：n2an1 ，当n3时，anan1a31，即an

2a1

1

1

1

1因为 ，所以T1 3 n

1

1

1

1n2Tn1222(n1)2n2 1n

212

1T1111n1 n

1 11

n1n

，即

1

ABCA1B1C1AA12A1CABCACB90A1（2）股定理求出O为中点，即可得证；1详解】A1CBCBCACA1CACACC1A1A1CACCBCACC1A1BCBCC1B1A1BCC1B11，A1O1，设COx，则C1O2x，CO2AO2AC2，AO2OC2CA2，AC2AC2CC2 1 1 1x212x)24x1ACA1CA1C1 AC2BBDAA1AA1DDAA1中点，AA1BB12BD2 AB2AC在Rt△ABC，AB2ACACACCM，连接C1MC1M∥A1C，C1MABCAMABCC1M(2AC)2AC则在Rt△AC1M中，AM2AC,C1MA1(2AC)2AC(2AC)2AC(2AC)2AC

BC (22)2(2)2(22)2(2)2(又ABCC1B1所以AB与平面BCCB所成角的正弦值为 13 1 40只小鼠均分为两组，分别为对照组（不加．XX单位：g（2×295%的把握认为药物对小鼠生长有抑制作用．Pk2k（1）EX（2（i）（ii）（2（i）1X的可能取值为0,12C0

C1

C2 则P(X0)2020 ，P(X1)2020 ，P(X2)2020 XEX019120219 2可得第11位数据为14.4，后续依次为17.3,17.3,18.4,20.421.523.223.6,，20位为23.22123.6m23.223.623.440(6614由（i）可得，K 6.4003.841，20202020%已知直线x2y10与抛物线C:y22px(p0)交于A,B两点，且|AB| （1）p（1）p（2）12（2）MNxmynMx1y1Nx2y2MFNF0mn的关系，以及MNF的面积表达式，再结合函数的性质即可求出其最小值．1AxA,yA,BxB,yB由x2y10y24py2p0y

4p,y

2py22

A y

yA

4yA2p2p6 4yA2F10MN的斜率不可能为零，MNxmynMx1y1Nx2y2，y2由xmy

y24my4n0y1y24my1y24n16m216n0m2n0MFNF0，所以x11x21y1y20，即my1n1my2n1y1y20，亦即m21yymn1yyn1201 y1y24my1y24n4m2n26n14m2nn120，n1n26n10n32

或n3 1n设点F到直线MN的距离为d1nxx y

yy1m21m216m21 4n26n1

11

1n11n1所以MNF的面积S1MNd1n1

n1n12n3

n3

当n3 时，MNF的面积 222212 mn的关系，一是为了减元，二是通f(xaxsinxx0,πcos3

2 若a8f(xf(xsin2xa（2）(,（1）求导,然后令tcos2x,讨论导数的符号即可（21

f(xsin2x,g(x)的最大值,0比较大小,得出a的分界点,f(x)a

cosxcos3x3sinxcos2xsincos6cos2x3sin2a acos4

32cos2cos4令cos2xt,则tf

(x)g(t)a 当a8, 8t2t (2t1)(4t(x)g(t) 当t01,xππ,f(x0 42 当t1,1,x0π,fx0 4 π

ππ所以f(x)在0,上单调递增,在,上单调递 4

422g(x)f(xsin

at22t g(x)f(x)2cos2xg(t)22

x1

2(2t1)a24t设(ta24t2

(t)4 4t32t(t)4 1a(,3]g(x(ta3g(x在0,上单调递减,g(x)g(0)0 2 a(3],f(xsin2x,符合题意2a(3 1

当t0,tt23t3

,所以 所以t(0,1,使得t0,即x0,,gx0 2 当tt,1,(t0,x0xg(x0g(x单调递增.x0x0g(x)g(0)0, 综上a的取值范围为( :本题采取了换元,注意复合函数的单调性tcosx在定义域内是减函数,若tcosx,当tt,1,(t0,x0xg( x2tP(2,1，直线ly1tsinA，B|PA||PB|4

