2023年普通高等學校招生全國統(tǒng)一考試（全國甲卷）語文一、現(xiàn)代文閱讀（36分）（一）論述類文本閱讀（3小題，9分 （二）實用類文本閱讀（3小題，12分220 （三）文學類文本閱讀（3小題，15分巴金 二、古代詩文閱讀（34分（一）文言文閱讀（5小題，20分母之言。其于昆弟，尤篤有愛。執(zhí)母喪，倚廬三年，．．ABCDEFGH （二）古代詩歌閱讀（2小題，9分 （三）名篇名句默寫（1小題，6分《鄒忌諷齊王納諫》中記載，齊王接受鄒忌的意見，廣開言路。一開始有很多人進諫，以至 三、語言文字運用（20分（5小題，20分 們能古相并。于代局，人尤是建治級人做的的今看來多成題，定心好是當，果決壞，就應了。”二教是一講，說“釜是，‘’是。破和沉是詞。釜’‘釜’意，就把砸；舟’‘舟’意，就把鑿。樣法動叫‘動’同做的具古叫釜現(xiàn)叫‘’同水運工，代‘’現(xiàn)叫船這是今匯演。古叫冠現(xiàn)叫‘子古叫‘’現(xiàn)叫鞋都這情。曹步有‘在中’句 里的釜和舟跟釜舟’的釜和舟思同。三教講比簡，說比少他樣：“羽河攻朝軍，河后把飯鍋碎把鑿，了己退，示進退決，于秦打了后大就‘釜舟個語示定大決，顧何牲意。” B.又吃魚又嫌C.前怕狼后怕 D.首尾不能兼解要點。要求闡釋準確，語言流暢，不超過60個字。 四、寫作（60分8002023年普通高等學校招生全國統(tǒng)一考試（全國甲卷）語文一、現(xiàn)代文閱讀（36分）（一）論述類文本閱讀（3小題，9分從重構上古史體系角度看，2070—80年代，夏鼐、蘇秉琦等系統(tǒng)地構建了新石器 【答案】1. 2. 3.【1【2【3（二）實用類文本閱讀（3小題，12分220220赫 【答案】4. 5.【4220赫茲”，據(jù)此看出，谷物幼苗根部發(fā)出的輕響是主動發(fā)出的聲音。【5B.“下面的發(fā)現(xiàn)可以作為證明第一段中心觀點的材料”錯。原文第一段“那么樹木會如何.D.“都使用了打比方和舉例子的說明方法”錯。原文第二段“樹木為了確保信息能夠快速220赫茲?！@些幼苗的頂端總會往聲源方向生長，這意【6（三）文學類文本閱讀（3小題，15分巴金 【答案】7.C 【7C.“依據(jù)是這些材料本身缺乏生命力”錯誤，由文中“詩應該給人以創(chuàng)造的喜悅，詩應該【8【9二、古代詩文閱讀（34分（一）文言文閱讀（5小題，20分ABCDEFGH 【答案】10. 11. 12.【10之”主謂賓齊全，B【12【13（二）古代詩歌閱讀（2小題，9分 【答案】14.A 15.①用動詞“垂”“謝”賦予“柳”“梅”動態(tài)的美感，運用虛寫的手【14【15（三）名篇名句默寫（1小題，6分 兩句詩“ “ ①.門庭若②.時時而間進 ③.停杯投箸不能食 ④.拔劍四顧心茫然 ⑤.忽如一夜春 ⑥.千樹萬樹梨花開（砌下落梅如雪亂，拂了一身還滿）三、語言文字運用（20分（5小題，20分 ‘ B.又吃魚又嫌C.前怕狼后怕 D.首尾不能兼解方法，擬出講解要點。要求闡釋準確，語言流暢，不超過60個字。 【答案】17. 18.語句：④；修改為：古人不能跟我們相提并論【17【18【19【20【21四、寫作（60分800勞動的生活模式，進入了節(jié)時省力、高能高速的全新狀態(tài)。隨著智能搜索引擎、OSO圖文人像識別、ChatGPT遺憾的是，在技術發(fā)展使時間變得充裕、人們對時間的使用更加自由的同時，現(xiàn)代人5GGSAppeVisonPro2023年普通高等學校招生全國統(tǒng)一考試（全國甲卷） {x|x3k,k {∣x{∣x D.若復數(shù)ai1ai2,aR，則a A.- B. C. D. A. B. C. D. ab1

