# Monday 13 May 2013 Afternoon

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2 2 Section A (36 marks) Find the equation of the line which is perpendicular to the line y = 2x - 5 and which passes through the point _ 4, i. Give your answer in the form y = ax + b. [3] 2 Find the coordinates of the point of intersection of the lines y = 3x - 2 and x + 3y =. [4] 3 (i) Evaluate _ 0. 2i -2. [2] 4 (ii) Simplify _ a i [3] 4 Rearrange the following formula to make r the subject, where r V = rr ( a + b) [3] 3 5 You are given that f( x) = x + kx When f( x ) is divided by (x - 2), the remainder is 8. Find the value of k. [3] 6 Find the coefficient of x 3 in the binomial expansion of _ 2-4xi 5. [4] 7 (i) Express 25 5 in the form 5 k. [2] 38 (ii) Simplify , giving your answer in the form a + b 5. [3] Express 3x 2-2x + 5 in the form a _ x - b i - c. Hence state the minimum value of y on the curve 2 y = 3x - 2x + 5. [5] 9 n -, n and n + are any three consecutive integers. (i) Show that the sum of these integers is always divisible by 3. [] (ii) Find the sum of the squares of these three consecutive integers and explain how this shows that the sum of the squares of any three consecutive integers is never divisible by 3. [3] OCR /0 Jun3

3 2 2 0 The circle _ x - 3i + _ y - 2i = 20 has centre C. 3 Section B (36 marks) (i) Write down the radius of the circle and the coordinates of C. [2] (ii) Find the coordinates of the intersections of the circle with the x- and y-axes. [5] (iii) Show that the points A _, 6i and B_ 7, 4i lie on the circle. Find the coordinates of the midpoint of AB. Find also the distance of the chord AB from the centre of the circle. [5] You are given that f( x) = _ 2x - 3i_ x + 2i_ x + 4i. (i) Sketch the graph of y = f( x). (ii) State the roots of f( x - 2) = 0. [3] [2] (iii) You are also given that g( x) = f( x) (A) Show that g( x) = 2x + 9x - 2x - 9. [2] (B) Show that g( ) = 0 and hence factorise g( x ) completely. [5] [Question 2 is printed overleaf.] OCR /0 Jun3 Turn over

6 2 Section A (36 marks) 2 3 (i) OCR 203

7 3 3 (ii) 4 5 OCR 203 Turn over

8 4 6 7 (i) 7 (ii) OCR 203

9 5 8 9 (i) 9 (ii) OCR 203 Turn over

10 6 Section B (36 marks) 0 (i) 0 (ii) OCR 203

11 7 0 (iii) OCR 203 Turn over

12 8 (i) (ii) (iii)(a) OCR 203

13 9 (iii)(b) OCR 203 Turn over

14 0 2 (i) 8 y y = x - 2 x y = x - 2 Fig. 2 OCR 203

15 2 (ii) OCR 203 Turn over

17 GCE Mathematics (MEI) Advanced Subsidiary GCE Unit 475: Introduction to Advanced Mathematics Mark Scheme for June 203 Oxford Cambridge and RSA Examinations

18 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 203

19 475 Mark Scheme June 203 Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M Method mark awarded 0, A0, A Accuracy mark awarded 0, B0, B Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations Meaning in mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph M dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working

20 475 Mark Scheme June 203 Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a. Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. b. An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader. c. The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. 2

22 475 Mark Scheme June 203 h. For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate s data. A penalty is then applied; mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. 4

23 475 Mark Scheme June 203 Question Answer Marks Guidance y = 0.5x + 3 oe www isw 3 B2 for 2y = x + 6 oe for 3 marks must be in form y = ax + b or M for gradient = oe seen or used 2 [3] 2 substitution to eliminate one variable M or multiplication to make one pair of coefficients the same; condone one error in either method and M for y = their m (x 4) or M for y = their mx + c and (4, ) substituted simplification to ax = b or ax b = 0 form, or equivalent for y M or appropriate subtraction / addition; condone one error in either method independent of first M (0.7, 0.) oe or x = 0.7, y = 0. oe isw A2 A each [4] 3 (i) M for 2 or for 0.04 oe or oe soi ie M for one of the two powers used correctly M0 for just with no other working 0.4 [2] 3 (ii) 8a 9 3 [3] B2 for 8 or M for and B for a soi ignore ± eg M for 2 3 ; M0 for just 2 5

