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# MATHEMATICS GUIDE MATH 1009

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Japanese University Entrance Examination Problems in Mathematics Edited by Ling-Erl Eileen T. Wu Mathematical Association of America Table of Contents Introduction .......................... ........................... ........................... ....................... 3 1990 University Entrance Center Examination (UECE) ...................... ...................... ................... 4 Results of Performance on 1990 UECE .......................... ........................... ......................... 7 Evaluation of 1990 University Entrance Center Examination (UECE) I. Opinions of and Evaluations by Senior High School Teachers 1. Preface .......................... ........................ ......................... ......................... 7 2. Content and Scope of the Exam Problems ........................ ....................... ...................... 8 3. Analysis of the Exam Problems ........................... ............................ ....................... 9 4. Summary ........................ ......................... ........................ ........................ 10 II. Analysis by Division of Research, Senior High School Division, Association of Japanese Mathematical Education 1. Guidelines for Exam Problems ........................... ........................... ........................ 11 2. On This Year’s Problems ............................ ............................ ........................... 11 3. Students’ Group Divisions and Performances ...................... ...................... ..................... 11 4. Content and Intent of Math I, Math II Problems, and Exam Results ........................ .................... 11 5. Analysis of Criticism and Opinions on Exam Problems ............................ ........................... 12 6. The Make-Up Exam ...................... ....................... ...................... .................... 13 7. Points to be Considered in Constructing Problems in the Future ....................... ......................... 13 Survey of UECE ........................ ........................ ......................... ........................ 13 Results of Survey on University Entrance Center Examination ................... .................... ................. 14 Conclusion ........................... ........................... ......................... ....................... 15 References .......................... ........................... ........................... ....................... 15 Appendix I: Individual University Examinations ....................... ......................... ..................... 15 Appendix II: Solutions to University Examinations ...................... ...................... ...................... 19 Published and distributed by The Mathematical Association of America This project was supported by the Alfred P. Sloan Foundation. Any opinions, findings, conclusions, or recommendations expressed herein do not necessarily reflect the views of the Alfred P. Sloan Foundation. °1993 by The Mathematical Association of America (Incorporated) 2 JAPANESE EXAM 3 Introduction 2. Secondary Exam This additional exam is given independently by each university at In Japan, higher education is considered to be one of the most im- a later date. Generally the examination dates for national and local portant factors in obtaining a good job. Consequently, one’s social public universities are divided into two types: status depends greatly on the level and quality of one’s education. (i) Two-stage type—two exam dates, usually two weeks apart, with Since Japan is a fairly affluent nation, most families can afford higher pre-determined allocation of entrants, about a 9 to 1 ratio. education for their children. However, with a limited number of pres- tigious universities, competition on the entrance exam is very keen. (ii) AB-date (or ABC-date) type—two (or three) exam dates, A-date Although the overall acceptance rate at four-year institutions in recent and B-date, (and C-date), usually one week apart, uniformly set years has been above 60%, the acceptance rate at national and local each year. public universities has been only 26%. As a result even at the ele- A-date coincides with First-stage date in (i). mentary school level some Tokyo students attend additional evening or Saturday classes to prepare for the entrance examination for the Thus there are four possible paths offering a student two chances to best private junior high schools (grades 7, 8, 9) and high schools take an entrance exam for a public university: (grades 10, 11, 12) in order to be well prepared for the university (a) A-date exam −→B-date exam entrance examination. The purpose of this report is to give samples of Japanese entrance (b) A-date exam −→Second-stage exam examinations in the field of mathematics along with the performance (c) First-stage exam −→B-date exam results of the 1990 examination. It is hoped that a better understand- (d) First-stage exam −→Second-stage exam. ing of the level of mathematics expected of Japanese students will give mathematics educators in the U. S. a basis for comparison when If a student passes the First-stage exam of a university and decides reviewing the U. S. secondary school mathematics curriculum and to enter the university, the