GCE Further Mathematics A Y531/01: Pure Core Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y531/01: Pure Core Advanced Subsidiary GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations 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 2021Y531/01 Mark Scheme October 2021 2 Annotations and abbreviations Annotation in RM assessor Meaning and  BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This question included the instruction: In this question you must show detailed reasoning.Y531/01 Mark Scheme October 2021 3 Question Answer Marks AO Guidance 1 (a) 8 – 2λ = –6 – 3µ and –11 + 5λ = 11 + µ B1 1.1a Forming 2 correct equations in λ and µ. Could be –2 + 3λ = 8 – µ Any two correct equations 8 – 2λ = –6 – 3µ –33 + 15λ = 33 + 3µ => –25 + 13λ = 27 M1 1.1 Attempt to solve (eg scaling one equation and adding or rewriting to a standard form for solution BC). Must reach an equation (possibly incorrect) with only one unknown. –2λ + 3µ = –14 5λ – µ = 22 3λ + µ = 10 λ = 4, µ = –2 A1 1.1 Both – 2 + 3×4 = 10 and 8 – –2 = 10 so they do intersect A1 2.4 Checking for consistency in 3rd equation and conclusion. Equation must be correct and both sides must be evaluated Allow eg 8 2 4 6 3 2 0 0 − × = − − × − = Might see λ = 4 substituted into last equation and then µ being found with this. ie – 2 + 3×4 = 8 – –2 alone is not sufficient for A1, need to see both sides becoming 10 x: 8 – 2×4 = 0 & –6 – 3×–2 = 0 y: –11 + 5×4 = 9 & 11 + –2 = 9 [4] (b) (0, 9, 10) B1 1.1 Allow as vector [1]Y531/01 Mark Scheme October 2021 4 Question Answer Marks AO Guidance 2 u = x + 1 B1 3.1a (u – 1)3 = u 3 – 3u 2 + 3u – 1 used in solution M1 1.1 Attempt to expand using binomial. 4 terms. Follow through on their u = x + 1 2x3 + 3x2 – 2x + 5 = 0 => 2(u 3 – 3u 2 + 3u – 1) + 3(u 2 – 2u + 1) – 2(u – 1) + 5 = 0 M1 1.1 Substituting into equation. Allow if no “= 0” here. Must have an attempt at expanding (u – 1)3 and (u – 1)2 Follow through on their u = x + 1 2u 3 – 3u 2 – 2u + 8 = 0 A1 2.5 Must be an equation For correct equation found using sums and products of roots allow SC2 (Method required was dictated in question) Only allocate marks using main scheme, or SC method [4] Question Answer Marks AO Guidance 3 3 + 5i is a root B1 1.2 Need to see statement that 3+5i is a root. May happen at end of question Attempt to expand (x – (3 + 5i))(x – (3 – 5i)) M1 1.1 Attempt to use the conjugate pair to derive a real quadratic May see (3 + 5i)(3 – 5i) = 9 + 25 = 34 and (3 + 5i) + (3 – 5i) = 6 instead of expansion = x2 – 6x + 34 so this must be a factor A1 2.2a x4 – 7x3 – 2x2 + 218x – 1428 = (x2 – 6x + 34)(x2 + ...x – 42) or (x2 – 6x + 34)(x2 – x + ...) M1 1.1 Attempt to factorise or divide resulting in x2 and one other term NB: This question required detailed reasoning (x2 – 6x + 34)(x2 – x – 42) A1 1.1 (x2 – x – 42) = (x – 7)(x + 6) => roots –6, 7 (and 3 + 5i) A1 1.1 3 + 5i may be mentioned as a root earlier in the solution [6]Y531/01 Mark Scheme October 2021 5 Question Answer Marks AO Guidance 4 (a) (i) Line drawn, perpendicular to line segment joining (0, -1) and (2,0) M1 1.1 Line needs to have negative gradient with |gradient| >1 and to intersect the y axis at a positive value If “shading out” is used then there needs to be an indication that the required region is below the line, such as “R” placed below line or “This region” written in etc. Region below line indicated as being the required region. A1 1.1 Exact perpendicularity not needed, but should be approximately perpendicular. [2] (a) (ii) m = –1/(½) = –2 M1 1.1 4x + 2y – 3 = 0 A1 1.1 Explicitly stated Note must be in required form ax+by+c=0 [2] (b) Circle centre (–1, 0) radius 3 or circle centre (0, 2) radius 2. M1 1.1 Radius can be implied by axis labels or tick-marks. Both circles correct A1 