Leaving Certificate 2015

Transcription

1 Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certificate 05 Marking Scheme Mathematics Higher Level

2 Note to teachers and students on the use of published marking schemes Marking schemes published by the State Examinations Commission are not intended to be standalone documents. They are an essential resource for examiners who receive training in the correct interpretation and application of the scheme. This training involves, among other things, marking samples of student work and discussing the marks awarded, so as to clarify the correct application of the scheme. The work of examiners is subsequently monitored by Advising Examiners to ensure consistent and accurate application of the marking scheme. This process is overseen by the Chief Examiner, usually assisted by a Chief Advising Examiner. The Chief Examiner is the final authority regarding whether or not the marking scheme has been correctly applied to any piece of candidate work. Marking schemes are working documents. While a draft marking scheme is prepared in advance of the examination, the scheme is not finalised until examiners have applied it to candidates work and the feedback from all examiners has been collated and considered in light of the full range of responses of candidates, the overall level of difficulty of the examination and the need to maintain consistency in standards from year to year. This published document contains the finalised scheme, as it was applied to all candidates work. In the case of marking schemes that include model solutions or answers, it should be noted that these are not intended to be exhaustive. Variations and alternatives may also be acceptable. Examiners must consider all answers on their merits, and will have consulted with their Advising Examiners when in doubt. Future Marking Schemes Assumptions about future marking schemes on the basis of past schemes should be avoided. While the underlying assessment principles remain the same, the details of the marking of a particular type of question may change in the context of the contribution of that question to the overall examination in a given year. The Chief Examiner in any given year has the responsibility to determine how best to ensure the fair and accurate assessment of candidates work and to ensure consistency in the standard of the assessment from year to year. Accordingly, aspects of the structure, detail and application of the marking scheme for a particular examination are subject to change from one year to the next without notice.

3 Contents Page Paper Model Solutions... Marking Scheme... Structure of the marking scheme... Summary of mark allocations and scales to be applied... Detailed marking notes... Paper Model Solutions... Marking Scheme... 5 Structure of the marking scheme... 5 Summary of mark allocations and scales to be applied... 5 Detailed marking notes Marcanna breise as ucht freagairt trí Ghaeilge... 6 []

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5 05. M9 Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certicate Examination 05 Mathematics Paper Higher Level Friday 5 June Afternoon :00 4:0 00 marks Model Solutions Paper Note: The model solutions for each question are not intended to be exhaustive there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his / her advising examiner. []

6 Instructions There are two sections in this examination paper. Section A Concepts and Skills 50 marks 6 questions Section B Contexts and Applications 50 marks questions Answer all nine questions. Write your answers in the spaces provided in this booklet. You may lose marks if you do not do so. You may ask the superintendent for more paper. Label any extra work clearly with the question number and part. The superintendent will give you a copy of the Formulae and Tables booklet. You must return it at the end of the examination. You are not allowed to bring your own copy into the examination. You will lose marks if all necessary work is not clearly shown. You may lose marks if the appropriate units of measurement are not included, where relevant. You may lose marks if your answers are not given in simplest form, where relevant. Write the make and model of your calculator(s) here: [4]

7 Section A Concepts and Skills 50 marks Answer all six questions from this section. Question (5 marks) Mary threw a ball onto level ground from a height of m. Each time the ball hit the ground it bounced back up to 4 of the height of the previous bounce, as shown. Height m (a) 0 Bounce Complete the table below to show the maximum height, in fraction form, reached by the ball on each of the first four bounces. Bounce 0 4 Height (m) (b) Find, in metres, the total vertical distance (up and down) the ball had travelled when it hit the ground for the 5 th time. Give your answer in fraction form m or S ( ( 4) m [5]

8 (c) If the ball were to continue to bounce indefinitely, find, in metres, the total vertical distance it would travel. 9 a r m [6]

9 Question Solve the equation x x 9x+ 0. Write any irrational solution in the form a+ b c, where a, b, c. (5 marks) f x x x x ( ) 9 + f x is a solution. (x ) is a factor () () x x x x x x + x x 9 + x 9x x + x x + x + or ( )( ) x x + Ax x x x + 9 x + Ax x x Ax + x x 9x + A A or x x x x x x x x Hence, other factor is x x ± ( ) 4()( ) ± 48 ± 4 x ± () Solutions: {, +, } [7]

10 Question f ( x) x + x 7, x Let (a) (i) Complete Table below. R. (5 marks) Table x f ( x) (ii) Use Table and the trapezoidal rule to find the approximate area of the region bounded by the graph of f and the x-axis. h A y + y + y + y + + y ( ) 5 square units ( ) n n 9 (b) (i) Find f ( x) dx. 9 ( + 7) x x dx x x + 7x ( ) ( ) (ii) Use your answers above to find the percentage error in your approximation of the area, correct to one decimal place % [8]

11 Question 4 (5 marks) (a) The complex numbers z, z and z are such that z, where i i. Write in the form a bi, +, and z z z + where a, b. + + z z z + i i i + + i + i i z + 5i 5 + i + 5i 5 i 5 + i 5 i 65 + i 6 z 5 + i 5+ i ( )( ) + 5i or i i i + i + i i i + i + i i i + i i + i 5 i i i z 5 i Let z a + bi 5 i a + bi 6 (5 i)( a + bi) 6+ (0) i 5a + 5bi ai + b 6+ (0) i (5a + b) + ( a + 5b) i [9]

