# MATHEMATICS (MEI) 4751 Introduction to Advanced Mathematics (C1)

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2 2 Section A (36 marks) 1 Find the equation of the line which is parallel to y = 3x + 1 and which passes through the point with coordinates (4, 5). [3] 2 (i) Simplify (5a 2 b) 3 2b 4. (ii) Evaluate ( 1 16 ) 1. (iii) Evaluate (16) 3 2. [2] [1] [2] 3 Make y the subject of the formula a = y 5. [3] c 4 Solve the following inequalities. (i) 2(1 x) > 6x + 5 (ii) (2x 1)(x + 4) < 0 [3] [2] 5 (i) Express in the form a 3. [2] (ii) Simplify Give your answer in the form b + c 2. [3] d 6 You are given that the coefficient of x 3 in the expansion of (5 + 2x 2 )(x 3 + kx + m) is 29, when x 3 + kx + m is divided by (x 3), the remainder is 59. Find the values of k and m. [5] 7 Expand ( x)4, simplifying the coefficients. [4] 8 Express 5x x + 6 in the form a(x + b) 2 + c. [4] 9 Show that the following statement is false. x 5 = 0 x 2 = 25 [2] OCR Jun10

3 3 Section B (36 marks) 10 (i) Solve, by factorising, the equation 2x 2 x 3 = 0. [3] (ii) Sketch the graph of y = 2x 2 x 3. [3] (iii) Show that the equation x 2 5x + 10 = 0 has no real roots. [2] (iv) Find the x-coordinates of the points of intersection of the graphs of y = 2x 2 x 3 and y = x 2 5x Give your answer in the form a ± b. [4] 11 y A B O x Fig. 11 Fig. 11 shows the line through the points A ( 1, 3) and B (5, 1). (i) Find the equation of the line through A and B. [3] (ii) Show that the area of the triangle bounded by the axes and the line through A and B is square units. [2] 32 3 (iii) Show that the equation of the perpendicular bisector of AB is y = 3x 4. [3] (iv) A circle passing through A and B has its centre on the line x = 3. Find the centre of the circle and hence find the radius and equation of the circle. [4] 12 You are given that f(x) = x 3 + 6x 2 x 30. (i) Use the factor theorem to find a root of f(x) = 0 and hence factorise f(x) completely. [6] (ii) Sketch the graph of y = f(x). [3] (iii) The graph of y = f(x) is translated by ( 1 0 ). Show that the equation of the translated graph may be written as y = x 3 + 3x 2 10x 24. [3] OCR Jun10

6 2 Section A (36 marks) 1 2(i) 2 (ii) 2 (iii) OCR 2010

7 3 3 4 (i) OCR 2010 Turn over

8 4 4 (ii) 5 (i) 5 (ii) OCR 2010

9 5 6 7 OCR 2010 Turn over

10 6 8 9 Section B (36 marks) 10 (i) OCR 2010

11 7 10 (ii) 10 (iii) 10 (iv) OCR 2010 Turn over

12 8 11 (i) 11 (ii) 11 (iii) OCR 2010

13 9 11 (iii) (continued) 11 (iv) OCR 2010 Turn over

14 10 12 (i) OCR 2010

15 11 12 (ii) 12 (iii) OCR 2010 Turn over

17 GCE Mathematics (MEI) Advanced Subsidiary GCE 4751 Introduction to Advanced Mathematics (C1) Mark Scheme for June 2010 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 pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, 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 2010 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile:

19 4751 Mark Scheme June 2010 SECTION A 1 y = 3x + c or y y 1 = 3(x x 1 ) y 5 = their m(x 4) o.e. y = 3x 7 or simplified equiv. 2 (i) 250a 6 b 7 (ii) 16 cao A1 2 1 allow for 3 clearly stated/ used as gradient of required line or (4, 5) subst in their y = mx + c; allow for y 5 = m(x 4) o.e. condone y = 3x + c and c = 7 or B3 www B1 for two elements correct; condone multiplication signs left in SC1 for eg a 6 + b 7 (iii) 64 2 condone ±64 for [±]4 3 or for only or for 3 ac = y 5 o.e. ac + 5 = y o.e. ( ac ) 2 [ y = ] + 5 o.e. isw for each of 3 correct or ft correct steps s.o.i. leading to y as subject or some/all steps may be combined; allow B3 for ( ) 2 [ y = ] ac+ 5 o.e. isw or B2 if one error 4 (i) 2 2x > 6x > 8x o.e. or ft x < 3/8 o.e. or ft isw or 1 x > 3x for collecting terms of their inequality correctly on opposite sides eg 8x > 3 allow B3 for correct inequality found after working with equation allow SC2 for 3/8 o.e. found with equation or wrong inequality 4 (ii) 4 < x < ½ o.e. 2 accept as two inequalities for one end correct or for 4 and ½ 5 (i) for 48 = 4 3 or 27 = 3 3 3

20 4751 Mark Scheme June (ii) www isw 3 B1 for 7 [B0 for 7 wrongly obtained] and B2 for or B1 for one term of numerator correct; if B0, then for attempt to multiply num and denom by k soi k = 12 attempt at f(3) m = 59 o.e. m = 4 cao A1 A1 A1 allow for expansion with 5x 3 + 2kx 3 and no other x 3 terms or for (29 5) / 2 soi must substitute 3 for x in cubic not product or long division as far as obtaining x 2 + 3x in quotient or from division m ( 63) = 59 o.e. or for k + m = 59 or ft their k x 2 x 2 x 16 x oe (must be simplified) isw 4 B3 for 4 terms correct, or B2 for 3 terms correct or for all correct but unsimplified (may be at an earlier stage, but factorial or n C r notation must be expanded/worked out) or B1 for 1, 4, 6, 4, 1 soi or for x [must have at least one other term] 8 5(x + 2) B1 for a = 5, and B1 for b = 2 and B2 for c = 14 or for c = 6 their ab 2 or for [their a](6/their a their b 2 ) [no ft for a = 1] 9 mention of 5 as a square root of 25 or ( 5) 2 = o.e. or x + 5 = 0 condone 5 2 = 25 or, dep on first being obtained, allow for showing that 5 is the only soln of x 5 = 0 Section A Total: 36 allow M2 for x 2 25 = 0 (x + 5)(x 5) [= 0] so x 5 = 0 or x + 5 = 0