（1）根据t1因为lxyABππx0t1cosy0t2所以PAPBt2t1 4，所以sin22πkπ，解得π1kπkZ ππ，所以3π 2由（1）可知，直线l的斜率为tan1，且过点2,1所以直线ly1x2xy30f(x2xaaa0yfx2，求a 22（1）xaxa讨论即可1xa,f(x2a2xax即3xa,xa,axa xa,f(x2x2aaxx3a,ax3a综上,不等式的解集为a3a 22x3a,xf(x)22x3a,xf(x的草图,f(x与坐标轴围成△ADO与

OAa ABa3a22,a2

2023年普通高等学校招生全国统一考试（全国甲卷）答卷前，考生务必将自己的姓名、准考证号填写在答题卡上回答选择题时，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考试结束后，将本试卷和答题卡一并交回12560分.在每小题给出的四个选项中，只有一项 51i32i2

2,3,A.

B. C.1 →

D.1已知向量a3,1,b2,2，则cosab,ab

2某校文艺部有4名学生，其中高一、高二年级各2名．从这4名学生中随机选2名组织校文艺汇演，则这2名学生来自不同年级的概率为（

C. D. A. B. C. D. A. B. C. D.FF, 为椭圆C

2P在CPF

0

B. C. D. e 曲线yx1在点1,2处的切线方程为 ye

ye

yex

yex 0)的离心率

，其中一条渐近线与圆(x

2

2

B两点，则|AB|

2

3

4在三棱锥PABC中，ABC是边长为2的等边三角形，PAPB2,PC A. C. D.fxe(x1)2．记a

f2,bf3,cf6，则 2 2 2bc

ba

cb

cayfxycos2x的图象向左平移yfx 6 直线y1x1的交点个数为 A. B. C. D.4520分 若fx(x1)2axsinxπ为偶函数，则a 3x2y若x，y满足约束条件2x3y3，则z3x2y的最大值 xy则球O的半径的取值范围是 70分.解答应写出文字说明、证明过程或演算步骤.17~21个试题考生都必须作答.22、23题为选考题，考生根据要求作答（一）60分记 的内角A,B,C的对边分别为a,b,c，已知b2c2

2求bc （（2）F为CMN为CFMFN0，求△MFN单位：g 19 （2（ⅰ）

n(ad

abcdacbdPK2k20fxaxsinxx0π

2 fxsinx0，求a已知直线x2y10与抛物线C:y22px(p0)交于A,B两点，AB （1）p（二）10分.22、23题中任选一题作答.如果多做，则按所做的第一题[4-4：坐标系与参数方程]（10分y1y1ABPAPB4[4-5：不等式选讲]（10分f(x2|xa|a,a0yfxx2，求a2023年普通高等学校招生全国统一考试（全国甲卷）答卷前，考生务必将自己的姓名、准考证号填写在答题卡上考试结束后，将本试卷和答题卡一并交回12560分.在每小题给出的四个选项1.设全集 A2,3,2,3,N25}NðUM{235}，51i32.2i2A.

C.1

D.151i3 51i 1(2i)(2 a3,1,b22，则

→

2 abab,abab 5211a5211

34,ab ，abab51312 → abab 17所以cosab,ab → aba 某校文艺部有4名学生，其中高一、高二年级各2名．从这4名学生中随机选2名组织校文艺汇演，则这2名学生来自不同年级的概率为（

C. D. 22名学生来自不同年级的基本事件有C1C142242 A. B. C. D.【分析】方法一：根据题意直接求出等差数列ann项和公式即a2a6a1da15d10a13d5a4a8a13da17d45d1a12

5

54d521020a2a62a410a4a845a45a89da8a41aad514 S55a320． A B. C. D.【详解】当k1A123B325k112k2时，判断框条件满足，第二次执行循环体，A358B8513k213k3A81321B211334k314当k4B34．FF为椭圆C:x2

2 的两个焦点，点PCPF

0 PF1

B. C. D.

0，所以

90∘

b2tan45∘11PFPFPF

2

0，所以

90c2514c2PF2PF2FF24216PF

2a

1 PF2PF22PFPF162PFPF20PF

2 曲线yx1在点 ye

ye

yex yex y

在点1eyekx1 2 x y

x

x2ky

x x yeex 所以曲线y 在点1,e处的切线方程为yexex 2 x2y2

0的离心率

(x2)2(y3)21交于A，B两点，则|AB|

2

3

4【详解】由e ，则a2b2

a2

5y2x22则圆心(2,3)到渐近线的距离d|223| 22所以弦长

4511在三棱锥PABC中，ABC是边长为2的等边三角形，PAPB2,PC A.