，且abc0，則cosac,bc

C. D. A. B. C. D.有50人報名足球俱樂部，60人報名乒乓球俱樂部，70人報名足球或乒乓球俱樂部，若已知某人報足球俱樂部，則其報乒乓球俱樂部的概率為（ A. B. C. D.“sin2sin21”是“sincos0”的 B.必要條件但不是充分條C.充要條 D.既不是充分條件也不是必要條 0)的離心率

，其中一條漸近線與圓(x

2

2

B兩點，則|AB|

2

4有五名志愿者參加社區(qū)服務，共服務星期六、星期天兩天，每天從中任選兩人參加服務，則恰有1人連續(xù)參加兩天服務的選擇種數(shù)為（ A. B. C. D.fxycos2xππyfxy1x1 6 B. C. D.在四棱錐PABCD中，底面ABCD為正方形，AB4,PCPD3,PCA45，則PBC的面 A.

B.

C.

D.x2y2FF為兩個焦點，O為原點，PcosFPF3，則|PO

A. C.

若y(x1)2axsinxπ為偶函數(shù)，則a 2 2x3yx，y滿足約束條件3x2yxy

，設z3x2y，則z的最大值 ，D為BC上一點，AD為BAC的平分線，AD 已知數(shù)列ana21Sn為ann2Snnan求an求數(shù)列an1n項和T2n XX單位：g（PPk2k已知直線x2y10與拋物線C:y22px(p0)交于A,B兩點，且|AB| （1）pf(xaxsinxx0,πcos3

2 若a8f(xf(xsin2xax2t|PA||PB|4f(x2xaaa0yfx2，求a2023年普通高等學校招生全國統(tǒng)一考試（全國甲卷）

A.{x|x3k,k{∣x

{∣xD.若復數(shù)ai1ai2,aR，則a A.- B. C. D.【詳解】因為ai1aiaa2iia2a1a2i22a

0a1 A. B. C. D.n1A123B325n112n2時，判斷框條件滿足，第二次執(zhí)行循環(huán)體，A358B8513n213n3A81321B211334n314n4B34． ab1

，且abc0，則cosac,bc

C. D. 【詳解】因為 ,所以 r,

abc →2,

a+b=- ,所以 a

2ab

112ab

ab如圖,設OA b,OC→由題知OAOB1OC

AB邊上的高OD

2,AD 2 所以CDCOOD

232 tanACD

AD1,cosACD c, c, 3 2

1 A B. C. D.【分析】根據(jù)題意列出關于q的方程,計算出q,S4q3q44q4q2,q3q24q40,即(q2)(q1)(q2)0q0,q2S4124815有50人報名足球俱樂部，60人報名乒乓球俱樂部，70人報名足球或乒乓球俱樂部，若 A. B. C. D.【詳解】報名兩個俱樂部的人數(shù)為50607040記“某人報足球俱樂部”為事件ABPA505PAB404 P(∣AP(

70.8“sin2sin21”是“sincos0”的 B.必要條件但不是充分條 D.既不是充分條件也不是必要條【詳解】當sin2sin21時，例如π,0但sincos0即sin2sin21推不出sincos0當sincos0sin2sin2(cos)2sin21，即sincos0能推出sin2sin21.sin2sin21是sincos0成立的必要不充分條件.x2y2

0)