student is required to proceed with regis- tration shortly after the exam result is known. Then the student loses expectation of student performance. applicant status for the Second-stage exam or accepted status for the The editor deeply appreciates the generous assistance of Professor B-date exam, whichever status the student held. Hiroshi Fujita of Meiji University, Tokyo, Japan in obtaining the data. Professor Fujita is a member of the committee currently reviewing Each university sets the relative weights of the two examination scores in its admissions decisions. The ratio of UECE score weights the Japanese national curriculum and is in charge of the Senior High to the individual university examination weights is typically about School Mathematics Division of the Council for Educational Cur- 40% to 60%, or 50% to 50%. Tokyo University is an extreme case, riculum. He has presented papers on Mathematics Education of the Senior Secondary Level in Japan at the National Council of Teachers setting the ratio of weights at 20% for the UECE score to 80% for of Mathematics (NCTM) meetings in the U.S., at the International the Tokyo University score. In 1990, about 6% of those admitted to the four-year national Congress on Mathematical Education (ICME) in Budapest, and at or local public universities gained admission by recommendations; other international conferences. The editor also would like to express some took only the UECE and some were exempt from both the special thanks to Professor Richard Askey, University of Wisconsin, for his thorough and helpful reading of the manuscript and for his UECE and the individual university examination. Although the recent improvement in secondary examination scheduling enables a student valuable comments and suggestions. The editor sincerely thanks the to take two national or local public university exams, most applicants Alfred P. Sloan Foundation for the financial support which made target an exam given on the first date or A-date because an exam this project possible. Also special appreciation is due Beverly Ruedi given on the second date tends to be more difficult and has a smaller at the MAA for her efficient and excellent work. To the publisher, Obunsha, the author gratefully acknowledges their assistance and co- allocation of admittees. According to the Ministry of Education, about 73% of university operation in granting permission to translate material from the ”1991 students were enrolled in private universities in 1987. In the past, Zenkoku Daigaku Nyuushi Mondai Seikai Suugaku (Kokkouritsu- JFSAT was primarily for the national and local public universities daihen)” [1991 University Entrance Examinations in Mathematics: Problems & Solutions—Edition for National & Local Public Uni- and only the individual university examination was required for pri- versities]. The editor attempted to give an accurate and readable vate universities. When UECE replaced JFSAT in 1990, one of the objectives of the UECE Committee was to make the examination translation, she accepts full responsibility for any errors that may easier for private as well as public universities to utilize. Sixteen appear here. of the 342 private universities joined all 132 national and local pub- The entrance examination for a national or local public university in Japan consists of two parts: lic universities to participate in 1990. Twenty-one private universities participated in UECE in 1991 and 31 participated in 1992. Currently, most private university applicants do not take the UECE. 1. Standardized Primary Exam—UECE Private universities admit a somewhat larger percentage of their The University Entrance Center Examination (UECE) is adminis- students by recommendations. In 1984 the rate was about 20%. In tered by a government agency, the Center for University Entrance Ex- 1991 at Asia University in Tokyo, 40 of the 400 admitted by recom- aminations. This examination is based on the high school curriculum mendations were exempt from taking the UECE and the university’s set forth by the Ministry of Education. Prior to 1990, this examina- entrance examination. Unlike national and local public universities, tion was called the Joint First Stage Achievement Test (JFSAT). All private universities’ entrance examination dates are not uniform, al- applicants for national or local public universities are required to takelowing applicants to take many private university examinations. In this examination. It is offered once a year, usually in mid-January, recent years students have applied to an average of 6 private univer- with the make-up examination scheduled one week later. sities. 4 Japanese University Entrance Examination Problems in Mathematics Even though the use of UECE in Japan is not yet as widespread as 1990 UECE in Mathematics that of the SAT in the U.S., it plays a similar role. See the following table for comparison data. Directions: Each problem contains several blanks. Blanks are repre- sented by bracketed, underlined numbers. Each blank must be filled 1987 Japan ∗ U. S.