2023年新高考全國II卷【語文+數(shù)學+英語】真題及答案解析

PAGE1頁/9絕密★2023年普通高等學校招生全國統(tǒng)一考試（新高考全國Ⅱ 一、現(xiàn)代文閱讀（35分）（一）I（5小題，19分1930 （二）II（4小題，16分社戲（節(jié)選

二、古代詩文閱讀（35分（一）文言文閱讀（本題共5小題，20分 （二）古代詩歌閱讀（本題共2小題，9分林逋 （三）名篇名句默寫（本題共1小題，6分 ， ， 三、語言文字運用（20分（一）I（2小題，7分是巋然不動，讓仙鶴①，以至于本應是勝利者的仙鶴，反而著急地叫了又叫……后操縱著小竹竿，皮影則甩手投足，舞槍弄棍，騎馬沖殺，無所不能，往往令觀眾②。（二）II（本題共3小題，13分?？墒亲罱?，樊女士發(fā)現(xiàn)，自從經(jīng)常戴上耳機聽著歌入睡以來，耳朵里開始有了“嗡嗡②？導致耳朵疼痛、耳屎變多等。有人覺得這都是小事，忍忍就過去了。但事實上，③是（）10文中畫波浪線部分有語病，請進行修改，使語言表達準確流暢?？稍鰟h少量詞語，四、寫作（60分800PAGE12頁/20絕密★2023年普通高等學校招生全國統(tǒng)一考試（新高考全國Ⅱ卷） 一、現(xiàn)代文閱讀（35分）（一）現(xiàn)代文閱讀I（本題共5小題，19分1930

【答案】1. 2. 3.【1故選B【2故選C3故選【4【5（二）II（416社戲（節(jié)選）

【答案】6. 7.【6故選D【7故選C?！?【9要的民俗文化活動，給人們帶來了節(jié)日般的快樂，豐富了人們的精神生活。二、古代詩文閱讀（35分（一）文言文閱讀（本題共5小題，20分 【答案】10.11. 12.【10故選B?！?2本題考查學生理解文章內(nèi)容故選A【13【14383，前秦帝苻堅率領大軍進駐壽陽，并臨淝水岸擺好陣勢，同晉將謝玄隔水對峙。（（二）古代詩歌閱讀（本題共2小題，9分林 【答案】15.C 16.①詩歌首聯(lián)“臥枕船舷歸思清”透露出愉悅閑適之心境，全詩景物都【15故選C【16（三）名篇名句默寫（本題共1小題，6分 ， ， ①.則遣從事以一少牢告廟 ①.請其矢 ①.小樓一夜聽春雨 ①. ①.星垂平野 ①.月涌大江（江天一色無纖塵皎皎空中孤月輪三、語言文字運用（20分（一）I（27是巋然不動，讓仙鶴①，以至于本應是勝利者的仙鶴，反而著急地叫了又叫……后操縱著小竹竿，皮影則甩手投足，舞槍弄棍，騎馬沖殺，無所不能，往往令觀眾②?！敬鸢浮?8①奈何不得②如醉如癡【18【19（二）II（313。可是最近，樊女士發(fā)現(xiàn)，自從經(jīng)常戴上耳機聽著歌入睡以來，耳朵里開始有了“嗡嗡②？導致耳朵疼痛、耳屎變多等。有人覺得這都是小事，忍忍就過去了。但事實上，③是（）10【答案】20. 21.①你只需要一副耳機②有什么限度③過度使用耳機的?！?0據(jù)以上分析看出B項與例句意義和用法相同。故選B?！?1耳機？”，此處應是一個問句，所以可填“有什么限度”。 【22四、寫作（60分800節(jié)日2023年普通高等學校招生全國統(tǒng)一考試（新高考全國Ⅱ卷）數(shù) 學1.在復平面內(nèi)，(1+3i)(3-i)對應的點位于 A.第一象限 B.第二象限 C.第三象限 D.第四象2.設集合A={0,-a}，B={1,a-2,2a-2}，若A?B，則a= A. B. C.3

D.某學校為了解學生參加體育運動情況，用比例分配的分層隨機抽樣方法作抽樣調(diào)查，名學生，則不同的抽樣結(jié)果共有（AC45

?C40

種 種若f(x)=(x+a)ln2x-1為偶函數(shù)，則a= 2x

C.2

D.x2x已知橢圓C

2=FFy=x+mC 3兩點，若

面積是△F2

面積的2倍，則m=

3

-2已知函數(shù)f(x)=aex-lnx在區(qū)間(1,2)上單調(diào)遞增，則a的最小值為

D.已知acosa=1+4

，則sina= 525-1+A.3--1+

C.3- 44記Sn為等比數(shù)列{an}的前n項和，若S4=-5，S6=21S2，則S8= A. B.

D.項符合題目要求。全部選對的得5分，部分選對的得2分，有選錯的得0PO，AB?APB=120?PA=2在底面圓周上，且二面角P-AC-O為45°，則 該圓錐體積為 B.該圓錐的側(cè)面積為42C.AC=2

D.△PAC3Oy=3

3(x-1)過拋物線Cy2=2px(p>0) p=MN=3 OMN為等腰三角若函數(shù)f(x)=alnx+b+c(a?0)既有極大值也有極小值，則 Abc>ab>

b2+8ac>010的概率為1-a10b(0<b<11的概率為1-b.考1次，三次傳輸是3次．收到的信號需要譯碼，譯碼規(guī)則如下：單次傳輸時，收到的信號1）.1，0，1l，0，1的概率為(1-a)(1-b11b(1-b)2+1-b當0<a<0.500的概率大于采用單次傳輸方0的概率三、填空題：本大題共4小題，每小題5分，共20a+b=2a-已知向量a，b滿足aa+b=2a-