(x2)2(y3)21交于A，B兩點，則|AB|

2

4【詳解】由e ，則a2b2

a2

5y2x22則圓心(2,3)到漸近線的距離d22所以弦長

4511則恰有1人連續(xù)參加兩天服務的選擇種數(shù)為（ A. B. C. D.假設a42人各參加星期六與星期天的社區(qū)服務，共有A212種方法，1人連續(xù)參加了兩天社區(qū)服務的選擇種數(shù)有51260種.fx為函數(shù)ycos2xπ向左平移π個單位所得函數(shù)，則yfx 6 y1x1的交點個數(shù)為 B. C. D.fxsin2xfxy1x1 fxy1x1的大小關系，從而精確圖像，由此得解 ycos2xππ 6 y

πcos2x66cos2x2sin2xfxsin2x y1x1顯然過01與10 fxy1x1 2x3π2x3π2x7πx3πx3πx7πfxy1x1

x3πf3πsin3π1y13π13π41 4 2

4 x3πf3πsin3π1y13π13π41 4

x7πf7πsin7π1y17π17π4 4

fxy1x1的交點個數(shù)為3 在四棱錐PABCD中，底面ABCD為正方形，AB4,PCPD3,PCA45，則PBC的面積為（ A.

B.

C.

D.【分析】法一：利用全等三角形的證明方法依次證得OCO，PDBPCA，從而得到PAPB，再在C中利用余弦定理求得PA，從而求得PB，由此在C法二：先在△PAC中利用余弦定理求得PA ，cosPCB1，從而求PAPC3，再利用空間向量的數(shù)量積運算與余弦定理得到關于PB,BPD的方程組，從而求得PB ，由此在PBC中利用余弦定理與三角形面積公式即可得解.ACBD交于OPO，則OACBD因為底面ABCD為正方形，AB4，所以ACBD42，則DOCO2 又PCPD3，POOP，所以PDOPCO，則PDOPCO，PCPD3ACBD42，所以PDBPCAPAPB在△PACPC3AC42PCA45PA2AC2PC22ACPCcosPCA329242 ，則PB

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

217故PA 217所以cosAPC 2172PA

17PAPCPAPCcosAPC173173 PBmBPD

1 1

PO

PAPC

PBPD，所以PAPC

PBPD PAPC2PAPCPBPD2PBPD則17923m2923mcosm26mcos110又在△PBDBD2PB2PD22PBPDcosBPD，即32m296mcos，m26mcos230②，兩式相加得2m2340，故PBm 故在PBCPC3PB17BC4PC2BC2 916 所以cosPCB 2PC 23 1cos22又0PCBπ，所1cos22所以PBC的面積為S1PCBCsinPCB13422 x2y2

F,

cosFPF

1， OP 5則|PO| A.

C.

【分析】方法一：根據(jù)焦點三角形面積公式求出△PF1F2P的坐標，從而得出OP的值；PFPF,PF2PF2 【詳解】方法一：設FPF20π

b2tanF1PF2b2 cos2sin2

1tan2

由cosF1PF2cos2cos2sin21tan25tan2a29b26c2a2b23

1FF 123 61y23 1 x29139

x2yx2y 3PFPF2a6PF2PF22PFPFFPFFF2

1即PF2PF2 PFPF12②，聯(lián)立 PFPF15,PF2PF221 PO 2PFPF

2PF1PF2

3 即PO

21 PFPF2a6PF2PF22PFPFFPFFF2

1即PF2PF2 PFPF12②，聯(lián)立①②，解得：PF2PF221 由中線定理可知，2OP2FF22PF2PF242，易知FF ，解得OP

30

1 1若y(x1)2axsinxπ為偶函數(shù)，則a 2 fπfπ，從而求得a2，再檢驗即可得解 2 yfxx12axsinxπx12axcosx 2 義域為R π π

所以f2f2，即a sa ，

則πa21212π，故a2 fxx122xcosxx21cosx，fxx21cosxx21cosxf又定義域為R，故fx為偶函數(shù)，所以a2

x2x3yx，y滿足約束條件3x2yxy

，設z3x2y，則z的最大值 y3xz過點Az 由2x3y3可得x3A(333x2y yzmax332315 2EF中點為OABBB1中點GMBB1C1C的NFGEGOMONMN，如圖，即R ON2MN則球心OON2MN

