清华大学留学生试卷
清华大学考博英语模拟试卷12(题后含答案及解析)
清华大学考博英语模拟试卷12(题后含答案及解析)题型有：1. Structure and V ocabulary 2. Cloze 3. Reading Comprehension 4. English-Chinese Translation 5. WritingStructure and V ocabulary1．The assessment center gives each applicant the opportunity to ______ whether they are suited to the work.A．exemplifyB．demonstrateC．exposeD．exhibit正确答案：B2．Fresh fruits and vegetables are generally less expensive when they are in________.A．saleB．needC．seasonD．time正确答案：C解析：“in sale”无此搭配．应为“on sale”，表示“出售，拍卖”，“in need”表示“需要的”。
“in season”表示“(水果等)应时的，时令的，“in time”表示“及时地，准时地”。
3．It is naive to expect (hat any society can resolve all the social problems it is faced with_______.A．for longB．in and outC．once for allD．by nature正确答案：C解析：once for all是“一劳永逸”。
A for long是“长久”；B．in and out是“里外”；D．by nature是“本质上”。
4．Mrs. Mary wore a string of beadsaround her neck.A．small pieces of goldB．small balls of precious stonesC．small pieces of woodD．small balls of glass正确答案：D解析：beads“有孔的小珠”与D项small balls of glass相符。
清华大学往年考试试卷真题
清华大学往年考试试卷真题清华大学是中国顶尖的高等学府之一，其考试试卷真题通常包含多个学科领域，以下是一个模拟的清华大学往年考试试卷真题的示例：清华大学数学分析考试试卷一、选择题（每题2分，共20分）1. 下列函数中，哪一个不是连续函数？A. f(x) = x^2B. f(x) = sin(x)C. f(x) = |x|D. f(x) = 1/x2. 函数f(x) = x^3 - 2x + 1在x=1处的导数是：A. 2B. 0C. -2D. 43. 积分∫(0,1) x^2 dx的值是：A. 1/3B. 1/2C. 2/3D. 1...（此处省略其他选择题）二、填空题（每题3分，共15分）1. 函数f(x) = 3x^2 + 2x - 1的极值点是_________。
2. 若f(x) = e^x，那么f'(x) = __________。
3. 函数y = ln(x)的定义域是_________。
...（此处省略其他填空题）三、简答题（每题10分，共30分）1. 证明函数f(x) = x^3在R上是单调递增的。
2. 解释什么是泰勒级数，并给出e^x的泰勒级数展开式。
3. 计算定积分∫(1, e) (x + 1/x) dx。
四、解答题（每题15分，共30分）1. 已知函数f(x) = x^2 - 4x + 4，求其在区间[0, 3]上的最小值。
2. 给定函数g(x) = sin(x) + cos(x)，求其在x=π/4处的导数，并解释其几何意义。
3. 解析下列微分方程：dy/dx = x^2 - y^2，初始条件为y(0) = 1。
五、附加题（10分）1. 讨论函数f(x) = x^3 - 6x^2 + 11x - 6在实数域R上的零点。
注意事项：- 请在答题纸上作答，不要在试卷上直接书写。
- 请保持答题纸整洁，字迹清晰。
- 选择题请用2B铅笔涂黑，填空题和解答题请用黑色签字笔书写。
祝考试顺利！请注意，以上内容仅为模拟示例，并非真实的清华大学考试试卷真题。
清华大学中学生标准学术能力(TDA)诊断性测试2024-2025学年高二上学期数学试卷和答案
标准学术能力诊断性测试2024年9月测试数学试卷（A 卷）本试卷共150分，考试时间90分钟.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.1.设,a b ∈R ，则“22log log a b >”是“1122b a ⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭”的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件2.集合(){}{}22ln 23,23,A x y x x B y y x x x A ==--==-+∈∣∣，则A B ⋂=R ð（）A.(),1∞-- B.()(],13,6∞--⋃C.()3,∞+ D.()[),16,∞∞--⋃+3.已知复数z 满足5z z ⋅=，则24i z -+的最大值为（）C. D.4.已知非零向量,a b 满足3a b = ，向量a 在向量b 方向上的投影向量是9b - ，则a 与b 夹角的余弦值为（） A.33 B.13 C.33- D.13-5.设函数()f x 的定义域为R ，且()()()()42,2f x f x f x f x -++=+=-，当[]1,2x ∈时，()()()2,303f x ax x b f f =+++=-，则b a -=（）A.9-B.6-C.6D.96.班级里有50名学生，在一次考试中统计出平均分为80分，方差为70，后来发现有3名同学的分数登错了，甲实际得60分却记成了75分，乙实际得80分却记成了90分，丙实际得90分却记成了65分，则关于更正后的平均分和方差分别是（）A.82，73 B.80，73 C.82，67D.80，677.已知()sin 404cos50cos40cos θθ-=⋅⋅ ，且ππ,22θ⎛⎫∈- ⎪⎝⎭，则θ=（）A.π3- B.π6- C.π6 D.π38.已知函数()2221x f x x =-++，则不等式()()2232f t f t +->的解集为（）A.()(),13,∞∞--⋃+ B.()1,3- C.()(),31,∞∞--⋃+ D.()3,1-二､多项选择题：本题共3小题，每小题6分，共18分.在每小题给出的四个选项中，有多项符合题目要求.全部选对得6分，部分选对但不全得3分，有错选的得0分.9.已知实数,,a b c 满足0a b c <<<，则下列结论正确的是（）A.11a c b c>-- B.a a c b b c +<+C.b c a c a b --> D.2ac b bc ab+<+10.已知函数()sin3cos3f x a x x =-，且()3π4f x f ⎛⎫≤⎪⎝⎭对任意的x ∈R 恒成立，则下列结论正确的是（）A.1a =±B.()f x 的图象关于点π,04⎛⎫ ⎪⎝⎭对称C.将()f x 的图象向左移π12个单位，得到的图象关于y 轴对称D.当π23π,1236x ⎡⎤∈-⎢⎥⎣⎦时，满足()2f x ≤-成立的x 的取值范围是π7π,3636⎡⎤-⎢⎥⎣⎦11.在长方体1111ABCD A B C D -中，已知4,2AB BC ==，13,AA M N =､分别为1111B C A B ､的中点，则下列结论正确的是（）A.异面直线BM 与AC 所成角的余弦值为7210B.点T 为长方形ABCD 内一点，满足1D T ∥平面BMN 时，1D T的最小值为5C.三棱锥1B B MN -的外接球的体积为14πD.过点,,D M N 的平面截长方体1111ABCD A B C D -所得的截面周长为+三､填空题：本题共3小题，每小题5分，共15分.12.若实数,x y 满足1232,34x y x y ≤+≤≤-+≤，则x y +的取值范围是__________.13.如图所示，在梯形ABCD 中，1,3AE AB AD =∥,3,BC BC AD CE =与BD 交于点O ，若AO x AD y AB =+ ，则x y -=__________.14.在四面体ABCD 中，3,,CD AD CD BC CD =⊥⊥，且AD 与BC 所成的角为30 .若四面体ABCD 的体积为2，则它的外接球表面积的最小值为__________.四､解答题：本题共5小题，共77分.解答应写出文字说明､证明过程或演算步骤.15.（13分）已知复数12213i z =-+=--.（1）若12z z z =，求z ；（2）在复平面内，复数12,z z 对应的向量分别是,OA OB ，其中O 是原点，求AOB ∠的大小.16.（15分）在ABC 中，角,,A B C 的对边分别是,,a b c ，且cos cos 1a C b A c -+=.（1）求角A ；（2）已知b D =为BC 边上一点，且2,BD BAC ADC ∠∠==，求AD 的长.17.（15分）如图所示，在四棱锥P ABCD -中，底面ABCD 为平行四边形，PA ⊥平面ABCD ，点Q 为PA 的三等分点，满足13PQ PA =.（1）设平面QCD 与直线PB 相交于点S ，求证：QS ∥CD ；（2）若3,2,60,AB AD DAB PA ∠==== ，求直线CQ 与平面PAD 所成角的大小.18.（17分）甲､乙两位同学进行投篮训练，每个人投3次，甲同学投篮的命中率为p ，乙同学投篮的命中率为()q p q >，且在投篮中每人每次是否命中的结果互不影响.已知每次投篮甲､乙同时命中的概率为15，恰有一人命中的概率为815.（1）求,p q 的值；（2）求甲､乙两人投篮总共命中两次的概率.19.（17分）已知函数()233x x f x a --=⋅+是偶函数，()246h x x x =-+.（1）求函数()e 2x y h a =-的零点；（2）当[],x m n ∈时，函数(()h f x 与()f x 的值域相同，求n m -的最大值.标准学术能力诊断性测试2024年9月测试数学（A卷）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.12345678A B C C D B A C二､多项选择题：本题共3小题，每小题6分，共18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6分，部分选对但不全的得3分，有错选的得0分.91011AD BC BD三､填空题：本题共3小题，每小题5分，共15分.12.21,55⎡⎤-⎢⎥⎣⎦13.11114.73π-四､解答题：本题共5小题，共77分.解答应写出文字说明､证明过程或演算步骤.15.（13分）解：（1）()() ()()12224i13i24i26i4i127i13i13i13i19i5 zzz-+---++-++ =====-+-+---5z∴==（2）依题意向量()()2,4,1,3OA OB=-=--于是有()()()214310OA OB⋅=-⨯-+⨯-=-OA OB====AOB∠为OA 与OB 的夹角，2cos2OA OBAOBOA OB∠⋅∴==-[]0,πAOB∠∈，3π4AOB∠∴=16.