2020届安徽省合肥七中、肥西农兴中学、合肥32中、合肥五中高三冲刺高考“最后一卷”英语试题
安徽省合肥七中、农兴中学、合肥32中、合肥五中2020届高三冲刺高考最后一卷英语试题及答案解析

绝密★启用前安徽省合肥七中、农兴中学、合肥32中、合肥五中2020届高三毕业班下学期冲刺高考最后一卷英语试题(时间:120分钟满分:150分)第一部分听力(共两节,满分30分)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,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What does the man like about the play?A. The story.B. The ending.C. The actor.2. Which place are the speakers trying to find?A. A hotel.B. A bank.C. A restaurant.3. At what time will the two speakers meet?A. 5:20.B. 5:10.C. 4:40.4. What will the man do?A. Change the plan.B. Wait for a phone call.C. Sort things out.5. What does the woman want to do?A. See a film with the man.B. Offer the man some help.C. Listen to some1great music.第二节(共15小题:每小题1.5分,满分22.5分)请听下面5段对话。
每段对话后有几个小题,从题中所给出的A、B、C三个选项种选出最佳选项,并标在试卷的相应位置。
听每段对话前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题给出5秒钟的作答时间。
每段对话读两遍。
听第6段材料,回答6、7题6. Where is Ben?A. In the kitchen.B. At school.C. In the park.7. What will the children do in the afternoon?A. Help set the table.B. Have a party.C. Light the candles.听第7段材料,回答第8、9题8. What are the two speakers talking about?A. A Family holiday.B. A business trip.C. A travel plan.9. Where did Rachel go?A. Spain.B. Italy.C. China.听第8段材料,回答第10至12题。
专题13 利用导数解决函数的极值、最值-学会解题之高三数学万能解题模板(2021版)【原卷版】

学习界的专题13 利用导数解决函数的极值、最值【高考地位】导数在研究函数的极值与最值问题是高考的必考的重点内容,已由解决函数、数列、不等式问题的辅助工具上升为解决问题的必不可少的工具,特别是利用导数来解决函数的极值与最值、零点的个数等问题,在高考中以各种题型中均出现,对于导数问题中求参数的取值范围是近几年高考中出现频率较高的一类问题,其试题难度考查较大.类型一利用导数研究函数的极值例1 已知函数f (x) =+ ln x ,求函数f (x)的极值.x【变式演练1】(极值概念)【西藏日喀则市拉孜高级中学2020 届月考】下列说法正确的是()A.当f '(x0 ) = 0 时,则f (x0 ) 为f (x) 的极大值B.当f '(x0 ) = 0 时,则f (x0 ) 为f (x) 的极小值C.当f '(x0 ) = 0 时,则f (x0 ) 为f (x) 的极值D.当f (x0 ) 为f (x) 的极值且f '(x0 ) 存在时,则有f '(x0 ) = 0【变式演练2】(图像与极值)【百师联盟2020 届高三考前预测诊断联考全国卷1】如图为定义在R 上的函数f (x)=ax3 +bx2 +cx +d (a ≠ 0)的图象,则关于它的导函数y =f '(x)的说法错误的是()A.f '(x)存在对称轴B.f '(x)的单调递减区间为⎛-∞,1 ⎫2 ⎪ ⎝⎭C.f '(x)在(1, +∞)上单调递增D.f '(x)存在极大值【变式演练3】(解析式中不含参的极值)【江苏省南通市2020 届高三下学期高考考前模拟卷】已知函数f (x)=(ax2 +x +1)e x ,其中e是自然对数的底数,a ∈R .(1)当a = 2 时,求f (x )的极值;(2)写出函数f (x )的单调增区间;(3)当a = 0 时,在y 轴上是否存在点P,过点P 恰能作函数f (x)图象的两条切线?若存在,求出所有这样的点;若不存在,请说明理由.【变式演练4】(解析式中含参数的极值)【四川省德阳市2020 届高三高考数学(理科)三诊】已知函数f (x )=ax - 2 ln x - 2 ,g (x )=axe x - 4x .(1)求函数f (x )的极值;(2)当a > 0 时,证明:g (x )- 2 (ln x -x +1)≥ 2 (ln a - ln 2 ).【变式演练5】(由极值求参数范围)【黑龙江省哈尔滨一中2020 届高三高考数学(理科)一模】已知函数学习界的007f ( x ) = x ln x -1 (m + 1) x2 - x 有两个极值点,则实数m 的取值范围为()2A . ⎛ - 1 , 0⎫B . ⎛-1, 1 -1⎫C . ⎛ -∞, 1 -1⎫ )D . (-1, +∞)e ⎪ e⎪ e⎪ ⎝ ⎭ ⎝⎭⎝⎭【变式演练 6】(由极值求其他)【四川省江油中学 2020-2021 学年高三上学期开学考试】已知函数f ( x ) = 1x 3 + ax 2 + bx (a , b ∈ R ) 在 x = -3 处取得极大值为 9.3(1) 求 a , b 的值;(2) 求函数 f (x ) 在区间[-4, 4] 上的最大值与最小值.类型二 求函数在闭区间上的最值万能模板内 容使用场景 一般函数类型解题模板第一步 求出函数 f (x ) 在开区间(a , b ) 内所有极值点;第二步 计算函数 f (x ) 在极值点和端点的函数值;第三步 比较其大小关系,其中最大的一个为最大值,最小的一个为最小值.