∗ with a single digit or sign. See the method shown in the following Population age 18 1,883,000 3,667,000 examples and answer in the specified space on the answer sheet: High school graduates 1,792,000 2,647,000 1. {1 },{2},{3},{4},... each represent values between0 and 9 or + or − signs. For example, to indicate −8 as the answer to {1 Proportion of the age group 95% 72% mark High school graduates {1} ° +01234567 8 9 enrolled in college 681,000 1,503,000 {2} − +01234567 ° 9 Proportion of HS grads 38% 57% Proportion of the age group 36% 41% 2. If the answer is a fraction, reduce the fraction to its lowest terms and indicate the sign in the numerator. For example, to indicate−2/9 Number of students who took: (1) as the answer to {3}{4}/{5},mark UECE (formerly JFSAT) 256,000 SAT 1,134,000 {3} ° +01 2 345678 9 Proportion of HS grads {4} − +01 ° 345678 9 enrolled in college 38% 75% {5} − +01 2 345678 ° Proportion of HS grads 14% 43% Proportion of the age group 14% 31% Mathematics A [Mathematics I] Applicants to universities or junior colleges 1,025,000 (100points,60minutes) (2) Accepted applicants 681,000 2,246,000 Section 1 (30 points) Acceptance rate 66% 1. Suppose the polynomial P(x) with integer coefficients satisfies the following conditions: The 1990 UECE in mathematics is divided into two sections: Sec- (A) If P(x) is divided by x − 4x +3, the remainder is 65x − 68. tion A consists of Mathematics I and Section B consists of either 2 Mathematics II, Industrial Mathematics, or Accounting/Statistics I,B) If P(x) is divided by x +6x − 7, the remainder is −5x + a. II. Depending on the student’s area of study, he or she will select Then we know that a = {1}. 2 the appropriate examination in Section B. Of the 327,543 applicants Let us find the remainder bx + c when P(x) is divided by x+ who took Section B in 1990, 327,034 took the Mathematics II, 52 4x − 21. took Industrial Mathematics, and 457 took Accounting/Statistics I, Condition (A) implies that {2b+c = {3}{4}{5} and a = {1}. II. Because of the significantly greater number of applicants takingondition (B) implies that {67} b + c = {8}{9}. It follows that Mathematics I and II, only those two examinations will be considered = {10} and c = {11}{12}{13}. in the following article. 2. Fill in the blanks in statements (A) through (D) with the appro- Ling-Erl Eileen T. Wu priate phrase [1], [2], [3] or [4] listed below: Menlo College (A) Given sets A, B, A ∪ B = A is {14 } for A ∩ B = B. 2 (B) For some integern, n being some multiple of 12 is {15} for n being a multiple of 12. (C) The center of the circle inscribed in triangle T coinciding with the center of the circle which circumscribes triangle T is {16 for triangle T to be an equilateral triangle. ∗ (D) Given real numbers a, b,and c, Data from Monbusho: Outline of Education in Japan, 1989. ∗∗ Data from the 1989–90 Fact Book on Higher Education by American Council on Education and Macmillan Co. and theDigest of Education Statis- |a + b + c| = |a| + |b| + |c| tics (Government Printing Office, Washington, DC, 1991). (1)The figure excludes approximately 138,000 of those who graduated prior17} for ab + bc + ca ≥ 0. [1] a necessary and sufficient condition t(2)987. First time freshman including H.S. graduates of other years. [2] a necessary but not sufficient condition Answers to Mathematics A [Mathematics I] 5 [3] a sufficient but not necessary condition Answers to Mathematics A [Mathematics I] [4] neither a sufficient nor a necessary condition (100 points) Section 2 (35 points) Section 1 1. a = {1} a =2 Let a be a constant. Consider the parabola 2 2 {2}b + c = {3}{4}{5} 3b + c =127 Ca : y = −x + ax + a . {6}{7}b + c = {8}{9} −7b + c =37 1. Since the coordinates of the vertax of C are ▯ ▯ b = {10} b =9 a ,{2}a2 , {1} {3} c = {11}{12}{13} c = 100 2 the vertex is on the curve y = {4}x . 2. {14} 1 2. Let ` be the line joining two points A (−1,1) and B (2,4).For the parabolaaCand the line ` to have a common point, the value of15} 2 a must be {16} 1 {7} a ≤ {5}{6} or a ≥ {8}. {17} 3 The coordinates of the common point of paraaola Che line ` are ({9}{10},{11}), when a = {5}{6} Section 2 and ▯ ▯▯ ▯ ▯ {12} {14} ▯ {7} 1. a ,{2}a2 a, a 2 , , when a = . {1} {3} 2 4 {13} {15} {8} Also,inorderfortheparabolC and the line segment AB to have {4}x 2 5x2 two distinct points of intersection, {16} 2. a ≤ {5}{6} a ≤− 1 {17} <a ≤ {18}. a ≥ {7}/{8} a ≥ 7/5 Section 3 (35 points) ({9}{10},{11}) ( −1,1) ▯ ▯ ▯ ▯ Consider the triangle ABC with coordinates A (0,3), B (−1,0) {12} ,14} 1,11 and C (2,1). {13} {15} 5 5 1. Thecenterofthecircumscribedcircleis {16}/{17} <a ≤ {18} 7/5 <a ≤ 2 ▯ ▯ {1 } {3} {2} {4} , and the radius is ▯ Section 3 {5} {6} ▯ ▯ ▯ ▯ . 1. {1} ,{3} 1, 5 {7} {2} {4} 4 4 Also, ▯ √ {8} {5} {6} 5 2 sin∠ABC = , {7} 4 {9} and the area of triangle ABC is {10}. {8} 4 From these conditions, we know that the radius of the inscribed{9} 5 circle is ▯ ▯ {10} 4 {11}{12} − {13}. {14} ▯ {11}{12} −▯ {13} √10 − √ 2 {14} 2 2. If point P moves along the sides of triangleABC,themaximum value of the distance between the origin O and point P is {15 the minimum value is 2. {15} 3 1 1 1 ▯ . ▯ √ {16}{17} {16}{17} 10 6 Japanese University Entrance Examination Problems in Mathematics Mathematics B [Mathematics II] (1) a = {1}, b = {2}{3} and the local maximum value of the (100points,60minutes) function f(x) is ▯ {4} {5} . Choose two of the following three sections: {6} (2) The value of the slope m of the tangent line at pointP on the curve y = f(x) is greater than or equal to {7tan θ Section 1 (50 points) ◦ ◦ ◦ (0 ≤ θ ◦ 180 ), then t◦e range o◦ values forθ is 0 ≤ θ < 1. In a circle with radius 2 and its center at lte origin O,e9}{10} or {11}{12}{13} ≤ θ < 180 . vertices of an inscribed hexagon beABCDEF, with the coordinatese volume of the solid generated by revolving the region bounded by the x-axis and the curve y = f(x) about the x- of A at (2,0), and with B in the first quadrant. (1) The components of the vector axis is {14}{15} AB +2DE − 3FA−→ {16}{17}{18}π. ▯ ▯ ▯ are {1}{2},{3}{4} {5} . 