= 已知直線lx-my+1=0

C(x-1)2+y2=4A，B“ABC8積為 m的一個 5f(x)=sin(wx+j)A，By=1y=f(x)2

=π，則f(π) 6記ABCABC的對邊分別為abcABC的面積為3，DBC中點，AD=1．（1）若?ADC=π，求tanB3（2）若b2+c2=8，求bc ?2an{a}b=? ?2an

，記SnTn分別為數(shù)列{an}，{bn}的前S4=32T3=16求{an}的通證明：當n>5Tn>Sncc的人判定為陽性，小于cp(c；誤診率是將未患病者判定為陽性的概率，記為q(c．假設數(shù)據(jù)在組內(nèi)均勻分布，以事件發(fā)（1）當漏診率p(c)=0.5％時，求臨界值c和誤診率q(c)（2）f(c)=p(c)+q(c)，當c?[95,105]f(c)的解析式，并求f(c)在區(qū)間[95,105]的最小值．A-BCDDA=DB=DCBD^CD?ADB=?ADC=60EBC的BC^DAFEF=DAD-AB-FC的中心為坐標原點，左焦點為(-250)，離心率為5C的左、右頂點分別為A1A2，過點(-40)的直C的左M，N22.（1）證明：當0<x<1x-x2<sinx<x（2）f(x)=cosax-ln(1-x2)，

x=0是f(x)的極大值a 1.在復平面內(nèi)，(1+3i)(3-i)對應的點位于 A.第一象限 B.第二象限 C.第三象限 D.第四象【詳解】因為(1+3i)(3-i)=3+8i-3i2=6+8i，則所求復數(shù)對應的點為(6,8)，位于第一象限.2.設集合A={0,-a}，B={1,a-2,2a-2}，若A?B，則a= A. B. C.3

D.【分析】根據(jù)包含關系分a-2=0和2a-2=0兩種情況討論，運算求解即可A?B若a-2=0，解得a=2，此時A={0-2}B={102}，不符合題意；若2a-2=0，解得a=1，此時A={0-1}B={1-10}，符合題意；a=1.3.某學校為了解學生參加體育運動的情況，用比例分配的分層隨機抽樣方法作抽樣調(diào)查，名學生，則不同的抽樣結(jié)果共有（A.C45

C20?C40 C C

種 種 【詳解】根據(jù)分層抽樣的定義知初中部共抽取60?400=40人，高中部共抽取60?200=20 根據(jù)組合公式和分步計數(shù)原理則不同的抽樣結(jié)果共有C40?C20 若f(x)=(x+a)ln2x-1為偶函數(shù)，則a= 2x

C.2

D.1f

3

=(-1+aln3，解得a=0a=0f(x)=xln2x-1(2x-1)(2x+1)>0x>1x<-1?則其定義域為?x??

2x 122x<-1?122?2(-x) 2x

?2x-1

2xf(-x)=(-x)ln2(-x)+1=(-x)ln2x-1=(-x)ln?2x+1

=x =f(x)2x f(x)為偶函數(shù).x2x已知橢圓C

2=FFy=x+mC 3兩點，若

面積是△F2

面積的2倍，則m=

3

-2【分析】首先聯(lián)立直線方程與橢圓方程，利用D>0，求出m范圍，再根據(jù)三角形面積比得到關于m的方程，解出即可.??y=x+?y=x+m與橢圓聯(lián)立?

y可得4x2+6mx+3m2-3=0

+y2=AB點，則D=36m2-4?4(3m2-3)>0，解得-2<m<2F1AB的距離d1,F2AB距離d2F1

2,0),F2

2,0)|-2+m2 2+|-2+m2 2+m2|-2+m2221SF2221

= =|-2+m|=2，解得m=- 或

SF

2+m +m 2 2已知函數(shù)f(x)=aex-lnx在區(qū)間(1,2)上單調(diào)遞增，則a的最小值為

D.f?(x)=aex-1?x

(1,2)上恒成立，再根據(jù)分參求最值即可求出f?(x)=aex-1?0在(12)上恒成立，顯然a>0xex?1 g(x)=xexx?(12)g?(x)=(x+1)ex>0，所以g(x)在(12)上單調(diào)遞增g(x)>g(1)=e，故e?1，即a?1=e-1a的最小值為e-1 已知acosa=1+4

，則sina= 525-1+A.3--1+

C.3- 4 4【詳解】因為cosa=1-2sin2a=

5，而a3-8(53-8(5=5-1 8.Sn為等比數(shù)列{an}的前n項和，若4=-5S6=21S2= A.B.D.nS4S8的關系即可解出；n項和的性質(zhì)求解．【詳解】方法一：設等比數(shù)列{an}的公比為q，首項為a1q=1S6=6a1=3?2a1=3S2q?S4=-5

6=

a(1-q411-1

=-5

a(1-q611-1

=

a1-1 1 1-由①可得，1+q2+q4=21q2=4S8

a(1-q811- 1

a(1-q411-1

?(1+q4)=-5?(1+16)=-85方法二：設等比數(shù)列{an}的公比為qS4=-5S6=21S2，所以q?-1S4=0，S2S4-S2S6-S4S8-S6成等比數(shù)列，所以(-5-

)2=S

+5)S=-1S=5

S2=-1S2S4-S2S6-S4S8-S6，即為-1-4-16S8+21，S8+21=-64S8=-85；S=5S=a+a+a+a=(a+a)(1+q2)=(1+q2)S

0 S4=-5矛盾，舍去．把握S4,S8的關系，從而減少相關量的求解，簡化運算項符合題目要求。全部選對的得5分，部分選對的得2分，有選錯的得0PO，AB?APB=120?PA=2在底面圓周上，且二面角P-AC-O為45°，則 該圓錐的體積為 B.該圓錐的側(cè)面積為42C.AC=2

