Matematika | Középiskola » Matematika angol nyelven emelt szintű írásbeli érettségi vizsga megoldással, 2012

ÉRETTSÉGI VIZSGA 2012. május 8 Azonosító jel: MATEMATIKA ANGOL NYELVEN EMELT SZINTŰ ÍRÁSBELI VIZSGA 2012. május 8 8:00 Az írásbeli vizsga időtartama: 240 perc Pótlapok száma Tisztázati Piszkozati NEMZETI ERŐFORRÁS MINISZTÉRIUM Matematika angol nyelven emelt szint írásbeli vizsga 0911 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 2 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: Important information 1. The exam is 240 minutes long, after that you should stop working 2. You may proceed to solve the problems in arbitrary order 3. In section II you are required to solve only four out of the five problems Please remember to enter the number of the question you have not attempted into the empty square below. Should there arise any ambiguity for the examiner about the question you ask not to be marked, it is question no. 9 that will not going to be assessed 4. You may work with any kind of

calculator as long as it is not capable of storing and displaying textual information and you may also consult any type of four digit mathematical table. The use of any other kind of electronic device or written source is forbidden. 5. Remember to show your reasonong when writing down the solutions; a major part of the score is given for this componenet of your work 6. Remember to include the substantial calculations in a clear manner 7. When refering to a theorem having a common name (e g Pythagoras’ Theorem, sine rule) that you have done at school you are not expected to state it meticulously: it is usually sufficient to put he theorem’s name. However, you are supposed to state clearly why and how does it apply. Any reference to any other theorem, however, is accepted only if it is stated precisely with all the conditions (no proof is needed) and you explain how it applies in the given situation. 8. Remember to answer each question (e g providing the result) also in text form 9.

You are supposed to work in pen; diagrams can still be drawn in pencil Anything outside the diagram and written in pencil cannot be marked by the examiner. If a solution or some part of a solution is crossed out then it is not going to be marked. 10. There is only one solution of each problem to be marked If you attempt a question more than once then you should cleary indicate the part you want to be marked. 11. Please, do not write anything in the shaded rectangular areas írásbeli vizsga 0911 3 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: I. 1. The sides a, b and c of a given triangle satisfy the following equations: c = 2b ; a 2 + b2 = 4 ; a 2 − b2 = 2 . a) b) c) Find the lengths of the sides of this triangle. Find the measure of the angles of this triangle. Find the radius of the inscribed circle of this triangle. You are expected to calculate the exact values of the respective results. írásbeli vizsga 0911 4 / 24 a) 4 points b)

5 points c) 4 points T.: 13 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 5 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: 2. a) A fair die is rolled twice and the two scores, in the order of their outcomes are entered for the digits a and b of the six digit number 8a567b , respectively. What is the probability that the digits of the six digit number hence obtained are all distinct? b) Four sets are given as follows: The elements of the set A are the two digit positive integers which are divisible by seven. The elements of the set B are the two digit positive multiples of 29. The elements of the set C are those two digit positive integers that are 11 less than a square number. The elements of the set D are those two digit positive integers which yield a square number when decreased by 13. b1) Find the number of elements of the set A ∪ C . b2) Find the number of elements of the set

B ∩ D . b3) Find those two digit positive integers that belong to exactly two of the above four sets. írásbeli vizsga 0911 6 / 24 a) 4 points b) 8 points T.: 12 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 7 / 24 2012. május 8 Matematika angol nyelven emelt szint 3. Azonosító jel: Identical pieces of two types of cheese are packed into two cylindrical boxes. Six pieces that are labelled by red stickers are put in one of the boxes as shown on the diagram and another six that are labelled by blue stickers are put in the other box. The two times six identical pieces completely fill the respective boxes. These boxes are now emptied and the twelve pieces are put on a table. Six of them are put back into one of the boxes with their labels up. How many ways are there to arrange these six pieces in the box? (Two arrangements are different if neither of them can be obtained by rotating the other one.) T.:

