S考试题及答案

安徽省2022年高考·英语·考试真题与答案解析————————————————————————————————————————第一部分听力(共两节，满分30分)做题时，先将答案标在试卷上。

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第一节(共5小题;每小题1.5分，满分7.5分)听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

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1. What does the man want to do?A. Have breakfast.B. Take a walk.C. Call his office.2. What was George doing last night?A.Having a meeting.B. Flying home.C. Working on a project.3. Why does the man suggest going to the park?A. It’s big.B. It’s quiet.C. It’s new.4. How does the woman sound?A. Annoyed.B. Pleased.C. Puzzled.5. Where is the man’s table?A. Near the door.B. By the window.C. In the corner.第二节(共15小题;每小题1.5分，满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟;听完后，各小题将给出5秒钟的作答时间。

广东省2022年高考·数学·考试真题与答案解析————————————————————————————————————————一、选择题1. 若集合，则（）{4},{31}M xN x x =<=≥∣M N = A. B. C. D. {}02x x ≤<123x x ⎧⎫≤<⎨⎬⎩⎭{}316x x ≤<1163x x ⎧⎫≤<⎨⎬⎩⎭【答案】D【分析】求出集合后可求.,M N M N ⋂【详解】，故，1{16},{}3M xx N x x =≤<=≥∣0∣1163M N x x ⎧⎫=≤<⎨⎬⎩⎭故选：D2. 若，则（）i(1)1z -=z z +=A. B. C. 1 D. 22-1-【答案】D【分析】利用复数的除法可求，从而可求.z z z +【详解】由题设有，故，故，21i1i i iz -===-1+i z =()()1i 1i 2z z +=++-=故选：D3. 在中，点D 在边AB 上，．记，则（）ABC 2BD DA =CA m CD n == ，CB=A. B. C. D. 32m n - 23m n -+ 32m n + 23m n + 【答案】B【分析】根据几何条件以及平面向量的线性运算即可解出．【详解】因为点D 在边AB 上，，所以，即，2BD DA =2BD DA =()2CD CB CA CD -=- 所以．CB =3232CD CA n m -=- 23m n =-+故选：B ．4. 南水北调工程缓解了北方一些地区水资源短缺问题，其中一部分水蓄入某水库.已知该水库水位为海拔时，相应水面的面积为；水位为海拔时，相应水面的面积1485m ．21400km ．1575m ．为，将该水库在这两个水位间的形状看作一个棱台，则该水库水位从海拔上21800km ．1485m ．升到）（）1575m ． 2.65≈A. B. C. D. 931.010m ⨯931.210m ⨯931.410m ⨯931.610m ⨯【答案】C【分析】根据题意只要求出棱台的高，即可利用棱台的体积公式求出．【详解】依题意可知棱台的高为(m)，所以增加的水量即为棱台的体积．157.5148.59MN =-=V 棱台上底面积，下底面积，262140.014010S ==⨯km m 262180.018010S '==⨯km m∴((66119140101801033V h S S =+=⨯⨯⨯+⨯'．(()679933320109618 2.6510 1.43710 1.410(m )=⨯+⨯≈+⨯⨯=⨯≈⨯故选：C ．5. 从2至8的7个整数中随机取2个不同的数，则这2个数互质的概率为（）A.B.C.D. 16131223【答案】D【分析】由古典概型概率公式结合组合、列举法即可得解.【详解】从2至8的7个整数中随机取2个不同的数，共有种不同的取法，27C 21=若两数不互质，不同的取法有：，共7种，()()()()()()()2,4,2,6,2,8,3,6,4,6,4,8,6,8故所求概率.2172213P -==故选：D.6. 记函数的最小正周期为T ．若，且的图象关()sin (0)4f x x b πωω⎛⎫=++> ⎪⎝⎭23T ππ<<()y f x =于点中心对称，则（）3,22π⎛⎫⎪⎝⎭2f π⎛⎫= ⎪⎝⎭A. 1B.C.D. 33252【答案】A【分析】由三角函数的图象与性质可求得参数，进而可得函数解析式，代入即可得解.【详解】由函数的最小正周期T 满足，得，解得，23T ππ<<223πππω<<23ω<<又因为函数图象关于点对称，所以，且，3,22π⎛⎫⎪⎝⎭3,24k k Z ππωπ+=∈2b =所以，所以，，12,63k k Z ω=-+∈52ω=5()sin 224f x x π⎛⎫=++ ⎪⎝⎭所以.5sin 21244f πππ⎛⎫⎛⎫=++= ⎪ ⎪⎝⎭⎝⎭故选：A7. 设，则（）0.110.1e ,ln 0.99a b c ===-，A. B. C. D. a b c <<c b a <<c a b<<a c b<<【答案】C【分析】构造函数，导数判断其单调性，由此确定的大小.()ln(1)f x x x =+-,,a b c 【详解】设，因为，()ln(1)(1)f x x x x =+->-1()111x f x x x'=-=-++当时，，当时，(1,0)x ∈-()0f x '>,()0x ∈+∞()0f x '<所以函数在单调递减，在上单调递增，()ln(1)f x x x =+-(0,)+∞(1,0)-所以，所以，故，即，1()(0)09f f <=101ln 099-<110ln ln 0.999>=-b c >所以，所以，故，所以，1()(0)010f f -<=91ln +01010<1109e 10-<11011e 109<故，a b <设，则,()e ln(1)(01)xg x x x x =+-<<()()21e 11()+1e 11xx x g x x x x -+'=+=--令，，2()e (1)+1x h x x =-2()e (21)x h x x x '=+-当时，，函数单调递减，01x <<-()0h x '<2()e (1)+1x h x x =-时，，函数单调递增，11x -<<()0h x '>2()e (1)+1x h x x =-又，(0)0h =所以当时，，01x <<-()0h x <所以当时，，函数单调递增，01x <<-()0g x '>()e ln(1)xg x x x =+-所以，即，所以(0.1)(0)0g g >=0.10.1e ln 0.9>-a c >故选：C.8. 已知正四棱锥的侧棱长为l ，其各顶点都在同一球面上.若该球的体积为，且36π，则该正四棱锥体积的取值范围是（）3l ≤≤A. B. C. D. 8118,4⎡⎤⎢⎥⎣⎦2781,44⎡⎤⎢⎥⎣⎦2764,43⎡⎤⎢⎥⎣⎦[18,27]【答案】C【分析】设正四棱锥的高为，由球的截面性质列方程求出正四棱锥的底面边长与高的关系，h 由此确定正四棱锥体积的取值范围.【详解】∵球的体积为，所以球的半径,36π3R =设正四棱锥的底面边长为，高为，2a h 则，,2222l a h =+22232(3)a h =+-所以，26h l =2222a l h =-所以正四棱锥的体积，42622411214()=333366936l l l V Sh a h l l ⎛⎫==⨯⨯=⨯-⨯- ⎪⎝⎭所以，5233112449696l l V l l ⎛⎫⎛⎫-'=-= ⎪ ⎪⎝⎭⎝⎭当，当，3l ≤≤0V '>l <≤0V '<所以当时，正四棱锥的体积取最大值，最大值为，l =V 643又时，，,3l =274V =l =814V =所以正四棱锥的体积的最小值为，V 274所以该正四棱锥体积的取值范围是.276443⎡⎤⎢⎥⎣⎦，故选：C.二、不定项选择题9. 已知正方体，则（）1111ABCD A B C D -A. 直线与所成的角为 B. 直线与所成的角为1BC 1DA 90︒1BC 1CA 90︒C. 直线与平面所成的角为 D. 直线与平面ABCD 所成的角为1BC 11BB D D 45︒1BC 45︒【答案】ABD【分析】数形结合，依次对所给选项进行判断即可.