（15分）解：（1）由正弦定理可得：cos sin cos sin cos 1sin a C b A C B A c C--+==()cos 1sin sin cos sin A C A C B ∴+=-，由()sin sin B A C =+可得：()cos sin sin sin cos sin A C C A C A C ⋅+=-+，cos sin sin sin cos sin cos cos sin A C C A C A C A C ⋅+=--，cos sin sin cos sin A C C A C∴⋅+=-sin 0C ≠ 可得：cos 1cos A A +=-，1cos 2A ∴=-，()0,πA ∈ ，2π3A ∴=（2）,BAC ADC BCA ACD ∠∠∠∠== ，BAC ∴ 与ADC 相似，满足：AC BC CD AC =，设CD x =，则有3x =解得：1,3x x ==-（舍去），即：1CD =2π3ADC BAC ∠∠== ，在ADC 中，由余弦定理可得：2222πcos 32AD CD AC AD CD+-=⋅⋅，即：211221AD AD +--=⨯⨯解得：1,2AD AD ==-（舍去），AD ∴的长为117.（15分）解：（1）证明：因为平面QCD 与直线PB 相交于点S ，所以平面QCD ⋂平面PAB QS=因为四边形ABCD 为平行四边形，AB ∴∥CD ，AB ⊄ 平面,QCD CD ⊂平面,QCD AB ∴∥平面QCDAB ⊂ 平面PAB ，平面QCD ⋂平面,PAB QS AB =∴∥QS ，AB ∥,CD QS ∴∥CD（2）过点C 作CH AD ⊥于点H ，PA ⊥ 平面,ABCD PA ⊂平面PAD ，所以平面PAD ⊥平面ABCD ，因为平面PAD ⋂平面ABCD AD =，且CH AD ⊥，CH ∴⊥平面PAD连接,QH CQH ∠∴是直线CQ 与平面PAD 所成的角因为点Q 为PA 的三等分点，232,223PA QA PA =∴==，在Rt DCH 中，333sin602CH =⋅= 在ACD 中，利用余弦定理可得：222223cos120,19223AC AC +-=∴=⨯⨯ ，在Rt QAC 中，222(22)1933QC QA AC =+=+=在Rt QCH 中，3312sin 233CH CQH CQ ∠===，可得π6CQH ∠=，即直线CQ 与平面PAD 所成的角等于π618.（17分）解：（1）设事件A ：甲投篮命中，事件B ：乙投篮命中，甲､乙投篮同时命中的事件为C ，则C AB =，恰有一人命中的事件为D ，则D AB AB =⋃，由于两人投篮互不影响，且在投篮中每人每次是否命中的结果互不影响，所以A 与B 相互独立，,AB AB 互斥，所以：()()()()P C P AB P A P B ==⋅()(()()(()()()P D P AB AB P AB P AB P A P B P A P B =⋃=+=⋅+⋅可得：()()1581115pq p q p q ⎧=⎪⎪⎨⎪-+-=⎪⎩解得：1335p q ⎧=⎪⎪⎨⎪=⎪⎩或3315,,,1533p p q p q q ⎧=⎪⎪>∴==⎨⎪=⎪⎩（2）设i A ：甲投篮命中了i 次；j B ：乙投篮命中了j 次，,0,1,2,3i j =，()30285125P A ⎛⎫== ⎪⎝⎭()2213223223365555555125P A ⎛⎫⎛⎫=⨯+⨯⨯+⨯= ⎪ ⎪⎝⎭⎝⎭()2223232323545555555125P A ⎛⎫⎛⎫=⨯+⨯⨯+⨯= ⎪ ⎪⎝⎭⎝⎭()3028327P B ⎛⎫== ⎪⎝⎭()2211221221433333339P B ⎛⎫⎛⎫=⨯+⨯⨯+⨯= ⎪ ⎪⎝⎭⎝⎭()2222112112233333339P B ⎛⎫⎛⎫=⨯+⨯⨯+⨯= ⎪ ⎪⎝⎭⎝⎭设E ：甲､乙两人投篮总共命中两次，则021120E A B A B A B =++由于i A 与j B 相互独立，021120,,A B A B A B 互斥，()()()()()()()()021*********P E P A B A B A B P A P B P A P B P A P B ∴=++=⋅+⋅+⋅8236454830412591259125271125=⨯+⨯+⨯=19.（17分）解：（1）()233x x f x a --=⋅+ 是偶函数，则()()f x f x -=，即11333399x x x x a a --⋅+=⋅+，()113309x x a -⎛⎫∴--= ⎪⎝⎭，由x 的任意性得119a =，即9a =()246h x x x =-+ ，()()()()()22e 2e 4e 618e 4e 12e 6e 2x xx x x x x y h a ∴=-=-⋅+-=-⋅-=-+，令()()e 6e 20x x -+=，则e 6x =或e 2x =-（舍去），即ln6x =，()e 2x y h a ∴=-有一个零点，为ln6（2）设当[],x m n ∈时，函数()f x 的值域为[],s t ，则函数()()h f x 的值域也为[],s t ，由（1）知()2933332x x x x f x ---=⋅+=+≥=当且仅当33x x -=，即0x =时等号成立，令()p f x =，则2p ≥，()2246(2)2h x x x x =-+=-+ 在区间[)2,∞+上单调递增，所以当[],p s t ∈时，()2,s h p ≥的值域为()(),h s h t ⎡⎤⎣⎦，即()()h s s h t t ⎧=⎪⎨=⎪⎩，则224646s s s t t t ⎧-+=⎨-+=⎩，即,s t 为方程246x x x -+=的两个根，解得23s t =⎧⎨=⎩，所以当[],x m n ∈时，()f x 的值域为[]2,3令()30x x λ=>，则()133,1x x y f x λλλ-==+=+>，3x λ= 在()0,∞+上单调递增，对勾函数1y λλ=+在()1,∞+上单调递增，由复合函数的单调性知，()f x 在()0,∞+上单调递增，()f x 是偶函数，()f x ∴在(),0∞-上单调递减令()3f x =，即333x x -+=，解得332x +=或332x =，即33log 2x +=或33log 2x -=，故n m -的最大值为3333535735log log log 222-+-=答案解析1.A【解析】由22log log a b >可得0a b >>，由1122b a⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭可得a b >，由a b >得不到0a b >>，故必要性不成立；由0a b >>可以得到a b >，故充分性成立，则“22log log a b >”是“1122b a ⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭”的充分不必要条件.2.B 【解析】集合(){}{}22ln 23230A x y x x x x x ==--=-->∣∣()(){}310{13},x x x x x x =-+>=<->∣∣或集合{}{}223,6B yy x x x A y y ==-+∈=>∣∣，{}()(]6,,13,6B y y A B ∞=≤∴⋂=--⋃R R ∣3.C【解析】复数z 满足5z z ⋅=，设22i,5z a b z z a b =+⋅=+=，()()2224i 24i (2)(4)z a b a b -+=-++=-++，则点()2,4-到圆225a b +=+=4.C【解析】设非零向量,a b 夹角为θ，向量a 在向量b 方向上的投影向量是39b - ，则cos ,39b a a b b θ⨯=-= ∣，解得3cos 3θ=-.5.D【解析】()()42f x f x -++= ，取()()1,312x f f =+=，()()()321211f f a b a b =-=-++=--，()()2f x f x +=- ，取()()0,2042x f f a b ===++，()()303,1423,2f f a b a b a +=---+++=-=- ，()()42f x f x -++= ，取2x =，则()21f =，则7b =，则729b a -=+=.6.B【解析】设更正前甲，乙，丙 的成绩依次为12350,,,,a a a a ，则12505080a a a +++=⨯ ，即507590655080a ++++=⨯ ，()222250(7580)(9080)(6580)807050a -+-+-++-=⨯ ，更正后平均分：()5016080908050x a =++++= ，()22222501(6080)(8080)(9080)807350s a ⎡⎤=-+-+-++-=⎣⎦ .7.A 【解析】()sin 40sin40cos cos40sin θθθ-=- 4cos50cos40cos 4sin40cos40cos θθ=⋅⋅=⋅⋅ 1cot40tan 4cos40θ⇒-=14cos40tan cot40θ-⇒=sin404sin40cos40cos40-=()sin 30102sin80cos40+-= 13cos102cos1022cos40+-=3313sin10cos10sin10cos102222cos40cos40--==()()sin 1060sin 50cos40cos40--===πππ,,223θθ⎛⎫∈-∴=- ⎪⎝⎭.8.C【解析】设()()21121x g x f x x =-=-++，()()2221112121x x x g x f x x x -⋅-=--=--+=--+++，()()2221102121x x x g x g x x x ⎛⎫⋅+-=-++--+= ⎪++⎝⎭，设()()1212121222,112121x x x x g x g x x x ⎛⎫⎛⎫>-=-+--+ ⎪ ⎪++⎝⎭⎝⎭()()()()()122121121222222021212121x x x x x x x x x x -⎛⎫=-+-=-+> ⎪++++⎝⎭，故()g x 为奇函数，且单调递增，()()()()()()22223212310230f t f t f t f t g t g t +->⇒-+-->⇒+->，()()()()()222302332g t g t g t g t g t +->⇒>--=-，故232t t >-，解得()(),31,t ∞∞∈--⋃+.9.AD【解析】A.0a b c <<<，可得a c b c -<-，故11a c b c>--，A 正确；B.设不等式成立，则()()a a c b c b b c b b b c++<++，可得ab ac ab bc +<+，即ac bc <，由0a b c <<<可得ac bc >，故假设不成立，B 错误；C.不妨假设211313210,,1332b c a c a b c a b --+--+=-<=-<=-<====--，故,C b c a c a b --<错误；D.设不等式成立，()()22,,,0ac b bc ab ac bc ab b a b c a b b a b c +<+-<--<-<<< ，()()a b c a b b -<-成立，故2ac b bc ab +<+成立，D 正确.10.BC【解析】A.