例 2 【河南省天一大联考 2020 届高三阶段性测试】已知函数 f ( x ) = ln x - x , g ( x ) = ax 2+ 2x (a < 0) .(1) 求函数 f( x ) 在⎡1 , e ⎤上的最值; ⎢⎣ e ⎥⎦(2) 求函数 h( x ) = f (x ) + g (x ) 的极值点.【变式演练 7】(极值与最值关系)【安徽省皖江联盟 2019-2020 学年高三上学期 12 月联考】已知函数 f ( x ) 在区间(a , b ) 上可导,则“函数 f ( x ) 在区间(a , b ) 上有最小值”是“存在 x 0 ∈(a ,b ) ,满足 f '(x 0 ) = 0 ”的⎨ 1 ()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【变式演练 8】(由最值求参数范围)【湖北省武汉市 2020 届高三下学期六月模拟】若函数⎧a ln x - x 2 - 2 (x > 0 )f ( x ) = ⎪x + + a (x < 0) 的最大值为 f (-1) ,则实数a 的取值范围为( )⎩⎪ xA . ⎡⎣0, 2e 2 ⎤⎦B . ⎡⎣0, 2e 3⎤⎦C . (0, 2e 2⎤⎦D . (0, 2e 3⎤⎦【变式演练 9】(不含参数最值)【安徽省江淮十校 2020-2021 学年高三上学期第一次联考】已知函数f (x ) = cos 2 x s in 2x ,若存在实数 M ,对任意 x 1 , x 2 ∈R 都有 f ( x 1 ) - f (x 2 ) ≤ M 成立.则 M 的最小值为()A.3 38B.32C.3 3 4D.2 3 3【变式演练 10】(含参最值)【重庆市经开礼嘉中学 2020 届高三下学期期中】已知函数f (x ) = (x - a - 1)e x -1 - 1x 2 + ax , x > 02(1) 若 f (x ) 为单调增函数,求实数 a 的值;(2) 若函数 f (x ) 无最小值,求整数 a 的最小值与最大值之和.【高考再现】1.【2018 年全国普通高等学校招生统一考试数学(江苏卷)】若函数 ƒ(x ) = 䂸x 3 — t x 䂸 + 1(t C R )在(t h + œ) 内有且只有一个零点,则 ƒ(x )在[ — 1h 1]上的最大值与最小值的和为.2【. 2018 年全国普通高等学校招生统一考试理科数学(新课标 I 卷)】已知函数 ƒ x = 䂸sinx + sin 䂸x ,则 ƒ x的最小值是 .3. 【2020 年高考全国Ⅱ卷理数 21】已知函数 f (x ) = sin 2x sin 2x .3 381 2 n (1) 讨论 f ( x ) 在区间(0,π) 的单调性;(2) 证明: f (x ) ≤ ;(3) 设 n ∈ N *,证明: sin 2x sin 22x sin 24x sin 22nx ≤ 3 . 4n4. 【2020 年高考天津卷 20】已知函数 f (x ) = x3+ k ln x (k ∈ R ) , f ' (x ) 为 f ( x ) 的导函数.(Ⅰ)当 k = 6 时,(i ) 求曲线 y = f ( x ) 在点(1, f (1)) 处的切线方程;(ii )求函数 g (x ) = f (x ) - f '(x ) + 9的单调区间和极值;x(Ⅱ)当 k - 3 时,求证:对任意的 x , x ∈[1, +∞) ,且 x> x , 有 f '( x ) + f ' (x ) > f (x 1 )- f (x 2 ) . 1 2 1 2 2x - x 1 25. 【2018 年全国卷Ⅲ理数】已知函数 ƒ x = 䂸+ x + tx 䂸 ln 1 + x — 䂸x .(1) 若 t = t ,证明:当— 1 ǹ x ǹ t 时,ƒ x ǹ t ;当 x Σ t 时,ƒ x Σ t ;(2) 若 x = t 是 ƒ x 的极大值点,求 t .6. 【2018 年全国普通高等学校招生统一考试文科】设函数 ƒ(x ) = [tx 䂸 — (3t + 1)x + 3t + 䂸]e x .(Ⅰ)若曲线 y = ƒ(x )在点(䂸h ƒ(䂸))处的切线斜率为 0,求 a ;(Ⅱ)若 ƒ(x)在 x = 1 处取得极小值,求 a 的取值范围.7. 【2018 年全国普通高等学校招生统一考试文科数学(天津卷)】设函数 ƒ(x )=(x — t 1)(x — t 䂸)(x — t 3),其中t 1h t 䂸h t 3 C R ,且t 1h t 䂸h t 3是公差为 d 的等差数列.(I )若t 䂸 = t h d = 1h 求曲线 y = ƒ(x )在点(t h ƒ(t ))处的切线方程;(II ) 若 d = 3,求 ƒ(x)的极值;4 4 (III ) 若曲线 y = ƒ(x) 与直线 y =— (x — t 䂸) — 6 3有三个互异的公共点,求d 的取值范围.【反馈练习】1.【2020 届高三 6 月质量检测巩固卷数学(文科)】若函数 f ( x ) = e x (-x 2 + 2x + a )在区间(a , a +1) 上存在最大值,则实数a 的取值范围为()⎛ -1 A ., -1 + 5 ⎫ B . (-1, 2)2 2 ⎪ ⎝ ⎭⎛ -1 C . 2 ⎫ , 2⎪⎛ -1 D .2⎫, -1⎪ ⎝ ⎭⎝⎭2. 【黑龙江省大庆市第四中学 2020 届高三下学期第四次检测】若函数 f (x ) = ae x- 1在其定义域上只有 3x个极值点,则实数a 的取值范围()⎛ e 2 ⎫⎛ e 2 ⎫ A . -∞, - ⎪ (1, +∞)⎝⎭ B . -∞, - ⎪⎝⎭C . ⎛-e , -1 ⎫ (1, +∞)D . ⎛-∞, - 1 ⎫4e 2 ⎪ e ⎪ ⎝ ⎭⎝ ⎭xx2 x3. 【湖北省金字三角 2020 届高三下学期高考模拟】已知函数 f ( x ) = e + - ln x 的极值点为1 ,函数 2g ( x ) = e x + x - 2 的零点为 x ,函数 h ( x ) = ln x的最大值为x ,则( ) 2 2x 3A. x 1 > x 2 > x 3B. x 2 > x 1 > x 3C. x 3 > x 1 > x 2D. x 3 > x 2 > x 14. 【湖北省宜昌一中、龙泉中学 2020 届高三下学期 6 月联考】已知函数(ff (e ) = 1,当 x >0 时,下列说法正确的是()ex )满足 x 2 f '(x ) + 2xf (x ) = 1+ ln x ,① f (x ) 只有一个零点;② f (x ) 有两个零点;- 5 + 5 - 5③ f (x) 有一个极小值点;④ f (x) 有一个极大值点A.