数学实验试题A(2010)

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电子科技大学《数学实验》2008-2009学年期末试题(含答案)

电子科技大学《数学实验》2008-2009学年期末试题(含答案)

电子科技大学二零零八到二零零九学年第二学期期末考试《数学实验》课程考试题A卷(120分钟) 考试形式:闭卷考试日期:2009年7月8日一、单项选择题(20分)1、三阶幻方又称为九宫图,提取三阶幻方矩阵对角元并构造对角阵用( )(A) diag(magic(3)); (B) diag(magic);(C) diag(diag(magic(3))); (D) diag(diag(magic))。

2、MATLAB命令P=pascal(3)将创建三阶帕斯卡矩阵,max(P)的计算结果是( )(A) 1 2 3 (B) 1 2 1 (C) 3 6 10 (D) 1 3 63、命令J=*1;1;1+**1,2,3+;A=j+j’-1将创建矩阵( )(A)123234345⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦; (B)234345456⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦(C)123123123⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦(D)111222333⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦4、data=rand(1000,2);x=data(:,1);y=data(:,2);II=find(y<sqrt(x)&y>x.^2);的功能是( )(A) 统计2000个随机点中落入特殊区域的点的索引值;(B) 统计1000个随机点落入特殊区域的点的索引值;(C) 模拟2000个随机点落入特殊区域的过程;(D) 模拟1000个随机点落入特殊区域的过程。

5、MATLAB计算二项分布随机变量分布律的方法是( )(A) binocdf(x,n,p); (B) normpdf(x,mu,s); (C)binopdf(x,n,p); (D) binornd(x,n,p)。

6、MATLAB命令syms e2;f=sqrt(1-e2*cos(t)^2);S=int(f,t,0,pi/2)功能是()(A) 计算f(x)在[0,pi/2]上的积分;(B) 计算f(t)不定积分符号结果;(C) 计算f(x)积分的数值结果;(D) 计算f(t)定积分的符号结果。

数学实验2009-2010第一学期期末试题A卷(附答案)

数学实验2009-2010第一学期期末试题A卷(附答案)

湛江师范学院2009年- 2010学年度第 一 学期考试科目: 数 学 实 验说明:把操作步骤(命令)及必要的运行结果写在纸上,操作过程中所得到的图形不用画在纸上。

仅有结果而无操作(命令)步骤的,该小题成绩以零分计算,请抄写工整。

(题目中,数据没有指定输出格式的,用默认格式,即小数格式)一、数值运算(每小题8分,共32分)已知有以下矩阵 ⎪⎪⎪⎭⎫⎝⎛-=314020112-H ⎪⎪⎪⎪⎪⎭⎫ ⎝⎛=1413012014260121A 1. 求矩阵A 的逆矩阵及与A 同阶的单位矩阵A=[1 2 1 0;6 2 4 1;0 2 1 0;3 1 4 1]; C=inv(A)B=eye(size(A))H=[-2 1 1;0 2 0;-4 1 3]; [V ,D]=eig(H)2. 求矩阵H 的特征值与特征向量,并判断H 是否可以相似对角化。

H=[-2 1 1;0 2 0;-4 1 3];[V ,D]=eig(H),det(V)3. 讨论下列向量组是否线性相关,并找出向量组的最大无关组,并将其余向量用该最大无关组线性表示;β1=(2 -1 3 5 ),β2=(4 -3 1 3); β3=(1 -1 -1 -1),β4=(3 -2 3 4) ,β5=(7 -6 -7 0);b=[2 -1 3 5;4 -3 1 3;1 -1 -1 -1;3 -2 3 4;7 -6 -7 0]; l=rank(b),m=rank(b(1:3,:)),n=rank(b([1 2 4],:))c=b([1 2 4],:)'; %转秩 d=b([3 5],:)'; h=c\d4. (1)求数值积分(假奇异积分)dxx x⎰11-2.0cos ;syms x; s=int(x^0.2*cos(x),-1,1);simplify(s) (2)二重积分⎰⎰+πππ20)cos sin (dxy x x y dysyms x y; s=int(int(y*sin(x)+x*cos(y),pi,2*pi),0,pi)二、符号运算(每小题8分,共32分)以下出现的字母均设为符号变量 1. 求二元函数的极限:33221)lim y x xy y x y x ++→-→(syms x y;f=(x^2*y+x*y^3)/(x+y)^3; limit(limit(f,x,-1),y ,2) 2. 设z=e 2xcos(3y),求yx z ∂∂∂2,并求π==∂∂∂y x y x z,12(1)>>syms x y; f=diff(exp(2*x)*cos(3*y),x); f=diff(f,y)(2)>> syms x y; f=diff(exp(2*x)*cos(3*y),x); f=diff(f,y),subs(x,1);subs(y ,pi)3. 求幂级数∑∞=+++012)12)(12n 1n n x (的和函数(注意最后结果要化简!)syms x n;g=symsum(1/((2*n+1)*((2*x+1)^(2*n+1))),n,0,inf);g=simple(g)4. 问k 为何值时,下面的方程组有非零解?请写出相应的MATLAB 命令x 1 -3x 3=0x 1+2x 2+kx 3=0 2x 1+kx 2 - x 3=0syms kA=[1 0 -3;1 2 k;2 k -1]; D=det(A); factor(D) ans= -(k+5)*(k-2)从上式分解可得:当k=-5 or k=2时,det(A)=0,从而有非零解。

