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The Mathematical Association of VictoriaSPECIALIST MATHEMATICSTrial written examination 12007Reading time: 15 minutesWriting time: 1 hour.............................................................. Student’s Name:QUESTION AND ANSWER BOOKStructure of bookNumber of questions Number of questionsNumber of marksto be answered8840Students are NOT permitted to bring mobile phones and/or any otherunauthorised electronic devices into the examination room.These questions have been written and published to assist students in their preparations for the 2007 Specialist Mathematics Examination 1. The questions and associated answers and solutions do not necessarily reflect the views of the Victorian Curriculum and Assessment Authority. The Association gratefully acknowledges the permission of the Authority to reproduce the formula sheet.This Trial Examination is licensed to the purchasing school or educational organisation with permission for copying within thatschool or educational organisation. No part of this publication may be reproduced, transmitted or distributed, in any form or by any means, outside purchasing schools or educational organisations or by individual purchasers, without permission.Published by The Mathematical Association of Victoria“Cliveden”, 61 Blyth Street, Brunswick, 3056Phone: (03) 9380 2399 Fax: (03) 9389 0399E-mail: office@.au Website: .au© MATHEMATICAL ASSOCIATION OF VICTORIA 2007Working spaceMAV SPECMATH EXAM 1/2007TURN OVERQuestion 1Given that P z z z z z 46450432=-+-+=]g and P i 20+=]g , find all the roots of P z 0=]g .3 marksInstructionsAnswer all questions in the spaces provided.A decimal approximation will not be accepted if an exact answer is required to a question.In questions where more than one mark is available, appropriate working must be shown.Unless otherwise indicated, the diagrams in this book are not drawn to scale.Take the acceleration due to gravity to have magnitude g m/s 2, where g = 9.8.MAV SPECMATH EXAM 1/2007Question 2u ~ and v ~ are vectors defined by cos sin u i j ~~~i i =+]]g g , sin cos v i j ~~~i i =+]]g g and 0211i r .a.Show that u ~ and v ~are unit vectors. 1 markb.Let a be the angle between the vectors u ~ and v ~. Express a in terms of i . 1 markc.Find a when 6i r=.1 markd.If 3i r=, find the vector resolute of v ~ in the direction of u ~. 2 marksMAV SPECMATH EXAM 1/2007TURN OVERQuestion 3 a.Expressx xx 22+ in partial fractions with integer numerators. 2 marksb. Hence show thatlog x xx dx b a2e 243++=--b l ywhere a and b are positive integers. Find the values of a and b .3 marksMAV SPECMATH EXAM 1/2007Question 4a.Show that, for x 03111, cos dx d x31=-]^_g h i 2 marksb. Hence , find the exact value of12161y3 marks7 MAV SPECMATH EXAM 1/2007TURN OVERQuestion 5An object of mass 2 kg falls from rest from a height of 50 metres. Its fall is opposed by an air resistance of magnitude of .v 0052 newton, where v is its velocity.a.Write an equation of motion for the falling object.1 markb.Show that dv dxg v v 40402=- 2 marksc.Hence, find the exact distance travelled for the object to reach a speed of 10 m/s.3 marksMAV SPECMATH EXAM 1/2007Question 6Let arctan f x x 4r=+]]g g , x R !.a.On the axes below, sketch the graph of f x ]g . On the sketch, clearly indicate theasymptotes and axes intercepts.3 marksb.Solve f x 125r=]g 1 markMAV SPECMATH EXAM 1/2007TURN OVERQuestion 7At time t seconds, a particle has position vectorcos sin sin cos t i t j t t r 3322~~~=+-+]]^]]^g g h g g h , where t 0$.a.Find its velocity vector v ~. 2 marksb.Find its maximum speed.3 marksc.Show that the particle never stops.1 markMAV SPECMATH EXAM 1/2007 10Question 8The position vector of a particle is given by 2sec tan r t i t j t ~~~+=]]]g g g where t 0$.a. Find the Cartesian equation of the path of the particle.2 marksb.Sketch the curve on the grid below, showing all important features.2 marks11 MAV SPECMATH EXAM 1/2007END OF QUESTION AND ANSWER BOOKc.Find the exact volume of revolution formed by rotating this curve between y 1= and y 2= about the y -axis.2 marksTotal 40 marksSPECIALIST MATHEMATICSWritten examinations 1 and 2FORMULA SHEETDirections to studentsDetach this formula sheet during reading time.This formula sheet is provided for your reference.© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2006Version 3 – March 2006REPRODUCED WITH PERMISSION MATHEMATICAL ASSOCIATION OF VICTORIA 2006SPECMATH 2Specialist Mathematics FormulasMensurationarea of a trapezium:12a b h +()curved surface area of a cylinder: 2π r hvolume of a cylinder: π r 2hvolume of a cone: 132πr h volume of a pyramid: 13Ah volume of a sphere: 433πr area of a triangle: 12bc A sin sine rule: a A b B cC sin sin sin ==cosine rule:c 2 = a 2 + b 2 – 2ab cos CCoordinate geometryellipse: x h ay k b−()+−()=22221hyperbola: x h ay k b−()−−()=22221Circular (trigometric) functionscos 2(x ) + sin 2(x ) = 11 + tan 2(x ) = sec 2(x ) cot 2(x ) + 1 = cosec 2(x )sin(x + y ) = sin(x ) cos(y ) + cos(x ) sin(y ) sin(x – y ) = sin(x ) cos(y ) – cos(x ) sin(y )cos(x + y ) = cos(x ) cos(y ) – sin(x ) sin(y ) cos(x – y ) = cos(x ) cos(y ) + sin(x ) sin(y )tan()tan()tan()tan()tan()x y x y x y +=+−1tan()tan()tan()tan()tan()x y x y x y −=−+1cos(2x ) = cos 2(x ) – sin 2(x ) = 2 cos 2(x ) – 1 = 1 – 2 sin 2(x )sin(2x ) = 2 sin(x ) cos(x ) tan()tan()tan ()2212x x x =−function sin –1cos –1tan –1domain [–1, 1][–1, 1]R range−ππ22,[0, !]−ππ22,Algebra (Complex numbers)z = x + yi = r (cos θ + i sin θ) = r cis θz x y r =+=22–π < Arg z ≤ πz 1z 2 = r 1r 2 cis(θ1 + θ2) z z r r 121212=−()cis θθz n = r n cis(n θ) (de Moivre’s theorem)© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2006REPRODUCED WITH PERMISSION MATHEMATICAL ASSOCIATION OF VICTORIA 20063SPECMATH Calculusd dxx nxn n()=−1∫=++≠−+x dxnx c nn n1111,d dxe aeax ax()=∫=+e dxae cax ax1d dxxxelog()()=1∫=+1xdx x celogd dxax a axsin()cos()()=∫=−+sin()cos()ax dxaax c1d dxax a axcos()sin()()=−∫=+cos()sin()ax dxaax c1d dxax a axtan()sec()()=2∫=+sec()tan()21ax dxaaxcd dxxsin−()=1()+>0c a,d dxxcos−()=1()+>0c a,d dxxxtan−()=+1211()∫+=+−aa xdxxac221tanproduct rule: ddxuv udvdxvdudx ()=+quotient rule: ddxuvvdudxudvdxv=−2chain rule: dydxdydududx=Euler’s method: If dydxf x=(),x0 = a and y0 = b, then x n + 1 = x n + h and y n + 1 = y n + h f(x n)acceleration: ad xdtdvdtvdvdxddxv ====22212constant (uniform) acceleration: v = u + at s = ut +12at2 v2 = u2 + 2as s =12(u + v)tTURN OVER© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2006 REPRODUCED WITH PERMISSION MATHEMATICAL ASSOCIATION OF VICTORIA 2006SPECMATH 4END OF FORMULA SHEETVectors in two and three dimensionsr i j k ~~~~=++x y z |r ~| = x y z r 222++= r ~1.r ~2 = r 1r 2 cos θ = x 1x 2 + y 1y 2 + z 1z 2!r r i j k ~~~~~==++d dt dx dt dy dt dzdt Mechanicsmomentum: p v~~=m equation of motion: R a~~=m friction: F ≤ µN© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2006REPRODUCED WITH PERMISSION MATHEMATICAL ASSOCIATION OF VICTORIA 2006。