12 5 a + b 6 (i) and a + 5 b 0 (ii) (i): 5 a + b 6 (ii): 5 a + 5b 0 6b 6 b From (ii): 5 b a a 5 z 5 + i n n (b) Let ω be a complex number such that ω, ω, and S + ω + ω + + ω. Use the formula for the sum of a finite geometric series to write the value of S in its simplest form. S + ω + ω + + ω a, r ω n ( ω n ) ( ) S ω 0 ω [0]

13 [] Question 5 (5 marks) (a) Solve the equation 6,. x x x + ( )( ), x x x x x x x x x x 4 6 : + x 9 6 : + x (b) Differentiate 6 x + x with respect to x. ( ) 6 6 ) ( + + x x x x x f ( ) 6 6 ) ( + + x x x f (c) Find the co-ordinates of the turning point of the function 6, 6. y x x x ) ( + x x f x x x ) 5 ( f ( ) 4 4 6, 5

14 Question 6 (a) (5 marks) Donagh is arranging a loan and is examining two different repayment options. (i) Bank A will charge him a monthly interest rate of 0 5%. Find, correct to three significant figures, the annual percentage rate (APR) that is equivalent to a monthly interest rate of 0 5%. F i P t ( + i) ( ) 4 8 % 0488 (ii) Bank B will charge him a rate that is equivalent to an APR of 4 5%. Find, correct to three significant figures, the monthly interest rate that is equivalent to an APR of 4 5%. F P + ( i) t ( + ) 045 i + i i 0 67 % []

15 (b) Donagh borrowed at a monthly interest rate of 0 5%, fixed for the term of the loan, from Bank A. The loan is to be repaid in equal monthly repayments over ten years. The first repayment is due one month after the loan is issued. Calculate, correct to the nearest euro, the amount of each monthly repayment. i( + i) t ( i ) t A P + ( ) 0 ( 005) or A A A A A A A [ ] 0 A []

16 Section B Contexts and Applications 50 marks Answer all three questions from this section. Question 7 (50 marks) A plane is flying horizontally at P at a height of 50 m above level ground when it begins its descent. P is 5 km, horizontally, from the point of touchdown O. The plane lands horizontally at O. P 0 5 km O 5 km Taking O as the origin, (x, f ( x ) ) approximately describes the path of the plane s descent where f () x 0004x x + cx+ d, 5 x 0, (a) (i) Show that d 0. and both x and f ( x ) are measured in km. f ( x) 0 004x x + cx + d f ( 0) d 0 d 0 (ii) Using the fact that P is the point ( 5, 05), or otherwise, show that c 0. f ( x) x x + cx f ( 5) 0 004( 5) ( 5) + c( 5) c 0 5 c 0 The plane lands horizontally at O f ( x ) 0 when x 0 f ( x) 0 007x x+ c f (0) c 0 c 0 (b) (i) Find the value of f ( x ), the derivative of f ( x ), when x 4. or f ( x) 0 004x x + cx + d f ( x) 0 007x x f ( 4) ( 4) ( 4) [4]

17 (ii) Use your answer to part (b) (i) above to find the angle at which the plane is descending when it is 4 km from touchdown. Give your answer correct to the nearest degree. tan θ f ( x) θ Angle of descent α 6497 (c) Show that ( 5, 0075) is the point of inflection of the curve y f ( x). f ( x) 0 007x x f ( x ) x x 5 f ( x) 0 004x x f ( 5) ( 5) ( 5, 0 075) ( 5) (d) (i) If (x, y) is a point on the curve y f ( x), verify that ( x 5, y + 05) is also a point on y f ( x). f ( x) 0 004x x f( x 5) 0 004( x 5) ( x 5) ( x x x ) ( x x ) x x + x y (ii) Find the image of ( x 5, y + 05) under symmetry in the point of inflection. Point: ( x 5, y+ 05) Point of inflection: ( 5, 0075) Change in x value: ( 5) ( x 5) x+ 5 Change in y value: 0075 ( y+ 05) y 0075 Image of point of inflection: x value: 5 + ( x + 5) x y value: ( y 0075) y (x, y) is image [5]

18 or Let ( x, y) be the image. x 5 + x y y, ( 5, 0 075), the point of inflection [6]

19 Question 8 (50 marks) An oil-spill occurs off-shore in an area of calm water with no currents. The oil is spilling at a rate of 40 6 cm per minute. The oil floats on top of the water. (a) (i) Complete the table below to show the total volume of oil on the water after each of the first 6 minutes of the oil-spill. Time (minutes) Volume ( 0 6 cm ) (ii) Draw a graph to show the total volume of oil on the water over the first 6 minutes. 4 0 Volume (0 6 cm ) Time (minutes) (iii) Write an equation for V(t), the volume of oil on the water, in cm, after t minutes. Line, slope 40 6, passing through (0, 0). 6 Vt ( ) 4 0 t ( ) (b) The spilled oil forms a circular oil slick millimetre thick. (i) Write an equation for the volume of oil in the slick, in cm, when the radius is r cm. V π r h π r (0 ) 0 π r cm [7]