2010高考全国Ⅰ数学试题与答案

2010高考全国Ⅰ数学试题与答案

2010年普通高等学校招生全国统一考试文科数学(必修+选修> 解读版参考公式:如果事件互斥,那么球的表面积公式如果事件相互独立,那么其中R表示球的半径球的体积公式如果事件A在一次实验中发生的概率是,那么次独立重复实验中事件恰好发生次的概率其中R表示球的半径一、选择题(1>(A> (B>- (C> (D>1.C【命题意图】本小题主要考查诱导公式、特殊三角函数值等三角函数知识【解读】(2>设全集,集合,,则A.B.C. D.2.C 【命题意图】本小题主要考查集合的概念、集合运算等集合有关知识【解读】,,则=(3>若变量满足约束条件则的最大值为(A>4 (B>3 (C>2 (D>13.B 【命题意图】本小题主要考查线性规划知识、作图、识图能力及计算能力.【解读】画出可行域<如右图),,由图可知,当直线经过点A(1,-1>时,z最大,且最大值为.<4)已知各项均为正数的等比数列{},=5,=10,则(A>(B> 7 (C> 6 (D>A4.A【命题意图】本小题主要考查等比数列的性质、指数幂的运算、根式与指数式的互化等知识,着重考查了转化与化归的数学思想.mmVxZudVti【解读】由等比数列的性质知,10,所以,所以(5>的展开式的系数是(A>-6 (B>-3 (C>0 (D>35.A. 【命题意图】本小题主要考查了考生对二项式定理的掌握情况,尤其是展开式的通项公式的灵活应用,以及能否区分展开式中项的系数与其二项式系数,同时也考查了考生的一些基本运算能力.mmVxZudVti【解读】的系数是 -12+6=-6(6>直三棱柱中,若,,则异面直线与所成的角等于(A>30° (B>45°(C>60° (D>90°6.C【命题意图】本小题主要考查直三棱柱的性质、异面直线所成的角、异面直线所成的角的求法.【解读】延长CA到D,使得,则为平行四边形,就是异面直线与所成的角,又三角形为等边三角形,(7>已知函数.若且,,则的取值范围是(A> (B>(C> (D>7.C【命题意图】本小题主要考查对数函数的性质、函数的单调性、函数的值域,考生在做本小题时极易忽视a的取值范围,而利用均值不等式求得a+b=,从而错选D,这也是命题者的用苦良心之处.mmVxZudVti【解读1】因为 f(a>=f(b>,所以|lga|=|lgb|,所以a=b(舍去>,或,所以a+b=又0<a<b,所以0<a<1<b,令由“对勾”函数的性质知函数在(0,1>上为减函数,所以f(a>>f(1>=1+1=2,即a+b的取值范围是(2,+∞>.mmVxZudVti【解读2】由0<a<b,且f(a>=f(b>得:,利用线性规划得:,化为求的取值范围问题,,过点时z最小为2,∴(C> mmVxZudVti<8)已知、为双曲线C:的左、右焦点,点P在C上,∠=,则A BC DA 1B 1C 1D 1O(A>2 (B>4 (C> 6 (D> 88.B 【命题意图】本小题主要考查双曲线定义、几何性质、余弦定理,考查转化的数学思想,通过本题可以有效地考查考生的综合运用能力及运算能力.mmVxZudVti 【解读1】.由余弦定理得cos ∠P =4【解读2】由焦点三角形面积公式得:4<9)正方体-中,与平面所成角的余弦值为 <A )<B )<C ) <D )9.D 【命题意图】本小题主要考查正方体的性质、直线与平面所成的角、点到平面的距离的求法,利用等体积转化求出D 到平面AC 的距离是解决本题的关键所在,这也是转化思想的具体体现.mmVxZudVti 【解读1】因为BB1//DD1,所以B 与平面AC 所成角和DD1与平面AC 所成角相等,设DO⊥平面AC,由等体积法得,即.设DD1=a,mmVxZudVti则,.所以,记DD1与平面AC所成角为,则,所以.【解读2】设上下底面的中心分别为;与平面AC所成角就是B与平面AC所成角,<10)设则<A)<B) (C> (D>10.C 【命题意图】本小题以指数、对数为载体,主要考查指数函数与对数函数的性质、实数大小的比较、换底公式、不等式中的倒数法则的应用.mmVxZudVti【解读1】 a=2=, b=In2=,而,所以a<b,c==,而,所以c<a,综上c<a<b.【解读2】a=2=,b=ln2=, ,; c=,∴c<a<b<11)已知圆的半径为1,PA、PB为该圆的两条切线,A、B为两切点,那么的最小值为(A> (B> (C> (D>11.D【命题意图】本小题主要考查向量的数量积运算与圆的切线长定理,着重考查最值的求法——判别式法,同时也考查了考生综合运用数学知识解题的能力及运算能力.mmVxZudVti 【解读1】如图所示:设PA=PB=,∠APO=,则∠APB=,PO=,,===,令,则,即,由是实数,所以,,解得或.故.此时.【解读2】设,换元:,【解读3】建系:园的方程为,设,<12)已知在半径为2的球面上有A、B、C、D四点,若AB=CD=2,则四面体ABCD的体积的最大值为mmVxZudVti(A> (B> (C> (D>12.B【命题意图】本小题主要考查几何体的体积的计算、球的性质、异面直线的距离,通过球这个载体考查考生的空间想象能力及推理运算能力.mmVxZudVti【解读】过CD作平面PCD,使AB⊥平面PCD,交AB与P,设点P到CD的距离为,则有,当直径通过AB与CD的中点时,,故.mmVxZudVti第Ⅱ卷注意事项:1.答题前,考生先在答题卡上用直径0.5毫M黑色墨水签字笔将自己的姓名、准考证号填写清楚,然后贴好条形码。