1.1 If M0A0 then SC1 for two circles with correct radii but centres (1, 0) and (0, –2) Correct region shaded or otherwise indicated A1 1.1 Region inside circle with radius 3 but outside circle with radius 2. [3]Y531/01 Mark Scheme October 2021 6 Question Answer Marks AO Guidance 5 (a) 5 12 5 12 1 0 13 13 13 13 0 1 12 5 12 5 13 13 13 13         − −   − =      =              AB M1 2.1 BC. AB or BA correct. Could see 1 1 1 0 5 12 5 12 13 0 1 12 5 12 5 13      − − −      =      5 12 5 12 13 13 1 0 13 13 12 5 0 1 12 5 13 13 13 13         − − −   − =       = ≠           −     BA AB so matrix multiplication is not commutative A1 2.2a BC. Other multiplication correct and conclusion [2] (b) Rotation about O M1 1.2 1 67.4° anticlockwise A1 1.1 or 1.18 rads [2] (c) (TB)–1 is a rotation about O by –67.4° anticlockwise (or 67.4° clockwise) M1 3.1a Correct inverse of their rotation TB. Could also be rotation of 292.6o anticlockwise So 1 cos( 67.4 ) sin( 67.4 ) sin( 67.4 ) cos( 67.4 ) −   − ° − − ° =     − ° − ° B 5 12 13 13 12 5 13 13     =       −   A1 1.1 or 1 0.385 0.923 0.923 0.385 −   =     − B (allow 0.384 for 0.385) NB: Question states “by considering the inverse transformation”. SC1 For correct inverse by other method. [2] (d) det B = 1 and det C = –3 M1 3.1a Could find BC and then find det(BC) = -3 So area of N =  1×–3 × 5 = 15 A1 3.2a Area must be 15, do not allow -15 or ±15 [2]Y531/01 Mark Scheme October 2021 7 Question Answer Marks AO Guidance 6 (a) ( 10) ( 10) 4 2 25 2 2 2 z − − ± − − × × = × M1 2.1 Correct substitution into formula. If formula quoted allow one slip. Or completing the square – one slip allowed. 5 5 i 2 2 z = ± A1 1.1 Allow 5 5i 2 z ± = or equivalent fractions NB: This question required detailed reasoning [2] (b) 3ω – 2 = 5i + 2iω => 3ω – 2iω = 2 + 5i M1 1.1 Expanding and rearranging Must rearrange to isolate � terms on one side and other terms on other side 2 5i (3 2i) 2 5i 3 2i ω ω + − = + ⇒ = − M1 1.1 Factorising and dividing by two term complex number NB: This question required detailed reasoning 2 5i 3 2i 6 4i 15i 10 3 2i 3 2i 9 4 ω + + + + − = × = − + + M1 2.1 Multiplying top and bottom by conjugate of bottom 4 19 i 13 13 ω = − + A1 1.1 Alternative method ω = a + bi => 3a + 3bi – 2 = 5i + 2ai – 2b M1 Assigning real and imaginary parts, to ω expanding and rearranging 3a – 2 = –2b and 3b = 5 + 2a M1 Comparing real and imaginary parts 9a – 6 + 10 + 4a = 0 4 13 ⇒ = − a M1 Using valid algebra to eliminate one unknown and finding the other 19 4 19 i 13 13 13 ⇒ = ⇒ = − + b ω A1 [4]Y531/01 Mark Scheme October 2021 8 Question Answer Marks AO Guidance 7 Basis Case: when n = 1: 23n – 3n = 23 – 3 = 8 – 3 = 5 which is divisible by 5. B1 2.1 At least one intermediate step must be shown Assume true for n = k ie 23k – 3k = 5p for some integer p M1 2.1 Must have statement in terms of some other variable than n 23(k + 1) – 3k + 1 = 23×23k – 3×3k = 8×(5p + 3k) – 3×3k M1 1.1 Uses laws of indices and then inductive hypothesis properly to eliminate either 23k or 3k (not both) or 8×23k – 3×( 23k – 5p) = 5×8p + 5×3k =5(8p + 3k) = 5q for some integer q and so this is also a multiple of 5 A1 2.2a AG. Further simplification to establish truth for k + 1 5(3p + 23k) So true for n = k ⇒ true for n = k + 1. But true for n = 1. So true for all integers n ≥ 1 A1 2.4 Clear conclusion for induction process, following a correct proof by induction. A formal proof by induction is required for full marks. [5]Y531/01 Mark Scheme October 2021 9 Question Answer Marks AO Guidance 8 (a) (t – 1)(6 – t(2 – 2t)) – (t – 1)((1 – t) – t(2 – 2t)) + (t – 1)((1 – t)(2 – 2t) – 6(2 – 2t)) M1 1.1 Correct process for expanding determinant. Fully expanded form: 2t3 + 7t2 – 14t + 5 (t – 1)[ (6 – t(2 – 2t)) - ((1 – t) – t(2 – 2t)) +((1 – t)(2 – 2t) – 