20 (ii) Find the rate, in cm per minute, at which the radius of the oil slick is increasing when the radius is 50 m. dv dt V dv dr cm per minute π rhwhere h 0 πrh cm dv 0 π r dr dr dr dv 4 0 dt dv dt 0 r π π (5000) 6 cm per minute (c) Show that the area of water covered by the oil slick is increasing at a constant rate 7 of 4 0 cm per minute. da A π r π r dr 6 da da dr 4 0 π r 4 0 cm dt dr dt 0 πr 7 per minute or r 6 (0) π (4 0 ) π (4 0 ) da dt 7 A r t t (d) The nearest land is km from the point at which the oil-spill began. Find how long it will take for the oil slick to reach land. Give your answer correct to the nearest hour. A 5 0 πr π(0 ) π0 cm 0 π0 π0 t minutes hours [8]

21 Question 9 (50 marks) The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function π f ( t) sin t 65, π where t is the number of days after March st and 65 t is expressed in radians. (a) Find the length of the day in Galway on June 5 th (76 days after March st ). Give your answer in hours and minutes, correct to the nearest minute. π f ( t) sin t 65 π f (76) sin hours 50 minutes (b) Find a date on which the length of the day in Galway is approximately 5 hours. π f ( t) sin t 5 65 π sin t π t t days after March is April 6. (c) Find f ( t), the derivative of f ( t ). π f ( t) sin t 65 π π f ( t) cos t π π cos t [9]

22 (d) Hence, or otherwise, find the length of the longest day in Galway. f (t) is a maximum when sin t 65 t hours π is a maximum of. or 95 π π f (t) 0 cos t cos π 0 65 t π π t t π f (9 5) sin π sin 7 hours (e) Use integration to find the average length of the day in Galway over the six months from March st to September st (84 days). Give your answer in hours and minutes, correct to the nearest minute. b 84 π b a 84 a 0 65 f ( x) dx sin t 65 π cos 84 t t π [ ] dt 84 ( ) ( ) 5 hours 5 minutes 0 [0]

23 Marking Scheme Paper, Section A and Section B Structure of the marking scheme Candidate responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide candidate responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. The scales and the marks that they generate are summarised in this table: Scale label A B C D E No of categories mark scales 0, 5 0,, 5 0,, 4, 5 0 mark scales 0, 0 0, 5, 0 0, 4, 8, 0 0,, 5, 8, 0 5 mark scales 0, 5 0, 7, 5 0, 5, 0, 5 0, 4, 7,, 5 0 mark scales 0, 0 0, 0, 0 0, 7,, 0 0, 5, 0, 5, 0 5 mark scales 0, 5 0,, 5 0, 8, 7, 5 0, 6,, 9, 5 0, 5, 0, 5, 0, 5 A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response correct response B-scales (three categories) response of no substantial merit partially correct response correct response C-scales (four categories) response of no substantial merit response with some merit almost correct response correct response D-scales (five categories) response of no substantial merit response with some merit response about half-right almost correct response correct response E-scales (six categories) response of no substantial merit response with some merit response almost half-right response more than half-right almost correct response correct response In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Thus, for example, in scale 0C, 9 marks may be awarded. Throughout the scheme indicate by use of * where an arithmetic error occurs. []

24 Summary of mark allocations and scales to be applied Section A Section B Question (a) (b) (c) Question Question (a) (b) Question 4 (a) (b) Question 5 (a) (b) (c) 5C 0C 0C 5E 5D 0C 5D 0C 0C 5B 0C Question 6 (a)(i)+(ii) 0C (b)( 5C Question 7 (a)(i) 5B (a)(ii) 5B (b)(i) 0C (b)(ii) 5B (c) 0D (d)(i) 5C (d)(ii) 0C Question 8 (a)(i) 5B (a)(ii) 5B (a)(iii) 5B (b)(i) 5B (b)(ii) 0D (c) 0C (d) 0C Question 9 (a) 0C (b) 0C (c) 0B (d) 0D (e) 0 []

25 Detailed marking notes NOTE: In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Rounding and units penalty to be applied only once in each section (a), (b), (c) etc. Throughout the scheme indicate by use of * where an arithmetic error occurs. Section A Question (a) Scale 5C (0,, 4, 5) Any term correct Note: (b) Any two terms correct Dividing by gets high partial credit at most. Correct decimal values high partial at most. Scale 0C (0, 4, 8, 0) NOTE: two solutions st solution Indicates addition of terms Recognises double distance after first hop Sum of all rises or drops or nd solution Indicates addition of terms Indicates Geometric Progression Correct Geometric Progression formula with correct substitution (c) Scale 0C (0, 4, 8, 0) Recognition of sum to infinity formula High Partial Credit Correct formula with correct substitution Sum of all rises or drops []

26 Question (a) Scale 5E (0, 5, 0, 5, 0, 5) Effort at finding root, i.e. f (), f ( ), etc. Low Mid Partial Credit: Finds one root correctly after division by incorrect factor Correct answers in decimal form from calculator with or without work High Mid Partial Credit: Tries division and gets x at very minimum Having got a quadratic equation with no remainder, fills in quadratic formula ± Note: If there is a remainder after division can only get maximum of 5 marks. [4]

27 Question (a)(i) and (ii) combined Scale 5D (0, 4, 7,, 5) Any one correct value Writes formula Mid Partial Credit: Correct table Correct formula for trapezoidal rule, and some correct substitution with h Completely incorrect table but applied correctly in a(ii) Correct table and 5 without work Note (): Answers in terms of h merit Mid Partial at most. Note (): Correct formula and some substitution gets High Partial. Note (): No formula and [ ( ) ] 0 gets High Partial. (b)(i) and (ii) combined Scale 0C (0, 4, 8, 0) Any correct integration Correct substitution of f (x) Correct % error formula Correct substitution of f (x) i.e. ( x + x 7) Correct integration with some correct substitution 97% Full Credit: Accept 8% without work for full credit. [5]