2010 AMC 10A 试题及答案解析

2010 AMC 10A 试题及答案解析

2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, in centimeters?SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as the width. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . The answer is .Problem 4A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is .Problem 5The area of a circle whose circumference is is . What is the value of ?SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to , which is equal to . The answer isTony works hours a day and is paid $per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then isand we know that he was for days, and for days. Thus, the answer is .Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is , which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is . What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds times more water than Logan's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is ()because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer isProblem 14Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?SolutionLet .Since ,Problem 15In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We can begin by first looking at Chris and LeRoy.Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.As Mike is a frog, his statement is false, hence there is at most one toad.As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and three frogs.Problem 16Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is.Solution 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has cubic inches. However, we can not just sum their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.Hence the total volume of the cuts is.Therefore the volume of the rest of the cube is.Solution 3We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.Each edge can be seen as a box, and each corner can be seen as a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?SolutionWe can solve this by breaking the problem down into cases and adding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 19Equiangular hexagon has side lengthsand . The area of is of the area of the hexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is .If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let be the probability that Isabella stops afterdrawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Theprobability of drawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequality is .Problem 24The number obtained from the last two nonzero digits of is equal to . What is ?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .If we divide by by taking out all the factors of in , we canwrite as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then his sequence contains numbers:Let be the smallest number for which Jim’s sequence has numbers. What is the units digit of ?SolutionWe can find the answer by working backwards. We begin with on the bottom row, then the goes to the right of the equal's sign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next row and for the row after that. However, at the fourth row, wesee that solving yields , in which case it would be incorrect since is not the greatest perfect square less than or equal to . So we make it a and solve . We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:Hence the solution is the last digit of , which is .。

数学实验考试试题

数学实验考试试题

数学实验考试试题一、选择题(每题 5 分,共 30 分)1、以下哪个软件常用于数学实验?()A ExcelB PhotoshopC WordD PowerPoint2、在数学实验中,要生成一组随机数,可以使用以下哪种方法?()A 手动输入B 使用随机数生成函数C 按照一定规律计算D 以上都不对3、进行数学建模时,以下哪个步骤是首先要做的?()A 收集数据B 提出假设C 建立模型D 模型求解4、用数学实验方法求解线性方程组,常用的方法是()A 消元法B 矩阵变换法C 迭代法D 以上都是5、要绘制一个函数的图像,以下哪个软件比较方便?()A MathematicaB 记事本C 计算器D 画图工具6、在数学实验中,误差分析的目的是()A 找出错误B 提高精度C 证明结果的正确性D 以上都是二、填空题(每题 5 分,共 30 分)1、数学实验的基本步骤包括:提出问题、()、建立模型、()、分析结果。