írásbeli vizsga 0911 8 / 24 12 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 9 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: 4. 1 1 1 1 ⋅ 3 ⋅ 5 ⋅ K ⋅ 2 n −1 , n ∈ N + . 7 7 7 7 Find the greatest natural number n for which a n > 49 −50 . a) Given is the sequence a n = b) Given is the sequence bn = 1 1 1 1 + 3 + 5 + K + 2 n −1 , n ∈ N + . 7 7 7 7 Determine the limit lim bn . n ∞ írásbeli vizsga 0911 10 / 24 a) 10 points b) 4 points T.: 14 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 11 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: II. You are expected to solve any four out of the problems 5 to 9. Write the number of the problem not selected in the blank square on page 3. 5. a) b) The four vertices of a rectangle are given in the cartesian system as A

( 0 ; 0 ), B ( 4 ; 0 ) , C ( 4 ; 1 ) and D ( 0 ; 1 ) . An interior point P ( x ; y ) of the rectangle is selected randomly. 1 1 What is the probability that y ≤ x + ? 3 2 Marci purchased 4 tombola tickets out of a total of 200 that were sold on the occasion of the Carnival. There were 10 prizes, altogether to be drawn A ticket cannot win more than one prize. b1) What is the probability that Marci wins exactly one prize on this tombola? b2) What is the probability that Marci wins something on this tombola? The calculations (the intermediate ones included) should be done correct to four decimal places. írásbeli vizsga 0911 12 / 24 a) 5 points b1) 5 points b2) 6 points T.: 16 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 13 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: You are expected to solve any four out of the problems 5 to 9. Write the number of the problem not selected

in the blank square on page 3. 6. The vertex of the graph of the quadratic function f : R R, f ( x ) = ax 2 + bx + c is V ( 4 ; 2 ) and it is also given that the point P ( 2 ; 0 ) is lying on the graph. a) Determine the coefficients a, b, and c. b) Write down the equation of the tangent to the graph at its point whose abscissa is equal to 3. c) Calculate the area bounded by the graph of f and the x-axis. írásbeli vizsga 0911 14 / 24 a) 6 points b) 5 points c) 5 points T.: 16 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 15 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: You are expected to solve any four out of the problems 5 to 9. Write the number of the problem not selected in the blank square on page 3. 7. Solve the following equation on the set of real numbers. ( 6 ⋅ 3log3 x ) log3 x ( ) = x2 log3 x − 6075 . T.: írásbeli vizsga 0911 16 / 24 16 points

2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 17 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: You are expected to solve any four out of the problems 5 to 9. Write the number of the problem not selected in the blank square on page 3. 8. A company has three branches, one in each of three cities, respectively. The average ages of the employees of the Kőszeg branch, those of the Tata branch and those of the Füred branch are 37 years, 23, years and 41 years, respectively. Three field trips were organized for the employees of the company. There were two branches going for each trip. There was no one else participating and each employee of the respective branches has, in fact, gone to the trip of their turn. The first trip was organized for the Kőszeg and the Tata branches and the average age of the participants was 29. The average age of the participants of the second trip ( those from the Kőszeg

and Füred branches) was 39.5 Finally, the average age of the participants of the third trip – they came from the Tata and the Füred branches -- was 33. What is the average age of the employees of this company? T.: írásbeli vizsga 0911 18 / 24 16 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 19 / 24 2012. május 8 Matematika angol nyelven emelt szint Azonosító jel: You are expected to solve any four out of the problems 5 to 9. Write the number of the problem not selected in the blank square on page 3. 9. An art gallery opened a new exhibition hall for children. The shape of the hall is a square based right pyramid whose internal dimensions are as follows: the edge of the base is 12 m long and the lateral edge is 10 m long. One of the exhibitor artists asked the contractors to stick a coloured band (a line) to hold the notices, all along the lateral walls, parallel to the edges of the base. The imaginary

plane through the colored band exactly halved the volume of the exhibition hall. a) Find the total length of the coloured band and also the height of the imaginary halving plane, above the ground level. On the occasion of the opening ceremony the sound engineer hanged the microphone from the apex of the hall in such a way that it was equally distant from each lateral wall as well as from the ground. b) Find the length of the hanging cable. You may neglect the size of the microphone and also that of the fastening. (Give your answers corect to the nearest centimeter.) írásbeli vizsga 0911 20 / 24 a) 9 points b) 7 points T.: 16 points 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 21 / 24 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 22 / 24 2012. május 8 Matematika angol nyelven emelt szint írásbeli vizsga 0911 Azonosító jel: 23 / 24 2012. május 8