21 4751 Mark Scheme June 2010 SECTION B 10 (i) (2x 3)(x + 1) x = 3/2 and 1 obtained 10 (ii) graph of quadratic the correct way up and crossing both axes crossing x-axis only at 3/2 and 1 or ft from their roots in (i), or their factors if roots not given crossing y-axis at 3 10 (iii) use of b 2 4ac with numbers subst (condone one error in substitution) (may be in quadratic formula) < 0 or 15 obtained 10 (iv) 2x 2 x 3 = x 2 5x + 10 o.e. x 2 + 4x 13 [= 0] M2 B1 B1 B1 B1 A1 for factors with one sign error or giving two terms correct allow for 2(x 1.5)(x + 1) with no better factors seen or ft their factors for x = 3/2 condone 1 and 2 marked on axis and crossing roughly halfway between; intns must be shown labelled or worked out nearby may be in formula or (x 2.5) 2 = or (x 2.5) = 0 oe (condone one error) or 15 seen in formula or (x 2.5) 2 = 3.75 oe or x = 2.5 ± 3.75 oe attempt at eliminating y by subst or subtraction or (x + 2) 2 = 17; for rearranging to 2 form ax + bx + c [ = 0] or to completing square form condone one error for each of 2 nd and 3 rd s use of quad. formula on resulting eqn (do not allow for original quadratics used) 2± 17 cao A1 or x + 2=± 17 o.e. 2nd and 3rd s may be earned for good attempt at completing square as far as roots obtained

23 4751 Mark Scheme June (iv) subst x = 3 into y = 3x 4 and obtaining centre = (3, 5) or using ( 1 3) 2 + (3 b) 2 = (5 3) 2 + (1 b) 2 and finding (3, 5) r 2 = (5 3) 2 + (1 5) 2 o.e. r = 20 o.e. cao eqn is (x 3) 2 + (y 5) 2 = 20 or ft their r and y-coord of centre 12 (i) trials of at calculating f(x) for at least one factor of 30 details of calculation for f(2) or f( 3) or f( 5) attempt at division by (x 2) as far as x 3 2x 2 in working correctly obtaining x 2 + 8x + 15 factorising a correct quadratic factor (x 2)(x + 3)(x + 5) 12 (ii) sketch of cubic right way up, with two turning points values of intns on x axis shown, correct ( 5, 3, and 2) or ft from their factors/ roots in (i) y-axis intersection at 30 A1 B1 A1 A1 A1 B1 B1 B1 or ( 1 3) 2 + (3 5) 2 or ft their centre using A or B condone (x 3) 2 + (y b) 2 = r 2 o.e. or (x 3) 2 + (y their 5) 2 = r 2 o.e. (may be seen earlier) M0 for division or inspection used or equiv for (x + 3) or (x + 5); or inspection with at least two terms of quadratic factor correct or B2 for another factor found by factor theorem for factors giving two terms of quadratic correct; M0 for formula without factors found condone omission of first factor found; ignore = 0 seen allow last four marks for (x 2)(x + 3)(x + 5) obtained; for all 6 marks must see factor theorem use first 0 if stops at x-axis on graph or nearby in this part mark intent for intersections with both axes or x = 0, y = 30 seen in this part if consistent with graph drawn

24 4751 Mark Scheme June (iii) (x 1) substituted for x in either form of eqn for y = f(x) correct or ft their (i) or (ii) for factorised form; condone one error; allow for new roots stated as 4, 2 and 3 or ft Section B Total: 36 (x 1) 3 expanded correctly (need not be simplified) or two of their factors multiplied correctly correct completion to given answer [condone omission of y = ] dep or for correct or correct ft multiplying out of all 3 brackets at once, condoning one error [x 3 3x 2 + 4x 2 + 2x 2 + 8x 6x 12x 24] unless all 3 brackets already expanded, must show at least one further interim step allow SC1 for (x + 1) subst and correct exp of (x + 1) 3 or two of their factors ft or, for those using given answer: for roots stated or used as 4, 2 and 3 or ft A1 for showing all 3 roots satisfy given eqn B1 for comment re coefft of x 3 or product of roots to show that eqn of translated graph is not a multiple of RHS of given eqn

25 OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre Qualifications (General) 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; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 2010

29 Reports on the Units taken in June ) (i) Only a small number of candidates failed to attempt this question. The vast majority used the factor theorem successfully to find the first factor, usually x 2. A great many were then able to complete the factorisation (mainly by first dividing, or using inspection to find the relevant quadratic factor, although some used the factor theorem again to find further factors). Most candidates seemed to realise that they needed to try factors of 30. A few candidates reached the correct factorisation but did not show evidence of using the factor theorem and did not earn the first two marks. (ii) (iii) The graphs of the cubic were generally good, with only occasional marks lost. As in 10(ii), the number of candidates who failed to show the y-intersection seemed to be fewer than previous years. A pleasing number of candidates gained all 3 marks here, although some had no idea of how to start. The most profitable method was to start from the revised factors and expand these to obtain the required function. Some who chose to replace x with (x 1) in the original equation came unstuck, more often by expanding (x 1) wrongly than in expanding the cubic term. 6

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