28 Question 4 (a) Scale 5D (0, 4, 7,, 5) NOTE: two solutions Some rationalisation Some relevant rearrangement. Mid Partial Credit: Gets zor in the form of z Correct use of conjugate in a + bi c + di + 5i 5 + i or One complex number correct Mid Partial Credit: Two complex numbers correct Correct use of conjugate in 5 i a + bi (b) Scale 0C (0, 4, 8, 0) Correct Geometric Progression formula Correct first term Correct ratio Values substituted in formula [6]

29 Question 5 (a) Scale 0C (0, 4, 8, 0) Indication of squaring Correct roots Note: must indicate required root (b) Scale 5B (0,, 5) Partial Credit: Any correct differentiation Indication of () (c) Scale 0C (0, 4, 8, 0) Differentiation equals 0 Finds x value. Note (): A linear equation from () gets low partial at most. Note (): Must put () 0 in (c) to get any marks. Note (): () only and () only.no credit [7]

30 Question 6 (a)(i) and (ii) combined Scale 0C (0, 4, 8, 0) Correct formula in either part Correct substitution in incorrect formula Any one section correct Note: Rate as 067% or gets High Partial. (b) Scale 5C (0, 5, 0, 5) NOTE: two solutions st solution Any correct step, i.e. correct formula Substitution in correct formula. or nd solution Correct equation. Listing some terms Some substitution Complete substitution and effort at evaluation. Note: If A and interchanged and remainder of work correct, may get High Partial credit. [8]

31 Section B Question 7 (a)(i) Scale 5B (0,, 5) Partial Credit: Recognises x 0 (a)(ii) Scale 5B (0,, 5) NOTE: two solutions st solution Partial Credit: Uses x 5 or f( x ) 05 Full credit: Begins with 0 and shows ( 5) 0 5 or similar or nd solution Partial Credit: Uses x 5 Gets f '( x) Uses f '( x) 0 when x 0 (b)(i) Scale 0C (0,, 7, 0) Any term correctly differentiated. Correct differentiation Full credit: is a correct answer (b)(ii) Scale 5B (0,, 5) Partial Credit: Recognition of connection between slope and Any right angled triangle (c) Scale 0D (0,, 5, 8, 0) Some correct differentiation of f '( x) Mention of f '( x) Mid Partial Credit: Correct f ''( x ) 0 Value of x substituted tan θ [9]

32 (d)(i) Scale 5C (0,, 4, 5) Some correct substitution Correct expansions (d)(ii) Scale 0C (0, 4, 8, 0) NOTE: two solutions st solution Work leading to change in x -value or y -value Correct change in x and y values or nd solution Uses ( x, y) as image, and no more Effort at calculating mid-point Question 8 (a)(i) Scale 5B (0,, 5) Partial Credit: One correct box (a)(ii) Scale 5B (0,, 5) Partial Credit: At least two points plotted No credit Bar chart (a)(iii) Scale 5B (0,, 5) Partial Credit: Incomplete equation for volume any function of t Attempt at finding slope (b)(i) Scale 5B (0,, 5) Partial Credit: Correct volume formula Converting mm to cm [0]

33 (b)(ii) Scale 0D (0,, 5, 8, 0) Mentions a relevant rate of change. Mid Partial Credit: dr dv dv Gets from and dt dr dt Writing down chain rule. Substitution of values (c) Scale 0C (0, 4, 8, 0) NOTE: two solutions st solution Mentions relevant rate of change. States chain rule i.e. or nd solution Effort to establish value of A Note: A in terms of t Must use calculus to get any credit. (d) Scale 0C (0, 4, 8, 0) r in centimetres Effort at expression of area Correct expression for time []

34 Question 9 (a) Scale 0C (0, 4, 8, 0) Uses t 76 Note: Correct substitution Using one error, but do not penalise again in (b) (b) Scale 0C (0, 4, 8, 0) Correct f (t) substituted. Note: Correct equation with t only Accept 5 or 6 substituted correctly and tested. (c) Scale 0B (0, 5, 0) Partial Credit: Any correct differentiation (note: 0 could be correct differentiation here) Note: Substituting 80 for one error (d) Scale 0D (0,, 5, 8, 0) both solutions f '( t) 0 Mid Partial Credit: Value of t Value of t substituted into f (t) f (t) maximum when sin θ Note: Accept 9 or 9 substituted and evaluated correctly for full marks. (e) Scale 0D (0,, 5, 8, 0) Correct expression in x or t Correct formula Correct limits Mid Partial Credit: Any correct integration Correct integration and effort at substitution Note: Integration with one error but finished correctly gets High Partial Credit. []

35 05. M0 Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certicate Examination 05 Mathematics Paper Higher Level Monday 8 June Morning 9:0 :00 00 marks Model Solutions Paper Note: The model solutions for each question are not intended to be exhaustive there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his / her advising examiner. []