D.△PAC33【詳解】依題意，?APB=120?，PA=2，所以OP=1,OA=OB 3(A1?π?(3

3)2?1=π，AB選項，圓錐的側(cè)面積為

3?2=23π，BC選項，設DAC的中點，連接ODPD32AC^ODAC^PD，所以?PDOP-AC-O的平面角，則?PDO=45?，所以OP=OD=32AD=CD

2AC=

1221222DPD

，所以SPAC=2?22 =2，D選項錯誤Oy=

3(x-1)過拋物線Cy2=2px(p>0) p=MN=3 OMN為等腰三角【詳解】Ay=

3(x-1)過點(10)，所以拋物線C:y2=2px(p>0)的p=1,p=22p=4，則A選項正確，且拋物線Cy2=4x2B選項：設M(x1,y1)N(x2,y2)y=由

3(x-1)y并化簡得3x2-10x+3=(x-3)(3x-1)=0?y2=?x=3x=1MN=x+x+p=3+1+2=16，B選項錯誤 CMN的中點為AMNA到直線l的距離分別為d1d2d因為d=1(d+d)=1(

NF)

12MN12 3即A到直線lMNMN為直徑的圓與直線l相切，C選項正確3Dy=

3(x-1)，即3x+y =03O到直線3x+y =0的距離為d 332所以三角形OMN1?16

3=43 y=-3(3-1)=-23y=-3?1-1?=23 ? OM

32+332+3

?1??1?3?+2?23??3??3所以三角形OMN不是等腰三角形，D選項錯誤.若函數(shù)f(x)=alnx+b+c(a?0)既有極大值也有極小值，則 ab>

b2+8ac>f(xf?xf?x在(0+?上有兩個變號零點，轉(zhuǎn)【詳f(x=alnx+b+

的定義域為(0+? ax2-bx-f(x) f(xf?x在(0+?a?0，因此方程ax2-bx-2c=0有兩x1x2，??Δ=b2+8ac>?于是?x+

=b>

，即有b2+8ac>0ab>0ac<0，顯然a2bc<0，即bc<0? ?xx=-2c>1 A錯誤，BCD正確.010的概率為1-a10b(0<b<11的概率為1-b.考1次，三次傳輸是3次．收到的信號需要譯碼，譯碼規(guī)則如下：單次傳輸時，收到的信號1）.1，0，1l，0，1的概率為(1-a)(1-b當0<a<0.500的概率大于采用單次傳輸方0的概率【分析】利用相互獨立事件的概率公式計算判斷AB；利用相互獨立事件及互斥事件的概率計算判斷C；求出兩種傳輸方案的概率并作差比較判斷D作答.它們相互獨立，所以所求概率為(1-b)(1-a)(1-b=1-a)(1-b)2，A對于B11，1，1l，0，1的事件，1110113個事件的積，它們相互獨立，所以所求概率為(1-b?b?1-b=b(1-b)2，B3它們互斥，由選項B知，所以所求的概率為C2b(1-b)2+1-b)3=1-b)21+2b3DC00P=1-a)21+2a，00P?=1-a，而0<a<0.5，P-P?=1-a)2(1+2a-1-a)=a(1-a)(1-2a)>0P>P?，D正確.三、填空題：本大題共4小題，每小題5分，共20a+b=2a-已知向量a，b滿足aa+b=2a-

= 33【分析】法一：根據(jù)題意結(jié)合向量數(shù)量積的運算律運算求解；法二：換元令c= r，a+a+b=2a-

2，即(a+b)2=(2a-b)22

a-r r ra+2a?b+

r=

r r4a?b+

，整理得

-2a?b=0a-b

3，即(a-b)2=r r ra-2a?b+

r=b=

，所以b 3

c=a-bc

3,a+b=c+2b,2a-b=2c+b 2由題(c+ 2

=(2c+

r，則

r r+4c?b+

r=

r r+4c?b+br

r =b

=c 3故答案為：3 【答案】2=1，而截去的正四棱錐的高為3，所以原正四棱錐的高為6 所以正四棱錐的體積為1?(4?4)?6=3231?(2?2)?3=43所以棱臺的體積為32-4=28(1?3?16+4(3

16?4)=28已知直線lx-my+1=08

C(x-1)2+y2=4A，B“ABC積為”的m的一個 5【答案】2（2-21-1中任意一個皆可以 AB，以及點CAB4-d【詳解】設點C到直線AB的距離為d，由弦長公式得AB=4-d

=1?d?

=8d=45或d=254-d1+11+1+4-d1+11+1+1+dm=?12

=45 1+1+

=25m=?252（2-21-1中任意一個皆可以 f(x)=sin(wx+j)A，By=1y=f(x)2

=π，則f(π) 6【答案】-2?

1 1 【分析】設A?x12?B?x22?，依題可得，x2-x1=6，結(jié)合sinx=2 w(x-x)=

w的值，再根據(jù)f?2π?=0以及f(0)<0

? f(x=sin?4x-2π?f(π)． 1 1 1 2【詳解】設A?x ?,B?x ?，由1 2

=6x2-x1=6由sinx=1x=π+2kπx=5π+2kπk?Z wx+j-(wx+j)=5π-π=2π，即w(

-x)=2π，\w=4

f?2π?=sin?8π+j?=08π+j=kπ，即j=-8π+kπk?Z．? ? f(x)=sin?4x-8π+kπ?=sin?4x-2π+kπ? fx

所以(

sin?4x-3π?f(x)=-sin?4x-3π? 又因為f(0)<0，所以f(x)=sin?4x-2π?，\f(π)=sin?4π-2π?= 3

-32記ABCABC的對邊分別為abcABC的面積為3，DBC中點，AD=1．若?ADC=π，求tanB3（2）若b2+c2=8，求bc（1）35（2）b=c=2（1）1，利用三角形面積公式求出a2，利用三角形面積公式求出aBC邊上的高，利用直角三角形求解作答.11：在ABC中，因為DBC?ADC=πAD=13=1AD?DCsin?=1AD?DCsin?ADC=1?1?1a2 33a=12=32在△ABD?ADB=2π，由余弦定理得c2=BD2+AD2-2BD?ADcos?ADB3即c2=4+1-2?2?1?(-1)=7，解得c2

7，則cosB 7+47+427?51-cos2sin1-cos2

211-1-7所以tanB=sinB 3 2：在ABC中，因為DBC?ADC=πAD=13=1=1AD?DCsin?ADC=1?1?1a2 33a=12=32

，解得a=4在ACD中，由余弦定理得b2=CD2+AD2-2CD?ADcos?ADB3即b2=4+1-2?2?1?1=3，解得b ，有AC2+AD2=4=CD2，則?CAD=π32C=πAAE^BCECE=ACcosC=3AE=ACsinC

3，BE=56所以tanB=AE 2

1：在△ABD與ACD?c2=1a2+1-2?1a?1?cos(π-???b2?