【详解】如图，连接、，因为，所以直线与所成的角即为直线与1B C 1BC 11//DA B C 1BC 1B C 1BC 所成的角，1DA 因为四边形为正方形，则，故直线与所成的角为，A 正确；11BB C C 1B C ⊥1BC 1BC 1DA 90︒连接，因为平面，平面，则，1AC 11A B ⊥11BB C C 1BC ⊂11BB C C 111A B BC ⊥因为，，所以平面，1B C ⊥1BC 1111A B B C B = 1BC ⊥11A B C 又平面，所以，故B 正确；1AC ⊂11A B C 11BC CA ⊥连接，设，连接，11A C 1111A C B D O = BO 因为平面，平面，则，1BB ⊥1111D C B A 1C O ⊂1111D C B A 11C O B B ⊥因为，，所以平面，111C O B D ⊥1111B D B B B ⋂=1C O ⊥11BB D D 所以为直线与平面所成的角，1C BO ∠1BC 11BB D D设正方体棱长为，则，，11C O =1BC =1111sin 2C O C BO BC ∠==所以，直线与平面所成的角为，故C 错误；1BC 11BB D D 30 因为平面，所以为直线与平面所成的角，易得，1C C ⊥ABCD 1C BC ∠1BC ABCD 145C BC ∠=故D 正确.故选：ABD10. 已知函数，则（）3()1f x x x =-+A. 有两个极值点 B. 有三个零点()f x ()f x C. 点是曲线的对称中心 D. 直线是曲线的切线(0,1)()y f x =2y x =()y f x =【答案】AC【分析】利用极值点的定义可判断A ，结合的单调性、极值可判断B ，利用平移可判断()f x C ；利用导数的几何意义判断D.【详解】由题，，令得或()231f x x '=-()0f x '>x >x <令得，()0f x '<x <<所以在上单调递减，在，上单调递增，()fx ((,-∞)+∞所以是极值点，故A 正确；x =因，，，(10f =>10f =->()250f -=-<所以，函数在上有一个零点，()f x,⎛-∞ ⎝当时，，即函数在上无零点，x≥()0f x f ≥>()f x ⎫∞⎪⎪⎭+综上所述，函数有一个零点，故B 错误；()f x 令，该函数的定义域为，，3()h x x x =-R ()()()()33h x x x x x h x -=---=-+=-则是奇函数，是的对称中心，()h x (0,0)()h x将的图象向上移动一个单位得到的图象，()h x ()f x 所以点是曲线的对称中心，故C 正确；(0,1)()y f x =令，可得，又，()2312f x x '=-=1x =±()(1)11f f =-=当切点为时，切线方程为，当切点为时，切线方程为，(1,1)21y x =-(1,1)-23y x =+故D 错误.故选：AC .11. 已知O 为坐标原点，点在抛物线上，过点的直线交C 于(1,1)A 2:2(0)C x py p =>(0,1)B -P ，Q 两点，则（）A. C 的准线为B. 直线AB 与C 相切1y =-C. D. 2|OP OQ OA ⋅>2||||||BP BQ BA ⋅>【答案】BCD【分析】求出抛物线方程可判断A ，联立AB 与抛物线的方程求交点可判断B ，利用距离公式及弦长公式可判断C 、D.【详解】将点的代入抛物线方程得，所以抛物线方程为，故准线方程为A 12p =2x y =，A 错误；14y =-，所以直线的方程为，1(1)210AB k --==-AB 21y x =-联立，可得，解得，故B 正确；221y x x y=-⎧⎨=⎩2210x x -+=1x =设过的直线为，若直线与轴重合，则直线与抛物线只有一个交点，B l l y l C 所以，直线的斜率存在，设其方程为，，l 1y kx =-1122(,),(,)P x y Q x y 联立，得，21y kx x y=-⎧⎨=⎩210x kx -+=所以，所以或，，21212Δ401k x x k x x ⎧=->⎪+=⎨⎪=⎩2k >2k <-21212()1y y x x ==又，||OP ==||OQ ==所以，故C 正确；2||||||2||OP OQ k OA ⋅===>=因为，，1||||BP x =2||||BQ x =所以，而，故D 正确.2212||||(1)||15BP BQ k x x k ⋅=+=+>2||5BA =故选：BCD 12. 已知函数及其导函数的定义域均为，记，若，()f x ()'f x R ()()g x f x '=322f x ⎛⎫- ⎪⎝⎭(2)g x +均为偶函数，则（）A. B. C. D. (0)0f =102g ⎛⎫-= ⎪⎝⎭(1)(4)f f -=(1)(2)g g -=【答案】BC【分析】转化题设条件为函数的对称性，结合原函数与导函数图象的关系，根据函数的性质逐项判断即可得解.【详解】因为，均为偶函数，322f x ⎛⎫- ⎪⎝⎭(2)g x +所以即，，332222f x f x ⎛⎫⎛⎫-=+ ⎪ ⎪⎝⎭⎝⎭3322f x f x ⎛⎫⎛⎫-=+ ⎪ ⎪⎝⎭⎝⎭(2)(2)g x g x +=-所以，，则，故C 正确；()()3f x f x -=(4)()g x g x -=(1)(4)f f -=函数，的图象分别关于直线对称，()f x ()g x 3,22xx ==又，且函数可导，()()g x f x '=()f x 所以，()()30,32g g x g x ⎛⎫=-=- ⎪⎝⎭所以，所以，()(4)()3g x g x g x -==--()(2)(1)g x g x g x +=-+=所以，，故B 正确，D 错误；13022g g ⎛⎫⎛⎫-== ⎪ ⎪⎝⎭⎝⎭()()()112g g g -==-若函数满足题设条件，则函数（C 为常数）也满足题设条件，所以无法确定()f x ()f x C +()f x 的函数值，故A 错误.故选：BC.三、填空题：本题共4小题，每小题5分，共20分．13. 的展开式中的系数为________________（用数字作答）．81()y x y x ⎛⎫-+ ⎪⎝⎭26x y 【答案】-28【分析】可化为，结合二项式展开式的通项公式求解.()81y x y x ⎛⎫-+ ⎪⎝⎭()()88y x y x y x +-+【详解】因为，()()()8881=y y x y x y x y x x ⎛⎫-++-+ ⎪⎝⎭所以的展开式中含的项为，()81y x y x ⎛⎫-+ ⎪⎝⎭26x y 6265352688C 28y x y C x y x y x -=-的展开式中的系数为-28()81y x y x ⎛⎫-+ ⎪⎝⎭26x y 故答案为：-2814. 写出与圆和都相切的一条直线的方程________________．221x y +=22(3)(4)16x y -+-=【答案】或或3544y x =-+7252424y x =-1x =-【分析】先判断两圆位置关系，分情况讨论即可.【详解】圆的圆心为，半径为，圆的圆心为，221x y +=()0,0O 122(3)(4)16x y -+-=1O (3,4)半径为，4，等于两圆半径之和，故两圆外切，5=如图，当切线为l 时，因为，所以，设方程为143OO k =34l k =-3(0)4y x t t =-+>O 到l 的距离，解得，所以l 的方程为，1d ==54t =3544y x =-+当切线为m 时，设直线方程为，其中，，0kx y p ++=0p>0k <，解得，14⎧=⎪⎪7242524k p ⎧=-⎪⎪⎨⎪=⎪⎩7252424y x =-当切线为n 时，易知切线方程为，1x =-故答案为：或或.3544y x =-+7252424y x =-1x =-15. 若曲线有两条过坐标原点的切线，则a 的取值范围是________________．()e xy x a =+【答案】()(),40,∞∞--⋃+【分析】设出切点横坐标，利用导数的几何意义求得切线方程，根据切线经过原点得到关于0x 的方程，根据此方程应有两个不同的实数根，求得的取值范围.0x a 【详解】∵，∴，()e x y x a =+(1)e xy x a '=++设切点为,则,切线斜率,()00,x y ()000e x y x a =+()001e x k x a =++切线方程为：,()()()00000e 1e x x y x a x a x x -+=++-∵切线过原点，∴,()()()00000e 1e x x x a x a x -+=++-整理得：,2000x ax a +-=∵切线有两条，∴,解得或,240a a =+> 4a <-0a >∴的取值范围是,a ()(),40,∞∞--⋃+故答案为：()(),40,∞∞--⋃+16. 已知椭圆，C 的上顶点为A ，两个焦点为，，离心率为．过2222:1(0)x y C a b a b+=>>1F 2F 12且垂直于的直线与C 交于D ，E 两点，，则的周长是1F 2AF ||6DE =ADE ________________．