36 Instructions There are two sections in this examination paper. Section A Concepts and Skills 50 marks 6 questions Section B Contexts and Applications 50 marks questions Answer all nine questions. Write your answers in the spaces provided in this booklet. You may lose marks if you do not do so. There is space for extra work at the back of the booklet. You may also ask the superintendent for more paper. Label any extra work clearly with the question number and part. The superintendent will give you a copy of the Formulae and Tables booklet. You must return it at the end of the examination. You are not allowed to bring your own copy into the examination. You will lose marks if all necessary work is not clearly shown. You may lose marks if the appropriate units of measurement are not included, where relevant. You may lose marks if your answers are not given in simplest form, where relevant. Write the make and model of your calculator(s) here: [4]

37 Section A Concepts and Skills 50 marks Answer all six questions from this section. Question (5 marks) An experiment consists of throwing two fair, standard, six-sided dice and noting the sum of the two numbers thrown. If the sum is 9 or greater it is recorded as a win (W). If the sum is 8 or less it is recorded as a loss (L). (a) Complete the table below to show all possible outcomes of the experiment. Die L L L L L L L L L L L L Die L L L L L W 4 L L L L W W 5 L L L W W W 6 L L W W W W (b) (i) Find the probability of a win on one throw of the two dice. 0 P ( W) (ii) Find the probability that each of successive throws of the two dice results in a loss. Give your answer correct to four decimal places. P 8 ( L,L,L) (c) The experiment is repeated until a total of wins occur. Find the probability that the third win occurs on the tenth throw of the two dice. Give your answer correct to four decimal places. ( ) P wins in P( wins, rd on 0th throw ) [5]

38 Question (5 marks) A survey of 00 shoppers, randomly selected from a large number of Saturday supermarket shoppers, showed that the mean shopping spend was The standard deviation of this sample was 0 7. (a) Find a 95% confidence interval for the mean amount spent in a supermarket on that Saturday. σ n 00 σ C. I. x ± ± 4 06 n We can be 95% confident that the mean amount spent was in the range 86 9 < μ < 94 5 (b) A supermarket has claimed that the mean amount spent by shoppers on a Saturday is 94. Based on the survey, test the supermarket s claim using a 5% level of significance. Clearly state your null hypothesis, your alternative hypothesis, and your conclusion. H : 0 Mean spend is 94 H Mean spend is not 94 : METHOD : x 90 45, σ 0 7, μ 94, n 00 x μ z 7 σ 07 n 7 > 96 Fail to reject null hypothesis ( Not enough evidence to reject the null hypothesis) or METHOD : M 94 is inside the confidence interval for the mean spend in the population 86 9 < μ < 94 5 worked out in part (i) etc. Fail to reject null hypothesis ( Not enough evidence to reject the null hypothesis) Or METHOD : C.I. based on a sample of 00 based on the claim is: < x < is inside this interval. Fail to reject null hypothesis ( Not enough evidence to reject the null hypothesis) [6]

39 (c) Find the p-value of the test you performed in part (b) above and explain what this value represents in the context of the question. ( < 7) P( z < 7) P z p-value: Meaning: If the mean amount spent really was 94, then the probability that the sample mean would be 9045 by chance is 87%. It is because this is more than a 5% chance that we do not reject the null hypothesis. [7]

40 Question (a) The co-ordinates of two points are A(4, ) and B(7, t ). (5 marks) The line l : x 4y 0 is perpendicular to AB. Find the value of t. t + t + Slope AB Slope 7 4 t + AB l t + 4 t 5 4 or AB : 4x + y + c 0 (4, ) 4x + y + c c 0 c 4(7) + ( t) 0 t 5 l 4 (b) Find, in terms of k, the distance between the point P(0, k ) and l. d (0) 4k k 5 (c) P(0, k ) is on a bisector of the angles between the lines l and l : 5x+ y 0 0. (i) Find the possible values of k. 8 4k k k 0 + k 5 ( 8 4k) ± 5( 0 + k) k 84 or 8k 84 k or k 48 4 (ii) If k > 0, find the distance from P to l. k d () 4 [8]

41 Question 4 (5 marks) Two circles s and c touch internally at B, as shown. (a) The equation of the circle s is ( x ) + ( y + 6) 60. Write down the co-ordinates of the centre of s. s c K B Centre: (, 6) Write down the radius of s in the form a 0, where a N. Radius: (b) (i) The point K is the centre of circle c. The radius of c is one-third the radius of s. The co-ordinates of B are (7, ). Find the co-ordinates of K. AK : KB : K, + + ( 5, 6) or Centre of s to B (translation) X ordinate goes up by 6 Y ordinate goes up by 8 ( 6 ) + 5 ( 8 ) 6 6 (ii) Find the equation of c. ( x 5) + ( y 6) ( 0) 40 (c) Find the equation of the common tangent at B. Give your answer in the form ax + by + c 0, where a,b, c Z Slope AB 7 6 Slope of tangent y x 7 Equation: ( ) x + y 4 0 [9]

42 Question 5 tan A + tan B (a) Prove that tan( A + B). tan Atan B (5 marks) tan ( A + B) sin cos ( A + B) ( A + B) sin AcosB + cos Asin B cos AcosB sin Asin B sin AcosB cos Asin B + cos Acos B cos Acos B cos AcosB sin Asin B cos AcosB cos AcosB tan A + tan B tan Atan B or sin A sin B + tan A + tan B cos A cos B tan A tan B sin Asin B cos Acos B sin Acos B + cos Asin B cos Acos B sin Acos B + cos Asin B cos Acos B sin Asin B cos Acos B sin Asin B cos Acos B sin( A + B) tan( A + B) cos( A + B) (b) Find all the values of x for which sin( x), 0 x 60, x in degrees. sin x x 60, 0, 40, 480, 780, 840 x 0, 40, 40, 60, 60, 80 or x60 +n(60 ), or x0 +n(60 ), x0 +n(0 ), or x40 +n(0 ), n0 x 0 or x 40 n x 40 or x 60 n x 60 or x 80 [40]