C. D.ABEPECEPEABCEABPECEPECPECEEPECE

3 ，PC PC2PE2CE2PECE所以V

BPEC

1S

AB11

321fxe(x1)2．记a

f2,bf3,cf6，则 2 2 2bcca

ba

cb g(x)(x1)2g(xx161 36

(

3)2429

16

70 2 6113640611 2 g(6g(3 因为611 2 6 24， 2 2)242843164384(32)0即61 2，所以g(6)g(2) g(2g(6g(3 yex为增函数，故acb，即bcayfx的图象由ycos2x的图象向左平移个单位长度得到，则 6 yfx的图象与直线y1x1的交点个数为 A. B. C. D.fxsin2xfxy1x1 fxy1x1的大小关系，从而精确图像，由此得解 ycos2xππ 6 ycos2xππcos2xπsin2xfxsin2x y1x1显然过01与10 fxy1x1 2x3π2x3π2x7πx3πx3πx7πfxy1x1

x3πf3πsin3π1y13π13π41 4 2

4 x3πf3πsin3π1y13π13π41 4

x7πf7πsin7π1y17π17π4 4

fxy1x1的交点个数为3 4520分 q1n项和公式和平方差公式化简即可求出公比q1则由8S67S3得86a173a1a10q1.q1时，因为8S67S3所以8

a1q61

7

a1q31即81q671q3，即81q31q371q3，即81q37q1若fx(x1)2axsinxπ为偶函数，则a fxx12axsinxπx12axcosxx2(a2)x1cosx a20，解得a2，3x2y若x，y满足约束条件2x3y3，则z3x2y的最大值 xyy3xz过点Az 由2x3y3可得x3A(333x2y yzmax332315在正方体ABCDA1B1C1D1中，AB4,O为AC1的中点，若该正方体的棱与球O的球面有公共点，则球O的半径的取值范围是 【答案】[222424242

2R43,R ，故Rmax 形，且OMNGH的对角线交点，连接MG，则MG R[2223故答案为：[22270分.解答应写出文字说明、证明过程或演算步骤.17~21题（一）60分记 的内角A,B,C的对边分别为a,b,c，已知b2c2

2求bc （2）（2）由（1）可知，只需求出sinA1a2b2c22bccosA

b2c2 2bc2，解得： 2

cos

cosacosBbcosAbsinAcosBsinBcosAsin sinAcosBsinBcos sin sinABsinB1， sinABsinABsinABsinB，即2cosAsinBsinB而0sinB≤1，所以cosA1，又0Aπ，所以sinA 3 故ABC

1bcsinA113 3△ (1)A1CABCA1CBCACBCBCACC1A1，ACC1A1BCC1B1；)得O为CC1A1CACxxA1O.1又因为ACB90ACBCBCACC1A1，2详解】A1CBCA1CACA1BABBCA1CACxA1C1x所以O为CCOC1AA1 ACAC,AC2AC2AA2 22即x2x2222AC2OCAC2OC1

20只分配到对照组，试验组的小白鼠饲养在高浓度臭氧环境，对照组的小g （2（ⅰ

n(adabcdacbdPK2k（2（i）（ii）（2（i）121.622.823.623.925.128.232.336.5)396220位为23.22123.6m23.223.623.4

40(6614 40(6614所以能有95的把握认为小白鼠在高浓度臭氧环境中与在正常环境中体重的增加量有差fxaxsinxx0π

2 fxsinx0，求a 2 a（1）a1fxfx，g00a0a0a<0法二：先化简并判断得sinx

sincos2

0a0a<0a01a1fxxsinxx0πcos2

2fx

cosxcos2x2cosxsinxsin

cos2x2sin2cos3xcos2x21cos2x

cos3xcos2x

cos cos3 cos3令tcosxx0π，所以tcosx0,1 cos3xcos2x2t3t22t3t22t22t2t12t1tt22t2t1因为t22t2t1210t10cos3xt30cos3xcos2x