【答案】13【分析】利用离心率得到椭圆的方程为，根据离心率得到222222213412043x y x y c c c+=+-=，即直线的斜率，进而利用直线的垂直关系得到直线的斜率，写出直线的方程：2AF DE DE，代入椭圆方程，整理化简得到：，利用x c =-22234120x y c +-=221390y c --=弦长公式求得，得，根据对称性将的周长转化为的周长，利用138c =1324a c ==ADE 2F DE △椭圆的定义得到周长为.413a =【详解】∵椭圆的离心率为，∴，∴，∴椭圆的方程为12c e a ==2a c =22223b a c c =-=，不妨设左焦点为，右焦点为，如图所示，∵222222213412043x y x y c c c+=+-=，即1F 2F ，∴，∴为正三角形，∵过且垂直于的直222AF a OF c a c ===，，23AF O π∠=12AF F △1F 2AF线与C 交于D ，E 两点，为线段的垂直平分线，∴直线斜率倒数为DE 2AF DE直线的方程：，代入椭圆方程，整理化简得到：DE x c =-22234120x y c +-=，221390y c --=判别式，()22224139616c c =+⨯⨯=⨯⨯∴，22264613cCD y =-==⨯⨯⨯=∴ ， 得， 138c =1324a c ==∵为线段的垂直平分线，根据对称性，，∴的周长等于DE 2AF 22AD DF AE EF ==，ADE 的周长，利用椭圆的定义得到周长为2F DE △2F DE △.222211*********DF EF DE DF EF DF EF DF DF EF EF a a a ++=+++=+++=+==故答案为：13.四、解答题本题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．17. 记为数列的前n 项和，已知是公差为的等差数列．n S {}n a 11,n n S a a ⎧⎫=⎨⎬⎩⎭13（1）求的通项公式；{}n a （2）证明：．121112na a a +++< 【答案】（1）()12n n n a +=（2）见解析【分析】（1）利用等差数列的通项公式求得，得到，利()121133n nS n n a +=+-=()23n n n a S +=用和与项的关系得到当时，,进而得：，利2n ≥()()112133n n n n n n a n a a S S --++=-=-111n n a n a n -+=-用累乘法求得，检验对于也成立，得到的通项公式；()12n n n a +=1n ={}n a ()12n n n a +=（2）由（1）的结论，利用裂项求和法得到，进而证得.121111211n a a a n ⎛⎫+++=- ⎪+⎝⎭【小问1详解】∵，∴,∴,11a =111S a ==111S a =又∵是公差为的等差数列，n n S a ⎧⎫⎨⎬⎩⎭13∴,∴,()121133n nS n n a +=+-=()23nn n a S +=∴当时，，2n ≥()1113n n n a S --+=∴,()()112133n n n n n n a n a a S S --++=-=-整理得：,()()111n n n a n a --=+即,111n n a n a n -+=-∴31211221n n n n n a a a a a a a a a a ---=⨯⨯⨯⋯⨯⨯，()1341123212n n n n n n ++=⨯⨯⨯⋯⨯⨯=--显然对于也成立，1n =∴的通项公式；{}n a ()12n n n a +=【小问2详解】()12112,11n a n n n n ⎛⎫==- ⎪++⎝⎭∴12111n a a a +++ 1111112121222311n n n ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=-+-+-=-< ⎪ ⎪ ⎪ ⎪⎢⎥++⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦ 18. 记的内角A ，B ，C 的对边分别为a ，b ，c ，已知．ABC cos sin 21sin 1cos2A BA B=++（1）若，求B ；23C π=（2）求的最小值．222a b c +【答案】（1）；π6（2）．5-【分析】（1）根据二倍角公式以及两角差的余弦公式可将化成cos sin 21sin 1cos2A BA B=++，再结合，即可求出；()cos sin A B B +=π02B <<（2）由（1）知，，，再利用正弦定理以及二倍角公式将化成π2C B =+π22A B =-222a b c +，然后利用基本不等式即可解出．2224cos 5cos B B+-【小问1详解】因为，即2cos sin 22sin cos sin 1sin 1cos 22cos cos A B B B BA B B B===++，()1sin cos cos sin sin cos cos 2B A B A B A BC =-=+=-=而，所以；π02B <<π6B =【小问2详解】由（1）知，，所以，sin cos 0B C =->πππ,022C B <<<<而，πsin cos sin 2B C C ⎛⎫=-=- ⎪⎝⎭所以，即有．π2C B =+π22A B =-所以222222222sin sin cos 21cos sin cos a b A B B B c C B+++-==．()2222222cos 11cos 24cos 555cos cos B BB BB-+-==+-≥-=-当且仅当的最小值为．2cos B =222a b c +5-19. 如图，直三棱柱的体积为4，的面积为．111ABC A B C -1A BC（1）求A 到平面的距离；1A BC （2）设D 为的中点，，平面平面，求二面角的正弦1AC 1AA AB =1A BC ⊥11ABB A A BD C --值．【答案】（1）（2【分析】（1）由等体积法运算即可得解；（2）由面面垂直的性质及判定可得平面，建立空间直角坐标系，利用空间向量BC ⊥11ABB A 法即可得解.【小问1详解】在直三棱柱中，设点A 到平面的距离为h ，111ABC A B C -1A BC则，111111111143333A A BC A A ABC A ABC A B BC C C B V S h V S A A V ---=⋅===⋅==解得，h =所以点A 到平面；1A BC 【小问2详解】取的中点E ,连接AE ,如图，因为，所以,1A B 1AA AB =1AE A B ⊥又平面平面，平面平面，1A BC ⊥11ABB A 1A BC 111ABB A A B =且平面，所以平面，AE ⊂11ABB A AE ⊥1A BC 在直三棱柱中，平面，111ABC A B C -1BB ⊥ABC由平面，平面可得，，BC ⊂1A BC BC ⊂ABC AE BC ⊥1BB BC ⊥又平面且相交，所以平面，1,AE BB ⊂11ABB A BC ⊥11ABB A 所以两两垂直，以B 为原点，建立空间直角坐标系，如图，1,,BC BABB 由（1）得，，AE =12AA AB ==1A B =2BC =则,所以的中点，()()()()10,2,0,0,2,2,0,0,0,2,0,0A A B C 1AC ()1,1,1D 则，,()1,1,1BD = ()()0,2,0,2,0,0BA BC ==设平面的一个法向量，则，ABD (),,m x y z = 020m BD x y z m BA y ⎧⋅=++=⎪⎨⋅==⎪⎩ 可取，()1,0,1m =-设平面的一个法向量，则，BDC (),,n a b c = 020m BD a b c m BC a ⎧⋅=++=⎪⎨⋅==⎪⎩ 可取，()0,1,1n =-r则，1cos ,2m n m n m n⋅===⋅所以二面角.A BD C --=20. 一医疗团队为研究某地的一种地方性疾病与当地居民的卫生习惯（卫生习惯分为良好和不够良好两类）的关系，在已患该疾病的病例中随机调查了100例（称为病例组），同时在未患该疾病的人群中随机调查了100人（称为对照组），得到如下数据：不够良好良好病例组4060对照组1090（1）能否有99%的把握认为患该疾病群体与未患该疾病群体的卫生习惯有差异？（2）从该地的人群中任选一人，A 表示事件“选到的人卫生习惯不够良好”，B 表示事件“选到的人患有该疾病”．与的比值是卫生习惯不够良好对患该疾病风险程度的(|)(|)P B A P B A (|)(|)P B A P B A 一项度量指标，记该指标为R ．（ⅰ）证明：；(|)(|)(|)(|)P A B P A B R P A B P A B =⋅（ⅱ）利用该调查数据，给出的估计值，并利用（ⅰ）的结果给出R 的估计(|),(|)P A B P A B 值．附，22()()()()()n ad bc K a b c d a c b d -=++++()2P K k ≥0.0500.0100.001k3.8416.63510.828【答案】（1）答案见解析（2）（i ）证明见解析；(ii)；6R =【分析】(1)由所给数据结合公式求出的值，将其与临界值比较大小，由此确定是否有99%2K 的把握认为患该疾病群体与未黄该疾病群体的卫生习惯有差异；(2)(i)根据定义结合条件概率公式即可完成证明；(ii)根据（i ）结合已知数据求.R 【小问1详解】由已知，222()200(40906010)=24()()()()50150100100n ad bc K a b c d a c b d -⨯-⨯==++++⨯⨯⨯又，，2( 6.635)=0.01P K ≥24 6.635>所以有99%的把握认为患该疾病群体与未患该疾病群体的卫生习惯有差异.【小问2详解】(i)因为，(|)(|)()()()()=(|)(|)()()()()P B A P B A P AB P A P AB P A R P B A P B A P A P AB P A P AB =⋅⋅⋅⋅所以()()()()()()()()P AB P B P AB P B R P B P AB P B P AB =⋅⋅⋅所以，(|)(|)(|)(|)P A B P A B R P A B P A B =⋅(ii)由已知，，40(|)100P A B =10(|)100P A B =又，，60(|)100P A B =90(|)100P A B =所以(|)(|)=6(|)(|)P A B P A B R P A B P A B =⋅21. 