2、常见的数学软件有()、()、Maple 等。

3、用数学实验方法研究函数的最值,可以通过()的方法来实现。

4、随机变量的数字特征包括()、()、方差等。

5、进行数据拟合时,常用的方法有()、()等。

6、数学实验中,数据的可视化可以帮助我们()、()。

三、简答题(每题 10 分,共 20 分)1、请简要说明数学实验与传统数学学习方法的区别。

答:传统数学学习方法通常侧重于理论推导和定理证明,通过纸笔计算和逻辑推理来解决数学问题。

而数学实验则是借助计算机软件和工具,通过实际操作和数据模拟来探索数学现象和解决问题。

在传统学习中,学生更多地依赖于抽象思维和逻辑推理,对于一些复杂的数学概念和问题,理解起来可能较为困难。

而数学实验可以将抽象的数学概念直观化,通过图像、数据等形式展现出来,让学生更容易理解和接受。

数学实验还能够让学生亲自参与到数学的探索过程中,培养学生的动手能力和创新思维。

同时,它也可以处理大规模的数据和复杂的计算,提高解决实际问题的效率。

全国大学生2010年数学竞赛A题

全国大学生2010年数学竞赛A题

12010高教社杯全国大学生数学建模竞赛题目(请先阅读“全国大学生数学建模竞赛论文格式规范”)A 题 储油罐的变位识别与罐容表标定通常加油站都有若干个储存燃油的地下储油罐,并且一般都有与之配套的“油位计量管理系统”,采用流量计和油位计来测量进/出油量与罐内油位高度等数据,通过预先标定的罐容表(即罐内油位高度与储油量的对应关系)进行实时计算,以得到罐内油位高度和储油量的变化情况。

许多储油罐在使用一段时间后,由于地基变形等原因,使罐体的位置会发生纵向倾斜和横向偏转等变化(以下称为变位),从而导致罐容表发生改变。

按照有关规定,需要定期对罐容表进行重新标定。

图1是一种典型的储油罐尺寸及形状示意图,其主体为圆柱体,两端为球冠体。

图2是其罐体纵向倾斜变位的示意图,图3是罐体横向偏转变位的截面示意图。

请你们用数学建模方法研究解决储油罐的变位识别与罐容表标定的问题。

(1)为了掌握罐体变位后对罐容表的影响,利用如图4的小椭圆型储油罐(两端平头的椭圆柱体),分别对罐体无变位和倾斜角为α=4.10的纵向变位两种情况做了实验,实验数据如附件1所示。

请建立数学模型研究罐体变位后对罐容表的影响,并给出罐体变位后油位高度间隔为1cm 的罐容表标定值。

(2)对于图1所示的实际储油罐,试建立罐体变位后标定罐容表的数学模型,即罐内储油量与油位高度及变位参数(纵向倾斜角度α和横向偏转角度β )之间的一般关系。

请利用罐体变位后在进/出油过程中的实际检测数据(附件2),根据你们所建立的数学模型确定变位参数,并给出罐体变位后油位高度间隔为10cm 的罐容表标定值。

进一步利用附件2中的实际检测数据来分析检验你们模型的正确性与方法的可靠性。

附件1:小椭圆储油罐的实验数据 附件2:实际储油罐的检测数据地平线 图1 储油罐正面示意图油位探针2油位探针地平线 图2 储油罐纵向倾斜变位后示意图(b) 小椭圆油罐截面示意图水平线1.2m(a) 小椭圆油罐正面示意图图4 小椭圆型油罐形状及尺寸示意图图3 储油罐截面示意图(b )横向偏转倾斜后正截面图地平线地平线垂直线油位探针(a )无偏转倾斜的正截面图油位探针。

(GCT)数学真题2010年 含答案解析.