2. 1, 1/2, 1/2, 1/4, 1/4, 1/4, 1/4,... is a sequence where 1/2 (2) If t is a real number, the magnitude of the vector k −→ −→ appears 2 times successively (k =1 ,2,3,... ). AB + tEF (1) Then the sum of the first 1000 terms is is a minimum when the value of t is {20}{21}{22} {6} {19} + {23} . 2 {7▯ (2) If the sum of the first n terms is 100,thenbecause and the minimum magnitude i{8}. {24}{25}{26} n =2 − {27}, 2. Let ABCD be a quadrilateral with −→ −→ n is a {28}{29} digit number provided102=0 .3010. BC =2 AD AB = CD = DA =2 Section 3 (50 points) −→ AD = a→ The numbers 1 through 9 are written individually on nine cards. Choose three cards from the nine, lettdng z denote the BA = b→ numbers of the cards arranged in increasing order. (1) Let M be the midpoint of CD.S∠BCM = {9}{10} ;◦ 1. There are {12} such x, y,az combinations. BM = ▯ {11}{12}. Also 2. The probability of having x,z all even is −→ {13} {15}−→ {3} BM = −a + b. (1) . {14} {16} {4}{5} (2) Let P be a point on AB,andQ be the point of intersection of PC and BM. Suppose PQ : QC =1:2 . Let us find AP : PB and BQ : QM.Ifweset 3. The probability of having x, y,and z be consecutive numbers is −→ −→ {6} . BP = tBA, {7}{8} we have −→ {17} −→ BQ = ( a + t b ). (2) 4. The probability of having x =4 is {18} Therefore, from(1) and (2), {9} {10}{11}. t ={19} . {20} 5. Possible values of x range from {12} to {13}.If k is an integer It follows that AP : PB = {21} : {22}, BQ : QM = {23} : {24} and such that {12 k ≤ {13}, the probability of x = k is {25}▯ BQ = {26} {27}{28}. ({14 − k)({15} − k. {16}{17}{18} Section 2 (50 points) The expected value ofx is 3 2 {19}. 1. The f√nction f(x)= √ + ax + bx has the local minimum {20} value −(2 3)/9 at x =1 / 3.Then, EVALUATION OF THE 1990 UECE 7 Answers to Mathematics B 4. {9} 5 [Mathematics II] (100 points) {10}{11} 42 Section 1 ▯ √ 5. {12}, {13} 1, 7 1. ({1}{2},{3}{4} {5})( −4,−4 3) ({14} − k)({15} − k) (9 − k)(8 − k) (8 − k)(9 − k) {6}/{7} 1/2 {16}{17}{18} 168 or 168 ▯ √ {8} 3 {19} 5 {20} 2 2. {9}{10} 60 ▯ √ {11}{12} 13 {13}−→ {15} → 3→ 1−→ a + b a + b {14} {16} 2 2 3. {17}/{18} 2/3 Results of Performance {19}/{20} 1/3 on 1990 UECE {21} : {22} 2:1 Number of Standard EXAMINATION Participants Average High Low Deviation {23} : {24} 4:5 MATHEMATICS A {25}▯ 4√ {27}{28} 13 Mathematics I 353,010 73.37 100 0 23.43 {26} 9 MATHEMATICS B Section 2 Mathematics II 327,034 64.27 100 0 22.62 1. {1} 0 Industrial Math. 52 40.87 91 0 23.69 {2}{3} −1 Accounting/Stat. I, I457 62.42 99 11 17.84 ▯ √ {4} {5} 2 3 {6} 9 {7}{8} −1 {9}{10} 90 {11}{12}{13} 135 {14}{15} 16 Evaluation of the 1990 University {16}{17}{18} 105 Entrance Center Examination (Direct translation of the text) 2. {19} 9 {20}{21}{22} 489 {23} 9 I. Opinions of and Evaluations by 2 2 n =2 {24}{25}{2− {27} n =2 100− 1 Senior High School Teachers {28}{29} 31 1. Preface Section 3 Discussions and research into establishing a more suitable univer- 1. {1}{2} 84 sity entrance examination to accommodate a great increase in the number and diversity of university applicants have allowed private {3} 1 universities to participate along with the national universities. 2. Accordingly, we revised the contents of the examinations. Origi- {4}{5} 21 nally, Math I and Math II combined were 100 minutes, 200 points; {6} 1 the UECE in mathematics is currently divided into two groups: Math 3. A[MathI]andMathB[MathII],eachexambeing60minuteslong {7}{8} 12 and worth 100 points. 8 Japanese University Entrance Examination Problems in Mathematics ThepurpoteUECi,asbt astedegreeof Section 3. Two-dimensional graphs and related expressions, trigono- mastery of the general fundamental learning established for the senior metric ratios high schools. Considering the various ways the UECE is utilized in selecting university entrants, fairness to the exam participants was 1. Find the center and the radius of the circle passing through three emphasized by posing the following questions: points A, B,and C in a plane. Also, find sin∠ABC and the area (1) whether exam problems (content, questions, score distributions, of 4ABC using either the law of sines or the law of cosines. Then find the radius of the inscribed circle. format, etc.) accurately assess the level of fundamental learning; (2) whether there is a wide difference in the degree of difficulty between the main exam and the make-up exam; 2. Find the maximum and minimum distances between the origin andapointonasideof 4ABC. (3) whether there is a great difference in the degree of difficulty among optional problem sections. Mathematics II With these criteria in mind, we will analyze the 1990 exam problems from the following points of view: Section 1. Vectors (1) whether the exam problems fulfill the purpose of the UECE 1. Coordinatized two-dimensional vectors exam; (1) Determine the coordinates of the given points and then find the (2) whether Math I, Math II are consistent with the guidelines set components of the vector. forth by the senior high schools; (2) Find the minimum value of the magnitude of the vector. (3) whether the content of exam problems tends to be esoteric; 2. Two-dimensional vectors (4) whether the degree of difficulty of optional problem sections for Math II is uniform in the three