FG2EGFG2EG22

1個交點，EF12.在ABC中，AB2，BAC60,BC ，D為BC上一點，AD為BAC的平分線，則AD ABcACbBCa方法一：由余弦定理可得，22b222bcos60°6，因為b0，解得：b1 12bsin60°12ADsin30°1ADbsin30° AD

3b1

2313

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

sin

sinB

6 2，sinC 2 因為 ，所以C45°，B180°60°45°75°又BAD30o，所以ADB75ADAB2．2．已知數(shù)列ana21Sn為ann2Snnan求an求數(shù)列an1n項和T2n （1）ann（2）n（2）n n

S1,n

,n1詳解】2Snnan，n22Sn1n1an12SnSn1nann1an12an化簡得：n2an1 ，當n3時，anan1a31，即an

2a1

1

1

1

1因為 ，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為中點，即可得證；1詳解】A1CBCBCACA1CACACC1A1A1CACCBCACC1A1BCBCC1B1A1BCC1B11，A1O1，設COx，則C1O2x，CO2AO2AC2，AO2OC2CA2，AC2AC2CC2 1 1 1x212x)24x1ACA1CA1C1 AC2BBDAA1AA1DDAA1中點，AA1BB12BD2 AB2AC在Rt△ABC，AB2ACACACCM，連接C1MC1M∥A1C，C1MABCAMABCC1M(2AC)2AC則在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與平面BCCB所成角的正弦值為 13 1 40只小鼠均分為兩組，分別為對照組（不加．XX單位：g（2×295%的把握認為藥物對小鼠生長有抑制作用．Pk2k（1）EX（2（i）（ii）（2（i）1X的可能取值為0,12C0

C1

C2 則P(X0)2020 ，P(X1)2020 ，P(X2)2020 XEX019120219 2可得第11位數(shù)據(jù)為14.4，后續(xù)依次為17.3,17.3,18.4,20.421.523.223.6,，20位為23.22123.6m23.223.623.440(6614由（i）可得，K 6.4003.841，20202020%已知直線x2y10與拋物線C:y22px(p0)交于A,B兩點，且|AB| （1）p（1）p（2）12（2）MNxmynMx1y1Nx2y2MFNF0mn的關系，以及MNF的面積表達式，再結合函數(shù)的性質即可求出其最小值．1AxA,yA,BxB,yB由x2y10y24py2p0y

4p,y

2py22

A y

yA

4yA2p2p6 4yA2F10MN的斜率不可能為零，MNxmynMx1y1Nx2y2，y2由xmy

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

或n3 1n設點F到直線MN的距離為d1nxx y

yy1m21m216m21 4n26n1

11

1n11n1所以MNF的面積S1MNd1n1

n1n12n3

n3

當n3 時，MNF的面積 222212 mn的關系，一是為了減元，二是通f(xaxsinxx0,πcos3

2 若a8f(xf(xsin2xa（2）(,（1）求導,然后令tcos2x,討論導數(shù)的符號即可（21

f(xsin2x,g(x)的最大值,0比較大小,得出a的分界點,f(x)a

cosxcos3x3sinxcos2xsincos6cos2x3sin2a acos4

32cos2cos4令cos2xt,則tf

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

ππ所以f(x)在0,上單調遞增,在,上單調遞 4

422g(x)f(xsin

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

x1

2(2t1)a24t設(ta24t2

(t)4 4t32t(t)4 1a(,3]g(x(ta3g(x在0,上單調遞減,g(x)g(0)0 2 a(3],f(xsin2x,符合題意2a(3 1

當t0,tt23t3

,所以 所以t(0,1,使得t0,即x0,,gx0 2 當tt,1,(t0,x0xg(x0g(x單調遞增.x0x0g(x)g(0)0, 綜上a的取值范圍為( :本題采取了換元,注意復合函數(shù)的單調性tcosx在定義域內是減函數(shù),若tcosx,當tt,1,(t0,x0xg( x2tP(2,1，直線ly1tsinA，B|PA||PB|4