①③B.①④C.②③D.②④5.【山东省潍坊市2020届高三6月高考模拟】已知函数f(x)的导函数f'(x)=x4(x-1)3(x-2)2(x-3),则下列结论正确的是()A.f (x)在x = 0 处有极大值B.f (x )在x = 2 处有极小值C. f (x)在[1, 3]上单调递减D.f (x )至少有3 个零点6.【云南省曲靖市2020 届高三年级第二次教学质量监测】已知实数a, b 满足0 ≤a ≤1,0 ≤b ≤ 1 ,则函数f (x)=x3 -ax2 +b2 x +1 存在极值的概率为()A.1B.3C.16 6 3D.37.【云南省红河自治州2019-2020 学年高三第二次高中毕业生复习统一检测】下列关于三次函数f ( x) =ax3 +bx2 +cx +d (a ≠ 0) ( x ∈R) 叙述正确的是()①函数f (x) 的图象一定是中心对称图形;②函数f (x) 可能只有一个极值点;③当x ≠-b时,f (x) 在x =x 处的切线与函数y = f (x) 的图象有且仅有两个交点;0 3a 0④当x ≠-b时,则过点(x, f (x))的切线可能有一条或者三条.0 3a 0 0A.①③B.②③C.①④D.②④8.【2020 届江西省分宜中学高三上学期第一次段考】已知e 为自然对数的底数,设函数f (x)=1 x2 -ax +b ln x 存在极大值点x ,且对于a 的任意可能取值,恒有极大值f (x )< 0 ,则下列结论2 0 0bb ( ) 中正确的是()A. 存在 x 0= ,使得f (x 0 ) < - 12eB. 存在 x 0= ,使得f (x 0 ) > -e 2C.b 的最大值为e 3D.b 的最大值为 2e 2ax 2⎛ 1 , 3⎫9. 【四川省内江市 2020 届高三下学期第三次模拟考试】函数f (x )= 2+(1﹣2a )x ﹣2ln x 在区间 2 ⎪⎝ ⎭内有极小值,则 a 的取值范围是()A . ⎛ -2, -1 ⎫B . ⎛-2, -1 ⎫3 ⎪2 ⎪ ⎝ ⎭⎝ ⎭C . ⎛ -2, - 1 ⎫ ⋃⎛ - 1 , +∞⎫D . ⎛ -2, - 1 ⎫ ⋃ ⎛ - 1 , +∞ ⎫ 3 ⎪ 3 ⎪ 2 ⎪ 2 ⎪ ⎝ ⎭ ⎝ ⎭⎝ ⎭ ⎝ ⎭10.【河北省衡水中学 2019-2020 学年高三下学期期中】已知函数 f (x ) =(x2- a )2- 3 x 2 -1 - b ,当时(从①②③④中选出一个作为条件),函数有 .(从⑤⑥⑦⑧中选出相应的作为结论,只填出一.组.即可)1 3 5 9① a ≤ - ② < a < ③ a = 1 ,-2 < b < 0 ④ a = 1 ,- < b < -2 或b = 0 ⑤4 个极小值点⑥1 个极小值点2 2 2 4⑦6 个零点⑧4 个零点1. 【福建省漳州市 2020 届高三高考数学(文科)三模】已知函数 f (x ) = ( x + 3) e x- 2m , m ∈ R .(1)若 m = 3,求 f ( x ) 的最值;2(2)若当 x ≥ 0 时, f (x - 2) + 2m ≥ 1 mx 2+ 2x +1 ,求 m 的取值范围.e 212. 【安徽省合肥七中、三十二中、五中、肥西农兴中学 2020 届高三高考数学(文科)最后一卷】已知函数 f (x ) = 1 x 2- 2x + a ln x , a > 1 . 2e(1) 讨论 f( x ) 的单调性;(2)若f (x )存在两个极值点x1 、x2 ,求f (x1 )+f (x2 )的取值范围.13.【2020 届安徽省芜湖市高三下学期教育教学质量监测】已知函数f (x)=ae x + 2e -x+(a - 2 )x .(1)若y =f (x )存在极值,求实数 a 的取值范围;(2)设1 ≤a ≤ 2 ,设g (x)= f (x)-(a + 2)cos x 是定义在⎛-∞,π ⎤上的函数.2 ⎥⎝⎦(ⅰ)证明:y =g'(x )在⎛-∞,π ⎤上为单调递增函数( g'(x)是y =g (x )的导函数);2 ⎥⎝⎦ (ⅱ)讨论y =g (x )的零点个数.14.【广东省惠州市2021 届高三上学期第一次调研】已知函数f (x) =x- ln(ax) .a(1)若a > 0 ,求f (x) 的极值;(2)若e x ln x +mx 2 +(1 -e x )x +m ≤ 0 ,求正实数m 的取值范围.15.【北京五中2020 届高三(4 月份)高考数学模拟】设函数f(x)=me x﹣x2+3,其中m∈R.(1)如果f(x)同时满足下面三个条件中的两个:①f(x)是偶函数;②m=1;③f(x)在(0,1)单调递减.指出这两个条件,并求函数h(x)=xf(x)的极值;(2)若函数f(x)在区间[﹣2,4]上有三个零点,求m 的取值范围.16.【辽宁省锦州市渤大附中、育明高中2021 届高三上学期第一次联考】已知函数f (x) =ae x - cos x -x(a ∈R).(1)若 a = 1 ,证明:f (x) ≥ 0 ;(2)若f (x) 在(0,π) 上有两个极值点,求实数 a 的取值范围.17.【西南地区名师联盟2020 届高三入学调研考试】已知函数f (x)=1x3 +bx2 +cx ,b 、c 为常数,且3学习界的007- 1< b < 1, f '(1) = 0 . 2(1)证明: -3 < c < 0 ;(2)若 x 是函数 y = f (x ) - cx 的一个极值点,试比较 f ( x - 4) 与 f (-3) 的大小. 0218.【山东省威海荣成市 2020 届高三上学期期中】某水产养殖公司在一片海域上进行海洋牧场生态养殖, 如图所示,它的边界由圆O 的一段圆弧 PMQ ( M 为此圆弧的中点)和线段 PQ 构成.已知圆O 的半径为12 千米, M 到 PQ 的距离为16 千米.现规划在此海域内修建两个生态养殖区域,养殖区域 R 1 为矩形 ABCD ,养殖区域 R 2 为 A M B ,且 A , B 均在圆弧上,C ,D 均在线段 PQ 上,设∠AOM =α.(Ⅰ)用α分别表示矩形 ABCD 和 A M B 的面积,并确定cos α的范围;(Ⅱ)根据海域环境和养殖条件,养殖公司决定在 R 1 内养殖鱼类,在 R 2 内养殖贝类,且养殖鱼类与贝类单位面积的年产值比为3 : 2 .求当α为何值时,能使年总产值最大.19.【江苏省南通市 2020 届高三下学期高考考前模拟卷】已知函数 f (x ) = ( x - a ) e x + b (a , b ∈ R ) .(1) 讨论函数 f( x ) 的单调性;(2) 对给定的 a ,函数 f( x ) 有零点,求b 的取值范围;(3)当 a = 2 , b = 0 时, F (x ) = f ( x ) - x + ln x ,记 y = F ( x ) 在区间⎛ 1 ,1⎫上的最大值为 m ,且4 ⎪ ⎝ ⎭m ∈[n, n + 1), n ∈Z ,求n 的值.20.【陕西省西安中学2020-2021 学年高三上学期第一次月考】已知函数f ( x) =x -1 -a ln x .(1)当 a = 1 时,求f(x)的最小值;(2)设m 为整数,且对于任意正整数n ,(1+1)(1+1) ⋅⋅⋅ (1+1) <m ,求m 的最小值.2 22 2n。