数学基础能力测试(25题,每题4分,共100分2.如果图1中给出了平面直角坐标系中直线l:y=ax+b的图像,那么坐标为(a,b的点在(.A.第1象限B.第Ⅱ象限C.第Ⅲ象限D.第Ⅳ象限3.若某单位员工的平均年龄为45岁,男员工的平均年龄为55岁、女员工的平均年龄为40岁,则该单位男、女员工人数之比为(.A.2:3B.3:2C.1:2D.2:14.如果图2中四边形ABCD顶点的坐标依次为A(-2,2,B(-1,5,C(4,3,D(2,1,那么四边形ABCD的面积等于(.A.16.5B.15C.13.5D.125.在实验室密闭容器*培育某种细菌,如果该细菌因每天的密度增长1倍,它在20天内密度增长到4百万株/m3,那么该细菌密度增长到百万株/m3时用了(天.A.2B.4C.8D.166.若图3中给出的函数y=x2+ax+a的图像与x轴相切,则a=(.A.0B.1C.2D.410.正三角形ABC中,D,E分别是AB,AC上的点,F,G分别是DE,BC的中点.已知BD=8厘米,CE=6厘米,则FG=(厘米.11.如图5所示,边长分别为1和2的两个正方形,放在同一水平线上,小正方形沿该水平线自左向右匀速穿过大正方形.设从小正方形开始穿人大正方形到恰好离开大正方形所用的时间为to,大正方形内除去小正方形占有部分之后剩下的面积为S(空白部分,则表示S 与时间t函数关系的大致图像为(.12.若某公司有10个股东,他们中任意6个股东所持股份的和都不少于总股份的50%,则持股最多的股东所持股份占总股份的最大百分比是(.A.25%B.30%C35%D.40%13.一个封闭透明的正四面体容器内装有水,正四面体的一个面放置在水平桌面时。

体内水面高度为四面体高h的,现将它倒置使原底面平行于水平桌瓯此时水面的高度与h 的比值为(.14.若从1,2,3,4,5,6,7,8,9,10这十个数中任意取3个不同的数,则它们能构成公比大于1的等比数列的概率是(.19.若a,b,c,d成等比数列,则函数y=A.有极大值,而无极小值B.无极大值,而有极小值C.有极大值,也有极小值D.无极大值,也无极小值逻辑推理能力测试(50题,每题2分,满分100分1.去年,有6000人死于醉酒,有4000人死于开车,但只有500人死于醉酒开车。

2010年全国高中数学联赛试题及答案详解(A卷)

2010年全国高中数学联赛一、填空题(每小题8分,共64分,)1. 函数x x x f 3245)(---=的值域是 .2. 已知函数x x a y sin )3cos (2-=的最小值为3-,则实数a 的取值范围是 .5.函数)1,0(23)(2≠>-+=a a a a x f x x 在区间]1,1[-∈x 上的最大值为8,则它在这个区间上的最小值是 .6. 两人轮流投掷骰子,每人每次投掷两颗,第一个使两颗骰子点数和大于6者为胜,否则轮由另一人投掷.先投掷人的获胜概率是 .7. 正三棱柱111C B A ABC -的9条棱长都相等,P 是1CC 的中点,二面角α=--11B P A B ,则=αsin .8. 方程2010=++z y x 满足z y x ≤≤的正整数解(x ,y ,z )的个数是 .二、解答题(本题满分56分)9. (16分)已知函数)0()(23≠+++=a d cx bx ax x f ,当10≤≤x 时,1)(≤'x f ,试求a 的最大值.10.(20分)已知抛物线x y 62=上的两个动点1122(,)(,)A x y B x y 和,其中21x x ≠且421=+x x .线段AB 的垂直平分线与x 轴交于点C ,求ABC ∆面积的最大值.11.(20分)证明:方程02523=-+x x 恰有一个实数根r ,且存在唯一的严格递增正整数数列}{n a ,使得 +++=32152a a a r r r .加 试2. (40分)设k 是给定的正整数,12r k =+.记(1)()()f r f r r r ==⎡⎤⎢⎥,()()l f r =(1)(()),2l f f r l -≥.证明:存在正整数m ,使得()()m f r 为一个整数.这里,x ⎡⎤⎢⎥表示不小于实数x 的最小整数,例如:112⎡⎤=⎢⎥⎢⎥,11=⎡⎤⎢⎥.3. (50分)给定整数2n >,设正实数12,,,n a a a 满足1,1,2,,k a k n ≤= ,记 12,1,2,,k k a a a A k n k +++== . 求证:1112n n k k k k n a A ==--<∑∑.4. (50分)一种密码锁的密码设置是在正n 边形12n A A A 的每个顶点处赋值0和1两个数中的一个,同时在每个顶点处涂染红、蓝两种颜色之一,使得任意相邻的两个顶点的数字或颜色中至少有一个相同.问:该种密码锁共有多少种不同的密码设置?。

