清华大学中学生标准学术能力测试THUSSAT2020年11月测试理科数学(一)试卷及答案
北京市清华大学2020届高三数学11月中学生标准学术能力诊断性测试试题理
北京市清华大学2020届高三数学11月中学生标准学术能力诊断性测试试题 理本试卷共150分，考试时间120分钟。
一、选择题：本大题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1.己知集合U ＝{－1，1，3，5，7，9}，A ＝{1，5}，B ＝{－1，5，7}，则U ð(A ∪B )＝A.{3，9}B.{I ，5，7}C.{－1，1，3，9}D.{－1，1，3，7，9}2.己知空间三条直线l ，m ，n ，若l 与m 垂直，l 与n 垂直，则A.m 与n 异面B.m 与n 相交C.m 与n 平行D.m 与n 平行、相交、异面均有可能3.复数z 满足|z －1|＝|z ＋3|，则|z|A.恒等于1B.最大值为1，无最小值C.最小值为1，无最大值D.无最大值，也无最小值4.某几何体的三视图如图所示(单位：cm) ，则该几何体的表面积(单位：cm 2)是A.16B.32C.44D.645.已知x ＋y>0，则“2|x|＋x 2>2|y|＋y 2”是“x>0”的A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件6.函数y ＝ln|x|·cos(2－2x)的图像可能是7·已知两个不相等的非零向量a ，b ，满足|a |与b －a 的夹角为60°，则|b |的取值范围是A.0,2⎛ ⎝⎭B.,12⎫⎪⎪⎣⎭C.2⎫+∞⎪⎪⎣⎭D.()1,+∞ 8.已知随机变量ξ的分布列为：则下列说法正确的是A.存在x ，y ∈(0，1)，E(ξ)>12 B.对任意x ，y ∈(0，1)，E(ξ)≤14C.对任意x ，y ∈(0，1)，D(ξ)≤E(ξ)D.存在x ，y ∈(0，1)，D(ξ)>14 9.设函数f(x)＝ax 3＋bx 2＋cx ＋d(a ，b ，c ，d ∈R 且a≠0)，若0<2f(2)＝3f(3)＝4f(4)<1，则f(1)＋f(5)的取值范围是A.(0，1)B.(1，2)C.(2，3)D.(3，4) 10.已知F 1，F 2分别为双曲线22221(0,0)x y a b a b-=>>的左、右焦点，若在双曲线右支上存在点P ，使得点F 2到直线PF 1的距离为a ，则该双曲线的离心率的取值范围是A.1,2⎛⎫ ⎪ ⎪⎝⎭ B.2⎛⎫+∞ ⎪ ⎪⎝⎭ C.( D.)+∞ 11.如图，在菱形ABCD 中，∠ABC ＝60°，E ，F 分别是边AB ，CD 的中点，现将△ABC 沿着对角线AC 翻折，则直线EF 与平面ACD 所成角的正切值最大值为3212.己数列{a n }满足a 1＝1，a n ＋1＝lna n ＋1n a ＋1，记S n ＝[a 1]＋ [a 2]＋···＋[a n ]，[t]表示不超过t 的最大整数，则S 2019的值为A.2019B.2018C.4038D.4037二、填空题：本大题共4小题，每小题5分，共20分。
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北京市清华大学2020届高三中学生标准学术能力诊断性测试(11月) 数学(文) Word版含答案
中学生标准学术能力诊断性测试2019年11月测试文科数学试卷(一卷)本试卷共150分，考试时间120分钟。
一、选择题：本大题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1.己知全集U ＝R ，集合A ＝{x|1x x-≥0}，B ＝{x|y ＝lg(3x －1)}，则A ∩(U ðB)＝ A.(0，1] B.(0，13] C.(13，1] D.(－∞，13] 2.己知a ∈R ，复数z ＝23a i i-+(i 为虚数单位)，若z 为纯虚数，则a ＝ A.23 B.23- C.6 D.－6 3.某单位200名职工的年龄分布情况如图所示，现要从中抽取25名职工进行问卷调查，若采用分层抽样方法，则40～50岁年龄段应抽取的人数是A.7B.8C.9D.104.下列函数中，在区间(0，＋∞)上单调递增的是A.y ＝3－xB.y ＝log 0.5xC.21y x =D.12x y x +=+ 5.已知抛物线y 2＝4x 的焦点为F ，直线l 过点F 与抛物线交于A 、B 两点，若|AF|＝3|BF|，则|AB|＝A.4B.92 C.132 D.1636.己知1tan()43πα-=-，则sin(2)2sin()cos()2παπαπα+--+= A.75 B.15 C.15- D.31257.设变量x 、y 满足约束条件20240240x y x y x y +-≥⎧⎪-+≥⎨⎪--≤⎩，且z ＝kx ＋y 的最大值为12，则实数k 的值为A.－2B.－3C.2D.38.在△ABC 中，角A ，B ，C 的对边分别为a ，b ，v ，若a ＝1，c ＝bsinA ＝asin(3π－B)，则sinC ＝9.某三棱锥的三视图如图所示，网格纸上小正方形的边长为l ，则该三棱锥外接球的表面积为A.27πB.28πC.29πD.30π10.函数13cos 6x y x e =-的大致图象是11.已知双曲线C ：22221(0,0)x y a b a b-=>>的右焦点为F ，直线l ：yx 与C 交于A ，B 两点，AF ，BF 的中点分别为M ，N ，若以线段MN 为直径的圆经过原点，则双曲线的离心率为A.3－12＋112.在△ABC 中，AB ＝8，AC ＝6，∠A ＝600，M 为△ABC 的外心，若AM AB AC λμ=+，λ，μ∈R ，则4λ＋3μ＝A.34B.53C.73D.83二、填空题：本大题共4小题，每小题5分，共20分。
THUSSAT2020年11月诊断性测试理科数学(一)卷试卷
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THUSSAT2023年11月诊断性测试数学答案
中学生标准学术能力诊断性测试2023年11月测试数学参考答案一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分．三、填空题：本题共4小题，每小题5分，共20分．13．1 14．2715 16．35四、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．（10分） （1）2sin sin c B C b B =cos ，所以由正弦定理得=B BC B Csin cos 2sin sin sin ， ∴=B 2cos 1，得=πB 3············································································ 3分−=−+=−==−++A B C A B A B1tan tan tan tan tan tan )(················ 5分（2）ABC ∆内切圆的面积为π，所以内切圆半径=r 1，由圆的切线性质得+−=∴=+−=c a b b c a 3···························· 7分 由余弦定理得=+−b c a ac 222，∴+=+−=+−c a ac a c ac 332222)((，将+=+a c 3代入，∴=+∴==π∆ac S ac ABC 238sin 31 ·············································· 10分（或∴++=+∴=⋅⋅++=∆a b c S r a b c ABC 2631)( ······················ 10分） 18．