()()sin3cos33sin 0,cos πf x a x x x ϕϕϕϕ⎛⎫=-=+=-=≤ ⎪⎝⎭()3π4f x f ⎛⎫≤ ⎪⎝⎭对任意x ∈R 恒成立，()f x ∴在3π4x =处取得极值，即3ππ3π42k ϕ⨯+=+，解得7π3ππ,sin 0,π,,sin 4422k ϕϕϕϕϕϕ=-+=-≤∴=-=-=- ，可求得1a =-，A 错误；B.()()3ππ3,0,44f x x f f x ⎛⎫⎛⎫=-=∴ ⎪ ⎪⎝⎭⎝⎭的图象关于点π,04⎛⎫ ⎪⎝⎭对称，B 正确；C.将()f x 的图象向左平移π12个单位，得到()π3ππ3331242g x x x x ⎛⎫⎛⎫=+⨯-=-=- ⎪ ⎪⎝⎭⎝⎭，函数图象关于y 轴对称，C 正确；D.()3π2342f x x ⎛⎫=-≤- ⎪⎝⎭，即3π1sin 342x ⎛⎫-≤- ⎪⎝⎭，7π3π11π2π32π646k x k ∴+≤-≤+，解得23π231π2ππ363363k x k +≤≤+，由题意知π23π,1236x ⎡⎤∈-⎢⎥⎣⎦，符合条件的k 的取值为1,0-，当1k =-时，π7π3636x -≤≤，均在定义域内，满足条件，当0k =时，23π31π3636x ≤≤，此时仅有23π36x =满足条件，所以满足()22f x ≤-成立的x 的取值范围为π7π23π,363636⎡⎤⎧⎫-⋃⎨⎬⎢⎣⎦⎩⎭，D 错误.11.BD【解析】A.MN ∥,AC BMN ∠∴为直线MN 与AC 所成角，在BMN 中，根据余弦定理可知222cos 2BM MN BN BMN BM MN∠+-=⋅，422BM MN BN ======，代入求得cos 10BMN A ∠=错误；B.取AD 的中点E ，取CD F ，取11A D 的中点S ，连接11,,,,EF D E D F AS SM ，SM ∥,AB AS ∥BM ，所以四边形ABMS 是平行四边形，AS ∥BM 且AS ∥11,D E D E ∴∥1BM D E ∴∥平面BMN ，同理可得1D F ∥平面BMN ，1DT ∥平面,BMN T ∈平面ABCD ，所以点T 的运动轨迹为线段EF ，在1ΔD EF 中，过点1D 作1D T EF ⊥，此时1D T 取得最小值，由题意可知，11D E D F EF ===，1111sin sin sin 105D EF BMN D T D E D EF ∠∠∠====，B 正确；C.取MN 的中点1O ，连接11B O ，则1111O N O M O B ==，过点1O 作1OO ∥1BB ，且111322OO BB ==，OM ∴为外接球的半径，在1Rt MB N 中，MN =，2R OM ∴==，34ππ,33V R C ∴==球错误；D.由平面11AA D D ∥平面11BB C C 得，过点,,D M N 的平面必与11,AA C C 有交点，设过点,,D M N 的平面与平面11AA D D 和平面11BB C C 分别交于,DO PM DO ∴∥,PM 同理可得DP ∥,ON 过点,,D M N 的平面截长方体1111ABCD A B C D -所得的截面图形为五边形DPMNO ，如图所示，以D 为坐标原点，以1,,DA DC DD 所在直线分别为,,x y z 轴建立空间直角坐标系，设,AO m CP n ==，则()()()()()0,0,0,2,0,,0,4,,1,4,3,2,2,3D O m P n M N ，()()()()0,2,3,1,0,3,2,0,,0,4,ON m PM n DO m DP n ∴=-=-== ，DP ∥,ON DO ∥PM ，()()2323m n n m ⎧=-⎪∴⎨=-⎪⎩，解得2m n ==，DO DP ∴==ON PM MN ====，所以五边形DPMNO 的周长为DO DP ON PM MN ++++==+，D 正确.12.21,55⎡⎤-⎢⎥⎣⎦【解析】令()()()()2323x y m x y n x y m n x m n y +=++-+=-++，2131m n m n -=⎧∴⎨+=⎩，解得()()2121,,235555m n x y x y x y ==-∴+=+--+，1232,34x y x y ≤+≤≤-+≤ ，则()()22441323,555555x y x y ≤+≤-≤--+≤-，24435555x y ∴-≤+≤-，即21,55x y ⎡⎤+∈-⎢⎣⎦.13.111【解析】建立如图所示的平面直角坐标系，设1AD =，则3BC =，()()()()220,0,3,0,,,1,,,33B C A m n D m n E m n ⎛⎫∴+ ⎪⎝⎭，所以直线BD 的方程为1n y x m =+，直线CE 的方程为()2329n y x m =--，联立两直线方程求得()()666655,,,,1,0,,11111111m n m n O AO AD AB m n +-⎛⎫⎛⎫∴=-==-- ⎪ ⎝⎭⎝⎭ ，6511,511m x my AO xAD y AB n ny -⎧=-⎪⎪=+∴⎨⎪-=-⎪⎩ ，解得651,,111111x y x y ==∴-=.14.73π-【解析】依题意，可将四面体ABCD 补形为如图所示的直三棱柱ADE FCB -，AD 与BC 所成的角为30 ，30BCF ∠∴= 或150，设,CB x CF y ==，外接球半径记为R ，外接球的球心如图点O ，11113sin 23324ABCD CBF V DC S xy BCF xy ∠⎛⎫∴=⋅⋅=⨯⨯== ⎪⎝⎭ ，解得8xy =，在2Rt OCO 中，2222222223922sin 4BF R OC OO CO BF BCF ∠⎛⎫⎛⎫==+=+=+ ⎪ ⎪⎝⎭⎝⎭，在BCF 中，由余弦定理可得2222cos BF BC CF BC CF BCF ∠=+-⋅⋅，要使外接球表面积最小，则R 要尽可能小，则BCF ∠应取30 ，(2222BF x y xy ∴=+≥-，当且仅当x y =时取等，(22min 99732444R BF xy ∴=+=+=-所以外接球表面积的最小值2min min 4π73πS R ==-.。
2024届北京市清华大学中学生标准学术能力诊断性测试1月测试英语试卷
2024届北京市清华大学中学生标准学术能力诊断性测试1月测试英语试卷一、阅读理解From hawk hikes to private sleepovers at the zoo, there is a great selection of animal-related experiences available to groups. Here are some top options to get closer to various wonderful wildlife.Chester ZooThe newest attractions here are the Madagascar Lemur Walkthrough experience, which gives visitors the opportunity to walk alongside ring-tailed and red-ruffed lemurs, and the interactive American Wetland Aviary, which is home to birds like scarlet ibises and flamingos. Group rates are available for parties of 15 or more and there are various catering options, including sit-down meals at the restaurant at the heart of the zoo.ZSL Whipsnade ZooUntil September 2022, it is offering groups of up to 60 the opportunity to experience a private Nature Night, on which they’ll get to explore the zoo privately after the public has left, take part in activities like quizzes, camp overnight, and get up early for a private tour along the green trail before it reopens to the public again.West Midland Safari ParkThe latest attraction at the park is the new African Walking Trail. Opened in May, the trail features three viewpoints that allow visitors to see the park’s African animals on foot. There’s also a four-mile drive-through safari area with red panda, penguin and lorikeet areas. Groups of ten plus, arriving in the same vehicle, can save more than 40%.Knowsley Safari ParkThe five-mile safari drive through the site takes you past free-roaming lions, rhinos and more than 100cheeky baboons. There’s a foot safari area, where the highlight is the Amur Tiger Trail with transparent walled viewing areas where you can get nose-to-nose with 450-pound tigers. Groups of 15 people and more, arriving in one vehicle, qualify for special ticket rates.1．Who is the passage intended for?A．Animal-loving students.B．Forest hiking fans.C．Group tour organizers.D．Wildlife preservationists.2．Visitors can experience private tours in ________.A．Chester ZooB．ZSL Whipsnade ZooC．West Midland Safari ParkD．Knowsley Safari Park3．From the passage, we know that ________.A．delicious meals are offered to tourists in the four parksB．private tours are available in the four parksC．all the parks can provide driving-through servicesD．