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mit高数试题

mit高数试题

MIT OpenCourseWare
18.01 Single Variable Calculus
Fall 2006
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Notation: 2.1 = Section 2.1 of the Simmons book; Notes G = section G of the Notes; 1A-3 = Exercise 1A-3 in Section E (Exercises) of the Notes (solved in section S) 2.4/13; 81/4 = in Simmons, respectively, section 2.4 Problem 13; page 81 Problem 4 Recitation 0. Wed. Sept. 6 Graphing functions. Read: Notes G, sections 1-4 Work: 1A-1a, 2a, 3abe, 6a, 7a Lecture 1. Thurs., Sept. 7 Derivative; slope, velocity, rate of change. Read: 2.1-2.4 Work: 1C-3abe, 4ab (use 3), 5, 6 (trace axes onto your answer sheet) Work: 1B-2, 1C-1a (start from the definition of derivative) Lecture 2. Fri. Sept. 8 Limits and continuity; some trigonometric limits Read: 2.5 (bottom p.70-73; concentrate on examples, skip the � − δ def’n) Read: 2.6 to p. 75; learn def’n (1) and proof “differentiable =⇒ continuous” at the end. Read: Notes C Work: 1D-1bcefg, 4a; 1C-2, 1D-3ade, 6a, 8a (hint: “diff =⇒ cont.”) Lecture 3. Tues. Sept. 12 Differentiation formulas: products and quotients; Derivatives of trigonometric functions.
数学建模英文版习题与答案