6(2 – 2t))] M1 1.1 Bringing (t – 1) or (t + 5) or (2t – 1) oe out as factor of the entire expression Factors may appear BC from no working (t – 1)(6 – 2t + 2t2 – 1 + t + 2t – 2t2 + 2 – 4t + 2t2 – 12 + 12t) = (t – 1)(2t2 + 9t – 5) = (t – 1)(2t – 1)(t + 5) A1 1.1 [3] (b) –5, ½, 1 B1 1.1 FT their complete factorisation of determinant into 3 linear factors. [1] (c) t = b2 + 2 M1 2.1 So that the system is Ar = c and so t ≥ 2 so cannot be –5, ½ or 1 therefore A–1 will exist (for all values of b) and so there will be a unique solution to the system for all values of b. A1 2.4 Complete reasoning must be seen for A1. Could test t = 1, ½, -5 in b2 = t – 2, and show that these do not give real values of b [2]Y531/01 Mark Scheme October 2021 10 Question Answer Marks AO Guidance 9 (a) 1 3 4 3 5 2 16 21 5 PQ       − −       = − = −                   − −  M1 2.1 Attempt to find the direction vector of the tunnel. Any non-zero multiple. 4 1 2 0 5 s t     −         − =         . M1 1.1 Use of  PQ.b = 0 in the solution. –4 – 2s + 5t = 0 => 2s = 5t – 4 => s = 2.5t – 2 A1 2.1 AG. Some intermediate work must be seen. [3] (b) M = 1 2��−−316 1 � + �−3 521�� = �−18 1 4 .5� B1 1.1 Position vector ordinates) of mid (or co -point - found 1 1 4 18.5 s t λ         = +             − r when z = 0 => –18.5 + λt = 0 M1 3.4 Using z = 0 and the equation of the line to find a ‘horizontal’ relationship between λ and t. Condone errors in, or omission of, x and y components. 18.5 t ⇒ = λ (so c = 18.5) A1 1.1 NB: Question can be answered just by considering the z coordinate. If done correctly and M1 A1 gained also allow B1 as implied. [3]Y531/01 Mark Scheme October 2021 11 Question Answer Marks AO Guidance (c) So we need to minimise 1 18.5 2.5 2 t t t       −     M1 3.3 Stating or implying that the length of the shaft is given by λb and using their λ / t relationship to reduce length of shaft to a form with only one variable. Or eg 1 18.5 0.4 0.8 0.4 0.8 s s s       +     + ( ) 1 (2.5 2 ) 1369 2 ( )2 2 4 y t t t = + − + = − + 1369 4 (7.25 10 5 t t − − 1 2 ) M1* 1.1 Finding expression for (squared) length of their vector May see ( ) 1 37 7.25 10 5 1 2 2 2 − + t t − − Or 39701 16 − 6845 2 �−1 + 6845 4 �−2 oe So to minimise set d 1369 d 4 yt = − = (10 10 0 t t − − 2 3 ) dep M1* 3.1a Correct method for minimisation of (squared) length of their vector (eg differentiating and setting to 0) Or attempt to complete the square in t–1. y t = − + 1369 4 (5( 1) 2.25 −1 2 ) 10 10 0 1 t t t − − 2 3 − = ⇒ = A1 2.2a So min when t–1 – 1 = 0, t = 1 So length of shaft = 1 18.5 0.5 1           or 1369 4 (7.25 10 1 5 1 − × + × − − 1 2 ) oe M1 3.4 Substituting their t into their form for length of shaft = 18.5 × 1.5 = 27.75 A1 1.1Y531/01 Mark Scheme October 2021 12 Question Answer Marks AO Guidance Alternate method: 1 3 1 1 3 5 4 2 16 21 18.5  −                = + =                         − − − a B1 4 0 2 1 2 0 4 2 2 5 1 0 0         −  − −            = − = =                               n × M1 Attempt to find normal to vertical plane containing tunnel 1 2 4 20 1 ( ) 4 2 10 20 2 0 5 20 1 k           − −     = − = =                                         b × M1 Attempt to find (multiple of) b by crossing their n with direction vector of tunnel. Need –18.5 + λ = 0 => λ = 18.5 M1 Using z = 0 to find λ May see multiple of b used eg –18.5 + 2λ = 0 So length of shaft = 1 1 18.5 21             M1 May see eg 1 2 19.5 4 9.25 1 13.25 18.5 2 0             = + =                   − r and then 19.5 1 13.25 4 0 18.5             −         − = 18.5 × 3/2 = 27.75 A1 [6] (d) So b is not parallel to the z-axis so the ventilation shaft does not go straight down. B1 3.2a Shaft not vertical [1]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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