2010全国高中数学联赛a卷加试试题及答案

2010全国高中数学联赛a卷加试试题及答案一、选择题(每题5分,共20分)1. 已知函数\( f(x) = ax^2 + bx + c \),其中\( a, b, c \)为常数,若\( f(1) = 0 \),则下列哪个选项是正确的?A. \( a + b + c = 0 \)B. \( a - b + c = 0 \)C. \( a + b - c = 0 \)D. \( a - b - c = 0 \)答案:A2. 一个圆的直径为10,那么它的面积是多少?A. 25πB. 50πC. 100πD. 25答案:B3. 一个等差数列的前三项分别为2,5,8,那么它的第五项是多少?A. 11B. 14C. 17D. 20答案:B4. 一个等比数列的前三项分别为2,6,18,那么它的第四项是多少?A. 54B. 42C. 36D. 54答案:A二、填空题(每题5分,共20分)5. 已知函数\( g(x) = 2x - 1 \),求\( g(3) \)的值。

答案:56. 一个三角形的三个内角分别为30°,60°,90°,那么这个三角形的类型是_______。

答案:直角三角形7. 一个数列的前四项为1,3,6,10,那么这个数列的通项公式是_______。

答案:\( n(n+1)/2 \)8. 已知一个矩形的长为10,宽为5,求它的周长。

答案:30三、解答题(每题10分,共60分)9. 已知等差数列\( \{a_n\} \)的前三项分别为1,4,7,求它的通项公式。

答案:\( a_n = 3n - 2 \)10. 已知等比数列\( \{b_n\} \)的前三项分别为2,8,32,求它的通项公式。

答案:\( b_n = 2^n \)11. 求函数\( h(x) = x^3 - 3x^2 + 2 \)的极值点。

答案:极小值点为\( x = 1 \),极大值点为\( x = 2 \)12. 已知一个圆的方程为\( (x - 2)^2 + (y - 3)^2 = 9 \),求通过原点且与该圆相切的直线方程。

2010数学真题卷及答案

2010年全国普通高等学校运动训练、民族传统体育专业单独统一招生考试数学一、选择题:每小题6分,共10小题,共60分.在每小题的四个选项中,只有一项是符合要求的.1.已知集合A ={x|x 2―1>0},B ={x|log 2x <0},则A ∩B 等于 ( )A .ØB .{x|x <-1}C .{x|x >1}D .{x|x <-1或x >1}2. 若不等式||x a -<1成立的充分条件是04<<x ,则实数a 的取值范围是( ) A. a ≥3B. a ≤3C. a ≥1D. a ≤13.函数)1(log 2-=x y 的反函数图像是 ( )A B4. 如图所示,∆OAB 是边长为2的等边三角形,直线x t =截这个三角形位于此直线左方的图形面积为y (见图中阴影部分)则函数y f t =()的大致图形为( )5.已知a 、b 是非零向量且满足(a -2b )⊥a ,(b -2a )⊥b ,则a 与b 的夹角是( )A .6π B .3π C .32π D .65π6椭圆22143x y +=的右焦点到直线y x =的距离是 ( )A.127. 过圆锥曲线C 的一个焦点F 的直线l 交曲线C 于A 、B 两点,且以AB 为直径的圆与F 相应的准线相交,则曲线C 为A. 双曲线B. 抛物线C. 椭圆D. 以上都有可能 8.若αααααcos sin cos 3sin ,2tan +-=则的值是( )A .31-B .-35C .31 D .35 9.直线052)3(057)3()1(2=-+-=-+-++y x m m y m x m 与直线垂直的充要条件是( )A .2-=mB .3=mC .31=-=m m 或D .23-==m m 或10.已知1(2)2x f x x ++=+,则1(2)f x -+= ( ) A.12x x -+ B.11x -+ C.211x x +-- D.21x x +-+二、填空题:每小题5分,共8小题,共计40分.将答案填在题中的横线上。

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………密………封………线………以………内………答………题………无………效……电子科技大学成都学院二零零九至二零一零学年第二学期期末考试《数学实验》课程考试题A卷(120分钟)考试形式:闭考试日期年月日课程成绩构成:平时10 分,期中0 分,实验30 分,期末60 分(本试卷100分)一、单项选择题(共30分)1.符号计算与一般数值计算有很大区别,它将得到准确的符号表达式。