已知点在双曲线上，直线l 交C 于P ，Q 两点，直线(2,1)A 2222:1(1)1x yC a a a -=>-,AP AQ 的斜率之和为0．（1）求l 的斜率；（2）若，求的面积．tan PAQ ∠=PAQ △【答案】（1）；1-（2．【分析】（1）由点在双曲线上可求出，易知直线l 的斜率存在，设，(2,1)A a :l y kx m =+，再根据，即可解出l 的斜率；()()1122,,,P x y Q x y 0AP BP k k +=（2）根据直线的斜率之和为0可知直线的倾斜角互补，再根据,AP AQ ,AP AQ即可求出直线的斜率，再分别联立直线与双曲线方程求出点tan PAQ ∠=,AP AQ ,AP AQ 的坐标，即可得到直线的方程以及的长，由点到直线的距离公式求出点到直线,P Q PQ PQ A PQ 的距离，即可得出的面积．PAQ △【小问1详解】因为点在双曲线上，所以，解得，即双曲线(2,1)A 2222:1(1)1x yC a a a -=>-224111a a -=-22a =22:12x C y -=易知直线l 的斜率存在，设，，:l y kx m =+()()1122,,,P x y Q x y 联立可得，，2212y kx m x y =+⎧⎪⎨-=⎪⎩()222124220k x mkx m ----=所以，，．2121222422,2121mk m x x x x k k ++=-=--()()22222216422210120m k m k m k ∆=++->⇒-+>所以由可得，，0AP BP k k +=212111022y y x x --+=--即，()()()()122121210x kx m x kx m -+-+-+-=即，()()()1212212410kx x m k x x m +--+--=所以，()()2222242124102121m mk k m k m k k +⎛⎫⨯+-----= ⎪--⎝⎭化简得，，即，()2844410k k m k +-++=()()1210k k m +-+=所以或，1k =-12m k =-当时，直线过点，与题意不符，舍去，12m k =-():21l y kx m k x =+=-+()2,1A 故．1k =-【小问2详解】不妨设直线的倾斜角为，因为，所以，,PA PB (),αβαβ<0AP BP k k +=παβ+=因为，所以，即，tan PAQ ∠=()tan βα-=tan 2α=-，解得，2tan 0αα-=tan α=于是，直线，直线，):21PA y x =-+):21PB y x =-+联立可得，，)222112y x x y ⎧=-+⎪⎨-=⎪⎩(23211002x x +-+-=因为方程有一个根为，所以，，2P x =P y =同理可得，Q x =Q y =所以，，5:03PQ x y +-=163PQ =点到直线的距离，A PQ d故的面积为．PAQ △11623⨯=22. 已知函数和有相同的最小值．()xf x e ax =-()lng x ax x =-（1）求a ；（2）证明：存在直线，其与两条曲线和共有三个不同的交点，并且从左y b =()y f x =()y g x =到右的三个交点的横坐标成等差数列．【答案】（1）1a =（2）见解析【分析】（1）根据导数可得函数的单调性，从而可得相应的最小值，根据最小值相等可求a.注意分类讨论.（2）根据（1）可得当时，的解的个数、的解的个数均为2，构建新1b >e x x b -=ln x x b -=函数，利用导数可得该函数只有一个零点且可得的大小关系，()e ln 2xh x x x =+-()(),f x g x 根据存在直线与曲线、有三个不同的交点可得的取值，再根据两类y b =()y f x =()y g x =b 方程的根的关系可证明三根成等差数列.【小问1详解】的定义域为，而，()e x f x ax =-R ()e '=-x f x a 若，则，此时无最小值，故.0a ≤()0f x '>()f x 0a >的定义域为，而.()ln g x ax x =-()0,+∞11()ax g x a x x'-=-=当时，，故在上为减函数，ln x a <()0f x '<()f x (),ln a -∞当时，，故在上为增函数，ln x a >()0f x '>()f x ()ln ,a +∞故.()min ()ln ln f x f a a a a ==-当时，，故在上为减函数，10x a <<()0g x '<()g x 10,a ⎛⎫ ⎪⎝⎭当时，，故在上为增函数，1x a >()0g x '>()g x 1,a ⎛⎫+∞ ⎪⎝⎭故.min 11()1ln g x g a a ⎛⎫==- ⎪⎝⎭因为和有相同的最小值，()e x f x ax =-()ln g x ax x =-故，整理得到，其中，11ln ln a a a a -=-1ln 1a a a-=+0a >设，则，()1ln ,01a g a a a a -=->+()()()222211011a g a a a a a --'=-=≤++故为上的减函数，而，()g a ()0,+∞()10g =故的唯一解为，故的解为.()0g a =1a =1ln 1a a a -=+1a =综上，.1a =【小问2详解】由（1）可得和的最小值为.e ()x x f x =-()ln g x x x =-11ln11ln 11-=-=当时，考虑的解的个数、的解的个数.1b >e x x b -=ln x x b -=设，，()e x S x x b =--()e 1x S x '=-当时，，当时，，0x <()0S x '<0x >()0S x '>故在上为减函数，在上为增函数，()S x (),0-∞()0,+∞所以，()()min 010S x S b ==-<而，，()e 0b S b --=>()e 2b S b b =-设，其中，则，()e 2b u b b =-1b >()e 20b u b '=->故在上为增函数，故，()u b ()1,+∞()()1e 20u b u >=->故，故有两个不同的零点，即的解的个数为2.()0S b >()e x S x x b =--e x x b -=设，，()ln T x x x b =--()1x T x x-'=当时，，当时，，01x <<()0T x ¢<1x >()0T x '>故在上为减函数，在上为增函数，()T x ()0,1()1,+∞所以，()()min 110T x T b ==-<而，，()e e 0b b T --=>()e e 20b b T b =->有两个不同的零点即的解的个数为2.()ln T x x x b =--ln x x b -=当，由（1）讨论可得、仅有一个零点，1b =ln x x b -=e x x b -=当时，由（1）讨论可得、均无零点，1b <ln x x b -=e x x b -=故若存在直线与曲线、有三个不同的交点，y b =()y f x =()y g x =则.1b >设，其中，故，()e ln 2x h x x x =+-0x >1()e 2x h x x'=+-设，，则，()e 1x s x x =--0x >()e 10x s x '=->故在上为增函数，故即，()s x ()0,+∞()()00s x s >=e 1x x >+所以，所以在上为增函数，1()1210h x x x'>+-≥->()h x ()0,+∞而，，(1)e 20h =->31e 333122()e 3e 30e e eh =--<--<故在上有且只有一个零点，且：()h x ()0,+∞0x 0311ex <<当时，即即，00x x <<()0h x <e ln x x x x -<-()()f x g x <当时，即即，0x x >()0h x >e ln x x x x ->-()()f x g x >因此若存在直线与曲线、有三个不同的交点，y b =()y f x =()y g x =故，()()001b f x g x ==>此时有两个不同的零点，e x x b -=1010,(0)x x x x <<此时有两个不同的零点，ln x x b -=0404,(01)x x x x <<<故，，，11e x x b -=00e x x b -=44ln 0x x b --=00ln 0x x b --=所以即即，44ln x b x -=44e x b x -=()44e 0x b x b b ----=故为方程的解，同理也为方程的解4x b -e x x b -=0x b -e x x b -=又可化为即即，11e x x b -=11e x x b =+()11ln 0x x b -+=()()11ln 0x b x b b +-+-=故为方程的解，同理也为方程的解，1x b +ln x x b -=0x b +ln x x b -=所以，而，{}{}1004,,x x x b x b =--1b >故即.0410x x b x x b =-⎧⎨=-⎩1402x x x +=。