sections; (1) Using the conditions given for the vectors, draw the graph and find the angle. Using the law of cosines, find the length of a side. (5) whether the problems provide sufficient information to evaluate Also, express the vector in terms of two linearly independent the student’s mathematical thinking ability and ability to perform → calculations; vectors a and b . (6) whether, for each problem, the level of difficulty, the method of Section 2. Progressions, derivatives, integrals, and trigonometric questioning, the distribution of points, format, and so forth are appropriate; functions (7) whether the problems reflect an improvement resulting from re- 1. Derivatives, integrals, and trigonometric functions viewing the past JFSAT and the Trial Center Exam administered (1) Determine the coefficients from the information given about the in 1988. minimum value of the function. Also determine the maximum value of the function. (2) Make use of the derivative to find the minimum value for the 2. Content and Scope of the slope. Find the range of the angle formed by the tangent and the Exam Problems positive x-axis. (3) Find the volume of the solid generated by revolving the region bounded by the curve and the x-axis about the x-axis. Mathematics I Section 1. Numbers and expressions, equations and inequalities 2. Progressions and logarithms (1) Consider the pattern of the progression, and make use of the sum formula for the geometric progression. This problem is beyond 1. Polynomials with integer coefficients and equations Using the two conditions given and the remainder theorem, divide the curriculum guidelines established for the senior high schools the integral expression by the second-degree expression. and requires special effort. (2) Given the sum, use the geometric progression to determine the 2. Proofs and statements number of terms n. Also, use the common logarithm to find the Covering combinations, integers, graphs and inequalities from Math digits of n. I, determine whether the condition given in each part is necessary, sufficient, or both. 3. Probability (1) Determine the value of the combination. Section 2. Functions, two-dimensional graphs and expressions (2) Determine the probability of choosing three numbers from {2,4,6,8}. 1. Find the coordinates of the vertex of a parabola and the equation of its locus. (3) Determine the probabilityof choosing three consecutive numbers. (4) Determine the probability of choosing three numbers, of which 4 is the smallest number. 2. Find the condition for having common points between the line joining the given two points and the parabola, and find the coordi- (5) If the smallest of three numbers is k,findtherangefor k.Also nates of the common points. Also, find the condition that the line find the probability of k being the smallest number and then find segment AB and the parabola have two distinct common points. the expected value of the smallest number. EVALUATION OF THE 1990 UECE 9 3. Analysis of the Exam Problems Section 3 1. In order to find the center of the circumscribed circle of 4ABC, Mathematics I solve a system of simultaneous equations by setting the equation of the circumscribed circle to be Section 1 1. This basic problem using the remainder theorem is typical for 2 2 x + y + ax + by + c =0 , a university entrance exam. If one knows the remainder theorem,or solve for the intersecting point of two perpendicular bisectors of is unthinkable that one should miss this problem, aside from errors in calculations. This problem is suitable for testing the level of thees. We suspect that there were students who did not follow the basic learning established for senior high schools. proper order in these questions and who used sin∠ABC in the next part to solve for the radius of the circumscribed circle. The radius of the inscribed circle can be obtained by the formula 2. The format for this problem on expressions and proofs is multiple-choice because it was assumed that more participants could answer the problems intuitively. S = rs. (1) In set theory,(A∪B) ⊃ B and A ⊃ (A∩B) are basic knowl- Depending on the textbook, the exact specification of this formula edge, but the senior high school textbooks do not cover this ines. detail. We think that many answered hastily without considering 2. The shortest distance between the originO and the point P can the proof. be found simply by using the formula for the distance between a (2) If n is a multiple oftn2,h n2 is a multiple of 12. This is obvious. To disprove the converse is not too difficult. point and a line. Since the forms of the formula vary depending on the textbook, this problem requires some work. Overall, it is a good (3) If T is an equilateral triangle, then the center of the insquestion. There is no objection to its level, distribution of points, or circle and that of the circumscribed circle coincide. This format. be proven with the knowledge gained from junior high school mathematics. We think many chose the answer [A] without con- sidering the proof. Mathematics II (4) Perhaps it did not occur to the students to square both sides of Section 1 |a + b + c| = |a| + |b| + |c| 1. (1) The positions of the vertices of the regular hexagon can be and even if it had occurred to them, it would have appeared thattained by using the trigonometric ratios or symmetric points. finding a counterexample to the converse required too much timee think it might have been better to