Matematika angol nyelven emelt szint Part I. Azonosító jel: number of the problem 1. 2. 3. 4. maximal score maximal score attained score 13 12 51 12 14 16 16 64 16 16 ← problem not selected Score on the written examination 115 Part II. date score attained teacher rounded to the next integer (pontszám egész számra kerekítve) integer score input for program (programba beírt egész pontszám) Part I. (I rész) Part II. (II rész) teacher (javító tanár) registrar (jegyző) date (dátum) date (dátum) írásbeli vizsga 0911 24 / 24 2012. május 8 ÉRETTSÉGI VIZSGA 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 0911 MATEMATIKA ANGOL NYELVEN EMELT SZINTŰ ÍRÁSBELI ÉRETTSÉGI VIZSGA JAVÍTÁSI-ÉRTÉKELÉSI ÚTMUTATÓ NEMZETI ERŐFORRÁS MINISZTÉRIUM Matematika angol nyelven középszint Javítási-értékelési útmutató

Important Information Formal requirements: 1. The papers must be assessed in pen and of different colour than the one used by the candidates. Errors and flaws should be indicated according to ordinary teaching practice. 2. The maximal score for each questions is printed in the first shaded rectangle next to the question. The score given by the examiner should be entered into the other rectangle. 3. In case of correct solutions it is enough to enter the maximal score into the corresponding rectangle. 4. In case of faulty or incomplete solutions, please indicate the corresponding partial score within the body mof the paper. Substantial requirements: 1. In case of some problems there are more than one marking schemes given However, if you happen to come accross with some solution different from those outlined here, please identify the parts equivalent to those in the solution provided in this booklet and do your marking accordingly. 2. The scores given in this booklet can be split further

Keep in mind, however, that any partial score can be an integer number only. 3. If the candidate’s argument is clearly valid and the answer is correct then the maximal score can be given even if the actual solution is less detailed than the one in this booklet. 4. If there is a calculation error or any other flaw in the solution, then the score should be deducted for the actual item only where the error has occured. If the candidate is going on correctly working with the faulty intermediate result and the problem has not suffered substantial damage due to the error, then the subsequent partial scores should be awarded. 5. If there is a fatal error within an item (these are separated by double lines in this booklet), then no points should be given in this item, even for formally correct steps. If, however, the wrong result obtained by invalid argument is used correctly throughout subsequent steps, maximal scores should be given for the parts remaining, unless the problem has changed

essentially due to the error. 6. If an additional remark or a measuring unit appears in brackets in this booklet then the solution is complete even if it does not appear in the candidate’s solution. 7. If there are more than one correct attempts to solve a problem, it is the one indicated by the candidate that can be marked. 8. You should not give any bonus points (points beyond the maximal score for a solution or for some part of a solution). 9. You should not reduce the score for erroneous calculations or steps unless its results are used by the candidate in the actual course of the solution. 10. There are only 4 questions to be marked out of the 5 in part II of this examination. Hopefully, the candidate has entered the number of the question not to be marked int he square area provided for this. Accordingly, this question should not be assessed even if there is some kind of solution contained in the paper. Should there be any ambiguity about the student’s request with respect to

the question not to be considered, it is the last one in this problem set, by default, that should not be marked. írásbeli vizsga 0911 2 / 17 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató I. 1. a) The values of a and b can be computed from the simultaneous system formed by the second and the third equations. Adding thesee two equations yields 2a 2 = 6 that is a = 3 (a > 0). 1 point For the value of b. b 2 = 4 − 3 = 1 that is b = 1 . c = 2 . (The sides of the triangle are units long, respectively.) 2 points For the value of a. 3 , 1 and 2 Total: 1 point For the value of c. 4 points 1. b) 2 1 point Since 12 + 3 = 22 , the triangle is right angled by the converse of 1 point Pythagoras’ theorem and the right angle is opposite to the longest side. 1 point 1 sin β = , therefore β = 30o , 1 point 2 1 point and thus α = 60o . Total: 5 points Remark:if the candidate identifies the halved regular triangle by its sides without