安徽省合肥七中、农兴中学、合肥32中、合肥五中2020届高三冲刺高考最后一卷英语答案解析

绝密★启用前安徽省合肥七中、农兴中学、合肥32中、合肥五中2020届高三毕业班下学期冲刺高考最后一卷英语试题参考答案解析一、听力1-5 CABBA 6-10 CBACB11-15 CAABC 16-20 CABAC二、阅读21.D 22.B 23.D【解析】这是一篇应用文。
deCaires Taylor's水下展览是一个集艺术和拯救海洋生物于一体的展览。
文章介绍了其中四处展览的特色。
1.细节理解题。
根据第一段的and with over 1,000 underwater artworks across the globe,deCaires Taylor has artfully highlighted the threats to our oceans while actively helping to create new life in them.(全球有超过1000件水下艺术品,deCaires Taylor巧妙地强调了对海洋的威胁,同时积极帮助创造新的生命)可知,deCaires Taylor 水下展览的共同特点是把艺术展览和拯救海洋生物巧妙地结合在一起。
D. Combining art with saving ocean creatures.(把艺术和拯救海洋生物结合在一起)符合以上说法,故选D项。
2.细节理解题。
根据Nassau, Bahamas部分下的Off the coast of Nassau sits the world's largest underwater sculpture, shallow enough for divers to view.(在拿骚的海岸附近,有世界上最大的水下雕塑,浅到足以让潜水者们可以看见)可知,在Nassau可以看到世界上最大的水下雕塑。
故选B项。
3.细节理解题。
根据Great Barrier Reef, Australia部分下的Expected to open to the public in early 2020 , deCaires Taylor's latest project aims to rehabilitate parts of the world's largest reef system.(预计在2020年对公众开放,deCaires Taylor's的最新项目是修复世界上最大的珊瑚礁的部分)可知,Museum of Underwater Art的目的是修复珊瑚礁。
精品解析:安徽省合肥七中、肥西农兴中学、合肥五中2020届高三冲刺高考理综化学试题(解析版)

合肥七中、肥西农兴中学、合肥三十二中、合肥五中2020届高三冲刺高考“最后一卷”理科综合化学试卷可能需要的相对原子质量:H 1 C 12 N 14 O 16 Si 28 S 32 Cl 35.5 K 39 Fe 56 Cu 64 1.煤、石油、天然气仍是人类使用的主要能源,同时也是重要的化工原料,我们熟悉的塑料、合成纤维和合成橡胶主要是以石油、煤和天然气为原料生产的。
下列说法中不正确的是A. 工业上可通过石油的热裂解生产苯、甲苯等芳香烃()B. 煤干馏的产品有焦炉煤气、煤焦油和焦炭C. 羊毛、蚕丝、棉花和麻等是天然的蛋白纤维或植物纤维D. 天然气是一种清洁的化石燃料,作为化工原料它主要用于合成氨和甲醇【答案】A【解析】【详解】A.工业上,通过石油的热裂解获得乙烯、1,3-丁二烯等短链不饱和烃,通过煤的干馏获得苯、甲苯等芳香烃,故A错误;B.煤干馏是隔绝空气加强热发生反应,产品主要为出炉煤气、煤焦油和焦炭,故B正确;C.羊毛、蚕丝、棉花和麻等在自然界中存在,属于天然纤维,故C正确;D.天然气广泛用于民用及商业燃气灶具、热水器,是一种清洁的化石燃料,天然气也可用作化工原料,以天然气为原料的一次加工产品主要有合成氨、甲醇、炭黑等,故D正确;答案选A。
2.设N A为阿伏加德罗常数的数值。
下列叙述正确的是()A. 将11.2LCl2溶于足量水,所得溶液中Cl2、Cl-、HClO和ClO-四种微粒总数为0.5N AB. 钾在空气中燃烧可生成多种氧化物,7.8g钾在空气中完全燃烧时转移的电子数为0.2N AC. 常温常压下,0.1mol NH3与0.1mol HCl充分反应后所得到的产物中含有的分子数仍为0.1N AD. 在2.8g晶体硅中所含的Si-Si键的数目为0.4 N A【答案】B【解析】【详解】A. 因为气体状况未知,所以无法计算氯气物质量,故A错误;B. 7.8g钾的物质的量为0.2mol,而钾反应后变为+1价,故0.2mol钾反应后转移电子为0.2N A,故B正确;C. NH 3和HCl 反应后生成的氯化铵为离子化合物,故无氯化铵分子,故C 错误;D. 硅晶体中每个硅原子与其周围四个硅原子形成四个共价键,但是每个共价键是两个硅原子共用的,所以1molSi 含有2mol 共价键,2.8g 晶体硅的物质的量是0.1mol ,则含有Si−Si 键的个数为0.2 N A ,故D 错误; 故选B 。
安徽省合肥七中、三十二中、五中、肥西农兴中学2020届高三高考数学(文科)最后一卷试题

