GCE Further Mathematics B (MEI) Y434/01: Numerical methods Advanced GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y434/01: Numerical methods Advanced GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations OCR (Oxford Cambrid... ge 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 2021Y434/01 Mark Scheme October 2021 2 Annotations and abbreviations Annotation in scoris 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank page Highlighting Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only 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 indicates that the instruction In this question you must show detailed reasoning appears in the question.Y434/01 Mark Scheme October 2021 3 Question Answer Marks AOs Guidance 1 (a) i 1.414214−√2 √2 or 1.4142142−2 2 oe soi 0.000000309449 isw 0.000000618898 isw M1 A1 A1 1.1a 1.1 1.1 ignore modulus signs to 2 sf or more to 2 sf or more [3] 1 (a) ii the second relative error is double the first relative error oe B1 2.2a [1] 1 (b) Ben is wrong because the spreadsheet stores 1.414214 to a higher precision than is displayed (and so when the square of this number is calculated, 2 is returned) isw B1 2.4 or 1.414214 is an approximation to √2 so 1.4142142 ≠ 2 oe [1] 2 (a) x f(x) Δ Δ² 1 ‒0.65 0.3 2 ‒0.35 1.82 2.12 3 1.77 1.82 3.94 4 5.71 1.82 5.76 5 11.47 M1 A1 1.1 1.1 finds 4 Δ values, allow one error all correct [2]Y434/01 Mark Scheme October 2021 4 Question Answer Marks AOs Guidance 2 (b) the second differences are constant oe B1 1.1 allow the 3rd differences are zero [1] 2 (c) ‒0.65 + 0.3(x ‒ 1) + 1.82×(�−1)(�−2) 2! [P2(x) =] 0.91x² ‒ 2.43x + 0.87 M1 A1 A1 1.1 1.1 1.1 must be correct form; allow 1 substitution error two of three terms correct all correct [3] 3 (a) sinh x² ‒ x³ ‒ 2 = 0 B1 1.1 must see = 0 [1] 3 (b) =IF(H5>0,G5,E5) B1 1.1 or =IF(H5<0,E5,G5) must see = [1] 3 (c) 1.48719×17.2899−2×‒0.77825 17.2899−−0.77825 oe awrt 1.50928 awrt 1.52603 M1 A1 A1 3.1a 1.1 1.1 may be implied by 1.509… NB f(1.50928) = ‒0.6111 to 4 sf [3]Y434/01 Mark Scheme October 2021 5 Question Answer Marks AOs Guidance 3 (d) the ratios are decreasing which suggests the convergence is (slightly) faster than 1st order the ratios are close to 1 which suggests the convergence is slow B1 B1 2.2b 2.2b allow between 1st and 2nd order do not allow eg not first order [2] 4 (a) 4.2472072‒4 0.1 or 4.0239468‒4 0.01 or 4.0023871‒4 0.001 or 4.0002386‒4 0.0001 2.472072 (with h = 0.1) 2.39468 (with h = 0.01) 2.3871 (with h = 0.001) 2.386 (with h = 0.0001) M1 A1 A1 A1 3.1a 1.1 1.1 1.1 use of forward difference method any two correct any three correct all four correct may be implied by one correct answer [4] 4 (b) comparison of last two estimates 2.39 is secure or 2.386 is possible M1 A1 1.1 2.2b if M0 allow SC1 for 2.39 is secure or 2.386 is possible regardless of justification [2] 5 (a) 48×0.5 soi £24 M1 A1 3.3 3.4 [2] 5 (b) consistent because 1.77 < 24 B1 2.4 allow consistent because error < mpe [1]Y434/01 Mark Scheme October 2021 6 Question Answer Marks AOs Guidance 5 (c) 52×0.495 £25.74 M1 A1 3.3 3.4 [2] 5 (d) this could happen if a large number of items eg cost less than £1 eg cost £1.99 or £2.99 etc eg more than 50p over the pound eg the mean error per item was 52.38p B1 3.5a [1] 5 (e) mpe = £0.99n B1 3.4 condone omission of units, allow 99n pence [1] 5 (f) expected error for Nina’s model is £0 since you would expect to round half the prices up and half down oe or expected error in Kareem’s model is ‒£0.495n since you would expect the average “chop” to be 49.5p oe so new model should be “estimated cost” + £0.495n B1 B1 2.4 3.5c U6Y434/01 Mark Scheme October 2021 7 Question Answer Marks AOs Guidance [2] 6 (a) 1 2 1 2 x x − + seen ( ) 2 1 0.5ln 1 1 2 1 2 n n n n n n n x x x x x x x + − + + = − − + oe soi M1 M1 2.1 1.1 may be implied by correct iterates condone omission of subscripts 13 2.0791668 1.7783346 1.7360141 1.7351281 1.7351277 1.735128 M1 A1 1.1 1.1 at least three further correct iterates derived from starting at 1 if M0 allow SC1 for 1.735128 from N-R method used with different �0 and at least 3 correct iterates shown correct to at least 5 sf where appropriate [4] 6 (b) M1 A1 2.4 1.1 tangent at (1,1) (1,1) to (3,0) [2]Y434/01 Mark Scheme October 2021 8 Question Answer Marks AOs Guidance 6 (c) N-R generally has 2nd order convergence whereas fixed point iteration generally has 1st order convergence B1 2.4 allow eg N-R converges faster allow eg fixed point iteration more likely to fail oe [1] 6 (d) ln(-0.403) is undefined (so the spreadsheet cannot compute a value ) B1 2.2a [1] 6 (e) 0.5 1.0739769 1.4524673 1.6245304 1.6932631 1.7194743 1.7293015 converges to β M1 A1 2.1 2.2a need to see at least 3 iterates correct to at least 5 sf [2]Y434/01 Mark Scheme October 2021 9 Question Answer Marks AOs Guidance 6 (f) 0.5 0.4764669 0.4528879 0.4293074 0.4057756 0.3823498 … 0.1116318 0.1111278 0.1110835 0.1110821 0.1110821 0.111082 M1 A1 1.1 2.2a at least 3 correct iterates derived from starting at 0.5 if M0 allow SC1 for 0.111082 from relaxation method used with different �0 and at least 3 correct iterates shown iterates correct to at least 5 sf [2] 7 (a) 1 16 isw or 0.0625 isw B1 2.2a [1] 7 (b) by comparison of T16 and T32 0.6 is certain or 0.63 is probable B1 2.2b [1]Y434/01 Mark Scheme October 2021 10 Question Answer Marks AOs Guidance 7 (c) r appears to be between 0.25 and 0.5 so order of convergence is between 1st and 2nd order B1 B1 2.2b 2.2b Alternative r > 0.25 so convergence slower than 2nd order r < 0.5 so convergence faster than 1st order B1 B1 [2] 7 (d) 2��+�� 3 or 4�2�−�� 3 soi = (2*O5 + N5)/3 or =(4*N6 ‒ N5)/3 M1 A1 1.1 1.1 must see = [2] 7 (e) awrt 0.62658745 awrt 0.00029 awrt 0.354 B1 B1 B1 1.1 1.1 1.1 [3]Y434/01 Mark Scheme October 2021 11 Question Answer Marks AOs Guidance 7 (f) S2n and difference from table used in extrapolation awrt 0.62658745 and awrt 0.00029 used 0.62658745 + 0.00029 × � 1−� awrt 0.62674355 to awrt 0.62675058 comparison with their S64 0.6267 is secure M1 A1 A1 A1 M1 A1 3.1a 1.1 1.1 1.1 3.2a 2.2b eg their 0.62658745 and their 0.00029 may see more dp for difference 0.35 ≤ r ≤ 0.36 or 0.62675 is possible; allow 0.626746 the last two A marks are only available if answers obtained from extrapolation to infinity from S64 If M0 allow SC2 for awrt 0.626607 obtained from 16×0.62658745−0.62629755 15 then SC1 for 0.627 obtained from comparison with S64 [6]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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