going into further details then still full score should be given. 1. c) The radius of the incircle can be calculated as the a ratio of the area and the semiperimeter r = p . 2 The area of the triangle is the half of the product of 1⋅ 3 3 the two legs: = . 2 2 3 3 ⎛ 3 −1 ⎞ 2 ⎜= ⎟. r= = 2 ⎟⎠ 1 + 2 + 3 3 + 3 ⎜⎝ 2 Total: írásbeli vizsga 0911 3 / 17 This point is due if this 1 point idea is clear from the solution. 1 point 2 points If the final result is written as an 4 points approximating decimal, then at most 3 points may be given. 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 2. a) Considering the experiment of rolling a fair die twice there are 36 (equally probable) ways to assign the values of a and b. The actual scores, however, must be from the set {1, 2, 3; 4}, by condition. (There are 4 possible values of a and 3 values only of b) therefore, there are 3 ⋅ 4 = 12 numbers of the required property. 12 1

The probability in question is hence = . 36 3 Total: 1 point 1 point 1 point 1 point 4 points 2. b) The four sets as the lists of their elements are as follows A = { 14; 21; 28; 35; 42; 49; 56; 63; 70; 77; 84; 91; 98 } ,. ( A = 13.) 1 point B = { 29; 58; 87 }, 1 point C = { 14; 25; 38; 53; 70; 89 }, 1 point D = { 13; 14; 17; 22; 29; 38; 49; 62; 77; 94} . 1 point b1) The cardinality of A ∪ C is 17. ( There are 1 point exactly two common elements of the 6-element set C and of the 13-element set A.) b2) The set B ∩ D has 1 element.(The single 1 point common element of the sets B and D is 29. b3) Below are listed those two digit positive integers which belong to exactly two of the above four sets: 2 points 29; 38; 49; 70; 77. Total: 8 points Remarks: 1. If the answers for questions b1) and b2) are wrong because the candidate has made some errors when listing the elements of the sets A, B, C and/or D but the operations with these faulty sets are performed correctly, then the

respective 1-1 points should be given. 2. If the answer for question b3) differs from the correct one by one element only then 1 point still may be given (instead of 2). 3. The score should not be reduced further if the answer for for question b3) is wrong because the candidate has made some errors when listing the elements of the sets A, B, C and/or D. 4. If, instead of listing the elements of the respective sets A-D the candidate proceeds differently then full score may be given for consistent reasoning only. Answers without reasoning may be given at most 3 points (instead of 8). írásbeli vizsga 0911 4 / 17 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 3. There are 2 ways to put back six pieces of the same colour. If both colours are represented: 1 red and 5 blue pieces can be arranged in 1 way only, (since the five blue ones are inevitably consecutive). 2 red ones and 4 blue ones can be arranged in 3 different ways, since 2 red

pieces are either next to each other or they are separated by one or two blue pieces. There are 4 ways to put back 3 red and 3 blue pieces, since the 3 red pieces are either consecutive, (and so are the blue ones then) or they are separated by 1-1 or 2-1 blue pieces, respectively. (There are two ways to do this in the latter case.) R R R R B B B B R R B B R B B R R B R B R 1 point 2 points 1 point Stating just 1 or 2 possibilities is worth 1 2 points point. Incomplete argument is worth 1 point. This 1 point is due for 1 point the correct answer only. Stating just 1 possibility is 0 point, 2 or 3 possibilities are 2 points worth 1 point. Incomplete argument is worth 1 point. R B B There are 3 ways to put back 4 red and 2 blue pieces just like in the 2 + 4 case. By the same symmetry there is just 1 way to put back 5 red pieces and 1 blue piece just like in the 1 + 5 case. These points may be given if the answer is wrong due to some error when calculating 1 point the cases 2+4

