GCE Further Mathematics B (MEI) Y435/01: Extra pure Advanced GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics B (MEI) Y435/01: Extra pure Advanced 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 2021Y435/01 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.Y435/01 Mark Scheme October 2021 3 Question Answer Marks AO Guidance 1 (a) DR z = f(2, y) = 8 + 4y – 2y2 M1 1.1 Deriving correct equation of graph of section. = 10 – 2(y – 1)2⇒ max at (1, 10) or (2, 1, 10) A1 1.1 Finding TP by completing the square, use of “–b/2a”, differentiation or mid-point between roots. Working must be shown. B1 1.1 ∩-shaped parabola which crosses horizontal axis twice. Condone incorrect variable names on axes (eg x-y for y-z). Crossing z-axis at 8, y-axis at 1 5 ± and showing (1,10) as a max A1 1.1 Coordinates of intercepts and max must be shown on graph or apparent in working. Allow decimal values (awrt –1.2 and 3.2) for the y-intercepts. z intercept must be shown as positive and max in 1st quadrant. However, scale is unimportant except that the negative y-intercept must be closer to O than the positive one. [4]Y435/01 Mark Scheme October 2021 4 (b) 2 z 3 2 x xy x ∂ = + ∂ B1 1.1 z x y 2 4 y ∂ = − ∂ B1 1.1 0 2 3 2 0 2 3 either 0 or or 3 2 zx x xy x x y y x ∂ = ∂ ⇒ + = ⇒ = = − = − M1 1.1 Setting a partial derivative to 0 and deriving condition(s) on x and/or y. Or 2 2 0 1 or 4 or 2 4 zy y x x y x y ∂ = ∂ ⇒ = = = ± 2 4 2 0, 4 9 3 9 z x y y y y y ∂ = = − ⇒ = ⇒ = ∂ M1 1.1 Substituting condition into other partial derivative equation to derive a non-zero value for x or y. 2 3 0, 2 6 0 6... z y x y x x x ∂ = = − ∂ ⇒ + = ⇒ = − 2 2 3 1 0, 4 1 3 0 6 2 z y x x x x x ∂ = = ∂ + = ⇒ = − or 32 0, 2 12 4 0 9 z x y x y y y ∂ = = ± ∂ ± = ⇒ = x = –6 A1 1.1 y = 9 z = –54 so (–6, 9, –54) A1 1.1 From correct working only. ... or 0 0 so (0, 0, 0) x y = ⇒ = A1 1.1 Derived from both z 0 x ∂ = ∂ and z 0 y ∂ = ∂ (could be by observation). If an extra SP is presented then A1 can be awarded for either SP correct and then A0. [7]Y435/01 Mark Scheme October 2021 5 2 (a) From Lagrange’s Theorem the order of any subgroup of G must be a factor of 8 and 6 is not a factor of 8 B1 2.4 Or “order of any subgroup of G (or an group of order 8) must be 1, 2 or 4 (or 8)” or “order of any subgroup must be a factor of the order of the group and 6 is not a factor of 8”. If referenced, Lagrange’s Theorem does not have to be quoted provided that it is applied. So B1 for eg “6 is not a factor of 8 so by Lagrange’s Theorem there can be no subgroup of G of order 6” but B0 for eg “By Lagrange’s Theorem there can be no subgroup of G of order 6”. [1] 2 (b) g2 (or g6) B1 2.2a May see eg gg or g◦g used here and/or throughout. Allow any multiplicative notation and any symbol for a binary operation. g6 (or g2) and no other B1 2.2a [2] 2 (c) e↔0 B1 2.2a Only needs to be seen once. g↔1, g2↔2, g3↔3, g4↔4, g5↔5, g6↔6, g7↔7 B1 2.2a Any one. g↔3, g2↔6, g3↔1, g5↔7, g6↔2, g7↔5 B1 2.2a Any other. g4↔4 does need not be seen again g↔5, g2↔2, g3↔7, g5↔1, g6↔6, g7↔3 and g↔7, g2↔6, g3↔5, g5↔3, g6↔2, g7↔1 B1 2.2a Other two. Ignore repeats. g4↔4 does need not be seen again Alternative method: e↔0 B1 Only needs to be seen once Either g↔1 or g↔3 or g↔5 or g↔7 M1 Giving all 4 possible isomorphism options for any generator of G (ie g, g3, g5 or g7) g↔1, g2↔2, g3↔3, g4↔4, g5↔5, g6↔6, g7↔7 A1 Completing the specification of any one isomorphism g↔3, g2↔6, g3↔1, g5↔7, g6↔2, g7↔5 and g↔5, g2↔2, g3↔7, g5↔1, g6↔6, g7↔3 and g↔7, g2↔6, g3↔5, g5↔3, g6↔2, g7↔1 A1 Other three. Ignore