43 Question 6 (a) Construct the centroid of the triangle ABC below. Show all construction lines. (Where measurement is used, show all relevant measurements and calculations clearly.) AC cm; BC 7 cm (5 marks) A 5 55 cm 5 55 cm B 5 85 cm 5 85 cm C or A B C [4]

44 (b) Prove that, if three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal line. Diagram: D A A E B E B F C F Given: AD BE CF, as in the diagram, with AB BC To Prove: DE EF Construction: Draw AE DE, cutting EB at E and CF at F Draw F B AB, cutting EB at B, as in diagram. Proof: B F BC (opposite sides in a parallelogram) AB (by assumption) BA E E F B (alternate angles) A E B F E B (vertically opposite angles) Δ ABE is congruent to Δ FBE (ASA) AE FE But AE DE and F E FE (opposite sides in a parallelogram) DE EF [4]

45 Section B Contexts and Applications 50 marks Answer all three questions from this section. Question 7 (40 marks) A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale). The sides of the plate, HK and PQ, are tangential to each circle. The larger circle has centre A and radius 4r cm. The smaller circle has centre B and radius r cm. The length of [HK ] is 8r cm and AB 0 7 cm. H 8r 4r K A 0 7 B r Q P (a) Find r, the radius of the smaller circle. (Hint: Draw BT KH, T AH. ) H T 8r r K A 0 7 B r Q ( r) + ( 8 ) ( 0 7) AT + BT AB r P 9r r + 64r r 0 cm [4]

46 (b) Find the area of the quadrilateral ABKH. ABKH BKHT + ΔABT cm ( 60)( 60) (c) (i) Find HAP, in degrees, correct to one decimal place. 60 tan HAB 60 HAB HAP 8 9 (ii) Find the area of the machine part, correct to the nearest cm. Area large sector HAP + area HABK + area sector KBQ 8 9 π ( 80) π ( 0) [44]

47 Question 8 (65 marks) In basketball, players often have to take free throws. When Michael takes his first free throw in any game, the probability that he is successful is 0 7. For all subsequent free throws in the game, the probability that he is successful is: 0 8 if he has been successful on the previous throw 0 6 if he has been unsuccessful on the previous throw. (a) Find the probability that Michael is successful (S) with all three of his first three free throws in a game. P ( S, S, S) (b) Find the probability that Michael is unsuccessful (U) with his first two free throws and successful with the third. P ( U,U,S) (c) List all the ways that Michael could be successful with his third free throw in a game and hence find the probability that Michael is successful with his third free throw. S, S, S U, U, S S, U, S U, S, S P P P P ( S, S, S) ( U,U,S) ( S, U,S) ( U,S, S) P [45]

48 (d) (i) Let p n be the probability that Michael is successful with his th n free throw in the game th (and hence ( p n ) is the probability that Michael is unsuccessful with his n free throw). Show that p n pn. ( ) p n+ P S,S + P(U,S) ( p ) p n p n n (ii) Assume that p is Michael s success rate in the long run; that is, for large values of n, we have pn+ pn p. Using the result from part (d) (i) above, or otherwise, show that p p p n pn p 0 8pn pn 0 75 p 0 8 n (e) For all positive integers n, let an p pn, where p 0 75 as above. an+ (i) Use the ratio to show that an is a geometric sequence with common ratio. a 5 n a n+ a n p p p p n p 5 n ( p ) p ( p ) 5 n n n n [46]

49 (ii) Find the smallest value of n for which p p n < a p n p n a p p ( 0 ) < ar 0 05 n ln 0 < ln n n ( ) ln n > 5 9 ln 0 n > 6 9 n 7 (f) You arrive at a game in which Michael is playing. You know that he has already taken many free throws, but you do not know what pattern of success he has had. (i) Based on this knowledge, what is your estimate of the probability that Michael will be successful with his next free throw in the game? Answer: 075 or p (ii) Why would it not be appropriate to consider Michael s subsequent free throws in the game as a sequence of Bernoulli trials? Events not independent [47]

50 Question 9 (a) (45 marks) Joan is playing golf. She is 50 m from the centre of a circular green of diameter 0 m. The diagram shows the range of directions in which Joan can hit the ball so that it could land on the green. Find α, the measure of the angle of this range of directions. Give your answer, in degrees, correct to one decimal place. 0 m 5 sin α 0 50 α 5 79 α 478 α 5 50 m α (b) At the next hole, Joan, at T, attempts to hit the ball in the direction of the hole H. Her shot is off target and the ball lands at A, a distance of 90 metres from T, where ATH 8. TH is 85 metres. Find AH, the distance from the ball to the hole, correct to the nearest metre. 90 m A T 8 85 m H AH (90)(85) cos AH 57 or [48]

51 90 m A T 8 X 85 m H Draw AX perpendicular to TH AX triangle ATX: sin8 AX TX cos8 TX XH 04 ( 58 7) ( 04 ) AH + AH 566 [49]