and/or 1+5. 1 point There are 14 different arrangements of six pieces 1 point altogether. Total: 12 points Remarks: 1. A clear diagram can be accepted as a correct argument 2. If the candidate correctly enumerates the number of possible arrangements in the cases of 6 red, 5 red – 1 blue, 4 red – 2blue, 3 red – 3 blue, respectively and then arguing by symmetry, multiplies the correct sum by 2, then the 3 red – 3 blue arrangements are counted twice: accordingly, the score should be reduced by 2 points. írásbeli vizsga 0911 5 / 17 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 4. a) 1 1 1 1 1 ⋅ 3 ⋅ 5 ⋅ K ⋅ 2 n−1 = 1+3+5+K+(2 n−1) 7 7 7 7 7 The exponent of 7 is the sum of the first n terms of an arithmetic progression, whose first term is 1 and common difference is 2. 1 an = (1+ 2 n−1) n 7 2 1 an = n 2 7 1 Now the inequality n 2 > 49 −50 should be solved (in 7 integers). 1 1 Since 49 = 7 2 , the task is n 2

> 100 . 7 7 an = 2 7 n < 7100 . Since the function x a 7 x is monotonically increasing n 2 < 100 . The highest square number below 100 is 81. The greatest natural number satisfying the conditions is 9. Total: 10 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point points 4. b) first solution The term bn is the sum of the first n terms of a 1 geometric progression whose first term is 7 1 and its common ratio is 2 . 7 The number lim bn is the sum (s) of the geometric n∞ series whose first term is b = These 2 points are due if this idea is clear from the solution. 1 point 1 1 and r = 2 . 7 7 1 b ⎛ 7 ⎞ Since r < 1 , s = = 7 ⎜= ⎟ . 1 − r 1 − 1 ⎝ 48 ⎠ 72 7 . The limit in question is 48 1 point 1 point Total: írásbeli vizsga 0911 1 point 6 / 17 4 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 4. b) second solution bn is the sum of the first n terms of a

geometric 1 progression whose first term is 7 1 and its common ratio is 2 . 7 1 1− n 1 1 1 1 1 49 bn = + 3 + 5 + K + 2 n−1 = ⋅ 7 7 7 7 7 1− 1 49 7 ⎛ 1 ⎞ bn = ⋅ ⎜1 − n ⎟ . 48 ⎝ 49 ⎠ 7 lim bn = n∞ 48 Total: írásbeli vizsga 0911 7 / 17 1 point These 2 points are due if this idea is clear from the solution. 1 point 1 point 1 point 4 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató II. 5. a) y 1 D A 1 C B x 1 1 ⎛ 1⎞ x + cuts the y-axis at the point ⎜ 0; ⎟ , 3 2 ⎝ 2⎠ ⎛3 ⎞ and the same line cuts the y = 1 line at the point ⎜ ; 1⎟ . ⎝2 ⎠ The favourable outcomes of the question are those points of the 1 1 rectangle which are lying below the straight line y = x + . 3 2 1 3 ⋅ 29 2 The area of this region is A f = 4 − 2 = . 2 8 (According to the definition of geometric probability) the 29 29 probability of the given event is p = 8 = (= 0.90625) 4 32 Total: The straight line y =

1 point 1 point 1 point 1 point 1 point 5 points 5. b1) first solution ⎛ 200 ⎞ ⎟⎟ ways Marci could buy his 4 tickets There are ⎜⎜ ⎝ 4 ⎠ out of the 200 tombola tickets. Out of the 200 tickets there are 10 winning ones and 190 that don’t win anything. Marci wins a single prize if and only if there is one among his four tickets from the winning ten and the remaining three are from the not winning 190. ⎛10 ⎞ ⎛190 ⎞ ⎟⎟ ways for this to happen. There are ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎝1⎠ ⎝ 3 ⎠ ⎛10 ⎞ ⎛190 ⎞ ⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ 1 ⎠ ⎝ 3 ⎟⎠ ⎝ The probability is: ≈ 0.1739 ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 4 ⎠ This point is due if this 1 point idea is clear from the solution. 2 points 1 point Total: írásbeli vizsga 0911 1 point 8 / 17 5 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 5. b1) first solution ⎛ 200 ⎞ ⎟⎟ ways to draw the 10 winning There are ⎜⎜ ⎝