安徽省合肥七中、三十二中、五中、肥西农兴中学2020届高三高考数学(文科)最后一卷试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.设全集{}1,2,3,4,5U =,集合{}1,3,5M =,集合{}3,4N =,则图中阴影部分所示的集合是( )A .{}1B .{}3,4C .{}2,3,4D .{}4 2.已知i 为虚数单位,复数231z i i =++,则复数z 的虚部是( ) A .i B .1 C .2i D .23.如图是某学校高三年级的三个班在一学期内的六次数学测试的平均成绩y 关于测试序号x 的函数图象,为了容易看出一个班级的成绩变化,将离散的点用虚线连接,根据图象,给出下列结论:①一班成绩始终高于年级平均水平,整体成绩比较好;②二班成绩不够稳定,波动程度较大;③三班成绩虽然多次低于年级平均水平,但在稳步提升.其中错误的结论的个数为( )A .0B .1C .2D .34.对于常数m 、n ,“0mn >”是“方程221mx ny +=的曲线是椭圆”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5.鲁班锁起源于中国古代建筑中首创的榫卯结构,相传由春秋时代鲁国工匠鲁班所作. 下图是经典的六柱鲁班锁及六个构件的图片,下图是其中一个构件的三视图,则此构件的体积为A .334000mmB .333000mmC .332000mmD .330000mm 6.已知α为锐角,且3cos(),cos 245παα+== A .2425 B .725 C .-2425 D .±24257.如图的程序框图的算法思路源于我国古代数学名著《数书九章》中的“中国剩余定理”.执行该程序框图,则输出的n 为( )A .50B .53C .59D .62 8.函数()52sin 33x x x x f x -+=-([,0)(0,])x ππ∈-的图象可能为( ) A .B .C .D .9.已知函数()y f x =是定义在R 上的奇函数,且满足(2)()0f x f x ++=,当[2,0]x ∈-时,2()2f x x x =--,则当[4,6]x ∈时,()y f x =的最小值为( ) A .8- B .1- C .0 D .110.已知1F ,2 F 是双曲线2222:1(0,0)x y E a b a b-=>>的左、右焦点,点M 在E 上,1MF 与x 轴垂直,211sin 4MF F ∠=,则E 的离心率为( )A B .32 C D .211.古希腊雅典学派算学家欧道克萨斯提出了“黄金分割”的理论,利用尺规作图可画出已知线段的黄金分割点,具体方法如下:(1)取线段2AB =,过点B 作AB 的垂线,并用圆规在垂线上截取112BC AB ==,连接AC ;(2)以C 为圆心,BC 为半径画弧,交AC 于点D ;(3)以A 为圆心,以AD 为半径画弧,交AB 于点E .点E 即为线段AB 的黄金分割点。若在线段AB 上随机取一点F ,则使得BE AF AE ≤≤的概率约为( )(参考数据2.236≈)A .0.236B .0.382C .0.472D .0.61812.一个等腰三角形的周长为10,四个这样相同等腰三角形底边围成正方形,如图,若这四个三角形都绕底边旋转,四个顶点能重合在一起,构成一个四棱锥,则围成的四棱锥的体积的最大值为( )A.81 B.27 C. D.二、填空题13.不等式组201030x y x y y +-≥⎧⎪-+≤⎨⎪-≤⎩表示的区域的面积为______.14.已知向量1,22AB ⎛= ⎝⎭,3122BC ⎛⎫= ⎪ ⎪⎝⎭,则ABC ∠=________. 15.若直线y kx b =+是曲线2x y e -=的切线,也是曲线1x y e =-的切线,则b =__________.16.如图,在平面四边形ABCD 中,2,sin 14CD DAC =∠=,2,33D B ππ==则四边形ABCD 的面积的最大值为_____.三、解答题17.如图,在直三棱柱111ABC A B C -中,AC BC ⊥,E 为11A C 的中点,1CE AC ⊥.(1)证明:CE ⊥平面11AB C ;(2)若1C E 1AA ,2AB BC =,求三棱锥1E AB C -的体积. 18.设{}n a 是等比数列,其前n 项的和为n S ,且22a =,2130S a -=.(1)求{}n a 的通项公式;(2)若48n n S a +>,求n 的最小值.19.某大学为调研学生在A ,B 两家餐厅用餐的满意度,从在A ,B 两家餐厅都用过餐的学生中随机抽取了100人,每人分别对这两家餐厅进行评分,满分均为60分. 整理评分数据,将分数以10为组距分成6组:[)0,10,[)10,20,[)20,30,[)30,40,[)40,50,[)50,60,得到A 餐厅分数的频率分布直方图,和B 餐厅分数的频数分布表:(Ⅰ)在抽样的100人中,求对A 餐厅评分低于30的人数;(Ⅱ)从对B 餐厅评分在[)0,20范围内的人中随机选出2人,求2人中恰有1人评分在[)0,10范围内的概率;(Ⅲ)如果从A ,B 两家餐厅中选择一家用餐,你会选择哪一家?说明理由.20.已知圆22(2):4F x y -+=,动点()(),0Q x y x ≥,线段QF 与圆F 相交于点P ,线段PQ 的长度与点Q 到y 轴的距离相等.(Ⅰ)求动点Q 的轨迹W 的方程;(Ⅱ)过点()2,4A 作两条互相垂直的直线与W 的交点分别是M 和N (M 在N 的上方,A ,M ,N 为不同的三点),求向量NM 在y 轴正方向上的投影的取值范围.21.已知函数()212ln 2f x x x a x =-+,1a e >. (1)讨论()f x 的单调性;(2)若()f x 存在两个极值点1x 、2x ,求()()12f x f x +的取值范围.22.在直角坐标系xOy 中,曲线1C的参数方程为21x y ⎧=⎪⎨=-+⎪⎩(t 为参数),以原点O 为极点,以x 轴正半轴为极轴,建立极坐标系,曲线2C 的极坐标方程为π2cos 4ρθ⎛⎫=+ ⎪⎝⎭. (1)判断曲线1C 与曲线2C 的位置关系;(2)设点(),M x y 为曲线2C 上任意一点,求2x y +的最大值.23.已知函数()|2|f x x a =-.(1)当2a =,求不等式()||6f x x +≤的解集;(2)设()|1|30f x x x +-+≤对[2,1]x ∈--恒成立,求a 的取值范围.参考答案1.D【分析】先理解韦恩图表达的集合的具体形式,再据此求集合的交集和补集即可。
2020届安徽省合肥七中、三十二中、五中、肥西农兴中学高三最后一卷数学(文)试题(解析版)

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
【答案】B
【解析】 时,如果 ,或 ,方程 不是椭圆;当方程 的曲线是椭圆时, ,则 成立,即可得出结论.