ask the participants to find the coordinates of pointB or point F before asking (1). and work. We feel that more students considered the problem intuitively. (2) Expressing the components of the vector In these types of multiple-choice questions, it is difficult to see the −→ −→ AB + tEF, logical thinking process. We would like to see an improvement made in the problem format; one which emphasizes the thought process the minimum value of the magnitude of this vector can be found that leads to the conclusion, rather than a format that encouragesfrom a second-degree function int. Because there is another way guessing the answer to the problem. We assume that this type of to solve the problem using the inner product, the problem should theorem-proof problem develops logical thinking. be designed to ensure fairness to all participants. Section 2 2. (1) BM can be obtained by the law of cosines or the Pythagorean theorem. Considering 1. This is a basic problem. We think that the short questions in the first part were easy and not confusing. The coordinates of the vertex BM→ can be obtained from the standard second-degree form. The equation of the locus can be found easily by eliminating the parameter a. as the midpoint vector of CD, BM can be found easily. Students easily solved this problem. This problem expresses the parabola ba Ca notation rarely used (2) From in textbooks. ▯ ▯ BQ =1 /3 2BP + BC −→ , 2. Parts {1} through {15} are textbook-level basic questions which one can find are considered easy to solve. For the remaining parts, it is important to keep in mind that {17} . {18} x2 − (a − 1)x − a +2=0 and − 1 ≤ x ≤ 2 Setting in order to have two distinct real solutions. This is a good question to BM = kBQ,−→ test graphing ability and mathematical thinking. We feel participants were unfamiliar with this type of problem; they either thought thatrom the fact that a and b are linearly independent, one can they did not have enough conditions or miscalculated somewhere. find the value of t.Fromthisfa{21}, {22}, {23},and {24} Overall, the number of questions and the distribution of points arean be found. In particular, when making use of equation (1) well-balanced. These are good standard questions. to find BQ, we think some students gave up in the middle of 10 Japanese University Entrance Examination Problems in Mathematics the process due to lack of graphic observation. Although it was5. From {12}, {13}, one can solve the problem if one understands stated in the problem that the point of the question. To find the probability of x = k,onecan −→ → generalize the method for 4. Since the students might not have been BA = b, accustomed to calculating combinations involving a letter (rather than a number), we consider this a bad problem. it might have been clearer if it had been specified that Section 3 is a set of good questions overall. −→ −→ −→ → AB = b because AD = a. Section 1 is a set of good questions which tie the graphs to the vectors. In terms of the level of difficulty, distribution of point4. Summary format, they are well-constructed questions which determine whether basic concepts are well understood. A summary of those points we felt strongly about on this year’s Center Examination and a record of our suggestions follows: Section 2 1. (1) All are basic problems. 1. This year’s averages are 73 for Math I and 64 for Math II, higher (2) We can see that an effort was made to demonstrate the steps in than the target scores of 60%. It has won much approval from senior high school teachers. Overall, there are many good, standard prob- a general problem dealing with differentiation and trigonometrilems with well-researched content. The problems were appropriate functions. for testing the degree of achievement in the senior high schools. (3) Here is a basic problem with calculations that are not very com- plex. Overall, (3) is a good question. However, if (1) is not 2. The content of Math I was sufficientlyconsistent with the content solved, then (2) and (3) cannot be solved, and the motive for in the learning guidelines for the senior high schools. In Math II, the asking (2) and (3) is lost. We hope to see a modification in thcontent of Sections 1 and 2 were adequate, but the level of Section 2.2 problem format. on grouped progressions was beyond the scope of the textbooks used. However, this type of application problem is useful in determining 2. (1) The steps for these problems were very difficult for the stuthe student’s innovative mathematical thinking ability. It is hoped dents. We would like to have first asked a more concrete ques- that an improved form of the question can be developed. tion such as “find the sum of the first 10 terms” to demonstrate the pattern. Eight points are assigned for answers {20rough 3. The problem on progressions crosses the entire range of Math I {23}. In order to give credit for steps in the solving process,and Math II. There were many good questions testing the student’s would have preferred to assign 4 points to the numerator and 4 ability to integrate their mathematical knowledge. We do not consider points to the denominator. the content obscure, but rather well-balanced. k−1 k−1 (2) If one observes that there are 2 terms of 1/(2 ) and that these terms sum to 1, the problem can be easily solved; how- 4. The optional problem sections in Math II were all expected to ever, it was probably a very hard question for those who were