10 ⎠ tickets out of the total of 200. Marci has 4 of the 200 tickets and and he has none of the remaining 196 ones. Therefore, Marci wins a single prize if and only if he has exactly one out of the 10 winning tickets and the remaining 9 winning tickets are all among the remaining 196 tickets non of which is owned by Marci. ⎛ 4 ⎞ ⎛196 ⎞ ⎟⎟ favourable outcomes. Therefore, there are ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎝1⎠ ⎝ 9 ⎠ ⎛ 4 ⎞ ⎛196 ⎞ ⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ 1 ⎠ ⎝ 9 ⎟⎠ ⎝ The probability is ≈ 0.1739 ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 10 ⎠ 1 point This point is due if this 1 point idea is clear from the solution. 2 points 1 point Total: 5 points 5. b2) first solution The probability of the complementary event is calculated. In our case this is the event that Marci did not win anything on the tombola. This can happen only if each of his 4 tickets are among the 190 not winning ones. ⎛ 200 ⎞ ⎟⎟ equally probable outcomes, There are ⎜⎜

⎝ 4 ⎠ ⎛190 ⎞ ⎟⎟ favourable ones. out of which there are ⎜⎜ ⎝ 4 ⎠ The probability that Marci did not win on the ⎛190 ⎞ ⎜⎜ ⎟⎟ 4 ⎠ (≈ 0.8132) tombola is hence ⎝ ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 4 ⎠ 1 point 1 point 1 point Therefore, the probabilty that Marci has won ⎛190 ⎞ ⎜⎜ ⎟ 4 ⎟⎠ ⎝ ≈ 0.1868 something on the tombola is 1 − ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 4 ⎠ Total: írásbeli vizsga 0911 1 point These 2 points are due if this idea is clear from the solution. 1 point 9 / 17 1 point 6 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 5. b2) second solution The given event occurs if and only if Marci wins 1, 2, 3 or 4 out of the 10 prizes. The probability that he has one winning ticket is ⎛10 ⎞ ⎛190 ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎝ 1 ⎠ ⎝ 3 ⎠ ≈ 0.1739 ⎛ 200 ⎞ ⎜⎜ ⎟⎟ 4 ⎝ ⎠ The probability that he has two winning tickets is ⎛10 ⎞ ⎛190

⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ≈ 0.0125 ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 4 ⎠ The probability that he has three winning tickets is ⎛10 ⎞ ⎛190 ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎝ 3 ⎠ ⎝ 1 ⎠ ≈ 0.0004 ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 4 ⎠ The probability that each of his 4 tickets are among ⎛10 ⎞ ⎛190 ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ 4 0 ⎠ ≈ 0,0000. the winning ones is ⎝ ⎠ ⎝ ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 4 ⎠ This point is due if this 1 point idea is clear from the solution. 1 point 1 point 1 point 1 point Therefore the probability that Marci has won some prizes on the tombola is the sum of the four 1 point probabilities just computed and it is 0.1868 Total: 6 points If the candidate is working in the event space of the second solution of question b1 by investigating if Marci owns 1, 2, 3 or 4 out of the 10 winning tickets then the probability of ⎛ 4 ⎞ ⎛196 ⎞ ⎛ 4 ⎞ ⎛196 ⎞ ⎛ 4 ⎞ ⎛196 ⎞ ⎛ 4 ⎞

⎛196 ⎞ ⎟ ⎜⎜ ⎟⎟ ⋅ ⎜⎜ ⎟ + ⎜ ⎟⋅⎜ ⎟ + ⎜ ⎟⋅⎜ ⎟ + ⎜ ⎟⋅⎜ 1 ⎠ ⎝ 9 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 8 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎜⎝ 7 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 6 ⎟⎠ ⎝ the question is computed as: . ⎛ 200 ⎞ ⎜⎜ ⎟⎟ ⎝ 10 ⎠ írásbeli vizsga 0911 10 / 17 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 6. a) first solution Given is the vertex, the equation of the graph of the 2 function f is y = a( x − 4) + 2 . P is also lying on the graph, therefore 4a + 2 = 0 , 1 yielding a = − . 2 1 1 2 Hence f ( x ) = − (x − 4) + 2 = − x 2 + 4 x − 6 , 2 2 and thus b = 4 , c = −6 . Total: 2 points 1 point 1 point 1 point 1 point 6 points 6. a) second solution The graph of f is a parabola, its equation can be written as y = ax 2 + bx + c . Substituting the coordinates of the vertex V ( 4 ; 2 ) 16a + 4b + c = 2 . (1) Substituting the coordinates of the given point P( 2 ; 0)