【详解】
当 时,方程 的曲线不一定是椭圆,
例如:当 时,方程 的曲线不是椭圆而是圆;
或者是 , 都是负数,曲线表示的也不是椭圆;
又当 时, ,且 是定义在 上的奇函数,
∴ 时, ,
∴当 时, ,
所以当 时,函数 的最小值为 .
故选B.
【点睛】
本题主要考查了函数周期性的判定及应用,以及函数的奇偶性的应用,其中解答中熟练应用函数周期性的判定方法,得出函数的周期是解答的关键,着重考查了推理与运算能力,属于基础题.
10.已知 , 是双曲线 的左、右焦点,点M在E上, 与x轴垂直, ,则E的离心率为( )
三、解答题
17.如图,在直三棱柱 中, , 为 的中点, .
(1)证明: 平面 ;
(2)若 , , ,求三棱锥 的体积.
【答案】(1)证明见解析;(2) .
【解析】(1)证明出 平面 ,可得出 ,再由 结合线面垂直判定定理可得出 平面 ;
(2)计算出 的面积,然后以点 为顶点,利用锥体的体积公式计算得出三棱锥 的体积.
2020年安徽省合肥七中、肥西农兴中学、合肥三十二中、合肥五中联考高考物理冲刺最后一卷

2020年安徽省合肥七中、肥西农兴中学、合肥三十二中、合肥五中联考高考物理冲刺最后一卷1.用金属铷为阴极的光电管,观测光电效应现象,实验装置示意如图甲所示,实验中测得铷的遏止电压U C与入射光频率ν之间的关系如图乙所示,图线与横轴交点的横坐标为5.15×1014Hz.已知普朗克常量ℎ=6.63×10−34J⋅s.则下列说法中正确的是()A. 欲测遏止电压,应选择电源左端为正极B. 当电源左端为正极时,滑动变阻器的滑片向右滑动,电流表的示数持续增大C. 增大照射光的强度,产生的光电子的最大初动能一定增大D. 如果实验中入射光的频率ν=7.00×1014Hz,则产生的光电子的最大初动能E k=1.2×10−19J2.如图所示为某质点做直线运动的v−t图象,其中在t1和t2时刻的瞬时速度分别为v1和v2,则质点在t1~t2运动过程中,下列说法正确的是()A. 质点的速度方向和加速度方向相同B. 质点的速度大小和加速度大小都逐渐变小C. 合力对质点做正功D. 质点运动的平均速度大于v1+v223.在沿平直轨道行驶的列车上,乘客使用斜靠在支架上的平板电脑进行阅读。
支架固定在水平桌面上,平板电脑和支架始终相对静止,其简化结构图如图所示。
在列车向前加速行驶过程中,关于平板电脑所受作用力,下列分析正确的是()A. 只受重力和支架弹力的作用B. 受重力、支架的弹力和摩擦力的作用C. 受支架的作用力方向水平向后D. 受支架的作用力方向竖直向上4.我国航天技术走在世界的前列,探月工程“绕、落、回”三步走的最后一步即将完成,即月球探测器实现采样返回,如图所示为该过程简化后的示意图,探测器从圆轨道1上的A点减速后变轨到椭圆轨道2,之后又在轨道2上的B点变轨到近月圆轨道3.已知探测器在轨道1上的运行周期为T1,运行速度为v1,在轨道3上运行速度为v3,O为月球球心,C为轨道3上的一点,AC与AO之间的最大夹角为θ,下列说法正确的是()A. 探测器在轨道2运行时的机械能大于在轨道1运行时的机械能B. 探测器在轨道2运行时的线速度v2大小满足v1<v2<v3C. 探测器在轨道3上运行时的周期为√sin3θT1D. 探测器在轨道2上运行和在圆轨道1上运行,加速度大小相等的位置有两个5.如图所示,边长为L的正方形金属线框放在垂直纸面向外的匀强磁场中,线框与磁场方向垂直,磁场的磁感应强度大小为B0,保持线框不动,将磁场的磁感应强度随时间均匀增大到B1;保持磁场的磁感应强度B1不变,再将线框绕对称轴(图中虚线)匀速转动转过60°,两个过程流过导线框截面的电荷量相等,则磁感应强度B0和B1的比值为()A. 2:3B. 3:2C. 1:2D. 2:16.带电荷量分别为+4q和−q的两点电荷组成了电荷系统,其电场线分布如图所示,图中实线为电场线,未标注方向,虚线上A、B、C、D等间距,根据图象可判断()A. B的电荷为−qB. 放在D点的试探电荷不受电场力作用C. A点的电势高于D点电势D. 若把一个带正电的试探电荷从A向B移动,电场力做正功,电势能减小7.如图所示,一根轻质弹簧放在光滑斜面上,其下端与斜面底端的固定挡板相连,弹簧处于自然伸长状态。
安徽省合肥七中、肥西农兴中学、合肥三十二中、合肥五中2020届高三冲刺高考“最后一卷”文综地理试题

合肥七中、肥西农兴中学、合肥三十二中、合肥五中2020届高三冲刺高考“最后一卷”文综地理试卷一、选择题下图是昆明市宜良县城通往靖安哨村的盘山公路。
该公路于上世纪九十年代修建,路基宽5米,因考虑到沿线主要为邻村土地,只好顺山梁而修,蜿蜒曲折,短短6.8公里共有68道拐,无意中创造了这惊艳世界的公路奇观。
后来,不仅吸引了游客,还吸引了一些顶级车赛在此举行。
据此完成下面小题。
1. 推测当时该公路选建在山脊上是考虑A. 山谷施工难度较大B. 少占耕地和节省资金C. 山谷地质灾害频发D. 沿途居民点分布密集2. 该公路能吸引赛车手的主要原因是A. 平直开阔,能见度较好B. 坡陡险峻,风景较壮观C. 地质坚硬,不容易塌陷D. 弯多密集,极具挑战性3. 该公路的修建,最初是为了A. 村民出行提供方便B. 通过旅游加快脱贫C. 吸引高新技术产业D. 增加矿产资源外运【答案】1. B 2. D 3. A【解析】【1题详解】与山脊部位相比,山谷地带更平坦,有利于公路修建,故A项错误;公路沿线主要为邻村土地,若沿山谷修建会占用邻村耕地,需要付出大量补偿金,从而加大修建成本,故B项正确;山谷有地质灾害,但其不是将公路选建在山脊的原因,故C项错误;图中显示山脊上居民点极少,故D项错误。
所以选B。
【2题详解】根据“蜿蜒曲折,短短6.8公里共有68道拐”的材料信息,可见该公顺山梁修建,弯多密集,迎合了赛车手追求自我挑战的心理需求,故选项A错误、D正确。
其他选项,如地质坚硬、坡陡险峻等,均不是吸引赛车手的原因,BC错误。
所以选D。
【3题详解】过去,靖安哨村到宜良县城需要绕道而行,路程长达数十公里,交通十分不便。
该公路的修建,最初就是为了村民出行提供方便,而不是为了增加矿产资源的外运和吸引高新技术产业,且当地经济发展水平低,公路等级低,故选项C、D错误,选项A正确;该公路修建后,因奇特景观吸引来了游客,旅游经济才渐渐发展起来,B错误。
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安徽省合肥七中、肥西农兴中学、合肥三十二中、合肥五中2020届高三冲刺高考“最后一卷”英语试题(时间:120分钟满分:150分)第一部分听力(共两节,满分30分)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,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What does the man like about the play?A. The story.B. The ending.C. The actor.2. Which place are the speakers trying to find?A. A hotel.B. A bank.C. A restaurant.3. At what time will the two speakers meet?A. 5:20.B. 5:10.C. 4:40.4. What will the man do?A. Change the plan.B. Wait for a phone call.C. Sort things out.5. What does the woman want to do?A. See a film with the man.B. Offer the man some help.C. Listen to some great music.第二节(共15小题:每小题1.5分,满分22.5分)请听下面5段对话。
每段对话后有几个小题,从题中所给出的A、B、C三个选项种选出最佳选项,并标在试卷的相应位置。
听每段对话前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题给出5秒钟的作答时间。
每段对话读两遍。
听第6段材料,回答6、7题6. Where is Ben?A. In the kitchen.B. At school.C. In the park.7. What will the children do in the afternoon?A. Help set the table.B. Have a party.C. Light the candles.听第7段材料,回答第8、9题8. What are the two speakers talking about?A. A Family holiday.B. A business trip.C. A travel plan.9. Where did Rachel go?A. Spain.B. Italy.C. China.听第8段材料,回答第10至12题。
10. How did the woman get to know about third-hand smoke?A. From young smokers.B. From a newspaper article.C. From some smoking parents.11. Why does the man say that he should keep away from babies?A. He has just become a father.B. He wears dirty clothes.C. He is a smoker.12. What does the woman suggest smoking parents should do?A. Stop smoking altogether.B. Smoke only outside their houses.C. Reduce dangerous matter in cigarettes.听第9段材料,回答第13至16题。
13. Where does Michelle Ray come from?A. A middle-sized city.B. A small town.C. A big city.14. Which place would Michelle Ray take her visitors to for shopping?A. The Zen Garden.B. The Highlands.C. The Red River area.15. What does Michelle Ray do for complete quiet?A. Go camping.B. Study in a library.C. Read at home.16. What are the speakers talking about in general?A. Late-night shopping.B. Holiday plan.C. Louisville.听第10段材料,回答第17至20题。
17. Why do some people say they never have dreams according to Dr. Garfield ?A. They forget about their dreams.B. They don't want to tell the truth.C. They have no bad experiences.18.Why did Davis stop having dreams?A. He got a serious heart attack.B. He was too sad about his brother's deathC. He was frightened by a terrible dream.19. What is Dr. Garfield’s opinion about dreaming?A. It is very useful.B. It makes things worse.C. It prevents the mind from working.20. Why do some people turn off their dreams completely?A. To sleep better.B. To recover from illnesses.C. To stay away from their problems.第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30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。