1a2+1-2?

，a3整理得1a2+2=b2+c2，而b2+c2=8，則a= 32又 =1?3?1?sin?ADC 3，解得sin?ADC=1，而0<?ADC<π，于 ?ADC=π2所以b

=2方法2：在ABC中，因為DBC中點，則2AD=AB+AC，又CB=AB-AC，于是4AD2+CB2=AB+AC)2+AB-AC)2=2(b2+c2=16，即4+a2=16，解得3a= 3又 =1?3?1?sin?ADC 3，解得sin?ADC=1，而0<?ADC<π，于 ?ADC=π2所以b

=2 ?2an{a}b=? ?2an

，記SnTn分別為數(shù)列{an}，{bn}的前S4=32T3=16求{an}的通證明：當n>5Tn>Sn（1）an=2n+3【分（1）設等差數(shù)列{an}的公差為d，用a1d表示Sn及Tn，即可求解作答（2）1，利用（1）的結(jié)論Snbn，再分奇偶結(jié)合分組求和法求出Tn，并與Sn作差比2，利用（1）Snbn，再分奇偶借助等差數(shù)列前n項和公式求出TnSn作差比較作答.1設等差數(shù)列{a}的公差為d，而

=?an-6,n=2k-1,k?N* ?2a,n= 則b1=a1-6,b2=2a2=2a1+2d,b3=a3-6=a1+2d-6?于?S4=4a1+6d=?于?T3=4a1+4d-12=

，解得a1=5d=2an=a1+n-1)d=2n+3所以數(shù)列{an}的通項公式是an=2n+32

=n(5+2n+3)=n2+4n，

=?2n-3,n=2k-1,k?N*? ?

?4n+6,n=當nbn-1+bn=2(n-1-3+4n+6=6n+1T=13+(6n+1)?n=3n2+7n (當n>5時T-S=32+7n-n2+4n=1n(n-1>0，因此T>S( 當n為奇數(shù)時，T= -

=3(n+1)2+7(n+1)-[4(n+1)+6]=3n2+5n-5

(當n>5時T-S=32+5n-5-n2+4n=1(n+2)(n-5>0，因此T>S( ?所以當n>5Tn>Sn?

=n(5+2n+3)=n2+4n，

=?2n-3,n=2k-1,k?N* 當n

?4n+6,n=T=(b+

+

)+(b+

+b)=-1+2(n-1)-3?n+14+4n+6?n=3n2+7 ，

(當n>5時T-S=32+7n-n2+4n=1n(n-1>0，因此T>S( 當n為奇數(shù)時，若n?3，則T=(b+

+b)+(b+

+

)=-1+2n-3?n+1+14+4(n-1)+6?n

=3n2+5n-5，顯然T=

=-1滿足上式，因此當nT=3n2+5n-5

(當n>5時T-S=32+5n-5-n2+4n=1(n+2)(n-5>0，因此T>S( 所以當n>5Tn>Sncc的人判定為陽性，小于cp(c；誤診率是將未患病者判定為陽性的概率，記為q(c．假設數(shù)據(jù)在組內(nèi)均勻分布，以事件發(fā)當漏診率p(c)=0.5％時，求臨界值c和誤診率q(c)（2）f(c)=p(c)+q(c)，當c?[95,105]f(c)的解析式，并求f(c)在區(qū)間[95,105]的最小值．（1）c=97.5q(c)=3.5?（2）f(c=?-0.008c+0.82,95?c?100，最小值為0.02?（1）根據(jù)題意由第一個圖可先求出c，再根據(jù)第二個圖求出c?97.5的矩形面積即根據(jù)題意確定分段點100，即可得出f(c)的解析式，再根據(jù)分段函數(shù)的最值求法1依題可知，左邊圖形第一個小矩形的面積為5?0.002>0.5%，所以95<c<100，所以(c-95)?0.002=0.5%，解c=97.5，q(c)=0.01?(97.5-95)+5?0.002=0.035=2當c?[95,100]f(c)=p(c)+q(c)=(c-95)?0.002+(100-c)?0.01+5?當c?100,105時，f(c)=p(c)+q(c)=5?0.002+(c-100)?0.012+(105-c)??f(c=?-0.008c+0.82,95?c?100?所以f(c)在區(qū)間[95,105]的最小值為0.02A-BCDDA=DB=DCBD^CD?ADB=?ADC=60EBC的BC^DAFEF=DAD-AB-F（2）33（1）BC^ADEBC^DA（2）AE^平面BCD，所以以點E為原ED,EB,EA所在直線分別為x,y,zABDABF的一個法向量，根據(jù)二面角的向量公式1AEDEEBCDB=DCDE^BCDA=DB=DC?ADB=?ADC=60，所以ACD與△ABD\AC=AB，從而AE^BC②，由①②，AE DE=E，AE,DE?平面ADE，所以，BC^平面ADE，而AD?平面ADE，所以BC^DA．22不妨設DA=DB=DC=2，BD^CD，\BC=22,DE=AE 2\AE2+DE2=4=AD2，\AE^DE，又BCD\AE^BCD．

AE^ BC=EDEBC?EEDEBEAxyzD(200A(00,2B(0,20E(000DABABF的一個法向量分別為n1=(x1y1z1)n2=(x2y2z2)D-AB-F平面角為q,AB=(0,2-2)，EF=DA=

20,2)F

20,2)AF=

2,0,0)，

2x1

2z1=0x=1，所以n=1,1,111?112y1

2z1=??