visitors can have access to walking trails in the four parksScientists regularly make vital new discoveries, but few can claim to have invented an entirely new field of science. Chemist Carolyn Bertozzi is one of them. Her discovery of biorthogonal chemistry (生物正交化学) in 2003 created a brand-new discipline of scientific investigation, which has enabled countless advances in medical science and led to a far greater understanding of biology at a molecular (分子的) level. On October 5, Bertozzi was awarded the Nobel Prize in Chemistry, jointly with two other professors. She is also the only woman to be awarded a Nobel Prize in science this year, after an all-male line-up in 2021.Bertozzi was the middle daughter of an MIT physics professor and a secretary. Few predicted that Bertozzi would be the most famous person in the family. While her academic performance was not bad in high school, she was fond of playing soccer. She end ed up being admitted to Harvard University. Despite her talent in soccer, she found it too time-consuming and quit the sport to devote herself to academics.But before becoming a rock star scientist, Bertozzi almost became an actual rock star. When she started at Harvard, she was tempted to major in music. That idea was “unpopular” with her parents, and she was timid about defying them. Instead, she chose the premed (医学预科的) trackthat included classes in math and sciences, and declared herself a biology major at the end of her first year of college.Her interest in music did not completely fall by the wayside, however. Bertozzi played keyboards and sang backup vocals for a hair metal band. Bertozzi, however, did not play with the band for long. Once the band’s practices and performances conflicted with her labs and classes, there was only one outcome.Plus, she’d soon have organic chemistry to think about a course which is infamous for weeding out pre-meds. Without any clear career ambitions up to that point, Bertozzi had been thinking about possibly becoming a doctor when, in her sophomore year (大二学年), she suddenly fell so head over heels in love with her chemistry course that she couldn’t tear herself away from her textbooks long enough to go out on Saturday nights. A torture to many was pure pleasure for her. Bertozzi changed her major from biology to chemistry a year later.Bertozzi has sometimes joked about her having missed out on her chance to follow Morello to LosAngeles. “I didn’t get on that bus, and my playing is now limited to ‘The Wheel's on the Bus Go Round,’ I’m waiting for my sons to get old enough to appreciate 1980s heavy metal!”4．Which of the following statements is TRUE according to the passage?A．Bertozzi is one of those scientists who made significant new discoveries.B．Bertozzi was the only female to win a Nobel Prize in science in 2021.C．Bertozzi played keyboards and sang backup vocals throughout her college years.D．Bertozzi initially planned to become a doctor.5．The underlined word in Para. 3 means ________.A．tell B．disobey C．approach D．threaten6．The organic chemistry course Bertozzi took was known to be ________.A．easy and enjoyableB．difficult to pass for pre-med studentsC．popular among hair metal band playersD．a required course for all college students7．What kind of person do you think Carolyn Bertozzi is?A．Brave and sympathetic.B．Athletic and critical.C．Humble and passionate.D．Talented and creative.Willie Sutton, a once celebrated American criminal, was partly famous for saying he robbed banks because “that’s where the money is.” Actually, museums are where the money is. In a single gallery there can be paintings worth more, taken together, than a whole fleet of jets. And while banks can hide their money in basements, museums have to put their valuables in plain sight.Nothing could be worse than the thought of a painting as important as The Scream, Edvard Munch’s impressive image of a man screaming against the backdrop of a blood-red sky, disappearing into a criminal underworld that doesn’t care much about careful treatment of art works. Art theft is a vast problem around the world. As many as 10,000 precious items of all kinds disappear each year. And for smaller museums in particular, it may not be a problem they can afford to solve. The money for insurance on very famous pictures would be budget destroyers even for the largest museums.Although large museums have had their share of embarrassing robberies, the greatest problem is small institutions. Neither can afford heavy security. Large museums attach alarms to their most valuable paintings, but a modest alarm system can cost $500,000 or more. Some museums are looking into tracking equipment that would allow them to follow stolen items once they leave the museums. But conservators are concerned that if they have to insert something, it might damage the object. Meanwhile, smaller museums can barely afford enough guards, relying instead