（1）根據(jù)t1因為lxyABππx0t1cosy0t2所以PAPBt2t1 4，所以sin22πkπ，解得π1kπkZ ππ，所以3π 2由（1）可知，直線l的斜率為tan1，且過點2,1所以直線ly1x2xy30f(x2xaaa0yfx2，求a 22（1）xaxa討論即可1xa,f(x2a2xax即3xa,xa,axa xa,f(x2x2aaxx3a,ax3a綜上,不等式的解集為a3a 22x3a,xf(x)22x3a,xf(x的草圖,f(x與坐標軸圍成△ADO與

OAa ABa3a22,a2

2023年普通高等學校招生全國統(tǒng)一考試（全國甲卷）答卷前，考生務必將自己的姓名、準考證號填寫在答題卡上回答選擇題時，選出每小題答案后，用鉛筆把答題卡上對應題目的答案標號涂黑考試結束后，將本試卷和答題卡一并交回12560分.在每小題給出的四個選項中，只有一項 51i32i2

2,3,A.

B. C.1 →

D.1已知向量a3,1,b2,2，則cosab,ab

2某校文藝部有4名學生，其中高一、高二年級各2名．從這4名學生中隨機選2名組織校文藝匯演，則這2名學生來自不同年級的概率為（

C. D. A. B. C. D. A. B. C. D.FF, 為橢圓C

2P在CPF

0

B. C. D. e 曲線yx1在點1,2處的切線方程為 ye

ye

yex

yex 0)的離心率

，其中一條漸近線與圓(x

2

2

B兩點，則|AB|

2

3

4在三棱錐PABC中，ABC是邊長為2的等邊三角形，PAPB2,PC A. C. D.fxe(x1)2．記a

f2,bf3,cf6，則 2 2 2bc

ba

cb

cayfxycos2x的圖象向左平移yfx 6 直線y1x1的交點個數(shù)為 A. B. C. D.4520分 若fx(x1)2axsinxπ為偶函數(shù)，則a 3x2y若x，y滿足約束條件2x3y3，則z3x2y的最大值 xy則球O的半徑的取值范圍是 70分.解答應寫出文字說明、證明過程或演算步驟.17~21個試題考生都必須作答.22、23題為選考題，考生根據(jù)要求作答（一）60分記 的內角A,B,C的對邊分別為a,b,c，已知b2c2

2求bc （（2）F為CMN為CFMFN0，求△MFN單位：g 19 （2（ⅰ）

n(ad

abcdacbdPK2k20fxaxsinxx0π

2 fxsinx0，求a已知直線x2y10與拋物線C:y22px(p0)交于A,B兩點，AB （1）p（二）10分.22、23題中任選一題作答.如果多做，則按所做的第一題[4-4：坐標系與參數(shù)方程]（10分y1y1ABPAPB4[4-5：不等式選講]（10分f(x2|xa|a,a0yfxx2，求a2023年普通高等學校招生全國統(tǒng)一考試（全國甲卷）答卷前，考生務必將自己的姓名、準考證號填寫在答題卡上考試結束后，將本試卷和答題卡一并交回12560分.在每小題給出的四個選項1.設全集 A2,3,2,3,N25}NeUM{235}，51i32.2i2A.

C.1

D.151i3 51i 1(2i)(2 a3,1,b22，則

→

2 abab,abab 5211a5211

34,ab ，abab51312 → abab 17所以cosab,ab → aba 某校文藝部有4名學生，其中高一、高二年級各2名．從這4名學生中隨機選2名組織校文藝匯演，則這2名學生來自不同年級的概率為（