2y2

2z2=n1n1n1n1?n2

y2=1n2=0,1,1

cosq

6，從而sinq23?23?

31-1-9D-AB-F的正弦值為33已知雙曲線C的中心為坐標原點，左焦點為(-250)，離心率為5C的左、右頂點分別為A1A2，過點(-40)的直C的左M，N2221（1）x-y1 （1）由題意求得ab（2）曲線的坐標分別寫出直線MA1與NA2的方程yx+2=-1x- Px=-1上1222

y=1(a>0,b>0)，由焦點坐標可知c= 5-22 5-225a則由e=c 可得a=2，b5a

c2-c2-22

y =y 2由(1)可得A1(-20)A2(20)，設M(x1y1)N(x2y2)0MNx=my-4，且-1<m<1 1x2-y2=聯(lián)立可得(4m2-1)y2-32my+48=0，且D=64(4m2+3>01 則y+y 32m,yy 4m2 1

4m2

2x+2

(x+2)，直線NA的方程為y

x-

(x-2) 聯(lián)立MA1與直線NA2的方程可得x+2=y2(x1+2)=y2(my1-2)=my1y2-2(y1+y2)+2x-

y1(my2-

my1y2-6m

-2

2

2 4m2 4m2

1=4m2

1=- 3m?4m2-1-6

4m2-1-6x+2=-1x=-1x=x- Px=-1上運動【點睛】關鍵點點睛求雙曲線方程的定直線問題，意在考查學生的計算能力，轉(zhuǎn)化能力和.22.（1）證明：當0<x<1x-x2<sinx<x（2）f(x)=cosax-ln(1-x2)，

x=0f(x)a【答（1）證明見詳解（2）(-?,

2)

2,【分析（1）F(x)=x-sinxx?(0,1)，G(x)=x2-x+sinxx?(0,1)，求導，（2）根據(jù)題意結(jié)合偶函數(shù)的性質(zhì)可知只需要研究f(x)在(0,1)上的論0<a2<2和a2?2，結(jié)合（1）中的結(jié)論放縮，根據(jù)極大值的定義分析求解（1）F(x)=x-sinxx?(0,1)F?(x)=1-cosx>0對"x?(0,1)F(x)在(0,1)上單F(x)>F(0)=0，x>sinx,x?(0,1)；構建G(x)=sinx-(x-x2)=x2-x+sinxx?(0,1)則G?(x)=2x-1+cosxx?(0,1)g(x)=G?(x)x?(0,1)g?(x)=2-sinx>0對"x?(0,1)恒成立，g(x)在(0,1)g(x)>g(0)=0，即G?(x)>0對"x?(0,1)恒成則G(x)在(0,1)上單調(diào)遞增，可得G(x)>G(0)=0，所以sinx>x-x2,x?(0,1)；綜上所述x-x2<sinx<x（2）令1-x2>0，解得-1<x<1f(x)的定義域為(-1,1)，a=0f(x)=-ln(1-x2)x?(-1,1)，y=-lnu在定義域內(nèi)單調(diào)遞減，y=1-x2在(-10)(0,1)上單調(diào)遞減，f(x)=-ln(1-x2)在(-10)上單調(diào)遞減，在(0,1)上單調(diào)遞增，x=0f(x)的極小值點，不合題意，所以a?0a?0時，令b=a> 因為f(x)=cosax-ln(1-x2)=cos(ax)-ln(1-x2)=cosbx-ln(1-x2)，f(-x)=cos(-bx)-ln?1-(-x)2?=cosbx-ln(1-x2)=f(x)， 所以函數(shù)f(x)在定義域內(nèi)為偶函數(shù)f?(x)=-bsinbx

x2

,x?(-1,1)（i）當0<b2?2時，取m=min?1,1?x?(0m)，則bx?(0,1)

x(b2x2+2-b2由（1）可得f?(x)=-bsin(bx) >-b2x 且b2x2>02-b2?0,1-x2>0

x2

x2 1-所以f?(x)

x(b2x2+2-b21-

0即當x?(0m)?(0,1)時f￠x)0，則f(x)在(0m)上單調(diào)遞增，結(jié)合偶函數(shù)的對稱性可知：f(x)在(-m,0)上單調(diào)遞減，x=0f(x)（ⅱ）當b2>2時，取x??01??(0,1)，則bx?(0,1) b f?(x)=-bsinbx-

<-b(bx-b2x2)- (-b3x3+b2x2+b3x+2-b2x2，

x2 1-構建h(x)=-b3x3+b2x2+b3x+2-b2x??01? b h?(x)=-3b3x2+2b2x+b3x??01? b h?(0)=b3>0h??1?=b3-b>0，則h?(x)>0對"x??01??b b 可知h(x)在?01?上單調(diào)遞增，且h(0)=2-b2<0h?1?=2>0 b ?b 所以h(x)在?0,1?內(nèi)存在唯一的零點n??0,1? b b x?(0n)時，則h(x)<0x>0,1-x2>0f?(x)

1-

(-b3x3+b2x2+b3x+2-b2)<0即當x?(0n)?(0,1)時f?(x)<0，則f(x)在(0n)上單調(diào)遞減，結(jié)合偶函數(shù)的對稱性可知：f(x)在(-n,0)上單調(diào)遞增，x=0f(x)22綜上所述：b2>2，即a2>2，解得a 或a< 22a的取值范圍為(-?