24 475 Mark Scheme June V r ( a b) oe www as final answer 3 M for dealing correctly with 3 M0 if triple-decker fraction, at the stage where it happens, then ft; and M for dealing correctly with π(a + b), ft condone missing bracket at rh end and M for correctly finding square root, ft their r 2 = ; square root symbol must extend below the fraction line M0 if ±... or r >... for M3, final answer must be correct [3] 5 f(2) = 8 seen or used M or long division oe as far as obtaining a remainder (ie not involving x) and equating that remainder to 8 (there may be errors along the way) k 20 = 8 oe A after long division: 2(k + 6) 20 = 8 oe A0 for just 2 5 instead of 32 unless 32 implied by further work [k =] 3 A [3] 6

25 475 Mark Scheme June www 4 B3 for 2560 from correct term (NB coefficient of x 4 is 2560) or B3 for neg answer following and then an error in multiplication or M2 for ( 4) 3 oe; must have multn signs or be followed by a clear attempt at multn; or M for 2 2 ( 4) 3 oe (condone missing brackets) or for 0 used or for seen for those who find the coefft of x 2 instead: allow M for 0 used or for seen ; and a further SC if they get 280, similarly for finding coefficient of x 4 as 2560 ignore terms for other powers; condone x 3 included; but eg = = 24 gets M2 only condone missing brackets eg allow M2 for x 3 5 C 3 or factorial notation is not sufficient but accept oe 0 may be unsimplified, as above M only for eg 0, 2 2 and 4x 3 seen in table with no multn signs or evidence of attempt at multn [lack of neg sign in the x 2 or x 4 terms means that these are easier and so not eligible for just a mark MR penalty] [4] 7 (i) oe or k = 7/2 oe 2 [2] M for 25 = 5 3 or soi M0 for just answer of 5 3 with no reference to 25 7

26 475 Mark Scheme June (ii) attempting to multiply numerator and denominator of fraction by 2 5 M some cands are incorporating the into the fraction. The Ms are available even if this is done wrongly or if is also multiplied by 2 5 denominator = 9 soi M must be obtained correctly, but independent of first M eg M for denominator of 9 with a minus sign in front of whole expression or with attempt to change signs in numerator A [3] 8 3(x 2) 2 7 isw or a = 3, b = 2 c = 7 www 4 B each for a = 3, b = 2 oe condone omission of square symbol; ignore = 0 and B2 for c = 7 oe or M for 7 or for 5 their a(their b) 2 3 may be implied by their answer or for their b 2 soi 3 7 or ft B B0 for (2, 7) may be obtained by starting again eg with calculus [5] 9 (i) 3n isw accept equivalent general explanation [] 8

27 475 Mark Scheme June (ii) at least one of (n ) 2 and (n + ) 2 correctly expanded M must be seen M0 for just n n 2 + n 2 + 3n B accept even if no expansions / wrong expansions seen comment eg 3n 2 is always a multiple of 3 so remainder after dividing by 3 is always 2 B dep on previous B B0 for just saying that 2 is not divisible by 3 must comment on 3n 2 term as well SC: n, n +, n + 2 used similarly can obtain first M, and allow final B for similar comment on 3n 2 + 6n + 5 allow B for [3] 0 (i) [radius =] 20 or 2 5 isw B B0 for 20 oe n n [centre =] (3, 2) B condone lack of brackets with coordinates, here and in other questions [2] 9

28 475 Mark Scheme June (ii) substitution of x = 0 or y = 0 into circle equation M or use of Pythagoras with radius and a coordinate of the centre eg or h = 20 ft their centre and/or radius equation may be expanded first, and may include an error bod intent allow M for (x 3) 2 = 20 and/or (y 2) 2 = 20 (x 7)(x + ) [=0] M no ft from wrong quadratic; for factors giving two terms correct, or formula or completing square used with at most one error completing square attempt must reach at least (x a) 2 = b following use of Pythagoras allow M for attempt to add 3 to [±]4 (7, 0) and (, 0) isw A accept x = 7 or (both required) ( 7) [ y ] oe 2 M no ft from wrong quadratic; for formula or completing square used with at most one error completing square attempt must reach at least (y a) 2 = b following use of Pythagoras allow M for attempt to add 2 to [±] 0,2 or , 2 isw A [5] accept 4 44 y oe isw 2 annotation is required if part marks are earned in this part: putting a tick for each mark earned is sufficient 0