（12分）（1）过D 作AB 的垂线交AB 于H 点，设==AC AB a ，则======−=BC BD a HD HB BD AH BH BA a 2,2,,1)，==AD ，由题意得，二面角−−C SA D 的平面角为∠CAD ， ········································· 2分∴∠===DA CAD DH cos ············································ 4分 （2）分别以AB ，AC ，AS 为x 轴，y 轴，z 轴建立直角坐标系，∴A B a 0,0,0,,0,0)()(，C a S h 0,,0,0,0,)()(，则−−λλλλE a h F a h ,0,1,0,,1)()()()(，SB SC =且==∴=λSB SCSE SF SE SF, ····················································· 6分 又,SAB SAC BSA CSA ∆∆∴∠=∠，那么SAESAF ∆∆，则=AE AF····························································································· 8分 故∆SEF 与∆AEF 都是等腰三角形，取EF 的中点G ，则SG 与AG 均垂直于EF ，111111,,1,,,,,,1G a a h SG a a h AG a a h λλλλλλλλλ⎛⎫⎛⎫⎛⎫−=−=− ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭222222)()(，平面⊥AEF 平面SBC 等价于SG AG ⊥ ················· 10分 2222a a SG AG h h h λλ∴⋅=+−−=λ4410)(，又SCB a h ∠=︒∴=∴=λ360,,2························ 12分 19．（12分） （1）1122,222n n n n a a a a ++=+∴−=−)(，即−=−+a a n n 22211)(，∴−a n 2}{是公比为21的等比数列 ······························································· 2分 ⎝⎭⎝⎭⎪⎪−=⋅∴=+⋅⎛⎫⎛⎫−−a a n n n n 2223,231111····················································· 4分（2）211211112312612222n n n n S a a a n n −⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫=+++=+++++=+−⎢⎥⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦············································································································· 6分⎝⎭⎝⎭⎪ ⎪∴−−=−⋅−−=⋅<⎛⎫⎛⎫S n S n n n nn222023266,266111， 即>∴−n n 26069,1取最小值14 ·································································· 8分 （3）n C n λ==<⎝⎭⎪⎛⎫++−C nC n n n n 33,122111)(，得>n 2， 即<+C C n n 1，有>>C C C 345，又===λλC C C 3,4123， 故C n }{中最大项为C C ,23 ······································································· 10分 又b m }{中最小值为∴−>λb C m n 3,7min max 2)()(，即−>λλ33472， ∴−⋅+>λλ3710)()(，又>∴>λλ30,7 ················································· 12分 20．（12分）（1）由题意可知：X 的所有可能取值为2.3，0.8，0.5，==⋅⋅=P X 2452.30.3134)( ······································································ 1分 =X 0.8包含的可能为“高低高”“低高高”“低低高”， ==⋅⋅+⋅⋅+⋅⋅=P X 2452452450.80.5114134114)( ·········································· 2分 ==−−=P X 0.510.30.50.2)( ··································································· 3分X 的分布列为：·············································································································· 4分 数学期望=E X 1.19)( ················································································ 5分 （2）设升级后一件产品的利润为Y ，Y 的所有可能取值为−−−a a a 2.3,0.8,0.5 ····················································· 6分⎝+⎭ ⎪=−=+⋅⋅=⎛⎫b b P Y a 3245102.31463)(························································· 7分⎝⎭⎝⎭−⎝⎭⎪ ⎪ ⎪=−=−⋅⋅++⋅⋅+−⋅⋅=⎛⎫⎛⎫⎛⎫bP Y a b b b 15624524524500.8134114114)( ············· 8分=−=−−=+−P Y a b b 101050.5163561)( ························································ 9分 =−⋅+−⋅+−⋅=+⋅−+−E Y a a a b a b b 101052.30.80.5 1.190.963561)()()()()( ············································································································ 11分()E Y E X b a >⇒⋅−>0.90)(，即：>>∈b a a 290,0.4110)(][（备注：21不写出不扣分） ······························· 12分 21．