2)

2,+?)1.當0<a2?2時，利用sinx<xx?(0,1)2.a2?2時，利用x-x2<sinxx?(0,1)，換元放縮2023年普通高等學校招生全國統(tǒng)一考試(新課標Ⅱ卷英語學本試卷共12頁。考試結(jié)束后,注意事項:1.答題前,考生先將自己的姓名、準考證號碼填寫清楚,將條形碼準選擇題必須使用2B鉛筆填涂;非選擇題必須使用0.5毫米黑色字跡的簽字筆書寫,字體工整、筆跡清楚。請按照題號順序在答題卡各題目的答題區(qū)域內(nèi)作答,超出答題區(qū)域書寫的答案無效;在草稿紙、試卷上答題無效。作圖可先使用鉛筆畫出,保持卡面清潔,不要折疊,不要弄破、弄皺,不準使用涂改液、修正帶、刮紙 聽力(1-20小題)在筆試結(jié)束后進行。 閱讀(共兩節(jié),滿分50分)第一節(jié)(共15小題;每小題2.5分,滿分37.5分閱讀下列短文，從每題所給的A、B、C、DAYellowstoneNationalParkoffersavarietyofrangerprogramsthroughoutthepark,andthroughouttheyear.Thefollowingaredescriptionsoftherangerprogramsthissummer.ExperiencingWildlifeinYellowstone(May26toSeptemberWhetheryou’rehikingabackcountrytrail(小徑),camping,orjustenjoyingthepark’samazingwildlifefromtheroad,thisquickworkshopisforyouandyourfamilyLearnwheretolookforanimalsandhowtosafelyenjoyyourwildlifewatchingexperience.MeetattheCanyonVillageJuniorRangerWildlifeOlympics(June5toAugustKidscantesttheirskillsandcomparetheirabilitiestotheanimalsofYellowstone.Stayforaslittleoraslongasyourplansallow.MeetinfrontoftheVisitorEducationCenter.CanyonTalksatArtistPoint(June9toSeptemberFromaclassicviewpoint,enjoyLowerFalls,theYellowstoneRiver,andthebreathtakingcolorsofthecanyon(峽谷)whilelearningaboutthearea’snaturalandhumanhistory.Discoverwhyartistsandphotographerscontinuetobedrawntothisspecialplace.MeetonthelowerplatformatArtistPointontheSouthRimDriveforthisshorttalk.PhotographyWorkshops(June19&JulyEnhanceyourphotographyskills—joinYellowstone’sparkphotographerforahands-onprogramtoinspirenewandcreativewaysofenjoyingthebeautyandwonderofYellowstone.6/19—Waterfalls&WideAngles:meetatArtistWhichofthefourprogramsbeginsthe B.JuniorRangerWildlifeC.CanyonTalksatArtistPoint. D.ExperiencingWildlifeinWhatistheshorttalkatArtistPointWorksoffamous B.ProtectionofwildC.Basicphotography D.HistoryofthecanyonWherewilltheparticipantsmeetfortheJuly10photographyArtist CCanyonVillage D.VisitorEducationBTurningsoil,pullingweeds,andharvestingcabbagesoundliketoughworkformiddleandhighschoolkids.Andatfirstitis,saysAbbyJaramillo,whowithanotherteacherstartedUrbanSprouts,aschoolgardenprogramatfourlow-incomeschools.Theprogramaimstohelpstudentsdevelopscienceskills,environmentalawareness,andhealthylifestyles.Jaramillo’sstudentsliveinneighborhoodswherefreshfoodandgreenspacearenoteasytofindandfastfoodrestaurantsoutnumbergrocerystores.“Thekidsliterallycometoschoolwithbagsofsnacksandlargebottlesofsoftdrinks,”shesays.“Theycometousthinkingvegetablesareawful,dirtisawful,insectsareawful.”Thoughsomeareinitiallyscaredoftheinsectsandturnedoffbythedirt,mostareeagertotrysomethingnew.UrbanSprouts’classesattwomiddleschoolsandtwohighschools,includehands-experimentssuchassoiltesting,flower-and-seeddissection,tastingsoffreshordriedproduce,andworkinthegarden.Severaltimesayear,studentscookthevegetablestheygrow,andtheyoccasionallymakesaladsfortheirentireschools.Programevaluationsshowthatkidseatmorevegetablesasaresultoftheclasses.“Wehavestudentswhosaytheywenthomeandtalkedtotheirparentsandnowthey’reeatingdifferently,”Jaramillosays.Sheaddsthattheprogram’sbenefitsgobeyondnutrition.Somestudentsgetsointerestedingardeningthattheybringhomeseedstostarttheirownvegetablegardens.Besides,workinginthegardenseemstohaveacalmingeffectonJaramillo’sspecialeducationstudents,manyofwhomhaveemotionalcontrolissues.“Theygetoutside,”shesays,“andtheyfeelsuccessful.”WhatdoweknowaboutAbbySheusedtobeahealth B.Shegrewupinalow-incomeC.Sheownsafastfood D.SheisaninitiatorofUrbanWhatwasaproblemfacingJaramilloatthestartoftheThekids’parentsdistrusted B.StudentshadlittletimeforherC.Somekidsdislikedgardenwork. D.TherewasnospaceforschoolWhichofthefollowingbestdescribestheimpactofthe B.C.Short- D.WhatcanbeasuitabletitlefortheRescuingSchool B.ExperiencingCountry D.ChangingLocalCReadingArt:ArtforBookLoversisacelebrationofaneverydayobject—thebook,representedhereinalmostthreehundredartworksfrommuseumsaroundtheworld.Theimageofthereaderappearsthroughouthistory,inartmadelongbeforebooksaswenowknowthemcameintobeing.Inartists’representationsofbooksandreading,weseemomentsofsharedhumanitythatgobeyondcultureandtime.Inthis“bookofbooks,”artworksareselectedandarrangedinawaythatemphasizestheseconnectionsbetweendifferenterasandculturesWeseescenesofchildrenlearningtoreadathomeoratschoolwiththebookasafocusforrelationsbetweenthegenerations.Adultsareportrayed描繪aloneinmanysettingsandposesabsorbedinavolume,deepinthoughtorlostinamomentofleisure.Thesescenesmayhavebeenpaintedhundredsofyearsago,buttheyrecordmomentswecanallrelateto.Booksthemselvesmaybeusedsymbolicallyinpaintingstodemonstratetheintellect(才智),wealthorfaithofthesubjectBeforethewideuseoftheprintingpress,booksweretreasuredobjectsandcouldbeworksofartintheirownright.Morerecently,asbookshavebecomeinexpensiveoreventhrowaway,artistshaveusedthemastherawmaterialforartworks—transformingcovers,pagesorevencompletevolumesintopaintingsandsculptures.Continueddevelopmentsincommunicationtechnologieswereoncebelievedtomaketheprintedpageoutdated.Froma21st-centurypointofview,theprintedbookiscertainlyancient,butitremainsasinteractiveasanybattery-powerede-reader.Toserveitsfunction,abookmustbeactivatedbyauser:thecoveropened,thepagesparted,thecontentsreviewed,perhapsnoteswrittendownorwordsunderlined.Andincontrasttoourincreasinglynetworkedliveswheretheinformationweconsumeismonitoredandtracked,aprintedbookstilloffersthechanceofawhollyprivate,“off-line”activity.WhereisthetextmostprobablytakenAnintroductiontoa B.AnessayontheartofC.Aguidebooktoa D.AreviewofmodernWhataretheselectedartworksWealthand B.HomeandC.Booksand D.WorkandWhatdotheunderlinedwords“relateto”inparagraph2 