普通话考试标准模拟试题及答案普通话考试标准模拟试题及答案普通话就是以北京语音为标准音，以北方话为基础方言，以典型的现代白话文著作为语法规范的现代民族共同语。

以下是店铺整理的普通话考试标准模拟试题及答案，欢迎学习!普通话考试标准模拟试题及答案篇1一、读单音节字词(100个音节，共10分，限时3.5分钟)贼(zéi) 列(liè) 枕(zhěn) 次(cì) 聋(lónɡ) 饼(bǐnɡ) 日(rì) 谨(jǐn) 裙(qún) 绢(juàn) 值(zhí) 冯(fénɡ) 炯(jiǒnɡ)咸(xián) 呆(dāi) 卤(lǔ) 僧(sēnɡ) 扭(niǔ) 肾(shèn) 抓(zhuā) 盆(pén) 战(zhàn) 耳(ěr) 基(jī) 丑(chǒu) 凝(nínɡ)免(miǎn) 外(wài) 穷(qiónɡ) 陋(lòu) 春(chūn) 昂(ánɡ) 喘(chuǎn) 滨(bīn) 娘(niánɡ) 方(fānɡ) 购(ɡòu) 仍(rénɡ)睡(shuì) 跟(ɡēn) 环(huán) 浮(fú) 擦(cā) 快(kuài) 滑(huá) 渺(miǎo) 疆(jiānɡ) 台(tái) 醒(xǐnɡ) 秘(mì)坑(kēnɡ) 善(shàn) 允(yǔn) 逛(ɡuànɡ) 甩(shuǎi) 照(zhào) 拨(bō) 叠(dié) 翁(wēnɡ) 床(chuánɡ) 舜(shùn) 肿(zhǒnɡ)俗(sú) 腭(è) 牌(pái) 骚(sāo) 雪(xuě) 批(pī) 洒(sǎ) 锌(xīn) 瑞(ruì) 锅(ɡuō) 垒(lěi) 休(xiū) 谈(tán) 目(mù)犬(quǎn) 榻(tà) 窝(wō) 举(jǔ) 纵(zònɡ) 黑(hēi) 瘸(qué) 掏(tāo) 挪(nuó) 惹(rě) 贝(bèi) 哑(yǎ) 奏(zòu) 席(xí)掐(qiā) 榆(yú) 餐(cān) 字(zì) 考(kǎo) 编(biān) 滚(ɡǔn) 叼(diāo) 法(fǎ)二、读多音节词语(100个音节，共20分，限时2.5分钟)倘(tánɡ)使(shǐ) 苍(cānɡ)翠(cuì) 强(qiǎnɡ)求(qiú) 蒙(měnɡ)古(ɡǔ)包(bāo) 从(cónɡ)而(ér) 粉(fěn)末(mò)儿(ér)旋(xuán)转(zhuǎn) 情(qínɡ)怀(huái) 合(hé)同(tonɡ) 财(cái)产(chǎn) 手(shóu)脚(jiǎo) 灭(miè)亡(wánɡ)起(qǐ)飞(fēi) 跨(kuà)越(yuè) 挂(ɡuà)念(niàn) 佛(fó)经(jīnɡ) 高(ɡāo)傲(ào) 柴(chái)火(huo) 亏(kuī)损(sǔn) 犯(fàn)罪(zuì) 决(jué)议(yì) 耽(dān)误(wu) 增(zēnɡ)加(jiā) 作(zuò)用(yònɡ) 难(nán)怪(ɡuài) 少(shǎo)女(nǚ) 个(ɡè)体(tǐ)上(shànɡ)下(xià)危(wēi)害(hài)荒(huānɡ)谬(miù)斥(chì)责(zé)撇(piē)开(kāi)砂(shā)轮(lún)儿(ér) 原(yuán)料(liào)东(dōnɡ)欧(ōu) 侵(qīn)略(lüè) 大(dà)多(duō)数(shù) 思(sī)想(xiǎnɡ) 本(běn)子(zi) 状(zhuànɡ)况(kuànɡ)柔(róu)软(ruǎn) 训(xùn)练(liàn) 药(yào)品(pǐn) 政(zhènɡ)党(dǎnɡ) 蒜(suàn)瓣(bàn)儿(ér) 定(dìnɡ)律(lǜ)英(yīnɡ)雄(xiónɡ) 人(rén)均(jūn) 没(méi)谱(pǔ)儿(ér)三、朗读短文(400个音节，共30分，限时4分钟) 作品30号其实你在很久以前并不喜欢牡丹，因为它总被人作为富贵膜拜。

浙江省嘉兴市2022年中考·英语·考试真题与答案解析一、完形填空阅读下面短文,然后从各题所给的四个选项中选出一个最佳答案。

When you hear the world hero, you may think of someone like Mahatma Gandhi or Nelson Mandela. But there are also "everyday" heroes-they may not be ___1___ to the public, but they do what they can to help ___2___ people's lives as well. Everyday heroes are normal people who are working to make the world a better place. ___3___ them is Robert Lee.Lee grew up in New York City. ___4___ his family was not rich, he learned from a young age the importance of ___5___ the waste of food. When in college, he was part of a student group that gave out leftover（剩余的）food to ___6___ people. This experience made him realize how ___7___ the problem of food wastage actually was.After he left university, Lee and a friend set up a（n）___8___ called Rescuing Leftover Cuisine（RLC）. Its purpose is to collect ___9___ food from restaurants and send it for free to shelters（收容所）and food kitchens. Volunteers ____10____ the food from restaurants around the city and hand it out to those in need.RLC operates ____11____. Using an app developed by Lee's team, restaurants ____12____ how much leftover food they have each day. Then volunteers are sent to fetch it. The ____13____ is usually short, so volunteers can just do their deliveries （递送）by ____14____ from the restaurants to the shelters. Lee believes this is a quick way of delivery.So far, Lee and his team have successfully ____15____ over 150,000 kilograms of food, serving almost 300,000 ____16____ to people who need them. Lee has shown that a ____17____ action can help people around us greatly. ____18____ may have a greater influence than we imagine. When Lee is asked _____19_____ his next plan is,he says there's still a lot to do. "It's just the ____20____. With more restaurants, who knows how much more we can do."1. A. safe B. close C. friendly D. well-known2. A. record B. share C. improve D. manage3. A. Under B. Beside C. Among D. Opposite4. A. If B. As C. When D. Though5. A. storing B. avoiding C. burning D. recycling6. A. sick B. blind C. lonely D. homeless7. A. easy B. serious C. funny D. strange8. A. club B. school C. business D. organization9. A. cold B. cheap C. unsold D. uncooked10. A. pick up B. go over C. pay for D. throw away11. A. badly B. quietly C. simply D. slowly12. A. report B. notice C. discuss D. consider13. A. visit B. break C. meeting D. distance14. A. flying B. walking C. skating