（12分）（1）AB AF BF ++=4211，即+++=AF BF AF BF 2211，又1212AF AF BF BF a a a +=+=∴=∴=2,442,2 ······························· 2分 又e =,1,1c a =∴=∴=c b 22， 故椭圆E 的方程是+=y x 2122······································································ 4分 （2）依题意知直线BC 的斜率存在，设直线=+BC y kx m :，代入+=y x 2122，得++=kx m x 2122)(，即⎝⎭⎪+++−=⎛⎫k x kmx m 22101222 ①，()22412240km k m m k =−+−=−++>⎝⎭⎪⎛⎫212222)(，即−+>k m 21022 ②，设B x y C x y ,,,1122)()(，则−A x y ,22)(，++=−k x x km 212212 ③，+=−k x x m 2112122 ④ ····················································· 6分2,,A F B 三点共线，F 1,02)(，直线AB 不与坐标轴垂直，−−∴=∴−+=−−+−x x x kx m x kx m y y 11,1112211212)()()()(， ∴++−+−=kx x m x x k x x m 220121212)()(，+++∴−+−=−k k k m km k m k m 222111202*********)(，∴−−+−−=km k km k m m k m 2222202222，∴=−∴m k 2,直线=−BC y k x :2)(， 由②得：−+>∴<kk k 22410,1222，=−BC x 12，点−F 1,01)(到直线−−=BC kx y k :20的距离=d ，∴==−∆S BC d k x x F BC 2213121 ································································· 8分 ()()121212x x x x x x −=+−=−⎝⎭++ ⎪ ⎪ ⎪−⎛⎫k k k k 221144441222222==，∴=≠∆S k F BC 01)，设+=>k t 2112，则=−kt 212，∴===∆S F BC1=·································································· 10分 所以当=t 413时，即=+==t k k 336,21,44122（符合题意），∆S F BC 1的最大值为4，所以当=±k 6时，∆F BC 1的面积取最大值为4············································································································ 12分22．（12分） （1）定义域为(),11,−∞−+∞)(，由题意知−−=x x x ax ln 12)()(，则−−=x x a 1ln 1)()(有三个不同的实数根，当>x 1时，令=−−g x x x 1ln 1)()()(，∴'=−+∴'g x x g x ln 11,)()()(在∈+∞x 1,)(上单调递增，又⎝⎭⎪'+=⎛⎫g e 101，∴g x )(在⎝⎭ ⎪∈+⎛⎫x e 1,11上单调递减，在⎝⎭ ⎪∈++∞⎛⎫x e 1,1上单调递增，⎝⎭⎪∴≥+=−⎛⎫g x g e e 111)( ·············································································· 2分当<−x 1时，令=−−−h x x x 1ln 1)()()(，()()++++∴'=−−+''=+=−+x x x x h x x h x x x 1111ln 1,112322)()()(， 又''−=h 30)(，∴'h x )(在−∞−,3)(上单调递减，在−−3,1)(上单调递增，∴'≥'−=+>h x h 32ln 20)()(，∴h x )(在−∞−,1)(上单调递增 ····································································· 5分又−=h 20)(，当→−−x 1时，→+∞h x )(，当→−∞x 时，→−∞h x )(， 当→+x 1时，→−g x 0)(，当→+∞x 时，→+∞g x )(，∴−<<a e01··························································································· 6分 （2）由已知可知−−=xx x a1ln1)()( 有3个零点x x x ,,123， 不妨设<<x x x 123，显然−<<−x 211，由（1）中函数g x )(性质且1111a >>−++−e ee e 11， ∴<<+<<x x e112123，只需证+>+x x e2223，即>+−x x e 2232 ···················································· 8分 又⎝⎭⎝⎭⎝⎭⎝⎭ ⎪ ⎪ ⎪ ⎪+−−=+−+−−−−<⎛⎫⎛⎫⎛⎫⎛⎫g x g x x x x x e e e 211ln 11ln 10122222222)()()(，上面不等式证明如下： 令⎝⎭⎝⎭⎝⎭ ⎪ ⎪ ⎪=+−+−−−−∈+⎛⎫⎛⎫⎛⎫ϕx x x x x x e e e 1ln 11ln 1,1,1221)()()(，⎝⎭⎝⎭ ⎪ ⎪'=−+−−−−∈+⎛⎫⎛⎫ϕx x x x e e ln 1ln 12,1,121)()(⎣⎦⎝⎭⎢⎥ ⎪=+−−−>⎛⎫⎡⎤x x e ln 11202)(，∴ϕx )(在⎝⎭ ⎪∈+⎛⎫x e 1,11上单调递增，⎝⎭⎪∴<+=⎛⎫ϕϕx e 101)( ··········································································· 10分又−>−−+−>+x x e e e1,2112122，⎝⎭⎝⎭ ⎪ ⎪∴+−<⎛⎫⎛⎫g x g x e 21122)(，又显然有<g x g x 23)()(，⎝⎭⎪∴+−<⎛⎫x x e 21123，∴+>+x x e2223， ∴++>x x x e2123 ·················································································· 12分。