have been of the same level of difficulty, but Section 2.2 included unaccustomed to this concept. Also, regarding the problems on a progression which students were not used to, making the problem the number of the digits, if one had not observed that difficult. There is considerable disparity between the levels of diffi- culty in Sections 1 and 2. For a higher level problem like 2.2, we log(2100 − 1) ≈ log2 100, would have liked to have seen more attention devoted to the ques- tioning method and to suggest such approaches as giving a concrete or if one were not accustomed to dealing with logarithms, we think this problem would be difficult to solve. Part 2 is a goodxample and asking the question in parts. application problem involving progressions and testing mathe- 5. The 1990 exam problems are consistently appropriate for eval- matical thinking ability; however, we think the problem is be- uating basic mathematical thinking and calculation ability. We see yond the scope of textbooks, and we would have liked to have seen a concrete example leading step-by-step to the solution. the intent of the UECE as testing the student’s mastery of essential mathematical material. For Math I, Section 1.2, we would like to see an additional true-false section, more testing of the problem-solving Section 3 thought process, and the elimination of ambiguous questions. 1. Basic problem. Because of the nature of probability, the answer to (1) is related to all parts in (2) and all parts thereafter. It 6. For JFSAT, the exam time for both Math I and Math II com- appropriate question. binedwas100minutes.ForUECE,thoughthetimeallottedforsim- ple problems was transferred to harder problems within each exam, 2. Basic problem. it seemed that 60 minutes each for Math I and Math II was not sufficient. In order to determine more accurately the student’s math- 3. If one observes that once x is determined, y, z are determined, ematical ability, we wonder if it is possible to cover the same material then it is a simple and basic problem. and shorten the rest period in order to lengthen each testing period to 80 minutes. 4. Selecting y and z is the same as finding the combination of selecting 2 cards from 5 cards which are marked 5, 6, 7, 8, and 9. 7. As for topics covered, the exam has been improved by the in- This is also a basic problem. clusion of problems on progressions, reflecting efforts to respond to EVALUATION OF THE 1990 UECE 11 earlier criticisms and suggestions. To devise fair, valid problems is ness mathematics. 327,034 students took Math II, with an average a difficult task, but we hope that efforts toward this goal will be score of 64.27. For each of the three sections in Math II, the number continued. of participants who selected the problem and the average scores are shown in the following table: Section 1 2 3 II. Analysis of 1990 UECE by Division of Number of Participants 301,190 263,030 89,832 Research, Senior High School Division, Association of Japanese Mathematical Average (50 Points) 36.11 27.96 31.03 Education 4. Content and Intent of Math I, 1. Guidelines for Exam Problems Math II Exam Problems, and Exam Results The University Entrance Center Examination (UECE) was given for The content of the problems was as follows: the first time this year. The previous JFSAT [Math I, Math II] was idt Mt A[at]dM t B[Mt hemIe Mathematics I being 60 minutes long and worth 100 points. The basic guidelines Section 1. Second-degree equations, necessary and sufficient condi- of the exam problems have not been changed; the UECE is designed tions to measure the student’s level of mathematical achievement in the senior high school. The exam is to be used by individual universities Section 2. Parabolas to select entrants. Section 3. Trigonometric ratios In order to evaluate the student’s achievement in basic mathemat- Mathematics II ics, we need to avoid ambiguous or misleading problems. The exam problems should be such that students could answer them solely on Section 1. Vectors the basis of their understanding of the textbooks and their ability to apply that understanding; the scores should accurately reflect the Section 2. Differentiation, integration, sequences [progressions] Section 3. Probability level of understanding of the problems. Since this exam is one crite- rion used for judging a university entrant, we must, of course, avoid problems which all students can solve and problems which none can solve. Particularly for those universities which do not require math The intent of each problem was as follows: in the secondary examination, we need to ensure that exam problems provide sufficient information for the selection of the entrants. Also, Mathematics I the degree of the difficulty of the optional problem sections in Math II should not be widely disparate. Section 1 1. Determine the student’s understanding regarding the quotient and 2. On This Year’s Problems the remainder of second-degree polynomials with integer coefficients. 1. In keeping with the above mentioned guidelines, we devised 2. Determine the degree of understanding regarding the logic of problems of the appropriate level by taking the following into ac- necessary and sufficient conditions. count: i. the results of the analysis of the previous JFSAT; Section 2 On the topic of the relationship between a parabola and a line, deter- ii. the report of the 1989 exam committee regarding points needing mine knowledge of the basic facts regarding second-degree functions to be addressed in future exams; iii. opinions and criticisms of senior high school teachers and the and their graphs. Association of Japanese Mathematical Education. Section 3 As for the degree of difficulty, we aimed to maintain an average scoreOn the topic of inscribed and circumscribed circles of a triangle, of around 60 points. determine knowledge of basic facts regarding trigonometric ratios. 