数学建模英文版习题与答案

数学建模第一章 Modeling Change1.Difference equation （差分方程） Example 1. A Savings CertificateSuppose that you deposited $1,000 into your saving account initially. The interest is paid each month at a rate of 1% per month, then the value of your account will be as follows ：Letn=number of months,n a =the value of the account after n monthsThen we have ,01.01n n n n a a a a =-=∆+ So the difference equation is n n n a a a 01.01+=+ We have the dynamical system(动态系统) model{1000,.......2,1,001.10,1===+a n a a n n*Suppose you withdraw(提取) $50 each month,then the change is5001.01-=-=∆+n n n n a a a aTherefore the model becomes{1000....2,1,0,5001.101==-=+a n a a n n1.2动态方程,1b ra a n n +=+r and b are constants.(1≠r )An equilibrium value(均衡值)（or fixed point(不动点)）of a dynamical system )(1n n a f a =+is a solution a to equation a=f(a),which means that,if a a =0,then all a a n = Thus in this case the equilibrium value is rb a -=1 1st month $1,000 0a 3a 2a 1a 4th month $1,030.30 3rd month $1,020.10 2nd month $1,010Nowbra a b ra a n n +=+=+1Therefore )(1a a r a a n n -=-+Set a a b n n -=, then n n rb b =+1,thus 0b r b n n =,i.e, )(0a a r a a n n -=- hence rbr b a r a n n -+--=1)1(0 In practice, we may write rbc r a n n -+*=1 write C to be determined by 0a 1.3差分方程组【求均衡点 实际意义 说明参数】 Example 1. A Car Rental CompanyA car rental company operates in Orlando and Tampa. A traveler will rent a car in one city and return the car in either of the cities. The company wants to know if there are sufficiently many cars in each city.Let On=number of.cars in Orlando after n days Tn=number of cars in Tampa after n days Then the model is{nn n n n n T O T T O O 7.04.03.06.011+=+=++ To find the equilibrium value:{TO T T O O 7.04.03.06.0+=+=So 4O=3T ，i.e.,if 730=O Total cars and 740=T Total cars,then n O and n T will be unchanged.第二章 The Modeling Process, Proportionality, and Geometric Similarity(几何相似)【写一定的假设（Assumption ）要明确合理】 We already know kx y x y =⇔∝,k is constant.We may also consider .,,ln ,2etc e y x y x y x ∝∝∝Also y=mx+b is a usual assumption,i.e.,x b y ∝-.Geometrically, it is a straight line,which is easy to spot. Example 2. Modeling a Bass （欧洲鲈鱼） Fishing DerbyA fishing club will hold a fishing contest. In order to be environment friendly, the fish will be released immediately after caught. How to determine the weight of a fish?Problem Identification: Determine the weight of a fish in terms of some easily measurable dimensions(度量)Assumption: All fishes are geometrically similar, and the density of a bass is constant.Thus weight W ∝volume V ∝length 3L .that is,3kL W =Model Refinement: We only assume that the cross sectional areas are similar and use another dimension – girth g.Assume effective V ≈ length ⨯average cross sectional areaNow effective length ∝L average cross sectional area ∝2gThus ∝W L 2g ,i.e.,2kLg W =第三章 Least-Squares Criterion:Minimize the sum of the squares of deviations.(最小二乘法)【怎样画散点图 画趋势线 怎样运用最小二乘法公式】 Fitting a Straight LineGiven a collection of data(i i y x ,),i=1,.....m,and a linear model y=ax+b Recall the deviation of the model y=f(x) at (i i y x ,) is )(i i x f y - Thus the least-squares criterion is to minimize 2121)())((∑∑==--=-=mi iimi iib ax y x f y S79LCrossTherefore we need to solve for a and b from0)1()(20)()(211=---=∂∂=---=∂∂∑∑==mi i i i mi i i b ax y b S x b ax y a S That is∑∑∑∑∑=+=+ii i i i i y mb a x y x b x a x )()()(2We get 最小二乘法公式（其中m 为数据个数）)(,)()(,)(22222截距斜率Intercept x x m x y x y x b Slope x x m y x y x m a i iii i i i i i i i i i ∑∑∑∑∑∑∑∑∑∑∑--=--=第四章 Experimental Modeling 【给出一个散点图再给出数据然后怎样变化可让散点图直一点】Thus if the original curve is （1）concave up:（凸）Then usey