52 (c) At another hole, where the ground is not level, Joan hits the ball from K, as shown. The ball lands at B. The height of the ball, in metres, above the horizontal line OB is given by h 6t + t + 8 where t is the time in seconds after the ball is struck and h is the height of the ball. K (i) O Find the height of K above OB. B h 6t + t + 8 t 0 h 8 m (ii) The horizontal speed of the ball over the straight distance [OB] is a constant 8 m s -. Find the angle of elevation of K from B, correct to the nearest degree. h 0 6t + t t 4 6t 0 ( )( ) t 4, t t 4 OB m 8 tan OBK OBK [50]

53 (d) At a later hole, Joan s first shot lands at the point G, on ground that is sloping downwards, as shown. A vertical tree, [CE], 5 metres high, stands between G and the hole. The distance, GC, from the ball to the bottom of the tree is also 5 metres. The angle of elevation at G to the top of the tree, E, is θ, where θ tan. The height of the top of the tree above the horizontal, GD, is h metres and GD d metres. (i) Write d and CD in terms of h. E h tanθ d d h G θ d D h 5 m CD 5 h 5 m C (ii) Hence, or otherwise, find h. d + CD 5 ( h) + (5 h) h + h+ h h h h 0, h 0 h 0 m or θ tan GED 6 45 CGE 6 45 CGD h sin h 5 h 0 m or h GCE 54 sin h 0 8 h 0 m 5 [5]

54 Marking Scheme Paper, Section A and Section B Structure of the marking scheme Candidate responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide candidate responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. The scales and the marks that they generate are summarised in this table: Scale label A B C D E No of categories mark scales 0,, 5 0,, 4, 5 0,,,4,5 0 mark scales 0, 5, 0 0, 4, 8, 0 0,, 5, 8, 0 5 mark scales 0, 5,, 5 0, 4, 7,, 5 0 mark scales 5 mark scales A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response correct response B-scales (three categories) response of no substantial merit partially correct response correct response C-scales (four categories) response of no substantial merit response with some merit almost correct response correct response D-scales (five categories) response of no substantial merit response with some merit response about half-right almost correct response correct response E-scales (six categories) response of no substantial merit response with some merit response almost half-right response more than half-right almost correct response correct response NOTE: In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Rounding and units penalty to be applied only once in each section (a), (b), (c) etc. Throughout the scheme indicate by use of * where an arithmetic error occurs. [5]

55 Summary of mark allocations and scales to be applied Section A Section B Question (a) (b)(i)+(ii) (c) Question (a) (b) (c) Question (a) (b) (c) 0C 0C 5C 0C 0D 5C 0D 0C 5D Question 7 (a) (b) (c)(i) (c)(ii) Question 8 (a) (b) (c) (d)(i) (d)(ii) (e)(i) (e)(ii) (f) 5C 5C 5C 5D 0C 0C 5D 5C 0B 5C 5C 5B Question 4 (a) (b)(i) (b)(ii) (c) Question 5 (a) (b) 5B 5C 0C 5C 5D 0D Question 9 (a) (b) (c)(i) (c)(ii) (d)(i) (d)(ii) 0C 0C 5B 0C 5B 5D Question 6 (a) 5C (b) Diag 5B Const 5B Proof 0C [5]

56 Section A Question (5 marks) (a) Scale 0C (0, 4, 8, 0) At least one other correct entry Partially correct table with at least 5 correct totals or couples Five or more correct entries including at least one other loss and one other win Table correctly completed with totals or couples but no indication of W or L (b)(i)(ii) Scale 0C (0, 4, 8, 0) Favourable outcomes identified (i) correct only ( 0 5,, 0 7, 0 8, 0 ) 6 8 (i) omitted or of no merit but (ii) 8 (c) Scale 5C (0,, 4, 5) Relevant binomial formula with some substitution Identifies p 7 or ( p) or ( p) or p Listing at least any two of the ten throws High Partial Credit Probability of two wins in nine throws [54]

57 Question (5 marks) (a) Scale 0C (0, 4, 8, 0) Relevant formula with or without substitution with further work n High Partial Credit σ 96 evaluated n (b) Scale 0D (0,, 5, 8, 0) One relevant step e.g. null hypothesis or alternative hypothesis stated Some work towards finding z Mention of ± 96 Mid Partial Credit: z calculated Either null or alternative hypothesis stated and relevant work towards finding z Confidence interval from (a) and either null or alternative hypothesis stated Confidence interval based on 00 (i.e , 98 06) and either null or alternative hypothesis stated. z calculated and compared to ± 96 but: o Not stating null hypothesis and / or alternative hypothesis correctly o Not accepting or rejecting hypothesis o Incorrect conclusion for hypothesis Incorrect use of 94 and confidence interval Incorrect use of and confidence interval (c) Scale 5C (0,, 4, 5) Effort at finding P(z < 7) p value correct Not contextualising answer correctly [55]

58 Question (5 marks) (a) Scale 0D (0,, 5, 8,0) Slope AB or l Mid Partial Credit: Both slopes found Slopes linked to perpendicularity (b) Scale 0C (0, 4, 8, 0) Relevant formula with some correct substitution High Partial Credit Substitution into formula fully correct (c) Scale 5D (0,,, 4, 5) Relevant formula with some correct substitution Mid Partial Credit: One value for k found Work indicating two values for k Both values of k Positive value for k evaluated and distance calculated [56]