THUSSAT2020年11月诊断性测试文科数学(一)卷试卷(含解析))
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C1
A1
B1
14. 若 sin ( + )sin ( − ) = − 1 ，则 sin2 − sin2 =
．
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( ) 15. 已知圆 C 与 y 轴相切于点 P 0, 3 ，与 x 轴正半轴交于两点 A, B,
则圆 C 的方程为
．
APB = 30 ， A
C B
（第 13 题图）
16. 在锐角 ABC 中，内角 A, B,C 所对的边分别为 a,b, c ，若 b + a = 4 cos C ，则 1 + 1 的
同编号不同的 3 个黄色乒乓球和 3 个白色乒乓球，现从 A、B 两个口袋中各摸出 2 个球．
（1）求摸出的 4 个球中有 3 个黄色乒乓球和 1 个白色乒乓球的概率；
（2）求摸出的 4 个球中黄球个数 的数学期望．
18. （12 分）已知数列 an 是等差数列，若 a1 = 2 ，且 a3, 2a 2 , 2a4 −1成等比数列，数列 bn 满足
1 − 1) 2
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f (x) = 1 + x2 (10x −1)2
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9. 将函数 y = sin 2x 的图象沿 x 轴向左平移 ( 0)个单位长度得到函数
2
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3
4
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5
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2. 已知双曲线方程： 2x2 − 3y2 = 1，则该双曲线的渐近线方程为
中学生标准学术能力诊断性测试(THUSSAT)2022年11月英语试卷(含答案)
中学生标准学术能力诊断性测试2022年11月测试英语试卷本试卷共150分，考试时间100分钟。
第一部分阅读理解（共两节，满分60分）第一节（共15小题；每小题3分，满分45分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项，并在答题卡上将该项涂黑。
AChina is vast and beautiful.It encompasses just about every type of landscape and wildlife species imaginable,from tropical jungle to Himalayan plateal1.These massive ecosystems are home to an incredible wealth of biodiversity,including rare plant life and popular species.Sanjiangyuan National ParkThe first national park to be considered as part of the new system is Sanjiangyuan——the source of the Yangtze,Yellow and Mekong(Lancang)rivers.Situated in Qinghai province in far northwest China,the park covers123,000sq km of high-altitude wilderness on the Tibetan Plateau.Protected species include snow leopards,wild yaks,and Tibetan gazelles and antelope.Wuyi Mountain National ParkBoth a Biosphere Reserve and a UNESCO World Heritage Site,Wuyi Shan is one of the world's largest subtropical primordial（原始的）forest systems.Its dramatic gorges and rivers are home to thousands of types of protected wildlife like migratory birds and rare amphibians（两栖动物）.And many visitors come to explore the archaeological remains of Han City as well as numerous11th-century Dadaist temples and shrines.Northeast China Tiger and Leopard National ParkSituated in the Changbai Mountains along China's border with Russia,this park covers the wild habitats of the big cats and has seen success in encouraging leopards and tigers to repopulate.Since2017,12Siberian tiger cubs and11leopard cubs have been born here.Nearby is Jingpo Lake National Geo-park near Mudanjiang in Heilongjiang province,with lakes,volcanic craters and ancient lava flows.Hainan Tropical Rainforest National ParkPreserving the tropical ecosystem of Hainan Island,China's southernmost province,this national park is home to20%of the country's amphibians and39%of its birds,as well as the world's rarest primate,the critically