ADive into Jason deCaires Taylor's underwater art exhibits from around the world.Creating the world's first underwater sculpture(雕像) park in Grenada in 2006, and with over 1,000 underwater artworks across the globe, de Caires Taylor has artfully highlighted the threats to our oceans while actively helping to create new life in them. Here are some of the underwater art exhibits.Molinere Underwater Sculpture ParkGrenada, West IndiesInstalled in 2006 after Molinere Bay suffered destructive damage from the 2004 Hurricane Ivan, these 75 underwater art pieces formed the world's first underwater sculpture park and was named one of National Geographic's 25 Wonders of the World. It now provides a new habitat for marine life, and it also draws divers and glass-bottomed boats away from more delicate reefs nearby.Ocean AtlasNassau, BahamasOff the coast of Nassau sits the world's largest underwater sculpture, shallow enough for divers to view. This 60-ton,16-foot tall statue of a young Bahamian girl appears to hold up the ocean, just like her Greek namesake (同名人物) Atlas, who suspended the heavens in Greek myth(神话) .The Rising TideLondon, U.K.Unlike most of deCaires Taylor's works, these sculptures can be seen from land,on the banks of the River Thames, emerging during low tide. Within sight of the Houses of Parliament, they are a nod to the Four Horsemen of the Apocalypse and are a reminder of rising sea levels and the denial of climate change.Museum of Underwater ArtGreat Barrier Reef, AustraliaExpected to open to the public in early 2020, deCaires Taylor's latest project aims to rehabilitate parts of the world's largest reef system. Works in the underwater museum include a partially-submerged(半淹没的)figure that changes color as the sea warms and which can be seen from shore, and even a submerged coral-covered greenhouse.21.What do deCaires Taylor's underwater art exhibits have in common?A.Becoming National Geographic's Wonders of the World.B.Sharing names with gods in Greek myth.C.Changing colors with water temperature.D.Combining art with saving ocean creatures.22.Where can you see the largest underwater sculpture?A.Grenada. B.Nassau. C.London.D.Great Barrier Reef. 23.Which exhibit is intended to save the world's largest reef system?A.Molinere Underwater Sculpture Park. B.Ocean Atlas.C.The Rising Tide. D.Museum of Underwater Art.BRebecca Munkombwe, an 11-year-old girl from Zimbabwe, is regarded as a hero for saving a 9-year-old friend from the jaws of a crocodile by jumping on the crocodile and gouging (挖) itseyes out.According to Zimbabwean media, Rebecca and her friends had just got back from a swim in a stream near their home village when they heard the screams corning from the water. She was shocked to see her 9-year-old friend Latoya Muwani being dragged into the water by a crocodile. While all the other children were running scared, Rebecca ran toward the water, jumped on top of the crocodile and started beating it with her bare fists. However, that didn’t seem to bother the crocodile at all, so she then used her fingers to gouge its eyes out until it loosened its grasp of Latoya. Once she was free, Rebecca swam with her to the bank.Surprisingly, the 11-year-old heroine managed to save her young friend without suffering any wounds, while Latoya was lucky to escape with just mild wounds that were later treated at a regional hospital. Latoya’s parents praised Rebecca’s courage and thanked her for saving their daughter’s life. “I was at work when I learnt that my daughter had been attacked by a crocodile while swimming. For a moment I thought of the worst before I learnt that she’d survived after being saved by Rebecca,” Latoya’s father, Fortune Muwani, said. “I don’t know how she managed to do that, but I’m grateful to her.”Local authorities confirmed the crocodile attack, adding that the number of such incidents was increasing. Apparently, the lack of easily accessible water sources is forcing women and children to use unprotected sources like this crocodile-infested (遍布鳄鱼的) stream.24. What made Rebecca Munkombwe a hero?A. Her catching a crocodile hare-handed.B. Her rescuing a drowning friend in time.C. Her helping a friend out of deadly danger.D. Her making friends with a wild crocodile.25. Which of the following words can best describe Rebecca?A. Brave and smart.B. Calm and careful.C. Honest and considerate.D. Courageous and patient.26. What message do Fortune Muwani’s words convey?A. He knew every detail about the rescuing process.B. Latoya’s conditions were far better than expected.C. Latoya’s being attacked by a crocodile sounded unbelievable.D. He regretted not accompanying Latoya while she was swimming.27. Why were crocodile attacks on the rise?A. Because people have poor awareness of safety.B. Because people enjoy getting close to animals in the wild.C. Because people have little knowledge of accessible water sources.D. Because people have to share limited water sources with wild animals.CBoomerang children who return to live with their parents after university can be good for families, leading to closer, more supportive relationships and increased contact between the generations, a study has found.The findings contradict research published earlier this year showing that returning adult children trigger a significant decline in their parents’ quality of life and wellbeing.The young adults taking part in the study were “more positive than might have been expected” about moving back home – the shame is reduced as so many of their peers are in the same position, and they acknowledged the benefits of their parents’ financial and emotional support. Daughters were happier than sons, often slipping back easily into teenage patterns of behaviour, the study found.Parents on the whole were more uncertain, expressing concern about the likely duration of the arrangement and how to manage it. But they acknowledged that things were different for graduates today, who leave university with huge debts and fewer job opportunities.The families featured in the study were middle-class and tended to view the achievement of adult independence for their children as a “family project”. Parents accepted that their children required support as university students and then as graduates returning home, as they tried to find jobs paying enough to enable them to move out and get on the housing ladder.