πfx

cos3

0在0,上恒成立 2 2详解】

sincos2

sinx0x

2 1sin2

cos3

cosx0x gxfxsinx0g0f0sin00，g0a11a0a0，a0时，因为sinx

sinx

cos2 sinx cos2x 又x0,π，所以0sinx1，0cosx1，

cos2 fxsinxsinx

sincos2

a<0时，由于0xπax0fxsinxax

sincos2

sinxsinx

sincos2

fxsinx0a0，所以a的取值范围为0.sin sinxcos2xsin sinxcos2x sin3因为sinxcos2x

cos2

cos2

cos2xx0π，所以0sinx10cosx1 故sinxsinx0在0π 2cos a0fxsinxsinx

sincos2

a<0时，由于0xπax0fxsinxax

sincos2

sinxsinxsin

sincos2

a0fxsinxax

sinxax

cos2sin3x 3sin2xcos2x2sin4 gxaxcos2x0x2gxa 3sin20cos202sin4

cos3g0a

a0若0xπgx0gx在0π 2 若0xπgx0g0gx0 所以在0πx0x0πgx0 2

2此时gx在0,x1上有gx0，所以gx在0,x1上单调递增，则在0x1gxg00，即fxsinx0，不满足题意；a0. 2 已知直线x2y10与抛物线C:y22px(p0)交于A,B两点，AB （1）p（2）F为CMN为CFMFN0，求△MFN（1）p（2）12（2）MNxmynMx1y1Nx2y2MFNF0mn的关系，以及MNF的面积表达式，再结合函数的性质即可求出其最小值．1AxA,yA,BxB,yB由x2y10y24py2p0yy4pyy2py22

AAB

yA

y 4y y 4yA2F10MN的斜率不可能为零，MNxmynMx1y1Nx2y2，y2由xmy

y24my4n0y1y24my1y24n16m216n0m2n0MFNF0，所以x11x21y1y20，即my1n1my2n1y1y20，亦即m21yymn1yyn1201 y1y24my1y24n4m2n26n14m2nn120，n1n26n10n32

或n3 1n设点F到直线MN的距离为d1nxx y

yy1m21m216m21 4n26n1

11

1n11n1所以MNF的面积S1MNd1n1

n1n12n3

n3

n3

时，MNF的面积 222212 mn的关系，一是为了减元，二是通[4-4：坐标系与参数方程]（10分y1y1

yABPAPB4（1）根据t1因为lxyABππx0

y0

PA

4，所以sin212πkπ，解得π1kπk ππ，所以3π 2由（1）可知，直线l的斜率为tan1，且过点2,1所以直线ly1x2xy30[4-5：不等式选讲]（10分f(x2|xa|a,a0yfxx2，求a 22（1）xaxa讨论即可1xa,f(x2a2xax即3xa,xa,axa xa,f(x2x2aaxx3a,ax3a综上,不等式的解集为a3a 22x3a,xf(x)22x3a,xf(x的草图,f(x与坐标轴围成△ADO与