D. shipping15. A. saved B. heated C. lent D. grown16. A. jobs B. meals C. rooms D. seats17. A. small B. secret C. direct D. social18. A. It B. He C. You D. They19. A. who B. when C. what D. where20. A. need B. time C. promise D. beginning二、阅读理解AFUTURE VIEWI think it will be great if they invent somethingso that we only have to sleep for one or two hours every day. Then we’d all have much more time to do things and to enjoy ourselves.Life would be better, I think, and everyone would do a lot more with their lives.MiguelIf I could choose anything, I’d go for a timemachine so that I could go back and do some things differently. Of course, It will be great if there that’s impossible— but wouldn't it be great if it was possible? I wish I could go back in time to when I was a kid and say some of the things in a different way!HannahWell, of course, the best thing is medicine for rally bad diseases（疾病）.A lot of scientists are working very hard to stop diseases like Ebola, a terrible disease that has killed many thousand people in Africa. So if they found a special treatment for Ebola, or a way of completely preventing it, life would be easier in so many places.Isabella It will be great if there is some kind of fuel(燃料）we can use in cars that won’t produce any pollution. I guess there are scientists right now trying to do that, and I hope they’ll succeed because the world would be a much cleaner place, wouldn’t it? CharlieI wish there was a machine that did homework! Wouldn’t tat be fantastic? But I guess teachers wouldn’t be very happy.Georgina21. Why does Hannah want to go back in time?A. To be young again.B. To live a better life.C. To do something for Africans.D. To say some words differently.22. Who might be worried about the pollution of the world?A. Miguel.B. Isabella.C. Charlie.D. Georgina.23. In which part of a newspaper may this passage appear?A. Health.B. Science.C. Travel.D. Education.BIs there a connection between music and language? According to recent studies, the answer is yes: Music helps develop certain language abilities in the brain.A study from Northwestern University shows that playing a musical instrument can improve a person's hearing ability. As a part of the study, two groups of people listened to a person talking in a noisy room. The people in the first group were musicians, while those in the second group had no musical training. The musicians were found to be able to hear the talking person more clearly.Musicians hear better,says study leader Nina Kraus, because they learn to pay attention to certain sounds. For example, when the violinists play in an orchestra（管弦乐队）, they must listen closely to what they are playing and ignore(忽视)the other sounds. In this way, musicians can hear certain sounds better, even in a room with lots of noise.Gotfried Schlaug, a doctor at Harvard Medical School, works with stroke(中风)patients.Because of their illness, these people cannot say their names, addresses, or other information normally However, they can till sing. Dr. Schlaug was surprised to find singing words helped his patients finally speak. Why does this work? Schlaug isn't sure. Music seems to be able to make different parts of the brain active, including the bad parts. In some way, this helps patients use those parts of the brain again.So music is not only enjoyable, but also good for us in many other ways. Playing an instrument or singing, says Nina Kraus, can help us do better in school and keep our brain sharp as we get older.24. What were the people in the first group asked to do in the study of Northwestern University?A. Listen to different kinds of music.B. Make some noise in an empty room.C. Give the second group musical training.D. Find out what the talking person was saying.25. The writer uses the example of violinists in Paragraph 3 to explain________.A. how musicians play in an orchestraB. how musicians ignore certain soundsC. why musicians have better hearing abilityD. why musicians are not afraid of loud noise26. Gottfried Schlaug helps his stroke