4a + 2b + c = 0. (2) The line x = 4 is the axis of symmetry of the parabola, therefore the mirror image R ( 6 ; 0) of the given point P through this line is also lying on the graph. Therefore, 36a + 6b + c = 0. (3) Solving the symultaneous system (1)-(2)-(3) yields 1 a = − ; b = 4; c = −6. 2 Total: írásbeli vizsga 0911 11 / 17 1 point 1 point 1 point 3 points 6 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 6. b) The slope of the corresponding tangent is the value of the derivative of f at x = 3 . f ′( x ) = − x + 4 yielding m = f ′(3) = 1 . The straight line y = x + d is passing through the point of the graph of f whose abscissa is 3, therefore 3 its second coordinate is f (3) = . 2 3 This implies d = − . 2 3 The equation of the tangent is y = x − . 2 Total: This point is due if this 1 point idea is clear from the solution. 1 point 1 point 1 point 1 point 5 points 6. c) The zeros of f are 2 and 6,

therefore the area is equal to 6 6 ⎞ ⎛ 1 A = ∫ f ( x )dx = ∫ ⎜ − x 2 + 4 x − 6 ⎟dx = 2 ⎠ 2 2⎝ 1 point 1 point 6 ⎡ 1 ⎤ = ⎢− x 3 + 2 x 2 − 6 x ⎥ = ⎣ 6 ⎦2 1 point ⎞ ⎛ 4 = (− 36 + 72 − 36 ) − ⎜ − + 8 − 12 ⎟ . ⎠ ⎝ 3 16 A= . 3 1 point 1 point Total: írásbeli vizsga 0911 12 / 17 5 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 7. This point should be given if the candidate is 1 point checking the solutions by substituting int he given equation. 1 point x > 0 by the definition of common logarithm. (3 ) = x (x ) = (x ) log 3 x log 3 x log 3 x log 3 x 2 2 log 3 x 1 point Let y = x log3 x (, where y > 0 ). 1 point The equation becomes 6 y = y − 6075 , that is y 2 − 6 y − 6075 = 0 . 1 point 2 One of the solutions is y1 = −75 , which does not yield any solution of the original equation. The other solution is y 2 = 81 , yielding x log 3 x = 81 , and

hence log 3 x log3 x = log 3 81 = 4. ( ) 1 point 1 point 1 point 2 points (By the corresponding law of common logarithms) (log 3 x )2 = 4 . 1 point If log 3 x = 2 , 1 point then x1 = 3 = 9 . 1 point If log 3 x = −2 , 1 point 2 1 . 9 Both values satisfy the original equation. then x 2 = 3 − 2 = 1 point Total: írásbeli vizsga 0911 13 / 17 1 point 16 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 8. Let the sizes of the branches from Kőszeg, Tata, and Füred be k, t and f, respectively, and denote the sum of the ages of the members of the corresponding branches by S k , S t and S f , respectively. Using the given data one can write down the following equations: S k = 37k ; These 2 points are due if 2 points this idea is clear from the solution. 1 point S t = 23 t ; 1 point S f = 41 f ; 1 point S k + S t = 29(k + t ) ; 1 point Two of these relations are sufficient to solve the 1 point problem, therefore

3 1 point points should be given for any two of them. S k + S f = 39.5(k + f ) ; S t + S f = 33(t + f ) . Substituting the first three relations into the following three equations, respectively 4 37k + 23t = 29(k + t ) , that is t = k . 3 5 37k + 41 f = 39.5(k + f ) , that is f = k 3 4 23t + 41 f = 33(t + f ) , that is t = f . 5 The mean age of the total group of employees is S k + St + S f years. k +t + f Isolating t and f in terms of k yields the following expression for the mean age of the total 4 5 92 205 37 k + 23 ⋅ k + 41 ⋅ k 37 + + 3 3 = 3 3 = 4 5 4 k+ k+ k 3 3 37 + 99 136 = = = 34 . 4 4 The mean age of the total group of employees is 34 years. Total: írásbeli vizsga 0911 14 / 17 1 point Two of these relations are sufficient to solve the problem, therefore 3 1 point points should be given for any two of them. 1 point If the candidate is solving 1 point the simultaneous system of three unknowns by introducing an auxiliary triple for the respective sizes of the groups