endangered Hainan gibbon（长臂猿）.The national park encompasses a number of existing reserves，where hiking trails lead around lakes and up peaks, offering visitors the chance to be immersed in the rainforest and spot rare creatures in the wild.1.Which species can't be found in Sanjiangyuan National Park?A.Wild yaks.B.Snow leopards.C.Giant pandas.D.Tibetan gazelles.2.What contributed to the varieties of protected wildlife in Wuyi Mountain National Park?A.Its pleasant climate.B.Its air pressure.C.Its economic development.D.Its geographical condition.3.Which of the following is True according to the passage?A.Sanjiangyuan National Park is situated in the west of China.B.Wuyi Mountain National Park is the largest subtropical park.C.You can go backpacking in Hainan Tropical Rainforest National Park.D.You can see volcanoes in Northeast China Tiger and Leopard National Park.BAn enthusiastic backyard bird watcher named Stuart Dahlquist spent years leaving out food for a family of crows,but he never expected to be given anything in return for the snacks.The56-year-old Seattle home owner first became acquainted with the bird family after he rescued two chicks that had fallen out of their nest in his front yard six years ago.Dahlquist told The Dodo that he had always delighted in listening to the baby birds sing to their parents during feeding time,so when he went outside and found the tiny crows on the ground,he knew he had to help.Despite the sharp sound of alarm from the adults,Dahlquist managed to scoop up the clicks and put them back in the nest.He even left out food and water at the base of the tree in case they fella second time.He then began regularly throwing bird food into his front yard,and the crows apparently took notice.One day when Dahlquist was preparing for his daily feeding，he was surprised to find a sprig（带叶小枝）that had been decorated with a soda can tab（拉环）.Not only that，it was left in the exact spot where he fed the crows.The next day,he was offered a second soda tab-decorated branch and he was amazed by the discovery."This isn't only generous,it's creative, it's art,”Dahlquist wrote on social media“My mind is blown.”Regardless of the crows'intent,Dahlquist says he has maintained a close relationship with the bird family since the incident."They'll follow along when I take my walks,landing on the wires along the way,”Dahlquist told The Dodo.“The adult male is very friendly and will fly sometimes within a few feet,diving down to say,'Here I am!"".4.What's the relationship between Dahlquist and the adult crows at first?A.They were acquaintances.B.They were friendly to each other.C.The adult crows took a cautious attitude.D.Dahlquist refused to feed the crow family.5.What does the underlined word in Paragraph3mean?A.BlankB.AbsentC.ShockedD.Intelligent6.What can we learn from the passage?A.Dahlquist is a warm-hearted journalist.B.The crows made gifts in return for Dahquist's help.C.Dahlquist first became familiar with the crow family at the age of51.D.Dahlquist put the food at the base of the tree because the crows lived there.7.What