on elderly staff.Thieves sometimes try using artworks as money for other underworld deals. The planners of the 2006 robbery of Russborough House near Dublin, who stole 18 paintings, tried in vain to trade them for Irish Republican Army members held in British prison. Others demand a ransom (赎金) from the museum that owns the pictures. Once thieves in Frankfurt, Germany, made off with two major works by J.M.W. Turner from the Tate Gallery in London. The paintings, worth more than $80 million, were recovered in 2012 after the Tate paid more than $5 million to people having “information” about the paintings. Though ransom is illegal in Britain, money for looking into a case is not, provided that police agree the source of the information is unconnected to the crime. All the same, where information money end s and ransom begins is often a gray area.8．Why do smaller museums face a greater challenge in preventing art theft?A．They lack experienced staff.B．They cannot afford high-tech security systems.C．They do not have valuable artworks.D．They lack interest in art conservation.9．What is the concern of conservators regarding the use of tracking equipment to prevent art theft?A．It might damage the artwork.B．It is too expensive for smaller museums.C．It is difficult to insert into the paintings.D．It is ineffective for valuable paintings.10．From Paragraph 4, we can learn that ________.A．the thieves demanded a ransom from the Tate GalleryB．the Tate Gallery regained the lost paintings illegallyC．the money paid was considered an information fee, not a ransomD．the police requested the Tate Gallery to pay the money11．The purpose of this passage is ________.A．to remind criminals to protect and preserve the paintingB．to give suggestions on how to avoid the crimes of art theftC．to urge museums to set up more advanced security systemsD．to make people aware of art theft and the necessity of good security systemsWho cares if people think wrongly that the Internet has had more important influences than the washing machine? Why does it matter that people are more impressed by the most recent changes?It would not matter if these misjudgments were just a matter of people’s opinions. However, they have real impacts, as they result in misguided use of scarce resources.The fascination with the ICT(Information and Communication Technology) revolution, represented by the Internet, has made some rich countries wrongly conclude that making things is so “yesterday” that they should try to live on ideas. This belief in “post-industrial society” has ledthose countries to neglect their manufacturing sector (制造业) with negative consequences for their economies.Even more worryingly, the fascination with the Internet by people in rich countries has moved the international community to worry about the “digital divide” between the rich countries and the poor countries. This has led companies and individuals to donate money to developing countries to buy computer equipment and Internet facilities. The question, however, is whether this is what the developing countries need the most. Perhaps giving money for those less fashionable things such as digging wells, extending electricity networks and making more affordable washing machines would have improved people’s lives more than giving every child a laptop computer or setting up Internet centres in rural villages, I am not saying that those things are necessarily more important, but many donators have rushed into fancy programmes without carefully assessing the relative long-term costs and benefits of alternative uses of their money.In yet another example, a fascination with the new has led people to believe that the recent changes in the technologies of communications and transportation are so revolutionary that now we live in a “borderless world”. As a result, in the last twenty years or so, many people have come to believe that whatever change is happening today is the result of great technological progress, going against which will be like trying to turn the clock back. Believing in such a world, many governments have put an end to some of the very necessary regulations on cross-border flows of capital, labour and goods, with poor results.Understanding technological trends is very important for correctly designing economic policies, both at the national and the international levels, and for making the right career choices at the individual level. However, our fascination with the latest, and our under valuation of what has already become common, can, and has, led us in all sorts of wrong directions.12．What are the effects of people’ misjudgments on the influences of new technology?A．It stimulates innovation.B．It affects their personal opinions.C．It influences their use of resources.D．It leads to improved technology.13．Why is the “digital divide” a concern related to the fascination with the Internet in rich countries?A．