C. D. 22名學生來自不同年級的基本事件有C1C142242 A. B. C. D.【分析】方法一：根據(jù)題意直接求出等差數(shù)列ann項和公式即a2a6a1da15d10a13d5a4a8a13da17d45d1a12

5

54d521020a2a62a410a4a845a45a89da8a41aad514 S55a320． A B. C. D.【詳解】當k1A123B325k112k2時，判斷框條件滿足，第二次執(zhí)行循環(huán)體，A358B8513k213k3A81321B211334k314當k4B34．FF為橢圓C:x2

2 的兩個焦點，點PCPF

0 PF1

B. C. D.

0，所以

90°

b2tan45°11PFPFPF

2

0，所以

90c2514c2PF2PF2FF24216PF

2a

1 PF2PF22PFPF162PFPF20PF

2 曲線yx1在點 ye

ye

yex yex y

在點1eyekx1 2 x y

x

x2ky

x x yeex 所以曲線y 在點1,e處的切線方程為yexex 2 x2y2

0的離心率

(x2)2(y3)21交于A，B兩點，則|AB|

2

3

4【詳解】由e ，則a2b2

a2

5y2x22則圓心(2,3)到漸近線的距離d|223| 22所以弦長

4511在三棱錐PABC中，ABC是邊長為2的等邊三角形，PAPB2,PC A.

C. D.ABEPECEPEABCEABPECEPECPECEEPECE

3 ，PC PC2PE2CE2PECE所以V

BPEC

1S

AB11

321fxe(x1)2．記a

f2,bf3,cf6，則 2 2 2bcca

ba

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

(

3)2429

16

70 2 6113640611 2 g(6g(3 因為611 2 6 24， 2 2)242843164384(32)0即61 2，所以g(6)g(2) g(2g(6g(3 yex為增函數(shù)，故acb，即bcayfx的圖象由ycos2x的圖象向左平移個單位長度得到，則 6 yfx的圖象與直線y1x1的交點個數(shù)為 A. B. C. D.fxsin2xfxy1x1 fxy1x1的大小關系，從而精確圖像，由此得解 ycos2xππ 6 ycos2xππcos2xπsin2xfxsin2x y1x1顯然過01與10 fxy1x1 2x3π2x3π2x7πx3πx3πx7πfxy1x1

x3πf3πsin3π1y13π13π41 4 2

4 x3πf3πsin3π1y13π13π41 4

x7πf7πsin7π1y17π17π4 4

fxy1x1的交點個數(shù)為3 4520分 q1n項和公式和平方差公式化簡即可求出公比q1則由8S67S3得86a173a1a10q1.q1時，因為8S67S3所以8

a1q61

7

a1q31即81q671q3，即81q31q371q3，即81q37q1若fx(x1)2axsinxπ為偶函數(shù)，則a fxx12axsinxπx12axcosxx2(a2)x1cosx a20，解得a2，3x2y若x，y滿足約束條件2x3y3，則z3x2y的最大值 xyy3xz過點Az 由2x3y3可得x3A(333x2y yzmax332315在正方體ABCDA1B1C1D1中，AB4,O為AC1的中點，若該正方體的棱與球O的球面有公共點，則球O的半徑的取值范圍是 【答案】[222424242

2R43,R ，故Rmax 形，且OMNGH的對角線交點，連接MG，則MG R[2223故答案為：[22270分.解答應寫出文字說明、證明過程或演算步驟.17~21題（一）60分記 的內角A,B,C的對邊分別為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為CC1A1CACxxA1O.1又因為ACB90ACBCBCACC1A1，2詳解】A1CBCA1CACA1BABBCA1CACxA1C1x所以O為CCOC1AA1 ACAC,AC2AC2AA2 22即x2x2222AC2OCAC2OC1