29 475 Mark Scheme June (iii) show both A and B are on circle B explicit substitution in circle equation and at least one stage of interim working required oe or clear use of Pythagoras to show AC and BC each = 20 (4, 5) B2 B each or M for 7 6 4, B2 from correct midpoint and centre used; B for 0 [5] M for (4 3) 2 + (5 2) 2 or or ft their centre and/or midpoint, or for the square root of this may be a longer method finding length of ½ AB and using Pythag. with radius; no ft if one coord of midpoint is same as that of centre so that distance formula/pythag is not required eg centre correct and midpt (3, ) annotation is required if part marks are earned in this part: putting a tick for each mark earned is sufficient

30 475 Mark Scheme June 203 (i) sketch of cubic the right way up, with two tps and clearly crossing the x axis in 3 places B no section to be ruled; no curving back; condone slight flicking out at ends but not approaching another turning point; condone some doubling (eg erased curves may continue to show); accept min tp on y-axis or in 3 rd or 4 th quadrant; curve must clearly extend beyond the x axis at both ends crossing/reaching the x-axis at 4, 2 and.5 B intersections must be shown correctly labelled or worked out nearby; mark intent accept curve crossing axis halfway between and 2 if 3/2 not marked intersection of y-axis at 24 B NB to find 24 some are expanding f(x) here, which gains M in iiia. If this is done, put a yellow line here and by (iii)a to alert you; this image appears again there [3] (ii) 2, 0 and 7/2 oe isw or ft their intersections 2 B for 2 correct or ft or for ( 2, 0) (0, 0) and (3.5, 0) or M for (x + 2) x (2x 7) oe or SC for 6, 4 and /2 oe [2] 2

31 475 Mark Scheme June 203 (iii) (A) correct expansion of product of 2 brackets of f(x) M need not be simplified; condone lack of brackets for M eg 2x 2 + 5x 2 or 2x 2 + x 6 or x 2 + 6x + 8 or allow M for expansion of all 3 brackets, showing all terms, with at most one error: 2x 3 + 4x 2 + 8x 2 3x 2 + 6x 2x 6x 24 may be seen in (i) allow the M; the part (i) work appears at the foot of the image for (iii)a, so mark this rather than in (i) correct expansion of quadratic and linear and completion to given answer A for correct completion if all 3 brackets already expanded, with some reference to show why 24 changes to 9 condone lack of brackets if they have gone on to expand correctly; condone +5 appearing at some stage NB answer given; mark the whole process [2] 3

32 475 Mark Scheme June 203 (iii) (B) g() = [=0] B allow this mark for (x ) shown to be a factor and a statement that this means that x = is a root [of g(x) = 0] oe B0 for just g() = 2() 3 + 9() 2 2() 9 [=0] attempt at division by (x ) as far as 2x 3 2x 2 in working M or inspection with at least two terms of quadratic factor correct M0 for division by x + after g() = 0 unless further working such as g( ) = 0 shown, but this can go on to gain last MA correctly obtaining 2x 2 + x + 9 A allow B2 for another linear factor found by the factor theorem factorising a correct quadratic factor M for factors giving two terms correct; eg allow M for factorising 2x 2 + 7x 9 after division by x + NB mixture of methods may be seen in this part mark equivalently eg three uses of factor theorem, or two uses plus inspection to get last factor; allow M for (x + )(x + 8/4) oe after and 8/4 oe correctly found by formula (2x + 9)(x + )(x ) isw A allow 2(x + 9/2)(x + )(x ) oe; dependent on 2 nd M only; condone omission of first factor found; ignore = 0 seen SC alternative method for last 4 marks: allow first MA for (2x + 9)(x 2 ) and then second MA for full factorisation [5] 4