“However,” the study says, “day-to-day tensions about the prospects of achieving different dimensions of independence, which in a few extreme cases came close to conflict, characterized the experience of a majority of parents and a little over half the graduates”.Areas of divergence included chores, money and social life. While parents were keen to help, they also wanted different relationships from those they had with their own parents, and continuing to support their adult children allowed them to remain close.28.What is the finding of the previous research?A.Boomerang children made their parents happier.B.The parents were looking forward to their children’s return.C.The parents’ quality of life became worse than before.D.Boomerang children never did any housework.29.What are college graduates’ attitudes towards returning home?A.They are ashamed of turning to their parents for help.B.They are glad that they could come back.C.They are doubtful about whether they should return.D.They are proud to be independent from the family.30.What is the reason for the “boomerang children” phenomenon?A.The children want to keep in closer touch with their parents.B.The parents want to provide support to their children.C.It is harder for the children to secure a satisfying job.D.There is more housework needed to be done by the children.31. The underlined word “divergence” in paragraph 7 may be best replaced by ________.A. disagreementB. harmonyC. responsibilityD. cooperationDA team of engineers at Harvard University has been inspired by Nature to create the first robotic fly. The mechanical fly has become a platform for a series of new high-tech integrated systems. Designed to do what a fly does naturally, the tiny machine is the size of a fat housefly. Its mini wings allow it to stay in the air and perform controlled flight tasks.“It’s extremely important for us to think about this as a whole system and not just the sum of a bunch of individual components,” said Robert Wood, the Harvard engineering professor who has been working on the robotic fly project for over a decade. A few years ago, his team got the go-ahead to start piecing together the components. “The added difficulty with a project like this is that actually none of those components are off the shelf and so we have to develop them all on our own,” he said.They engineered a series of systems to start and drive the robotic fly. “The seemingly simple system which just moves the wings has a number of interdependencies on the individualcomponents, each of which individually has to perform well, but then has to be matched well to everything it’s connected to,” said Wood. The flight device was built into a set of power, computation, sensing and control systems. Wood says the success of the project proves that the flying robot with these tiny components can be built and manufactured.While this first robotic fly is linked to a small, off-board power source, the goal is eventually to equip it with a built-in power source, so that it might someday perform data-gathering work at rescue sites, in farmers’ fields or on the battlefield. “Basically, it should be able to take off, land and fly around,” he said.Wood says the design offers a new way to study flight mechanics and control at insect-scale. Yet, the power, sensing and computation technologies on board could have much broader applications. “You can start thinking about using them to answer open scientific questions, you know, to study biology in ways that would be difficult with the animals, but using these robots instead,” he said. “So there are a lot of technologies and open interesting scientific questions that are really what drives us on a day to day basis.”32.Which of the following statements was the difficulty engineers met while making the robotic fly?A.They did not have sufficient fund.B.No ready-made components were available.C.There was no model in their mind.D.It was hard for them to assemble the components.33.What do we know about the robotic fly?A.The robotic fly has been put into wide application.B.The robotic fly consists of a flight device and a control system.C.The robotic fly can collect information from many sources.D.The robotic fly can fly well with the cooperation of individual components. 34.Which of the following can be learned from the passage?A.The robotic fly can replace animals in some experiments.B.Animals are not allowed in biological experiments.C.The robotic fly is designed to learn about insects.D.There used to be few ways to study how insects fly.35.Which of the following might be the best title of the passage?A.The Development of Robotic FlyB.Robotic Fly Promotes Engineering ScienceC.Harvard’s Efforts in Making Robotic FlyD.Robotic Fly Imitates Real Life Insect第二节(共5小题;每小题2分,满分10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。