It leads to competition between rich and poor countries.B．It results in a lack of access to technology in developing countries.C．It increases the cost of computer equipment in rich countries.D．It promotes global digital cooperation.14．From Paragraph 4, we know that ________.A．donating for technology is always the better optionB．the author does not provide opinions on this matterC．donating for technology and basic needs should be balancedD．donating for basic needs should be prioritized over technology15．What is the passage mainly about?A．Significance of information and communication technology.B．Serious consequences of over-emphasizing high technology.C．Technological trends guiding economic policy making.D．How to use donation money in the new age.There’s a Symphony Just Below the Surface — Can You Hear It?Imagine it’s your birthday, and your friends and family pool their money to get you the best gift you can imagine: tickets for fabulous seats to see your favorite musical act. But what if you got to the venue and something terrible had just happened to you? 16 . Even while facing the prospect of extreme difficulty in your life, you are so thrilled to see your favorite group that fora couple of hours, you can put all of that behind you.17 . That is the ability to suspend our fears and worries and focus on what we love. In the example of the concert, we know that when the music ends, we may go back to our concerns, but while it’s playing, there is nothing we can do about them, so we might as well just give in.Life always has its music, and we don’t need to be front-row center at a concert to hear it. Throughout our lives, no matter what else is going on, a melody is present. But we are often so focused on the present moment that we fail to hear the melody. 18 .We can become magnificent listeners to life, with enough practice. And let’s face it, this is something we were born to do, so the skill is there, waiting for us to employ it. We can tap into the music, and when we do find ourselves distracted from it, we can use consciousness to bring us right back. It is as simple as saying, “OK, I’m distracted again; I am going to start listening again.”19 .Life is always playing music, but we have to listen, and we listen by being present. We can do this. 20 . When we do this, we’ll discover that the symphony inside of us is magnificent.A．As humans, we have been given a wonderful giftB．These feelings may last several minutes or even last several hoursC．In a word, wisdom and patience are the things that listening to the music of life requires D．Soon, we will find that we have to redirect ourselves less and less, and we hear the music more and moreE．You’d broken your knees, say, or you learned of a failure of examF．The noise of our worry drowns out all the other things we might otherwise hear and enjoy G．We just need to realize and engage with the music of life that is always playing二、完形填空When Alex Lin was 11 years old, he read a(an) 21 article in the newspaper, which said that people were 22 old computers in backyards, throwing TVs into streams, and dumping(丢弃) cell phones in the garbage. This was dangerous because e waste contains harmful 23 that can leak into the environment, getting into crops, animals, water supplies—and people.Alex was really worried and decided to make it next project for WIN—the Westerly Innovations Network. Alex and six of his friends had 24 this organization to help solve community problems two years before.But what could they do about this project with e-waste? The team spent several weeks gathering information about the harmful chemicals in e waste and their 25 on humans. They learned how to dispose (处置) of e-waste 26 and how it could be recycled. Then, they sent out a survey and found only one in eight knew what e-waste was, let alone how to dispose of it.Alex and his friends went into 27 . They advertised in the local newspaper and 28 notices to students, asking residents to bring their 29 electronics to the school parking lot. The drive lasted two days, and they 30 over 9,500 kilograms of e waste.The next step was to set up a long-term e-waste drop-off center for the town. After some research, they’d learned that reusing is the best way to 31 electronic devices and it is seven times more 32 than recycling. So, they began learning to refurbish(翻新) computers themselves and distributed them to students who didn’t have their own. In this way, they could help students in the area and protect the environment at the same time.For a 33 solution to e waste, the drop off center wasn’t enough. Laws would have to be passed. In 2016,WIN helped 34 for an e waste bill in their town, which required companies that manufactured or sold electronics to take back e waste. The bill clearly 35 the dumping of e waste.Because of the work of WIN, more and more people, like Alex and his team, are getting the message about safe disposal of e-waste. As Alex says, “Today’s technology should not become tomorrow’s harmful garbage.21．