or lny to squeeze the tail downward,OR use 2x or 3x to stretch the tail to theright(2) concave down:(凹)：Then use 2y or 3y to stretch the tail up-ward,OR usex or lnx to squeeze the tail to the left第五章 Simulation Modeling 【给一个随机现象描述模拟过程（random number 随机数） 按概率来分 用公式语言描述结果】Monte Carlo Fair Dice Algorithm Flip of a Fair Coin （抛硬币）： Head Tail0 0.51Let x be a random number in [0,1],define掷骰子第九章. Graphs of Functions as Models （量纲分析） Mass(质量) M Momentum （动量） 1-MTLLength(长度) LWork （功） 22-T ML Velocity （速度）1-LTDensity （密度） 3-ML Acceleration （加速度） 2-LT Viscosity （摩擦系数） 11--T ML Specific weight(重量) 22--T ML Pressure （压力） 21--T ML Force （力） 2-MTL Surface tension(张力) 2-MT Frequency （频率） 1-T Power （功率） 32-T ML Angular velocity （角速度） 1-T Rotational inertia （惯性） 2MLAngular acceleration （角加速度） 2-TTorque （转力距） 22-T ML Angular momentum （角动量） 12-T MLEntropy （能量） 22-T ML Energy （能量）22-T MLHeat22-L MLExample 1. Drag Force on a SubmarineWe are interested in the drag force experienced by a submarine. The main factors are Fluid velocity v,Characteristic dimension r (the length),Fluid density ρ,Fluid viscosity μ.Thus the model is f(D,v,r,ρ,μ)=0We haveD v rρμ2-MTL1-LT L 3-ML11--TMLTo find dimensionless(量纲) products 1)()()()()(11312=-----edcba TMLMLLLTMLTWe haveChoose a and e as free variables,then(1)a=1,e=0:b= -2,d=-1,c= -2,thusρρ221221rvDrDv==∏---(2)a=0,e=1:b= -1,d= -1,c= -1,thusρμμρvrrv==∏---1112Note that21∏is the Reynolds numberHence we have the model )(21∏=∏h,this is )(22ρμρvrhrvD=Suppose we use the model to test the drag force with rrm101=第十章Graphs of Functions as Models【军备竞赛能源危机】军备竞赛 Observations:(1) y is increasing, that is, y'>0. (2) y is concave up, that is, y''≤ 0. (3) If x=my, then y=y0 /sm.We propose the continuous model10,/0<<=s S y y y x ，Similary 10,/0<<=t t x x xy （S,t 为各自的生存率） (1) Change in 0y :If X increases its civil defense, then 0y and y' both increase. Therefore the curve y=f(x) shifts upward and has a larger slope than before.On the other hand, if missiles of Y are more effective, then 0y and y' decrease. Therefore the curve shifts downward and has a smaller slope than before.(2) Change in s:If missiles of Y are well protected, then s increases and y' decreases. Therefore the curve y=f(x) rotates downward and has a smaller slope than before.On the other hand, if the technology and weapon effectiveness of X ’s missiles is improved, then s decreases and y' increases. Therefore the curve rotates upward and has a larger slope than before.(3) Change in exchange ratio e=x/y:If X uses multiple warheads, then e increases. Therefore the curve y=f(x) rotates upward and has a larger slope than before.能源危机（供求曲线）SupposeS(q) = p* + α(q – q*), D(q) = p* – β(q – q*).After a tax of t, the new supply curve is S'(q).The new supply curve isS'(q) = p* + t + α(q – q*).To find the new equilibrium: S'(q) = D(q), that is, p* + t +α(q – q*) = p* – β(q – q*). Thusq1 = q* – t /(α+β) p1 = p* + βt /(α+β). Hence the price increase is p1 – p* = βt /(α+β). Thus,When D(q) is very steep, consumers will pay a larger portion of the tax; When S(q) is very steep, the industry will pay a larger portion of the tax.第十一章 Modeling with a Differential Equation 【画解的曲线（积分曲线）】 Example: Sketch solution curves (integral curves)：)2)(1('-+=y y y Equilibrium: y = – 1, y = 2Equilibrium point y* is stable ify(t) →y* when y0 is close to y*Therefore the equilibrium y* = –1 is stable but y* = 2 is unstable.Example: Sketch solution curves (integral curves)第十二章Modification: If there is no competition, the model is{y k ym dt dy x k xa dt dx )1()1(21-=-=Logistic modelThen the model with competition is{ymnxkymdtdyxabykxadtdx)1()1(21--=--=。