patients speak by________.A. playing music for themB. getting them to sing wordsC. teaching them to play instrumentsD. joining the bad parts of their brain27. What is the main purpose of the passage?A. To introduce different ways of playing music.B. To give some advice about making our brain active.C. To show the relationship between music and language abilities.D. To share the results of recent studies on treating stroke patients.CIt's late evening in the middle of the winter. The snow is deep, but off I go into the wild.Suddenly, I hear it: the howl（嚎叫)of wolves. Excitedly, I walk faster in the direction of the howling.I've been working as a volunteer wolf tracker in Yellowstone National Park for over a year It's my job to follow wolves to record information about them. Like wolves,however, I don't work alone. I'm one of a team of people who have been in a project to help reintroduce wolves to the park.Humans have always feared wolves. From fairy tales to movies, the wolf is always the bad guy —a danger to humans and other animals. In the 1920s, the organizers of the park shared this view about wolves and made the decision to drive them away from the park. As a result, the number of elk （麋鹿）in the park increased. The elk started looking for food near river banks since they didn't have to worry about wolves. This, in turn, stopped trees from growing along the rivers, doing great harm to the environment there.The return of wolves has already made a big difference. To begin with, the elk soon learned to stay away from the rivers, so new trees have been able to grow there. The trees become home to new wildlife, such as birds, and provide food for more animals.The park is now becoming more and more popular with those who want to see wolves in their natural environment. It has also helped the local people make more money. Tourists visit the area, stay in the local hotels, and eat in the local restaurants.Looking at the wild animals and the beautiful sceneries around me, I am amazed at the changes the wolf has made to Yellowstone. And happily, it is playing a good guy in this story.28. According to the passage, one of a wolf tracker’s jobs is to________.A. catch wolves in the parkB. feed wolves in the wildC. stop wolves from killing elkD. collect information about wolves29. Why did the organizers drive the wolves away from the park in the 1920s?A. The wolves were always howling at night.B. The elk in the park were in danger of dying out.C. The wolves had done great harm to the environment.D. The park was thought to be unsafe with wolves around.30. Paragraphs 4 and 5 are mainly about the _______ that wolves have brought to Yellowstone.A. risksB. troublesC. changesD. competitions31. What does the writer mean by the underlined sentence in the last paragraph?A. Wolves are important to Yellowstone.B. Yellowstone has been home to wolves.C. The wolf and the elk live in peace now.D. The number of wolves is growing fast.DWhenever I’m invited to somebody’s house for the first time, I’ll take a close look at their bookshelves. What they keep on their shelves, in my eyes, is a reflection(反映) of their personal tastes and interests. The books on a bookshelf tell the story of a life. My own shelves are a kind of roadmap through my life, telling a chronological tale. First up come my favorite childhood reads, like Watership Down and His Dark Materials. The collection then moves on to teenage things I read in school — Nineteen Eighty-Four, The Handmaid’s Tale and A Clockwork Orange. Finally, we come out of full-time education and enter the real world, and this part is a reflection of my professional(职业的) life, from the first exciting days of work experience to the present.If you pointed to any book on my shelves, I could name the year I got it, where I was working, and probably where I was living at the time.