OAa ABa3a22,a2

2023年普通高等学校招生全国统一考试(全国甲卷)第一部分听力(30分第一节(5小题;1.51.5分10WheredoestheconversationprobablytakeInthebook B.Intheregister C.InthedormWhatistheweatherlike B. C.WhatdoesthemanwanttodoontheDosome B.Havea C.GoWhatarethespeakerstalkingAnew B.Achangeoftheir C.AformerWhatdoweknowabout B.He’s C.He’s第二节(15小题;1.522.5分5A、B、C三个选项中5秒钟;5秒钟的作答时间，每段对话或独白读两遍。66、7题。WhichofthefollowingdoesthewomanThe B.Thesitting C.TheWhatdoesthewomansuggesttheydoGotoanother B.Seesomeother C.Visitthe78、9WhatisthemanHe’smakingaphoneHe’schairingaHe’shostingaWhatmakesMrs.JohnsonworriedaboutherdaughterinLackofmedicalInconvenienceofPoortransportation81012Whatpositiondoesthemanapply B.An C.AnWhichaspectofthecompanyappealstotheThecompany B.Thefree C.ThecompetitiveWhatisdifficultforthemantodealInterpersonalrelationships.B.Quality-quantity C.Unplanned91316HowdoesRobertsoundwhenspeakingofhisbeinga B. 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C.第二部分阅读理解(共两节,40分第一节(15小题:230分A、B、CD四个选项中,WheretoEatinBangkokisahighlydesirabledestinationforfoodlovers.IthasaseeminglybottomlesswellofdiningHerearesomesuggestionsonwheretostartyourBangkokeatingOfferingThaifinedining.NahmprovidesthebestofBangkokculinary(烹饪的)experiences.It’stheonlyThairestaurantthatranksamongthetop10oftheword’s50bestrestaurantslistHeadChefDavidThompsonwhoreceivedaMichelinstarforhisLoodon-basedThairestaurantofthesamename,openedthisbranchintheMetropolitanHotelin2010.IssaysStameseIssayaSiameseClubisintematoionallyknownThaicheflanKittichai’sfirstflagshipBangkokrestaurant.ThemenuinthisbeautifulcolonialhouseincludestraditionalThaicuisinecombinedwithmoderncookingmethods.Bo.Bo.tanhasbeenmakingwavesinBangkok’sculinarysencesinceitopenedin2009.Servinghard-to-findThaidishesinanelegantatmosphere,therestaurantistruetoThaicuisine’sroots,yetstillmanagestoaddaspecialtwist.Thisplaceisgoodforacandlelitdinneroraworkmeetingwithcolleagueswhoappreciatefinefood.Forthoseextremelyhungrythere’salargesetmenu.Earningfirstplaceonthelates“Asia’s50bestrestaurants”listprogressiveIndianrestaurantGagganisoneofthemostexcitingvenues(场所)toarriveinBangkokinrecentyears.Thebesttableinthistwo-storycolonialThaihomeoffersawindowrightintothekitchen,whereyoucanseechefGagganandhisstaffinaction.Culinarytheateratitsbest.WhatdoNahmandIssayaSiameseClubhaveinTheyadoptmoderncooking B.TheyhavebranchesinC.Theyhavetop-class D.TheyarebasedinWhichrestaurantoffersalargesetA. B.Bo. C.IssayaSiamese D.WhatisspecialaboutIthiresstafffrom B.ItputsonaplayeveryC.Itserveshard-to-findlocal D.ItshowsthecookingprocesstoTerriBoltonisadabhandwhenitcomestoDIY(do-it-yourself).Skilledatputtingupshelvesandpiecingtogetherfurniture,sheneverpayssomeoneelsetodoajobshecandoherself.ShecreditstheseskillstoherlategrandfatherandbuilderDerekLloyd.Fromtheageofsix,Terri,now26,accompaniedDerektoworkduringherschoolholidays.Aday’sworkwasrewardedwith£5inpocketmoney.Shesays:“I’msureIwasn’tmuchofahelptostartwithpaintingtheroomsandputtingdowntheflooringthroughoutthehouse.Ittookweeksandiswasbackbreakingwork,butIknowhewasproudofmyskills.”TerriwhonowrentsahousewithfriendsinWandsworthSouthWestLondonsaysDIYalsosavesherfromlosinganydepositwhenatenancy租期comestoanendSheadds:“I’vemovedhousemanytimesandIalwaysliketopersonalisemyroomandputuppictures.So,it’sbeenusefultoknowhowtocoverupholesandrepaintaroomtoavoidanychargeswhenI’vemovedout.”WithmillionsofpeoplelikelytotakeonDIYprojectsoverthatcomingweeks,newresearchshowsthatmorethanhalfofpeopleareplanningtomakethemostofthelong,warmsummerdaystogetjobsdone.Theaveragespendperprojectwillbearound£823.Twothirdsofpeopleaimtoimprovetheircomfortwhileathome.Twofifthwishtoincreasethevalueoftheirhouse.ThoughDIYhastraditionallybeenseenasmalehobby,theresearchshowsitiswomennowleadingthecharge.Whichisclosestinmeaningto“adabhand”inparagraphAAn 