ALEVELMATH9709_s09_qp_3

ALEVELMATH9709_s09_qp_3

ALEVELMATH9709_s09_qp_3UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONSGeneral Certi?cate of Education Advanced LevelREAD THESE INSTRUCTIONS FIRSTIf you have been given an Answer Booklet,follow the instructions on the front cover of the Booklet.Write your Centre number,candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use a soft pencil for any diagrams or graphs.Do not use staples,paper clips,highlighters,glue or correction?uid.Answer all the questions.Give non-exact numerical answers correct to3signi?cant?gures,or1decimal place in the case of angles in degrees,unless a different level of accuracy is speci?ed in the question.The use of an electronic calculator is expected,where appropriate.You are reminded of the need for clear presentation in your answers.At the end of the examination,fasten all your work securely together.The number of marks is given in brackets[]at the end of each question or part question.The total number of marks for this paper is75.Questions carrying smaller numbers of marks are printed earlier in the paper,and questions carrying larger numbers of marks later in the paper.1Solve the equation ln 2+e?x =2,giving your answer correct to2decimal places.[4] 2xThe diagram shows the curve y=√(1+2tan2x)for0≤x≤1 4π√(1+2tan2x)d x,giving your answer correct to2decimal places.[3] (ii)The estimate found in part(i)is denoted by E.Explain,without furthercalculation,whether another estimate found using the trapezium rule with six intervals would be greater than E or less than E.[1]3(i)Prove the identity cosec2θ+cot2θ≡cotθ.[3] (ii)Hence solve the equation cosec2θ+cot2θ=2,for0≤θ≤360?.[2] 4The equation x3?2x?2=0has one real root.(i)Show by calculation that this root lies between x=1and x=2.[2](ii)Prove that,if a sequence of values given by the iterative formulaxn+1=2x3n+25When(1+2x)(1+ax)23,simplifying the coef?cient.[4]6The parametric equations of a curve arex=a cos3t,y=a sin3t,where a is a positive constant and0d xin terms of t.[3] (ii)Show that the equation of the tangent to the curve at the point with parameter t isx sin t+y cos t=a sin t cos t.[3](iii)Hence show that,if this tangent meets the x-axis at X and the y-axis at Y,then the length of XY is always equal to a.[2]7(i)Solve the equation 2+(2√3)i ?4=0,giving your answers in the form x+i y,where x and yare real.[3] (ii)Sketch an Argand diagram showing the points representing the roots.[1] (iii)Find the modulus and argument of each root.[3] (iv)Show that the origin and the points representing the roots are the vertices of an equilateral triangle.[1] 8(i)Express100d t=110x√(1?x2)for x≥0and its maximum point M.The diagram shows the curve y=x2(i)Find the exact value of the x-coordinate of M.[4](ii)Show,by means of the substitution x=sinθ,that the area A of the shaded region between the curve and the x-axis is given byπA=12sin22θdθ.[3](iii)Hence obtain the exact value of A.[4]。

数学实验试题

数学实验试题

………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）一二三四合计一、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MATLAB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MATLAB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，的图形；（B）绘出函数F在[–，的图形；（C）绘函数F的导函数在[0，的图形；（D）绘函数F的导函数在[–，的图形4．在MATLAB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数据，计算对应的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．MATLAB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）将无理数取三位近似；（B）将无理数取四位近似数（C）将无理数取三位近似数；（D）将有理数取四位近似数8．MATLAB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）MATLAB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，（）（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………密………封………线………以………内………答………题………无………效……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。