2. Math I has three sections that were required of all students; in Mathematics II Math II the student has a choice of two of the three sections. We made sure that we gave problems in diverse areas. Section 1 1. Determine the student’s understanding of and ability to apply the 3. Students’ Group Divisions and Performance basic facts regarding two-dimensional vectors. For this year’s exam, there were 353,010 students who took Math A Section 2 with an average score of 73.37. There were 327,543 students who took Math B; of those, 52 took engineering math and 457 took busi- 1. Determine the student’s understanding of the basic facts related to differentiation and integration of a third-degree polynomial. 12 Japanese University Entrance Examination Problems in Mathematics 2. Test the student’s thinking ability and understanding of geomet- 1. Calculate a position vector and its magnitude. The percentage of ric progressions using a sequence of fractions. correct answers was high, about 75%. 2. This part tests the application of the law of cosines for triangles 3. Test the student’s ability to calculate the probability of a com- bination of given numbers. and the ability to find the position vector of an interior point and the ratio of the magnitudes of vectors. The percentage of correct answers moved downward from 90% to 45% in this part. The following indicates the percentage of correct answers, incor- Section 2 rect or incomplete answers, and no answers for each problem. 1. (1) Find the function from its minimum value and then find its maximum value. Mathematics I (2) Bound the slope of the tangent, and find the range of the angle Section 1 formed by the tangent and the positive x-axis. (3) Find the volume of a solid of revolution. It is a standard integra- 1. One would know how to solve this basic remainder theorem tion problem. The scoring rate of Part 1 was 75%. problem by factoring three second-degree expressions. The percent- age of correct answers for each part was high, above 80%. 2. (1) This problem is to note the pattern of the sequence. It is useful to express the number of terms using a geometric progression and 9 10 2. This problem tests the theoretical thought process involving nec- note that 2 = 512 < 1000 < 2 =1024 . essary and sufficient conditions. {16 can be solved by looking at the (2) This problem asks one to consider the converse. It was devised to graph. The percentage of correct answers was high: 81.38%. How- test the student’s thinking and application abilities. The percent- ever, the percentage of correct answers on other parts were unex- age of correct answers was unexpectedly low (26% for the first pectedly low (57.4% for {14 }, 60.48% for {15}, 49.22% for {17}); part and 14.7% for the second part) and many left this problem perhaps students were not used to finding counterexamples. blank. Section 3 Section 2 Find the probability of picking three cards out of 9 cards. Parts 1, This section tests basic understanding and application ability related 2, 3, and 4 were basic, and the percentage of correct answers ranged to second-degree curves. The percentage of correct answers for Part from 63% to 87%. Well done. Because Part 5 used the letter k, 1 was 85%. Though overall performance was good, we think the there were many wrong answers and blank answers; the percentage students were not used to eliminating parameters. Part 2 investigates of correct answers was cut down to 30%. The overall average of the situation when a parabola and a line intersect. With the exception Section 3 was 31.03 (out of 50 points possible), which is considered of the last inequality, the percentage of correct answers was above high for a probability problem. 85%, an impressive performance. The average for Section 2 was 27.84 (out of 35 points), the best among the three sections. Section 3 5. Analysis of Criticism and Opinions on Exam Problems This is a standard problem involving two-dimensional graphs, ex- pressions, and trigonometric ratios. The scoring rate was about 67%. (1) Overall this year’s exam problems were found to be basic, and In Part 1 the problem is to find the center and the radius of 4ABC’s suitable to test the achievement level of learning steps in the senior high schools. circumscribed circle and to find the radius of the inscribed circle. A surprisingly low percentage, 59%, of the students were able to find (2) On this year’s Math II, the problem on sequences in Section the center of the circumscribed circle. 2, Part 2 was good; others said it was beyond the scope of the We think that the great difference between the percentage of textbooks of the senior high schools. It is regrettable but, because correct answers for finding the area of 4ABC (76%) and the per- of the limit on the number of blanks, we could not set the question centage of correct answers for finding the radius of the inscribed up in steps. circle (about 55%) was because some participa

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