5WhydidTerri’sgrandfathergiveher£5aA.Forabirthday B.AsatreatforherC.TosupportherDIY D.ToencouragehertotakeupaHowdidTerriavoidlosingthedepositonthehousesheBymakingitlooklike B.ByfurnishingitC.Bysplittingtherentwitha D.BycancellingtherentalWhattrendinDIYdoestheresearchItisbecomingmore B.Itisgettingmoretime-C.Itisturningintoaseasonal D.ItisgainingpopularityamongIwasabout13whenanunclegavemeacopyofJosteinGaarder’sSophie’sWorld.Itwasfullofideasthatwerenewtome,soIspentthesummerwithmyheadinandoutofthatbookItspoketomeandbroughtmeintoaworldofphilosophy(哲学).ThatloveforphilosophylasteduntilIgottocollege.NothingkillstheloveforphilosophyfasterthanwhothinktheyunderstandFoucault,Baudrillard,orConfuciusbetterthanyou—andthentrytoexplainEricweiner’sTheSocratesExpress:InSearchofLifeLessonsfromDeadPhilosophersreawakenedmyloveforphilosophy.Itisnotanexplanationbutaninvitationtothinkandexperiencephilosophy.Weinerstartseachchapterwithasceneonatrainridebetweencitiesandthenframeseachphilosopher’sworkinthecontext(背景)ofonethingtheycanhelpusdobetterTheendresultisareadinwhichwelearntowonderlikeSocrates,seelikeThoreau,listenlikeSchopenhauer,andhavenoregretslikeNietzsche.This,morethanabookaboutundestandingphilosophy,isabookabourlearningtousephilosophytoimprovealife.Hemakesphilosophicalthoughtanappealingexercisethatimprovesthequalityofourexperiences,andhedoessowithplentyofhumorWeinerentersintoconversationwithsomeofthemostimportantphilosophersinhistoryandhebecomespartofthatcrowdintheprocessbydecoding解读theirmessagesandaddinghisownTheSocratesExpressisafun,sharpbookthatdrawsreadersinwithitsapparentsimplicityandgraduallypullsthemindeeperthoughtsondesire,loneliness,andaging.Theinvitationisclear:Weinerwantsyoutopickupacoffeeorteaandsitdownwiththisbook.Iencourageyoutotakehisoffer.It’sworthyourtime,eveniftimeissomethingwedon’thavealotof.Whoopenedthedoortophilosophyforthe B.EricC.Jostein D.AcollegeWhydoestheauthorlistgreatphilosophersinparagraphTocompareWeinerwithTogiveexamplesofgreatTopraisetheirwritingTohelpreadersunderstandWeinersWhatdoestheauthorlikeaboutTheSocratesItsviewsonhistoryarewell-ItsideascanbeappliedtodailyItincludescommentsfromItleavesanopenWhatdoestheauthorthinkofWeinersObjectiveandDaringandSeriousandhardtoHumorousandGrizzlybears,whichmaygrowtoabout2.5mlongandweighover400kg,occupyaconflictedcorneroftheAmericanpsyche-werevere(敬畏themevenastheygiveusfrighteningdreams.AskthetouristsfromaroundtheworldthatfloodintoYellowstoneNationalParkwhattheymosthopetosee,andtheiranswerisoftenthesame:agrizzlybear.“Grizzlybearsarere-occupyinglargeareasoftheirformerrange,”saysbearbiologistChrisServheen.Asgrizzlybearsexpandtheirrangeintoplaceswheretheyhaven’tbeenseeninacenturyormore,they’reincreasinglybeingsightedbyhumans.ThewesternhalfoftheU.S.wasfullofgrizzlieswhenEuropeanscame,witharoughnumberof50,000ormorelivingalongsideNativeAmericans.Bytheearly1970s,aftercenturiesofcruelandcontinuoushuntingbysettlers,600to800grizzliesremainedonamere2percentoftheirformerrangeintheNorthernRockies.In1975,grizzlieswerelistedundertheEndangeredSpeciesAct.Today,thereareabout2,000ormoregrizzlybearsintheU.S.TheirrecoveryhasbeensosuccessfulthatU.S.FishandWildlifeServicehastwiceattemptedtodelistgrizzlies,whichwouldloosenlegalprotectionsandallowthemtobehunted.Botheffortswereoverturnedduetolawsuitsfromconservationgroups.Fornow,grizzliesremainlisted.Obviouslyifprecautions预防aren’ttakengrizzliescanbecometroublesome,sometimeskillinganimalsorwalkingthroughyardsinsearchoffood.Ifpeopleremovefoodandattractantsfromtheiryardsandcampsites,grizzlieswilltypicallypassbywithouttrouble.Puttingelectricfencingaroundchickenhousesandotherfarmanimalquartersisalsohighlyeffectiveatgettinggrizzliesaway.