kind of person do you think Dahlquist is?A.GenerousB.CreativeC.HumorousD.StrictCArt is a powerful tool for telling a scientific story.With many scientific fields dealing with the strangest of the animal kingdom—like creatures with a handful of eyes and bodies so different from our own—art can help us experience these hard-to-imagine parts of the natural world and shed light on new scientific discoveries.Alia Payne,an interactive art professor came to the National Museum of Natural History to work with the live jellyfish collections.She always got the same question from visitors,"How do jellyfish stings work?"She had the scientific answer for them but found it difficult to explainwithout a clear visual.That's when a light bulb went off in Payne's mind.She built a3D model of one of the stinging cells that visitors could touch and interact with."I've always loved art for education,"said Payne."We learn more easily when we have something to play and interact with.”While art helps society experience science,it also helps scientists understand the natural world.Dr.Karen Osborn,a zoologist at the museum,specializes in the striking creatures of the ocean's largest habitat—the mid-water.Midwinter creatures are equipped with all sorts of unique features to get them through life in their extreme habitat,and Osborn's research helps us understand how these animals came to be.“So I started leaning photography,”Osborn explained.“It was really important because these animals don't look so great when they're preserved in a jar in the museum's collections."Artistic creativity provides an opportunity to showoff the bright colors,attractive body plans and cool adaptations of these creatures to the museum's scientists,visitors and the world.In turn, art connects scientists to their own creativity and aids in their scientific process and the communication of their discoveries.By weaving together science with imagination and storytelling,art helps highlight the beauty of the ocean's creatures—even those fit for horror movies,and connects people to their innate curiosity for the strangeness of the animal kingdom.8.What can be inferred from Paragraph1?A.Art can deal with the strangest animals.B.Art is an essential way of explaining science.C.We couldn't tell a scientific story without art.D.It is hard to imagine the art of the natural world.9.What do Alia and Karen have in common?A.Both of them are biologists.B.Both of them are interested in photography.C.Both of them take advantage of art for science.D.Both of them make many discoveries of animals.10.Which of the following is NOT the effect of art?A.It can connect people to nature.B.It can help scientists show off themselves.C.It can promote the creativity of the scientists.D.It can illustrate the ocean's creatures beautifully.11.What can be the best title of the passage?A.Art has two sides.B.Two great scientists.C.Science is important to art.D.Why science needs art.DIf you go back about30years and ask some people if they're familiar with the word“emoji”, the chances are that they'll laugh at your