59 Question 4 (5 marks) (a) Scale 5B (0,, 5) Partial Credit: Centre or radius (b)(i) Scale 5C (0,, 4, 5) Formula for ratio with some correct substitution Effort at setting up translation Substitution into ratio formula fully correct One ordinate only found Correct answer without supporting work (b)(ii) Scale 0C (0, 4, 8, 0) Identifies centre Identifies radius Equation of circle formed but error in substitution (c) Scale 5C (0,, 4, 5) Slope AB or slope of tangent Some correct substitution into relevant formula Equation of line fully substituted [57]

60 Question 5 (5 marks) (a) Scale 5D (0, 4, 7,, 5) Tan function in terms of Sine and Cosine Mid Partial Credit: sin (A+B) or cos (A+B) expanded Numerator or denominator in fraction form (method ) Numerator and denominator divided by cosa.cosb (Method ) Both numerator and denominator expressed in form of a single fraction (Method ) (b) Scale 0D (0,, 5, 8, 0) One value for x Mid Partial Credit: One value for x Two or more values for x Three or more values for x Question 6 (5 marks) (a) Scale 5C (0,, 4, 5) Some relevant calculation One side bisected One midpoint indicated One median drawn (b) Diagram / Given : Scale 5B (0,, 5) Partial Credit: Effort at Diagram or Given Construction: Scale 5B (0,, 5) Partial Credit: Construction attempted (diagram and/or description) Proof: Scale 0C (0, 4, 8, 0) More than one critical step omitted but still some substantial work of merit Proof completed with one critical step omitted [58]

61 Section B Question 7 (40 marks) (a) Scale 5C (0, 5,, 5) BT drawn correctly Pythagoras formula with some correct substitution Recognising ATB 90 Pythagoras formula fully substituted (b) Scale 5C (0, 5,, 5) Indicates two areas Effort at area of rectangle only Effort at area of triangle only Area of triangle correct Area of rectangle correct (c)(i) Scale 5C (0,, 4, 5) 60 tan HAB or equivalent in sin or cos 60 HAB in degrees. (c)(ii) Scale 5D (0,,, 4, 5) Effort at area of one region Mid Partial Credit: Area of one sector with correct substitution Area of two sectors with substitution correct in both. [59]

62 Question 8 (65 marks) (a) Scale 0C (0, 4, 8, 0) One correct probability Identifies all three probabilities correctly Three probabilities multiplied of which two are correct (b) Scale 0C (0, 4, 8, 0) One correct probability Identifies all three probabilities correctly Three probabilities multiplied of which two are correct (c) Scale 5D (0, 4, 7,, 5) Lists one new way Mid Partial Credit: Full listing only One new probability Sum of three probabilities Identifies all four probabilities correctly (d) (i) Scale 5C (0,, 4, 5) Indicates P(S, S) and/or P(U, S) or equivalent Substitution into equation for p n + (d) (ii) Scale 0B (0, 5, 0) Partial Credit: Partial substitution into equation (e)(i) Scale 5C (0,, 4, 5) a n + in terms of p and p n+ a n + in terms of p, p n, and p n+ an a n + substituted a n [60]

63 (e)(ii) Scale 5C (0,, 4, 5) a in numerical form n ar substituted a7 evaluated without checking a 6 (f) Scale 5B (0,, 5) Partial Credit: (i) correct only or (ii) correct only Question 9 (45 marks) (a) Scale 0C (0, 4, 8, 0) Effort at expressing sine function in terms of 5 and 50 Finds third side of triangle and makes effort to find an angle Half angle found (b) Scale 0C (0, 4, 8, 0) Cosine Rule with some correct substitution Effort at calculating AX or TX Cosine Rule substituted correctly Finds AX and formulates for TX (or vice versa) (c)(i) Scale 5B (0,, 5) Partial Credit: t 0 indicated Accept h 8 m without work (c)(ii) Scale 0C (0, 4, 8, 0) h 0 indicated OB found for positive value for t (d)(i) Scale 5B (0,, 5) Partial Credit: h d CD 5 h [6]

64 (d)(ii) Scale 5D (0,,, 4, 5) Pythagoras with some correct substitution. Effort at evaluating θ Mid Partial Credit: Pythagoras correctly substituted tan evaluated Quadratic equation expanded correctly CD sin CGD with CGD calculated GC sin GD GCE with GCE calculated GC [6]

65 Marcanna breise as ucht freagairt trí Ghaeilge (Bonus marks for answering through Irish) Ba chóir marcanna de réir an ghnáthráta a bhronnadh ar iarrthóirí nach ngnóthaíonn níos mó ná 75% d iomlán na marcanna don pháipéar. Ba chóir freisin an marc bónais sin a shlánú síos. Déantar an cinneadh agus an ríomhaireacht faoin marc bónais i gcás gach páipéir ar leithligh. Is é 5% an gnáthráta agus is é 00 iomlán na marcanna don pháipéar. Mar sin, bain úsáid as an ngnáthráta 5% i gcás iarrthóirí a ghnóthaíonn 5 marc nó níos lú, e.g. 98 marc 5% 99 bónas 9 marc. Má ghnóthaíonn an t-iarrthóir níos mó ná 5 marc, ríomhtar an bónas de réir na foirmle [00 bunmharc] 5%, agus an marc bónais sin a shlánú síos. In ionad an ríomhaireacht sin a dhéanamh, is féidir úsáid a bhaint as an tábla thíos. Bunmharc Marc Bónais [6]

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