A．alarming B．terrifying C．embarrassing D．inspiring 22．A．carrying B．burying C．taking D．destroying 23．A．subjects B．restrictions C．bacterial D．chemicals 24．A．developed B．recognized C．formed D．restored 25．A．affects B．effects C．consequences D．attempts 26．A．properly B．instantly C．constantly D．gradually 27．A．enthusiasm B．action C．behavior D．energy 28．A．distributed B．contributed C．established D．conducted 29．A．unexpected B．unwanted C．useless D．meaningless 30．A．obtained B．collected C．ordered D．donated 31．A．break down B．take in C．expose to D．deal with 32．A．efficient B．economical C．effective D．beneficial 33．A．lasting B．physical C．original D．crucial 34．A．push B．delay C．accept D．pass 35．A．prevents B．permits C．predicts D．forbids三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
清华北大真题集
清华北⼤真题集2015年清华⼤学领军计划测试（物理）注意事项：1.2016清华领军计划测试为机考，全卷共100分，考试时间与数学累积120分钟：2.考题全部为不定项选择题，本试卷为回忆版本，故有些问题改编为填空题。
1、在α粒⼦散射实验中，以下1到5五个区域哪个可能是中⼼原⼦存在的区域？2、质量为m ，电阻为R 的圆环在如图的磁场中下落，稳定时速度为v 。
求匀速下落时电动势，有以下两种计算⽅案。
⽅法⼀：由受⼒平衡RvL B mg 22=有结论 mgRv =ε⽅法⼆：由功能关系 mgv P R = RP R 2ε=有结论 mgvR =ε问：关于以上哪种⽅法说法正确的是（）A.都正确B.都不正确C.只有⽅案⼀正确D.只有⽅案⼆正确3、理想⽓体做kV p =的准静态过程，已知定容⽐热V C 和R ，求该过程的⽐热C4、如图所⽰，光滑且不计电阻的导轨上有⼀⾦属棒，⾦属棒电阻为R ，初速度为s m v /10=，空间中有恒定的垂直于导轨平⾯的磁场，磁感应强度为B ，当⾦属棒减速到10v 时，⽤时s 1，速度识别器最低记录是s m /001.0，求总共记录的该导体棒运动时间为多少？5、⾼为H出平抛⼀物体，同时在其正下⽅⽔平地⾯斜抛⼀物体，⼆者同时落到同地，则斜抛物体的射⾼为。
6、有⼀厚度为D的透明玻璃砖，⼀束⽩光以⼊射⾓60°⾓射⼊。
（1）求最早射出⾊光的折射率（玻璃折射率最⼩值为m inn）（2）若⽩⾊只有红黄绿三种颜⾊（并给出折射率）问那种⾊光最先射出？7、⼩磁铁在铝制空⼼杆中运动（⽆裂缝、有裂缝、有交错的矩形裂孔），则先落地的⼀个是哪⼀个？8、均匀带电半圆环，⼀半带正电，⼀半带负电，电荷密度为λ，求P点的场强和电势。
9、⼀个⼈在岸上以速度v⽔平拉船，岸⾼度为h，绳⼦与河夹⾓为θ。
此时船的速度和加速度为？2015年清华⼤学暑期夏令营测试（物理）本试卷共100分，考试⽤时90分钟。
注意事项：1.答卷前，考⽣务必将⾃⼰的姓名、准考证号、考点名称填写在答题卡上，并在规定位置粘贴考试⽤条形码。
高考清华大学语文试卷
第一部分：现代文阅读一、阅读下面的文字，完成下面小题。
【甲】浩渺星空，无数星河在宇宙深处闪烁，其中最亮的那颗，当属太阳。
太阳是太阳系的中心，是地球上一切生命的能量来源。
太阳在诞生之初，经过无数次的核聚变，形成了光和热，照亮了地球，温暖了万物。
然而，太阳并非永恒不变，它的寿命大约只有100亿年。
当太阳的生命走到尽头，它将变成一颗红巨星，然后成为一颗白矮星，最终消失在宇宙的深处。
【乙】人类对太阳的研究从未停止。
从古代的太阳崇拜到现代的太阳观测，人类对太阳的认识不断深化。
近年来，随着科技的进步，人类对太阳的了解更加全面。
通过卫星观测、望远镜观测等方式，科学家们发现了太阳黑子、太阳风等许多太阳现象。
这些现象不仅揭示了太阳的内部结构，也对地球的气候变化、无线电通信等方面产生了重要影响。
【丙】太阳对于地球的重要性不言而喻。
太阳光和热为地球上的生物提供了生存的基本条件。
没有太阳，地球将是一个寒冷、黑暗的世界。
此外，太阳还影响着地球的气候、季节和天气。
太阳辐射的变化会导致地球气候的变化，进而影响人类的生产和生活。
【丁】那么，我们该如何更好地利用太阳的能量呢？目前，人类已经掌握了太阳能电池、太阳能热水器等太阳能利用技术。
这些技术将太阳光转化为电能或热能，为人类的生活提供了便利。
未来，随着科技的不断发展，太阳能的利用将更加广泛，有望成为人类的主要能源之一。
【小题1】下列对文中加点词语的解释，不正确的一项是（）A. 永恒不变：永远不改变。
B. 深化：向更深的程度发展。
C. 深刻：达到事情或问题的本质。
D. 全面：各个方面的、完整的。
【小题2】下列对文章内容的理解和分析，不正确的一项是（）A. 文章以“浩渺星空，无数星河在宇宙深处闪烁，其中最亮的那颗，当属太阳”开篇，引出说明对象“太阳”。
B. 文章从太阳的诞生、发展、影响等方面进行了说明，条理清晰。
C. 文章通过举例子、列数字等方式，使说明内容更加生动形象。
D. 文章结尾展望了太阳能的利用前景，表明了作者对太阳能的期待。
2023清华大学考博英语模拟题1(考卷附答案)
2012清华大学考博英语模拟题1（考卷）一、阅读理解（每题2分，共40分）Section A下列每篇文章后面有5个问题，请根据文章内容回答问题。
Passage 1Questions:1. When did the concept of sustainable development first appear in the world?2. Which organization proposed the concept of sustainable development?3. What is the main idea of sustainable development?4. What does sustainable development emphasize?5. According to the passage, what is the relationship between sustainable development and the three aspects of economy, society and environment?Passage 2Questions:1. What is the main idea of this passage?2. What are the advantages of the Internet mentioned in the passage?3. What are the disadvantages of the Internet mentioned in the passage?4. How do countries address the issues of Internet security?5. Why is it important to safeguard Internet security?Section B下列每篇文章后面有5个问题，请根据文章内容回答问题。
清华大学数学试卷高三
一、选择题（每题5分，共50分）1. 下列各数中，绝对值最小的是（）A. -3B. -2C. 0D. 12. 若函数f(x) = 2x + 3，则f(-2)的值为（）A. -1B. 1C. 3D. 73. 已知数列{an}的通项公式为an = 3n - 2，则第10项an的值为（）A. 28B. 29C. 30D. 314. 在△ABC中，若∠A = 30°，∠B = 45°，则∠C的度数为（）A. 105°B. 120°C. 135°D. 150°5. 下列不等式中，正确的是（）A. 2x > xB. x + 1 > xC. x^2 > xD. x^2 > -x6. 已知等差数列{an}的首项为a1，公差为d，则第n项an的表达式为（）A. an = a1 + (n - 1)dB. an = a1 - (n - 1)dC. an = a1 + ndD. an = a1 - nd7. 下列函数中，在定义域内单调递增的是（）A. f(x) = x^2B. f(x) = -x^2C. f(x) = x^3D. f(x) = -x^38. 若等比数列{an}的首项为a1，公比为q，则第n项an的表达式为（）A. an = a1 q^(n - 1)B. an = a1 / q^(n - 1)C. an = a1 q^nD. an = a1 / q^n9. 下列数列中，是等差数列的是（）A. 1, 3, 5, 7, ...B. 2, 4, 8, 16, ...C. 1, 2, 4, 8, ...D. 1, 3, 6, 10, ...10. 若函数f(x) = ax^2 + bx + c的图像开口向上，则a的取值范围是（）A. a > 0B. a < 0C. a ≥ 0D. a ≤ 0二、填空题（每题5分，共25分）11. 若等差数列{an}的首项为a1，公差为d，则第n项an的表达式为______。
2023年清华大学强基计划数学真题试卷
2023年清华大学强基计划数学试卷
1、有六面旗，两面蓝，两面红，两面黄，除颜色外完全相同，从这些旗子中去除若干面（至少一面），从上到下悬挂在同一个旗杆上，可以组成一个信号序列，则不同的信号序列共有多少种？
2、已知a，x，k∈R，ln(x+a)−kax=0对任意的a∈R恒成立，求k 的最小值
3、11个黑球，9个红球，依次取出，剩下全是一种颜色就结束，求最后只剩下红球的概率。
4、三个复数的模分别为1，5，5√2，且这三个复数实部虚部均为整数，则这三个复数的积有多少个可能值。
5、椭圆x2
4+x2
3
=1，F为左焦点，A，B为椭圆上两点且FA=5，FB=8，
求直线AB的斜率k的范围。
6、数列x x满足x1=3
2
，x x+1=x x x，求使该数列{x x}有极限的x的最大值。
7、4
x+1+9
x+2
+16
x+3
=(4x+5)(2−x)有几个正实数解？
8、已知点M(8，1)，过点N(1，0)的直线L上有一个动点P，则|xx|+ 2|xx|的最小值为（）
9、两个人甲和乙，数字为2∼30之间的共29个自然数，现找出两个不同的数，把其和告诉甲，把其和告诉乙。
甲说：“虽然我不知道是哪两个数，但是肯定乙也不知道”，再问乙，乙说：“本来我不知道，但是听到甲说这句话，现在为我知道了”，甲听到乙说他知道了，然后就说：“现在我也知道了”，那么这两个数是多少？