20只分配到對照組，試驗組的小白鼠飼養(yǎng)在高濃度臭氧環(huán)境，對照組的小g （2（ⅰ

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

40(6614 40(6614所以能有95的把握認為小白鼠在高濃度臭氧環(huán)境中與在正常環(huán)境中體重的增加量有差fxaxsinxx0π

2 fxsinx0，求a 2 a（1）a1fxfx，g00a0a0a<0法二：先化簡并判斷得sinx

sincos2

0a0a<0a01a1fxxsinxx0πcos2

2fx

cosxcos2x2cosxsinxsin

cos2x2sin2cos3xcos2x21cos2x

cos3xcos2x

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

πfx

cos3

0在0,上恒成立 2 2詳解】

sincos2

sinx0x

2 1sin2

cos3

cosx0x gxfxsinx0g0f0sin00，g0a11a0a0，a0時，因為sinx

sinx

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

cos2 fxsinxsinx

sincos2

a<0時，由于0xπax0fxsinxax

sincos2

sinxsinx

sincos2

fxsinx0a0，所以a的取值范圍為0.sin sinxcos2xsin sinxcos2x sin3因為sinxcos2x

cos2

cos2

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

sincos2

a<0時，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此時gx在0,x1上有gx0，所以gx在0,x1上單調遞增，則在0x1gxg00，即fxsinx0，不滿足題意；a0. 2 已知直線x2y10與拋物線C:y22px(p0)交于A,B兩點，AB （1）p（2）F為CMN為CFMFN0，求△MFN（1）p（2）12（2）MNxmynMx1y1Nx2y2MFNF0mn的關系，以及MNF的面積表達式，再結合函數(shù)的性質即可求出其最小值．1AxA,yA,BxB,yB由x2y10y24py2p0yy4pyy2py22

AAB

yA

y 4y y 4yA2F10MN的斜率不可能為零，MNxmynMx1y1Nx2y2，y2由xmy

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

或n3 1n設點F到直線MN的距離為d1nxx y

yy1m21m216m21 4n26n1

11

1n11n1所以MNF的面積S1MNd1n1

n1n12n3

n3

n3

時，MNF的面積 222212 mn的關系，一是為了減元，二是通[4-4：坐標系與參數(shù)方程]（10分y1y1

yABPAPB4（1）根據(jù)t1因為lxyABππx0

y0

PA

4，所以sin212πkπ，解得π1kπk ππ，所以3π 2由（1）可知，直線l的斜率為tan1，且過點2,1所以直線ly1x2xy30[4-5：不等式選講]（10分f(x2|xa|a,a0yfxx2，求a 22（1）xaxa討論即可1xa,f(x2a2xax即3xa,xa,axa xa,f(x2x2aaxx3a,ax3a綜上,不等式的解集為a3a 22x3a,xf(x)22x3a,xf(x的草圖,f(x與坐標軸圍成△ADO與