在MATLAB命令窗口中键入命令syms x,y1=x^2;y2=sqrt(x);;int(y1-y2,x,0,1),屏幕显示的结果是()(A)y1 =x^(1/2) (B)ans= 1/3;(C)y2 =x^2;(D)ans= -1/32.在MA TLAB命令窗口中键入命令B=[8,1,6;3,5,7;4,9,2];B*B(:,1)。

结果是()(A)ans= (B)ans= (C)ans= (D)ans=91 67 67 6767 91 67 6767 67 91 673.在MA TLAB命令窗口中,键入命令syms x,F=1/(2+cos(2x));ezplot(diff(F)),结果是()(A)绘出函数F在[0,π]的图形;(B)绘出函数F在[–2π,2π]的图形;(C)绘函数F的导函数在[0,π]的图形;(D)绘函数F的导函数在[–2π,2π]的图形4.在MA TLAB命令窗口中,键入命令binopdf(3,6,1/2),结果是(A)ans= (B) ans= (C) ans= (D) ans=15/64 5/16 3/32 15.用赋值语句给定x MATLAB表达式是()(A)sqrt(7sin(3+2x)+exp(3)log(3)) (B)sqrt(7*sin(3+2*x)+exp(3)*log(3))(C)sqr(7*sin(3+2*x)+e^3*log(3)) (D)sqr(7sin(3+2x)+ e^3log(3))6.MA TLAB语句strcat(int2str(2010),'年是', s,'年')的功能是()(A)将数据2010转换为符号;(B)将数据2010与符号变量合并;(C)将几个符号变量合并为一个;(D)将符号变量转换为数值变量;7.语句L=sqrt(pi); x=fix(1000*L)/1000的功能是()(A)将无理数π取三位近似;(B)将无理数π取四位近似数(C)将无理数π取三位近似数;(D)将有理数π取四位近似数8.MA TLAB绘三维曲面需要构建网格数据,语句[x,y]=meshgrid(-2:2)返回数据中()(A)x是行向量,y是列向量;(B)x是列向量,y是行向量;(C)x是行元素相同的矩阵;(D)x是行向量相同的矩阵9.下面有关MATLAB函数的说法,哪一个是错误的()(A)函数文件的第一行必须由function开始,并有返回参数,函数名和输入参数;(B)MA TLAB的函数可以有多个返回参数和多个输入参数;(C)在函数中可以用nargout检测用户调用函数时的输入参数个数;(D)如果函数文件内有多个函数,则只有第一个函数可以供外部调用;10.设a,b,c表示三角形的三条边,表达式a+b>c|a+c>b|b+c>a,()………密………封………线………以………内………答………题………无………效……(A )是三条边构成三角形的条件; (B )是三条边不构成三角形的条件; (C )构成三角形时逻辑值为真; (D )不构成三角形时逻辑值为假二、程序阅读理解(24分)1.用plot 命令绘制平面曲线 v0=515;alpha=45*pi/180;T=2*v0*sin(alpha)/9.8;%第二行 t= T*(0:16) /16; x=v0 *t *cos(alpha);y=v0 *t *sin(alpha) -0.5*9.8*t.^2; plot(x,y,x,y,’r*’) Xmax=x(17)(1)对下面有关程序的功能的说法确切的是( ) (A )以515为初速度的抛射曲线的绘制;(B )以515为初速度的抛射曲线的绘制以及计算射程;(C )以515为初速度以45度为发射角的抛射曲线的绘制以及计算射程; (D )以515为初速度以45度为发射角的抛射曲线的绘制。

(2)第二行语句的功能是( ) (A )计算射程;(B )计算飞行时间;(C )计算航点的横坐标; (D )计算航点的纵坐标。

2.关于“牟合方盖”的实验程序如下h=2*pi/100;t=0:h:2*pi;r=0:0.05:1;x=r'*cos(t);y=r'*sin(t);z=sqrt(1-x.^2); %第三行 meshz(x,y,z),axis off colormap([0 0 1]) view(-47,56),hold onx1=cos(t);y1=sin(t);z1=abs(sin(t)); plot3(x1,y1,z1,'ro');(1)下面有关程序的功能的说法确切的是( )(A )绘圆柱面x 2 + y 2 = 1, x 2 + z 2 = 1的交线;(B )绘圆柱面x 2 + y 2 = 1, x 2 + z 2 = 1所围区域的边界曲面;(C )绘圆柱面x 2 + y 2 = 1, x 2 + z 2 = 1的交线及所围区域的边界曲面;(D )绘圆柱面x 2 + y 2 = 1, x 2 + z 2 = 1的交线及所围区域的边界曲面的上半部分。