question.Nowadays,however,emojis seem to have entered almost everyone's daily life.Why use emojis?The first reason why people claim they use emojis is due to their simplicity and the relative ease with which they have the potential to transfer ideas and thoughts.A recent study shows that people find it easier to convey ideas online rather than in ing emojis greatly makes that easier since it makes the message or emotion more concise（简洁的）.It's easier to make and send a"sad face"rather than state that you're sad.Linguists and people who study trends often guess if social media platforms are changing the way that we see the world.Due to their wide use and simplicity,social scientists have considered emojis the basis of an"international language".This arises from the fact that emojis are generally consistent worldwide and are understood by all and frequently used.In cross-cultural communication,emojis are consistent in all social media platforms.Hence,brands that want to expand overseas integrate emojis into their marketing schemes. It's safe to say that a great number of users on social plat form sare people who grew alongside the Internet.Famous brands and companies are switching their main marketing strategies to target this new market.Like all things that live on the Internet,there comes a time when they go out of date.However,this might not be the case with emojis.Ever since their birth,emojis have served as a gateway for clearer and more concise expression of digital speech.Hence,they will stick around for a long time.12.Which of the following is not the feature of emojis according to the passage?A.SimpleB.ConciseC.UniversalD.Traditional13.Why are emojis considered the basis of an"international language"?A.They have various meanings.B.They are frequently and widely used.C.They change the way we see the world.D.They promote cross-cultural communication.14.Who are the main marketing target of the famous brands?A.Senior citizens.B.Babysitters.C.The young generation.D.College students.15.What's the author's attitude towards the future of emojis?A.UncertainB.PositiveC.IndifferentD.Pessimistic第二节（共5小题；每小题3分，满分15分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
中学生标准学术能力诊断性测试2020年11月高三理科综合试卷 (word)
中学生标准学术能力诊断性测试 2020 年 11 月测试理科综合试卷（一卷）本试卷共 300 分，考试时间 150 分钟。
可能用到的相对原子质量：H 1 C 12 N 14 O 16 Fe 56Co 59一、选择题：本题共 13 小题，每小题 6 分，共 78 分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1．下列叙述中可以正确区分出自养生物和异养生物的是A．只有异养生物需要环境中的化合物B．细胞呼吸为异养生物所特有的代谢C．只有异养生物的细胞中含有线粒体D．只有自养生物能依靠无机养分生活2．下列关于氨基酸和蛋白质的叙述错误的是A．蛋白质功能的不同可能取决于氨基酸的种类、数目和排序不同B．改变蛋白质的一个氨基酸可能改变整个蛋白质的空间结构C．氨基酸仅通过脱水缩合的方式就可以形成具有生物学功能的蛋白质D．脊椎动物中血红蛋白一条多肽链的氨基酸差异越小，亲缘关系越近3．同一个生物体内的细胞存在着形态、结构和功能上的差异，但所有体细胞的细胞核中的遗传物质都是相同的，其原因是A．细胞是生命活动的基本单位，细胞核是细胞生命活动的控制中心B．细胞核是遗传信息库，具有控制细胞代谢和遗传的功能C．经过有丝分裂将复制完的 DNA 精确均分的结果D．DNA 主要存在于细胞核内，细胞核是遗传物质贮存和复制的场所4．将 T2噬菌体感染大肠杆菌后立即合成的 RNA 分离出来，分别与 T2噬菌体和大肠杆菌的 DNA 进行分子杂交，结果发现这种 RNA 只能与 T2的 DNA 杂交形成杂种链，而不能和大肠杆菌的 DNA 杂交，下列说法错误的是A．分子杂交依据碱基互补配对原则B．新合成的 RNA 是以 T2噬菌体的 DNA 为模板合成的C．大肠杆菌被感染后，其自身的 RNA 合成停止D．新合成的 RNA 可以作为大肠杆菌合成自身蛋白质的模板5．多巴胺是脑神经细胞分泌的一种神经递质，使人产生兴奋愉悦的情绪，多巴胺发挥作用后由转运载体运回突触前神经元。