OAa ABa3a22,a2

2023年普通高等學校招生全國統(tǒng)一考試(全國甲卷)第一部分聽力(30分第一節(jié)(5小題;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第二節(jié)(15小題;1.522.5分5A、B、C三個選項中5秒鐘;5秒鐘的作答時間，每段對話或獨白讀兩遍。66、7題。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. C.WhatwasRobertlikebeforehewas9yearsHehadwild B.Heenjoyed C.HelovedWhatdidRobert’sfather WhathelpedRobertbecomea B.Listeningto C.Reading101720WherewasOpenTchaikovskyCompetitionheldinAIn B.In C.InWhatdoesMaximsayaboutthecompetitionheattendedatItinspiredmanyyoungItwasthemusiceventofhisItwasalife-changingWhichkindofmusicaretheyoungplayersrequiredtoRock B.Pop C.ClassicalWhatdoesMaximvaluemostinyoungplayers’A B. C.第二部分閱讀理解(共兩節(jié),40分第一節(jié)(15小題:230分A、B、CD四個選項中,WheretoEatinBangkokisahighlydesirabledestinationforfoodlovers.IthasaseeminglybottomlesswellofdiningHerearesomesuggestionsonwheretostartyourBangkokeatingOfferingThaifinedining.NahmprovidesthebestofBangkokculinary(烹飪的)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(場所)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(哲學).ThatloveforphilosophylasteduntilIgottocollege.NothingkillstheloveforphilosophyfasterthanwhothinktheyunderstandFoucault,Baudrillard,orConfuciusbetterthanyou—andthentrytoexplainEricweiner’sTheSocratesExpress:InSearchofLifeLessonsfromDeadPhilosophersreawakenedmyloveforphilosophy.Itisnotanexplanationbutaninvitationtothinkandexperiencephilosophy.Weinerstartseachchapterwithasceneonatrainridebetweencitiesandthenframeseachphilosopher’sworkinthecontext(背景)ofonethingtheycanhelpusdobetterTheendresultisareadinwhichwelearntowonderlikeSocrates,seelikeThoreau,listenlikeSchopenhauer,andhavenoregretslikeNietzsche.This,morethanabookaboutundestandingphilosophy,isabookabourlearningtousephilosophytoimprovealife.Hemakesphilosophicalthoughtanappealingexercisethatimprovesthequalityofourexperiences,andhedoessowithplentyofhumorWeinerentersintoconversationwithsomeofthemostimportantphilosophersinhistoryandhebecomespartofthatcrowdintheprocessbydecoding解讀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預防aren’ttakengrizzliescanbecometroublesome,sometimeskillinganimalsorwalkingthroughyardsinsearchoffood.Ifpeopleremovefoodandattractantsfromtheiryardsandcampsites,grizzlieswilltypicallypassbywithouttrouble.Puttingelectricfencingaroundchickenhousesandotherfarmanimalquartersisalsohighlyeffectiveatgettinggrizzliesaway.“Ourhopeistohaveaclean,attractant-freeplacewherebearscanpassthroughwithoutlearningbadhabits,“saysJamesJonkel,longtimebiologistwhomanagesbearsinandaroundMissoula.HowdoAmericanslookatTheycausemixedfeelingsinTheyshouldbekeptinnationalTheyareofhighscientificTheyareasymbolofAmericanWhathashelpedtheincreaseofthegrizzlyTheEuropeansettlers’Theexpansionofbears’TheprotectionbylawsinceThesupportofNativeWhathasstoppedtheU.S.FishandWildlifeServicefromdelistingTheoppositionofconservationThesuccessfulcomebackofThevoiceoftheThelocalfarmers’WhatcanbeinferredfromthelastFoodshouldbeprovidedforPeoplecanliveinharmonywithAspecialpathshouldbebuiltforTechnologycanbeintroducedtoprotect第二節(jié)(5小題;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(滿足)mayseemlikethemost“feelgood”optionatthetime,butpsychologyresearchsuggestswaitingforthingsactuallymakesushappierinthelongrun.Andtheonlywayforustogetintothehabitofwaitingistopractice. .Putoffwatchingyourfavoriteshowuntiltheweekendorwait10extraminutesbeforegoingforthatcake.You’llsoonfindthatthemorepatienceyoupractice,themoreyoustarttoapplyittoother,moreannoyingsituations. Somanyofushavethebeliefthatbeingcomfortabelistheonlystatewewilltolerate,andwhenweexperiencesomethingoutsideofourcomfortzone,wegetimpatientaboutthecircumstances.Youshouldlearntosaytoyourself,“ .”You’llthengraduallybecomemorepatient.FindyourStartwithsmallAccepttheAllthisaddsuptoastateofItcanalsohelpuspracticemoreThisismerelyuncomfortable,notThey’reallsituationswherewecouldusealittleextra第三部分語言知識運用(45分)第一節(jié)(201.530分)A、B、CD四個選項中,選出可以填入空白處的最佳Manyyearsago,IboughtahouseintheGarfagnana,wherewestillgoeverysummer.Thefirsttime there,weheardthechugchug-chugofamotorbike itswaydownthehilltowardus.Itwas calledMario,comingto usaboxcontainingsometomatoesandabottleofwine.Itwasaverynice forhimtomake.Butwhenwelookedatthetomatoes,wewere becausetheyweresomisshapen:notatalllikethenice,round, 27 thingsyougetinasupermarket.Andthewinewascloudy,inafunnyoldbottlewithnolabel(標簽)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