“Ourhopeistohaveaclean,attractant-freeplacewherebearscanpassthroughwithoutlearningbadhabits,“saysJamesJonkel,longtimebiologistwhomanagesbearsinandaroundMissoula.HowdoAmericanslookatTheycausemixedfeelingsinTheyshouldbekeptinnationalTheyareofhighscientificTheyareasymbolofAmericanWhathashelpedtheincreaseofthegrizzlyTheEuropeansettlers’Theexpansionofbears’TheprotectionbylawsinceThesupportofNativeWhathasstoppedtheU.S.FishandWildlifeServicefromdelistingTheoppositionofconservationThesuccessfulcomebackofThevoiceoftheThelocalfarmers’WhatcanbeinferredfromthelastFoodshouldbeprovidedforPeoplecanliveinharmonywithAspecialpathshouldbebuiltforTechnologycanbeintroducedtoprotect第二节(5小题;210分Here’sariddle:Whatdotrafficjams,longlinesandwaitingforavacationtostartallhaveinThere’soneanswer. IntheDigitalAge,we’reusedtohavingwhatweneedimmediatelyandrightaiourfingertips.However,researchsuggeststhatifwepracticedpatience,we’dbeawholelotbetteroff.Hereareseveraltricks.Practicegratitude感激Thankfulnesshasalotofbenefits:Researchshowsitmakesushappier,lessstressedandevenmore .“Showingthankfulnesscanfosterself-control,”saidYeLi,researcherattheUniversityofMakeyourselfInstantgratification(满足)mayseemlikethemost“feelgood”optionatthetime,butpsychologyresearchsuggestswaitingforthingsactuallymakesushappierinthelongrun.Andtheonlywayforustogetintothehabitofwaitingistopractice. .Putoffwatchingyourfavoriteshowuntiltheweekendorwait10extraminutesbeforegoingforthatcake.You’llsoonfindthatthemorepatienceyoupractice,themoreyoustarttoapplyittoother,moreannoyingsituations. Somanyofushavethebeliefthatbeingcomfortabelistheonlystatewewilltolerate,andwhenweexperiencesomethingoutsideofourcomfortzone,wegetimpatientaboutthecircumstances.Youshouldlearntosaytoyourself,“ .”You’llthengraduallybecomemorepatient.FindyourStartwithsmallAccepttheAllthisaddsuptoastateofItcanalsohelpuspracticemoreThisismerelyuncomfortable,notThey’reallsituationswherewecouldusealittleextra第三部分语言知识运用(45分)第一节(201.530分)A、B、CD四个选项中,选出可以填入空白处的最佳Manyyearsago,IboughtahouseintheGarfagnana,wherewestillgoeverysummer.Thefirsttime there,weheardthechugchug-chugofamotorbike itswaydownthehilltowardus.Itwas calledMario,comingto usaboxcontainingsometomatoesandabottleofwine.Itwasaverynice forhimtomake.Butwhenwelookedatthetomatoes,wewere becausetheyweresomisshapen:notatalllikethenice,round, 27 thingsyougetinasupermarket.Andthewinewascloudy,inafunnyoldbottlewithnolabel(标签)onit.Thesecan’tbeany ,wethought.Butwewere hiskindness,sowe them.Whatwediscoveredisthatit’s tojudgewhatyoueatonlybyits .Thosetomatoes thatremindedmeoftheonesmyuncleusedtogrowwhenIwasachild.Nowadayssupermarkettomatoes perfectbuttasteofwater.Nobody’sgoingtohavea memoryofthose.It’sasurprisetheyhaven’tmanagedtogrowsquareonessothattheycan themeasily.Mario’swinemayhavebeencloudyandcomeoutofanoldbottle,butitwas It’sgoodtoeatthingsatthecorrecttime,whenthey’re ,andascloseaspossibletowherethey .WhatMariohad uswasthetasteoftheGarfagnana.21.A.B.C.D.22.A.B.C.D.23.A.B.C.D.24.A.B.C.D.25.A.B.C.D.26.A.B.C.D.27.A.B.C.D.28.A.B.C.D.29.A.sympatheticB.thankfulC.cautiousD.interested30.A.B.C.D.31.A.B.C.D.32.A