B.C. D.Whatdoestheauthorwanttosaybymentioningthee-Theprintedbookisnottotallyoutofdate.BTechnologyhaschangedthewayweC.Ourlivesinthe21stcenturyareD.PeoplenowrarelyhavethepatiencetoDAscitiesballoonwithgrowth,accesstonatureforpeoplelivinginurbanareasisbecominghardertofind.Ifyou’relucky,theremightbeapocketparknearwhereyoulive,butit’sunusualtofindplacesinacitythatarerelativelywild.Pastresearchhasfoundhealthandwellnessbenefitsofnatureforhumans,butanewstudyshowsthatwildnessinurbanareasisextremelyimportantforhumanwell-being.Theresearchteamfocusedonalargeurbanpark.Theysurveyedseveralhundredpark-goers,askingthemtosubmitawrittensummaryonlineofameaningfulinteractiontheyhadwithnatureinthepark.Theresearchersthenexaminedthesesubmissions,coding(編碼)experiencesintodifferentcategories.Forexample,oneparticipant’sexperienceof“Wesatandlistenedtothewavesatthebeachforawhile”wasassignedthecategories“sittingatbeach”and“l(fā)isteningtowaves.”Acrossthe320submissions,apatternofcategoriestheresearcherscalla“naturelanguage”begantoemerge.Afterthecodingofallsubmissions,halfadozencategorieswerenotedmostoftenasimportanttovisitors.Theseincludeencounteringwildlife,walkingalongtheedgeofwater,andfollowinganestablishedtrail.Namingeachnatureexperiencecreatesausablelanguage,whichhelpspeoplerecognizeandtakepartintheactivitiesthataremostsatisfyingandmeaningfultothem.Forexample,theexperienceofwalkingalongtheedgeofwatermightbesatisfyingforayoungprofessionalonaweekendhikeinthepark.Backdowntownduringaworkday,theycanenjoyamoredomesticformofthisinteractionbywalkingalongafountainontheirlunchbreak.“We’retryingtogeneratealanguagethathelpsbringthehuman-natureinteractionsbackintoourdailylives.Andforthattohappen,wealsoneedtoprotectnaturesothatwecaninteractwithit,”saidPeterKahn,aseniorauthorofthestudy.WhatphenomenondoestheauthordescribeatthebeginningofthePocketparksarenow B.WildnatureishardtofindinC.Manycitiesare D.PeopleenjoylivingclosetoWhydidtheresearcherscodeparticipantsubmissionsintoTocomparedifferenttypesofpark-goers. B.ToexplainwhytheparkattractsC.Toanalyzethemainfeaturesofthepark. D.Tofindpatternsinthevisitors’WhatcanwelearnfromtheexamplegiveninparagraphWalkingisthebestwaytogainaccesstoYoungpeoplearetoobusytointeractwithThesamenatureexperiencetakesdifferentforms.DThenaturelanguageenhancesworkperformance.WhatshouldbedonebeforewecaninteractwithnatureaccordingtoLanguage B.EnvironmentalC.Public D.Intercultural第二節(jié)(共5小題;每小題2.5分,滿分12.5分閱讀下面短文從短文后的選項中選出可以填入空白處的最佳選項。選項中有兩Asanartistwhosharesherjourneyonsocialmedia,I’moftenaskedbycuriousfollowershowtobeginanartjourney.Unfortunately,thereisnomagiclistIcanoffer.Idoremember,though,whatitwasliketobeacompletebeginner.SoI’veputtogethersomegoodtipsforstartinganartjourney.·StartsmallIsuggestusingasketchbook素描本forsmallstudiesThesesmallstudiesprovideinspirationandmaybeaspringboardformorecomplexworksinthefuture.16You’llwanttolookbackonyourjourneytoseehowfaryou’vecome.·Paintoftenandpaintfromlife.There’snobetterwaytoimprovethantoputinthosebrushmiles.Whetheryoupaintstilllifes,portraits,orlandscapes,paintfromlifeasmuchaspossible.·Continuallychallengeyourselftotrysomethingnew.18Artisticgrowthcanbeabitpainful.Welcometotheclub；we’veallbeenthere.Ilovetakingonchallenges.Ioncetookupachallengetocreateapaintingeverydayforamonthandposttheworksonline.·19Seekingandacceptingconstructivefeedback反饋iscrucialtogrowth.Ipostmyworkonsocialmediaand,inturn,havemetsomeofthekindestpeople.Theymakemefeelvaluedandrespected,nomattermylevelofartisticability.Thejourneyyou’reonwon’tfollowastraightpath.20Pushthrough,giveittimeandputintheeffort.Youwillharvesttherewardsofanartisticlife.GetoutofyourcomfortMakecareerplansandsetDon’tthrowawayyourbeginnerShareyourworkifyoufeelcomfortabledoingYou’llhitroadblocks,andyou’llfeeldiscouragedatEvaluateyourperformanceand,ifneeded,redefineyourYou’lldevelopthatpaintingmusclememorythatonlycomeswith第三部分語言運用(共兩節(jié),滿分30分第一節(jié)(共15小題;每小題1分,滿分15分,InAprillastyear,IsawapostonthePNP(PilotsNPaws)websitefromafamilyinTopeka.TheyhadtomovetoVirginiabuttheywereonaverytight .Theycouldnotaffordtopayfor fortheirdog,Tiffy,and wantedtotakeherwiththem.Itjust thatIwasplanninganotherPNPflightwithanotherpilot,Karen,who25totakeTiffyfromKansasCitytoVirginia.WhatIwastodowasflytoTopeka26WhenImetTiffy’sowners,theyseemedvery27.George,thehusband,wastryingtobecalm,butIcouldtellthiswas28forhim,havingtoleavehisdogtoa29andtrustthateverythingwould30.Aftersomegoodbyes,IaskedGeorgeandhiswifetohelpme31Tiffyintotheplane.IpromisedtotakecareofTiffyand32themassoonaswegottoKansasTheflightwas33,andTiffywasagreatpassenger.Thenextday, withKarenandmadeitbacktoGeorgeinVirginiawithinafewdays.Hewasso andsentmeanicee-mailwithpictures.ItfeltgreattoknowthatIhadhelpedbringthisfamilytogetheragain.21.A.B.C.22.A.B.C.23.A.B.C.24.A.B.C.D.25.A.B.C.D.26.A.seeB.lookC.handD.pick27.A.B.C.D.28.A.B.C.D.29.A.B.C.30.A.speedB.workC.comeD.take31.A.B.C.D.A. B. C. D.A.unnecessary B.unexpected C.unavoidable A. B. C. D.A.thankful B.generous C.proud 第二節(jié)(共10小題;每小題1.5分,滿分15分,WheneverItellpeoplethatIteachEnglishattheBerlinZoo,Ialmostalwaysgetaquestioninglook.Behindit,thepersonistryingtofigureoutwhoexactlyIteach…theanimals?SinceJune2017,rightbeforethe (arrive)ofthetwonewpandas,MengMengandJiaoQing,Ihavebeenhelpingthepandakeepersatthezootofeelmorecomfortableand (confidence)speakingEnglish.AndwhodotheyspeakEnglish Notthepandas,eventhough languageusedforthemedicaltraininginstructionsisactuallyEnglish.Theytalktothefloodofinternationaltouristsandto (visit)Chinesezookeeperswhooftencometocheckonthepandas,whichareonloanfromChina.Theyalsoneedtobereadytogive (interview)inEnglishwithinternationaljournalists.Thisis theyneedanEnglishtrainer.So,whataretheylearning? (basic),howtodescribeapanda’slife.It’sbeenanhonortowatchthepandaprogrammedevelop toseethepandassettleintotheirnewhome.Asalittlegirl,I (wish)tobeazookeeperwhenIgrewup.Now,I’mlivingoutthatdreamindirectlybyhelpingthepandakeepersdotheirjobinEnglish.第四部 寫作(共兩節(jié),滿分40分第一節(jié)(滿分15分 Ryan準備將學生隨機分為兩人一組，讓大家課后練習口語，你80DearI’mLiHuafromClass第二節(jié)(滿分25分