33 475 Mark Scheme June (i) y = 2x + 3 drawn accurately M at least as far as intersecting curve twice ruled straight line and within 2mm of (2, 7) and (, ) (.6 to.7, 0.2 to 0.3) B intersections may be in form x =..., y =... (2. to 2.2, 7.2 to 7.4) B 2 (ii) 2x 3 x 2 [3] M or attempt at elimination of x by rearrangement and substitution if marking by parts and you see work relevant to (ii), put a yellow line here and in (ii) to alert you to look may be seen in (i) allow marks; the part (i) work appears at the foot of the image for (ii) so show marks there rather than in (i) = (2x + 3)(x 2) M condone lack of brackets implies first M if that step not seen = 2x 2 x 6 oe A for correct expansion; need not be simplified; NB A0 for 2x 2 x 7 = 0 without expansion seen [given answer] implies second M if that step not seen after 2x 3 seen x isw oe M A [5] use of formula or completing square on given equation, with at most one error isw eg coordinates; after completing square, accept 57 or 4 6 better completing square attempt must reach at least [2](x a) 2 = b or (2x c) 2 = d stage oe with at most one error 5

34 475 Mark Scheme June (iii) x k x 2 and attempt at rearrangement M x k x k M for simplifying and rearranging to zero; condone one error; collection of x terms with bracket not required 2 eg M bod for 2 2 or M for x 2 2kx 2k 0 x k x k b 2 4ac = 0 oe seen or used M = 0 may not be seen, but may be implied by their final values of k [k =] 0 or 4 as final answer, both required A SC for 0 and 4 found if 3 rd M not earned (may or may not have earned first two Ms) [4] eg obtained graphically or using calculus and/or final answer given as a range 6

35 475 Mark Scheme June 203 Appendix: revised tolerances for modified papers for visually impaired candidates (graph in 2(i) with 6mm squares) 2 (i) y = 2x + 3 drawn accurately M at least as far as intersecting curve twice ruled straight line and within 3 mm of (2, 7) and (, ) (.6 to.8, 0.2 to 0.3) B intersections may be in form x =..., y =... (2. to 2.3, 7. to 7.4) B [3] if marking by parts and you see work relevant to (ii), put a yellow line here and in (ii) to alert you to look 7

36 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB 2EU OCR Customer Contact Centre Education and Learning Telephone: Facsimile: For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, CB 2EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 203

37 OCR Report to Centres June Introduction to Advanced Mathematics (C) General Comments Candidates coped well with this question paper. As commented in previous reports, many candidates are well-drilled in many of the techniques required for this unit, but applying them in unfamiliar situations (such as question 7) is found to be much more difficult. In general, the standard of algebra was quite good. However, candidates do not always see the need to use brackets correctly. Those who omitted brackets penalised themselves in some questions by then failing to expand brackets correctly or, in question 6, failing to cube the 4. Brackets were formally needed in the answer to question 3, for instance. Candidates are far happier dealing with whole numbers and quadratic equations that factorise, than they are with fractions, decimals and the use of quadratic formula. In the latter case, when this was necessary, some candidates 'gave up'. Those attempting the completing the square method with fractions were rarely successful. Comments on Individual Questions Section A ) In finding the equation of the line, most candidates obtained full marks. The main mistake was to use a gradient of 2, due to confusion between perpendicular and parallel. There was a significant number of arithmetic errors especially in coping with negative signs and the fraction. 2 2) In the main, this question was completed well. Some candidates found the arithmetic challenging, especially if rearranging x + 3y = to substitute in for y, with the resulting need to cope with fractions. A slight majority choose the substitution method rather than elimination. A few neglected to find y having found x. 3) In evaluating ( 0.2) 2, many stopped after evaluating as (or, sadly often, as 0.04 ). Those who converted to fractions first were more successful in reaching In the second part, the majority found the power of a correctly, but the 6 proved more challenging. A surprising number did 3 6 = 2 to obtain 2a ) There were many good answers in rearranging the formula. Most candidates managed at least one mark; some triple-decker fractions or the use of signs were seen. The π and the (a + b) sometimes became separated. The radius was sometimes considered to be ±, and the > sign was used on more than one occasion. It was encouraging to see very few penalties incurred due to a poor square root symbol.

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