(2)关于第三行语句错误的解释是( )(A )z 是矩形域上曲顶柱面高度值; (B )z 是与y 同型的矩阵; (C )z 是圆域上曲顶柱面高度值; (D )z 是与x 同型的矩阵3.中国农历年由天干(10干)和地支(12支)相配而成,计算农历年的MATLAB 程序如下year=input('input year:=');0.511.522.532000400060008000………密………封………线………以………内………答………题………无………效……S1='辛壬癸甲乙丙丁戊己庚'; S2='酉戍亥子丑寅卯辰巳午未申';k1=mod(year,10); k2=mod(year,12); if k1==0,k1=10;end %第五行 if k2==0,k2=12;end s1=S1(k1); s2=S2(k2);strcat(int2str(year),'年是', s1,s2,'年')(1)输入2010,实验程序的结果将给出( ) (A )2010年是丁亥年;(B )2010年是庚寅年;(C )2010年是戊子年;(D )2010年是丙戍年 (2)第五行语句的功能是( )(A )当年份是12的倍数时定位为地支12 (B )当年份是12的倍数时定位为天干12; (C )当年份是10的倍数时定位为地支10;(D )当年份是10的倍数时定位为天干104 非负函数 y =f(x)在有限区间上的图形为上半平面的一条曲线,曲线绕x 轴旋转时,产生以x 为对称轴的旋转曲面,其体积 syms a b xf=exp(a*x)*sin(b*x); f1=subs(f,a,-0.2); f2=subs(f1,b,0.5);V=pi*int(f2*f2,x,0,2*pi) double(V)t=(0:20)*pi/10; theta=t;r=f2(t);x=t'*ones(size(t)); %第九行 y=r'*cos(theta); %第十行 z=r'*sin(theta); %第十一行 mesh(z,y,x)colormap([0 0 0]) axis offview(-17,54)(1) 关于程序的功能确切的说法( ) (A)计算曲线段 f(x)=exp(a x)sin(b x), 绕X 轴旋转的旋转曲面体积 (B)计算曲线段 f(x)=exp(-0.2x)sin(0.5x), 绕X 轴旋转的旋转曲面体积 (C) 计算曲线段 f(x)=exp(a x)sin(b x), 绕X 轴旋转的旋转曲面体积并绘图 (D )计算曲线段 f(x)=exp(-0.2x)sin(0.5x), 绕X 轴旋转的旋转曲面体积并绘图(2)(A) (B ) (C ) (D) 三、程序填空(32分)1.维维安尼(Viviani )体是圆柱体( x – R /2)2 + y 2 ≤R 2/4被球面x 2 + y 2 + z 2 = R 2所割下的立体。

下面的实验程序功能是取R =2求体积上半部分,先利用符号计算处理重积分并转换为数值数据,再用蒙特卡罗方法计算体积做对比。

完成下面程序填空220[()]V f x dxππ=⎰02x π≤≤02x π≤≤02x π≤≤02x π≤≤()f x =()f x y f =y f =………密………封………线………以………内………答………题………无………效……syms x y;f=sqrt(4-x^2-y^2); y1=-sqrt(2*x-x^2);y2= ① S1=int(f,y,y1,y2); S2=int(S1,x,0,2)V= ② P=rand(10000,3);X=2*P(:,1);Y=2*P(:,2)-1;Z=2*P(:,3);II=find((X-1).^2+Y.^2<=1&Z<=sqrt(4-X.^2-Y .^2)); V1=8*length(II)/10000 2.对于二重积分 ,D 是直线y= x – 2和抛物线 y 2 = x 所围区域。

下面数学实验程序的 功能是利用MATLAB 符号计算方法将二重积分处理为二次积分进行计算(选择先对x 积分后对y 积分的秩序),然后将计算结果的符号表达式转换为双精度实数,syms x y; f=x*y*y;x1=y*y;x2= ① S1=int(f,x,x1,x2);S2=int(S1,y,-1,2);S= ②3.某年A 、B 两城镇人各有1000人,以后每年A 镇人口10%迁往B 镇; B 镇人口15%迁往A 镇。

则有迁移矩阵L ,L 有两个互异特征值λ1,λ2,属于两个特征值的特征向量分别为:α 1,α 2,记初始人口分布X 0=[1000,1000]T ,则存在不全为零的数c 1,c 2使得 X 0 = c 1α 1 + c 2α 2(解此方程组求出c 1,c 2)。

下面实验程序主要功能是利用L n X 0 = c 1λ1n α 1+ c 2λ2n α 2计算人口变化。

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