Li 閱讀下面材料,根據(jù)其內(nèi)容和所給段落開頭語續(xù)寫兩段,WhenIwasinmiddleschool,mysocialstudiesteacheraskedmetoenterawritingcontest.Isaidnowithoutthinking.Ididnotlovewriting.MyfamilycamefromBrazil,soEnglishwasonlymysecondlanguage.WritingwassodifficultandpainfulformethatmyteacherhadallowedmetopresentmypaperonthesinkingoftheTitanicbyactingoutaplay,whereIplayedalltheparts.Noonelaughedharderthanhedid.So,whydidhesuddenlyforcemetodosomethingatwhichIwassuretofail?His“BecauseIloveyourstories.Ifyou’rewillingtoapplyyourself,Ithinkyouhaveagoodshotatthis.”Encouragedbyhiswords,Iagreedtogiveitatry.IchosePaulRevere’shorseasmysubjectPaulReverewasasilversmith銀匠inBostonwhorodeahorseatnightonApril18,1775toLexingtontowarnpeoplethatBritishsoldierswerecoming.Mystorywouldcomestraightfromthehorse’smouth.Notabrilliantidea,butfunny;andunlikelytobeanyoneelse’schoice.Whatdidthehorsethink,ashespedthroughthenight?Didhegettired?Havedoubts?Didhewanttoquit?Isympathizedimmediately.Igottired.Ihaddoubts.Iwantedtoquit.But,likeRevere’shorse,Ikeptgoing.Iworkedhard.Icheckedmyspelling.Iaskedmyoldersistertocorrectmygrammar.IcheckedoutahalfdozenbooksonPaulReverefromthelibrary.IevenreadafewofWhenIhandedintheessaytomyteacher,hereadit,laughedoutloud,andsaid,“Great.Now,writeitagain.”Iwroteitagain,andagainandagain.WhenIfi