and 2 points gets the correct result but he does not show that the answer does not depend on actual choice of these auxiliary numbers then 2 1 point points should be deducted. 1 point 16 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 9. a) E x J G D M A 10 I K H C F B 12 Using the notations of the diagram the pyramids GHIJE and ABCDE are similar. V ABCDE = 2 , therefore the ratio of the corresponding VGHIJE AB FE 3 segments (e. g = = 2. GH KE AB 12 GH = 3 = 3 (≈ 9.524 ) 2 2 48 4 ⋅ GH = 3 (≈ 38.10 ) 2 The total length of the coloured band is 38.10 m By Pythagoras’ theorem in the right triangle ABD one gets BD = 12 2 and FB = 6 2 By Pythagoras’ theorem in the right triangle FBE one ( ) 7 ≈ 5.29) L 1 point 2 points 1 point 1 point gets (FE ) = 10 2 − 6 2 . 1 point FE = 28 = 2 1 point 2 2 ( KE = 3 28 2 ⎛ 2 7 ⎞ ⎜= ⎟ ⎜ 3 2 ≈ 4.2 ⎟ ⎝ ⎠ 1 point 3 ⎞ 28 ⎛ 2 −1 ⎟ ⎜ = 28

⋅ ≈ 1 . 09 3 3 ⎟ 2 ⎜⎝ 2 ⎠ The halving plane is 1.09 m high above the ground level. Total: FK = FE − KE = 28 − írásbeli vizsga 0911 15 / 17 1 point 9 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 9. b) first solution E N 2 2 r P r 6 6 O r M 6 F The microphone should be put into the centre O of the inscribed sphere of the pyramid. The segments EL and EM on the diagram are the altitudes of the respective lateral faces. By Pithagoras’ theorem in the right triangle ELC (EL )2 = 102 − 62 , yielding EL = 8 . Since the lenghts of the tangents to a circle (sphere) from an external point are all equal, MF = MN = 6 , NE = 2 . By Pithagoras’ theorem in the right triangle OEN (OE )2 = (ON )2 + E 2 . ( ) 2 28 − r = r 2 + 2 2 6 (≈ 2.27 ) r= 7 The distance of the microphone from the apex E is EO = EF − OF ≈ 5.29 − 227 = 302 meters Total: írásbeli vizsga 0911 16 / 17 6 L 1 point 1

point 1 point 1 point 1 point 1 point 1 point 7 points 2012. május 8 Matematika angol nyelven emelt szint Javítási-értékelési útmutató 9. b) seecond solution The microphone should be put at the point O. Denote its distance from the faces of the pyramid by x (meters). Using the notations of the diagram EL is the altitude of the lateral face EBC. Connecting the vertices of the pyramid ABCDE with O it is divided into five pyramids. Write down the volume of the pyramid ABCDE as the sum of the respective volumes of these five pyramids. The pyramids ABEO, BCEO, DCEO and ADEO are congruent, therefore their volumes are equal. (1) V ABCDE = V ABCDO + 4 ⋅ VBCEO AB 2 ⋅ EF 144 ⋅ 28 = = 48 ⋅ 28 . 3 3 AB 2 ⋅ x 144 ⋅ x VABCDO = = = 48 x . 3 3 T ⋅x V BCEO = BCE . The length of the altitude of the 3 triangle BCE perpendicular to the side BC can be computed in the right triangle BEL as 1 point 1 point VABCDE = EL = 10 2 − 6 2 = 8 . BC ⋅ EL 12 ⋅ 8 ABCE = = = 48.

2 2 T ⋅ x 48 ⋅ x = 16 x . Hence VBCEO = BCE = 3 3 Plugging the respective expressions obtained for the volume into equation (1) yields 48 ⋅ 28 = 48 x + 4 ⋅ 16 x = 112 x , 6 7 ≈ 2.27 (m) 7 The distance of the microphone from the apex E is EO = EF − OF ≈ 5.29 − 227 = 302 meters Total: 1 point 1 point 1 point 1 point yielding x = írásbeli vizsga 0911 17 / 17 1 point 7 points 2012. május 8