__▲It is an easy way to get to know about his hopes, dreams and interests. And a big part of that story is how we arrange(排列) our books, and get on with them. It’s such a personal thing, different from each other. Some want to be with their books in every room, others have them on shelves, behind glass or simply lining their floors. Each home has a story to tell through the way they live with their books.It’s often said that what we read says something about who we are. The way we fill our bookshelves might also send a message about ourselves. From the pile of bookson your bedside table to your well-organized shelves, whether we notice it or not, our book collections are telling a story before a single page has been turned.32. The books on the writer’s own shelves are organized according to _________.A. their pricesB. their writersC. his personal interestsD. the periods of his life33. Which of the following is the best sentence to fill in the blank in Paragraph 4?A. Every book is worth reading.B. Each bookshelf holds a special dream.C. Book collections tell so much about a person.D. The books a person reads depend on his choices.34. The writer seems to agree that ____________.A. we’d better not put our books on the floorB. the way we arrange our books shows our tastesC. the books at our bedside are usually more importantD. we can know about a person by reading the message he sends35. It can be learned from the passage that the writer is probably__________.A. a book loverB. a story tellerC. a shelf collectorD. a fashion designerEIt was last January, when I’d just finished a charity(慈善) and was on the train back home, that I put down my phone and started writing thank-you notes to people who had helped.When I got off the train, I felt amazingly good. The next day, I wrote more thank-yous and the same feeling of happiness hit me again. I suddenly had the idea: Why not keep on doing this for every day of the year?To keep on task, I decided to pick out a different theme for each month. January was charity.February would be neighbors, I decided. And I thought of a number of names rightaway: the owner of our local bookstore, who let me and my little son in before the store opened and offered to play his favorite songs; our babysitter, who dropped off a bag of old board games for our kids to play; …While writing the notes, I realized how often I had spent my time on the phone moving from app to app, appreciating(欣赏) other people’s lives. Writing thank-you notes allowed me the time to do something different, paying more attention to my own life.In the following months, I wrote to my friends, doctors, teachers and parenting role models. In July, my “food”month, I wrote to Julie, who used to cook at my favorite restaurant. It went like this:Dear Julie,I’ve been finding myself missing you lately. Thank you for hosting and cooking beautiful and thoughtful food. Jake and I will never forget when you sent out biscuits shaped into the number VI for our sixth anniversary (周年纪念日）. We talk about it every year.Thank you. We miss you.Love,GinaI was happy to receive a note back from her. Julie replied,“I don’t think I’ve ever received such a touching letter before. I’m going through a hard time right now, and this helps.”On December 31, I wrote my last card — to Jake, my husband, and our two kids. And I took a picture of us, so I could remember the feeling welling up inside me. Gratitude.36. The underlined word “this” in Paragraph 2 means ________.A. writing thank-you notesB. putting the phone awayC. taking the train back homeD. raising money for charity37. How did Julie most probably feel after reading the note from Gina?A. Shy and surprised.B. Pleased and relaxed.C. Nervous and excited.D. Moved and encouraged.38. The theme of the December notes might be “________”.A. photosB. familyC. holidaysD. community39.What did Gina get by doing her task according to the passage?A. Many thanks from her neighbors.B. More time to appreciate her own life.C. More help from people around her.D. Happy moments to be with her friends.40. The best title for the passage can be ________.A. My Thank You YearB. An Amazing TripC. Happiness inside UsD. Important People in Our Life三、词汇运用A. 用方框中所给词语的适当形式填空,每词仅用一次。