10、p，q都为质数，p整除7p+1，q整除7p+1，有多少组p，q
11、正整数a，b，c，x，y，z满足：ax=b+c，by=c+a，cz=a+b，则xyz 的可能值有（）。
清华大学语文考试试卷真题
1. 下列哪个成语的意思与“不耻下问”相同？A. 知无不言B. 勤学好问C. 博学多才D. 诲人不倦2. 下列哪个选项不是唐代诗人白居易的诗句？A. 离离原上草，一岁一枯荣。
B. 举头望明月，低头思故乡。
C. 春风得意马蹄疾，一日看尽长安花。
D. 春风又绿江南岸，明月何时照我还？3. 下列哪个选项是《红楼梦》中的人物？A. 林黛玉B. 贾宝玉C. 萧红D. 鲁迅4. 下列哪个选项不是《三国演义》中的故事？A. 赤壁之战B. 三顾茅庐C. 长坂坡之战D. 水浒传5. 下列哪个选项是《水浒传》中的108位好汉之一？A. 宋江B. 武松C. 李逵D. 关羽二、填空题（每题2分，共10分）1. 《诗经》是我国最早的一部诗歌总集，收录了自______年至______年的诗歌。
2. “______，不亦说乎？”出自《论语》，这句话表达了孔子对学习的看法。
3. “青青子衿，悠悠我心。
但为君故，沉吟至今。
”这句诗出自《诗经》中的______。
4. 《西游记》是我国四大名著之一，主要讲述了唐僧师徒四人取经的故事。
其中，孙悟空、猪八戒、沙僧分别代表了______、______、______。
5. 《红楼梦》的作者是______，书中描绘了贾、王、史、薛四大家族的兴衰历程。
三、阅读题（每题10分，共30分）1. 阅读下面的文言文，回答问题。
《陈情表》节选臣密言：臣以险衅，夙遭闵凶。
生孩六月，慈父见背；行年四岁，舅夺母志。
祖母刘悯臣孤弱，躬亲抚养。
臣少多疾病，九岁不行，零丁孤苦，至于成立。
既无伯叔，终鲜兄弟，门衰祚薄，晚有儿息。
外无期功强近之亲，内无应门五尺之僮。
茕茕孑立，形影相吊。
而刘氏日薄西山，气息奄奄，人命危浅，朝不虑夕。
臣无祖母，无以至今日；祖母无臣，无以终余年。
祖孙二人，更相为命，是以区区不能废远。
臣密今年四十有四，祖母刘今年九十有六，是臣尽节于陛下之日长，报养刘之日短也。
乌鸟私情，愿乞终养。
臣密言：臣之辛苦，非独蜀之人士及二州牧伯所见明知，皇天后土，实所共鉴。
2009年清华大学外国留学生本科入学考试.docx答案
2009年清华⼤学外国留学⽣本科⼊学考试.docx答案2009年清华⼤学外国留学⽣本科⼊学考试通识试题⼀、选择题：（6分）1、热带有（B）个季节。
A、1B、2C、3D、42、克⾥奥佩特拉Cleopatra是埃及（C）王朝的君主。
A、波旁B、都铎C、托勒密D、孔雀E、伊丽莎⽩3、三星堆位于中国的（A）省。
A、四川B、⼭西C、陕西D河南E、浙江4、“⽕把节”是（D）族的传统节⽇。
A、藏族B、壮族C、汉族D、彝族E、回族5、美国的国务卿相当于中国的（B）A、总理B、国防部长C、外交部长D、国务委员E、政协主席6、（D）是⾮洲最⼤的⽯油⽣产国。
A、刚果B、南⾮C、埃及D、尼⽇利亚E、埃塞俄⽐亚⼆、填空题：（8分）1、法国⼤⾰命发⽣在1789年，7⽉14⽇巴黎⼈民攻占了巴⼠底狱。
2、“民有、民治、民享”的纲领性⼝号是林肯在美国葛底斯堡Gettysburg演说中提出来的。
3、《兰亭序》是中国东晋“书圣”王羲之的代表作。
4、欧洲最长的河流是伏尔加河。
5、H2SO4是硫酸的化学式。
⽐重最轻的⾦属是锂。
6、E=mc2是科学家爱因斯坦在狭义相对论⾥提出来的质能公式。
三、分析题1、现有125⼈，其中有40⼈说韩国语，60⼈说英语，80⼈说汉语，说两种语⾔的有50⼈，说三种语⾔的有10⼈。
问不说这些语⾔的有⼏⼈。
答：125-（40+60+80-50-10X2）=152、张⽼头有七个⼦⼥，从⽼⼤到⽼七分别是A 、B 、C 、D 、E 、F 、G 。
⽬前知道如下情况：(1)A 有3个妹妹； (2)B 有1个哥哥; (3)C 是⼥的，她有2个妹妹；（4）D 有2个弟弟；（5）E 有2个姐姐；（6）F 也是⼥的，和G 没有妹妹。
请根据这些条件，推算出张⽼头的七个⼦⼥，谁是男的，谁是⼥的。
答：A 、B 、E 、G 是男的，C 、D 、F 是⼥的。
2010年⼀、选择题：（6分）1、物理学中每秒内振动的次数——( 1 )来描述物体振动的快慢。
清华大学期末考试试卷语文
语文一、选择题（每题2分，共20分）1. 下列词语中，字形、字音完全正确的是（）A. 踟蹰满志B. 欣欣向荣C. 脍炙人口D. 满目琳琅2. 下列句子中，没有语病的是（）A. 为了提高学生的综合素质，学校决定举办一场丰富多彩的文艺汇演。
B. 随着科技的发展，我们的生活变得越来越便捷，但同时也带来了许多新的问题。
C. 通过这次社会实践，我深刻体会到了理论与实践相结合的重要性。
D. 虽然天气很冷，但他还是坚持去参加晨跑。
3. 下列词语中，不属于成语的是（）A. 一丝不苟B. 独树一帜C. 一马当先D. 一帆风顺4. 下列句子中，修辞手法运用正确的是（）A. 月亮升上了天空，像一块银盘悬挂在夜空中。
B. 老师用鼓励的目光注视着我们，仿佛在说：“你们是最棒的！”C. 小明跑得像兔子一样快。
D. 夜晚的星空，像一幅美丽的画卷。
5. 下列句子中，标点符号使用错误的是（）A. 我国拥有五千年的悠久历史，是世界上文明古国之一。
B. 她的笑声，如同清泉一般，让人感到心旷神怡。
C. 他问：“今天天气怎么样？”D. 你看，那边的山，多像一只雄狮！二、填空题（每空2分，共20分）6. 下列诗句中，出自唐代诗人白居易的是：（）7. 下列词语中，形近字组词正确的是：（）8. 下列句子中，缺少主语的是：（）9. 下列句子中，关联词使用错误的是：（）10. 下列句子中，句子成分搭配不当的是：（）三、简答题（每题10分，共30分）11. 请简要分析下列诗句的意境：（5分）山重水复疑无路，柳暗花明又一村。
12. 请结合实际，谈谈你对“读书破万卷，下笔如有神”这句话的理解。
（10分）13. 请简要说明下列词语的词性及其在句子中的作用：（10分）① 精彩纷呈② 勤能补拙四、作文（40分）14. 请以“梦想的力量”为题，写一篇不少于500字的作文。
注意：文章应立意明确，结构完整，语言流畅，内容充实。
清华大学考博英语模拟试卷13(题后含答案及解析)
清华大学考博英语模拟试卷13(题后含答案及解析)题型有：1. Structure and V ocabulary 2. Cloze 3. Reading Comprehension 4. English-Chinese Translation 5. WritingStructure and V ocabulary1．Because of the massive oil spillage in the gulf, both the plant and animal lives in the area are in______.A．destinyB．amenityC．jeopardyD．tragedy正确答案：C解析：各项的意思是：destiny命运，天命；amenity娱乐设施；jeopardy危险，危难；tragedy惨事，灾难；悲剧。
根据句意判断，答案是C。
知识模块：词汇2．The membership card entitled him____certain privileges in the club.A．onB．inC．atD．to正确答案：D解析：entitle sb．to sth．是习惯用法，表示“使某人有权利(或资格)得到……”。
3．Shy people never ______ set out to attract attention of other people.A．willinglyB．voluntarilyC．decidedlyD．deliberately正确答案：B解析：willingly 自愿地，欣然地。
voluntarily自发地。
decidedly果断地。
deliberately故意地。
4．His plan was ______ by the committee.A．rejectedB．dejectedC．objectedD．projected正确答案：A5．It is strictly ______ that access to confidential documents is denied to all but a few.A．securedB．forbiddenC．regulatedD．determined正确答案：B解析：本题意为“拒绝一些人接触机密文件是被严格禁止的”。
2022年清华大学领军计划自主招生数学试题(问卷)
2022年清华大学领军方案自主招生数学试题〔问卷〕2022年清华大学领军方案自主招生数学试题说明：本试卷共30小题，共100分。
在每题给出的四个选项中，有一个或多个选项是符合题目要求的。
选不选对，得总分值；选对但不全的，得局部分；有选错的，得0分。
112?2??? 〔〕，那么 ?isin1?z1?z23313 A.0 B.1 C. D.222.设{an}为等差数列，p,q,k,l为正整数，那么“p?q?k?l〞是“ap?aq?ak?al〞的〔〕1.设复数z?cosA.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件 3.设A,B是抛物线y?x上的两点，O是坐标原点，假设OA?OB，那么〔〕 A.|OA|?|OB|?2 B.|OA|?|OB|?22C.直线AB过抛物线y?x的焦点D.O到直线AB的距离小于等于1 4.设函数f(x)的定义域为(?1,1)，且满足：①f(x)?0,x?(?1,0)；②f(x)?f(y)?f(22x?y),x,y?(?1,1)。
那么f(x)为〔〕 1?xyA.奇函数 B.偶函数C.减函数 D.有界函数 5.如图，直线y?kx?m与曲线y?f(x)相切于两点，那么F(x)?f(x)?kx有〔〕A.2个极大值点B.3个极大值点C.2个极小值点D.3个极小值点 6.?ABC的三边长分别为a,b,c。
假设c?2,?C??3，且sinC?sin(B?A)?2sin2A?0，那么〔〕A.b?2aB.?ABC的周长为2?232323 D.?ABC的外接圆半径为 332x7.设函数f(x)?(x?3)e，那么〔〕 A.f(x)有极小值，但无最小值 B.f(x)有极大值，但无最大值6 C.假设方程f(x)?b恰有一个实根，那么b?3e6D.假设方程f(x)?b恰有三个不同实根，那么0?b?3e2222228.A?{(x,y)|x?y?r},B?{(x,y)|(x?a)?(y?b)?r}，AB?{(x1,y1),(x2,y2)}，那么〔〕C.?ABC的面积为222A.0?a?b?2r B.a(x1?x2)?b(y1?y2)?022C.x1?x2?a,y1?y2?b D.a?b?2ax1?2by19.非负实数x,y,z满足4x?4y?z?2z?3，那么5x?4y?3z的最小值为〔〕 A.1B.2C.3D.4 10.设数列{an}的前n项和为Sn，假设对任意正整数n，总存在正整数m，使得Sn?am，那么〔〕 A.{an}可能为等差数列 B.{an}可能为等比数列 1222C.{an}的任意一项均可写成{an}的两项之差D.对任意正整数n，总存在正整数m，使得an?Sm11.运动会上，有6名选手参加100米比赛，观众甲猜想：4道或5道的选手得第一名；观众乙猜想：3道的选手不可能得第一名；观众丙猜想：1，2，6道选手中的一位获得第一名；观众丁猜想：4，5，6道的选手都不可能获得第一名。