repeats. g4↔4 does need not be seen again [4]Y435/01 Mark Scheme October 2021 6 3 (a) 3 3 0 det( ) 0 2 2 1 3 4 λ λ λ λ − − = − − A I M1 1.1a Formation of appropriate determinant soi. = − − − − × − − × (3 )[(2 )(4 ) 2 3] 3(0 2 1) λ λ λ oe M1 1.1 Attempt to expand determinant. Allow one slip. May see eg expansion by 1st col: (3 )[(2 )(4 ) 6] 1(6 0) − − − − + − λ λ λ Or other formulation eg: ((3 )(2 )(4 ) 6 0) (0 6(3 ) 0) λ λ λ λ − − − + + − + − + = –λ3 + 9λ2 – 20λ + 12 = 0 A1 1.1 Must be an equation. ISW. [3] 3 (b) 1, 2 and 6 substituted into (a) equation to verify B1 1.1 eg checking trace is insufficient. [1] 3 (c) 3a + 3b = a or 2a or 6a and 2b + 2c = b or 2b or 6b and a + 3b + 4c = c or 2c or 6c M1 1.1 Correctly forming 3 equations in 3 unknowns for one of their eigenvalues. May see explicit choice of eg c = 1 to form 3 equations in 2 unknowns. Or formation of appropriate determinant eg 2 4 4 . 0 1 3 λ λ − − − − − i j k λ = 1: 2a = –3b, b = –2c or λ = 2: c = 0, a = –3b or λ = 6: a = b, c = 2b M1 1.1 Attempt to solve equations for at least one of their eigenvalues leading to two unknowns in terms of 3rd. Attempt to expand determinant (might be in terms of λ) eg 2 8 7 6 2 . 2 λ λ λ  − +     −    Can be inferred by 2 correct coefficients. 3 3 1 2 or 1 or 1 1 0 2       −             −             A1 1.1 or any non-zero multiple. 3 3 1 2 and 1 and 1 1 0 2       −             −             A1 1.1 or any non-zero multiple. [4]Y435/01 Mark Scheme October 2021 7 3 (d) 3 3 1 2 1 1 1 0 2   −   −     M1 3.1a Forming matrix of their eigenvectors, E. 1 3 3 1 2 6 4 1 2 1 1 5 5 5 oe 10 1 0 2 1 3 3 −     − − −         − = − −     A1FT 3.1a BC. Finding inverse of their matrix of eigenvectors. May be in decimal form: 0.2 0.6 0.4 0.5 0.5 0.5 0.1 0.3 0.3   − −     − −   1 0 0 1 0 0 0 2 0 0 2 0 0 0 6 0 0 6 n n n           =         B1 3.1a Matrix of eigenvalues must be consistent with matrix of eigenvectors. Allow 1n. 3 3 1 1 0 0 2 6 4 1 2 1 1 0 2 0 5 5 5 10 1 0 2 0 0 6 1 3 3 n n     −   − −           −   − −         M1 3.1a Forming EΛ nE–1. Can be awarded if Λ n incorrect or uncalculated but eigenvectors must be in same order as eigenvalues. 1 0 0 2 6 4 0 2 0 5 5 5 0 0 6 1 3 3 2 6 4 5 2 5 2 5 2 6 3 6 3 6 n n n n n n n n    − −       − − =        − −     − × − × ×     × × M1 1.1 Proper attempt to multiply either the first two or the last two (of 3) in the correct order (with or without 1 10 ). or 3 3 1 1 0 0 2 1 1 0 2 0 1 0 2 0 0 6 3 3 2 6 2 2 6 1 0 2 6 n n n n n n n   −        − =          − ×     −     ×   1 1 3 3 1 2 6 4 1 2 1 1 5 2 5 2 5 2 10 1 0 2 6 3 6 3 6 6 15 2 6 18 15 2 3 6 12 15 2 3 6 1 4 5 2 6 12 5 2 3 6 8 5 2 3 6 10 2 2 6 6 6 4 6 n n n n n n n n n n n n n n n n n n n n n + +   −   − −      − − × − × × =         × ×   − + × + − + × + × − × + ×     − × + − × + × − + × + ×     − + × − + +   A1 1.1 or 3 3 2 6 2 6 4 1 2 2 6 5 5 5 10 1 0 2 6 1 3 3 n n n n n   − ×      − −   −   − − =       ×     etc. Condone 6×6n unsimplified. [6]Y435/01 Mark Scheme October 2021 8 4 (a) CF: un+2 – 3un+1 – 10un = 0 and un = αrn ⇒r2 – 3r – 10 = 0 M1 1.1a Deriving the auxiliary equation (allow one sign error). ⇒r = 5 or r = –2 CF is α5n + β(–2)n A1FT 1.1 FT correct roots of their AE to form CF (do not ISW). Condone missing brackets around –2 unless misused. Trial function: un = an + b B1 1.1a Correct form. Other forms eg an2 + bn + c are allowable provided a = 0 derived. a(n + 2) + b – 3[a(n + 1) + b] – 10(an + b) = 24n – 10 M1 1.1 Substituting their form correctly into recurrence relation. ⇒ (a – 3a – 10a) = 24 and 2a + b – 3a – 3b – 10b = –10 M1 1.1 Deriving two equations in a and b using a correct method (eg comparing coefficients) a = –2 and b = 1 so GS is un = 1 – 2n + α5n + β(–2)n A1 1.1 Full form of GS, including un =, must be seen. cao [6] 4 (b) Either: n = 0 => 1 + α + β = 6 or: n = 1 => 1 – 2 + 5α – 2β = 10 M1 1.1 Substituting n = 0 or n = 1 in their GS to derive an equation in α & β. This mark can be awarded if one of their equations is wrong. α + β = 5 and 5α – 2β = 11 => 2α + 2β = 10 => 7α = 21 M1 1.1 Deriving 2 equations from substituting n = 0 & 1, at least one correct for their GS, and attempting to solve . Attempt to solve can be implied by correct answer or valid algebra but incorrect answer with no working M0 α = 3 and β = 2 so un = 1 – 2n + 3×5n + 2×(–2)n A1FT 1.1 FT from their GS. Allow nonembedded values if GS seen in (a). Do not ISW. [3] 4 (c) From recurrence relation: u2 = 3u1 + 10u0 + 24×0 – 10 = 3×10 + 10×6 – 10 = 80 From particular solution: u2 = 1 – 2×2 + 3×52 + 2×(–2)2 = 1 – 4 + 75 + 8 = 80 B1 2.5 Both expressions properly seen (ie it must be clear that candidates are correctly using two different methods to find u2). [1]Y435/01 Mark Scheme October 2021 9 4 (d) vn = + + 1 2 5 2 − − pn n 3 2 ( p p )n n ( ) M1 3.1a Writing the limit to be deduced vn in a form which enables . If p < 5 then vn→∞ while if p > 5 then vn→ 0 as n→∞ B1 2.1 Convincing argument. FT for GS of the form: c – dn + αsn + βtn (where s t > ). At most one of c and d is 0. s and t are not equal and both not 0. Both α and β are not 0. Either s >1 or t >1 (or both). p = 5 A1 2.2a FT. p = s (must be a number). A0 If s = –t. q = 3 A1 2.2a FT. q = α(must be a number). A0 If s = –t. If M0 then SC2 for p = 5, q = 3. [4]Y435/01 Mark Scheme October 2021 10 5 g 2x x ∂ = ∂ or g 2y y ∂ = ∂ or g 4z z ∂ = ∂ M1 3.1a g(x, y, z) = x2 + y2 + 2z2 and surface is g = 126. Finding one correct partial derivative. May be rewritten as 1 1 2 2 z x y y x = = − − f ( , ) 63 2 2 but condone ±. 2 2 4 x g y z     ∇ =       is the normal to the tangent plane at each point. A1 3.1a Finding the normal vector. x2zy21 g z     −     ∇ = −       −     oe 2 0 . 2 . 0 4 4 1 x g y z z         ∇ = =             n M1 3.1a Dotting normal with normal to x-y plane. 2 0 2 0 cos 3 4 1 x y z π         =             M1 2.2a Expressing dot product in other form using correct value of angle. 2 2 2 2 2 2 2 2 2 1 (2 ) (2 ) (4 ) 1 2 1 2 4 126 2 4 2 126 2 x y z x y z z z z = + + × × = + + × = − + = + M1 1.1 Using cos , 1 3 2 π = forming magnitude of both normals and reducing to form a bz + 2 oe (could be done after squaring). or a bx by + + 2 2 (could see eg 2 2 x y + =108 oe after equating to 4z and eliminating z). 2 2 2 2 2 126 2 4 126 2 16 14 126 9 3 0 3 which is the equation of z z z z z z z z z Π. + = ⇒ + = ⇒ = ⇒ = ⇒ = ± ≥ ⇒ = A1 3.2a Not ± in final answer. [6]Y435/01 Mark Scheme October 2021 11 6 (a) 2 3 1 1 1 ( 1) ( 1)( 2) ( 1)( 2)( 3) 1 1 1 ( 1) ( 1) ( 1) q q q q q q q q q + + + + + + + + + < + + + + + +   M1 2.1 Correct statement that given series is less than an infinite GP (could be eg 1 12 ... q q + + or 2 1 1 ... 3 3 + + ). 1 11 1 1 q + q = − + A1 2.1 FT on their 1 a − r . 1 1 q q 1 1 = = + − A1 2.1 AG. Intermediate step must be seen. [3] 6 (b) 1 q 1 1 q ≥ ⇒ ≤ M1 2.2a But S S 1 1; q < ⇒ < clearly S > 0 so 0 1 < < S so S ∉ . A1 2.2a AG. S > 0 must be stated but need not be justified. Since 0 1 1 q < ≤ and S 1 q < then 0 1 < < S and ∴ ∉ S  . [2] 6 (c) 0 0 1 ! e e ! ( 1)! ! ! r r p q q p q r q r ∞ ∞ = = = = ⇒ = = − ∑ ∑ M1 3.1a Multiplying both sides by q! No need to mention q ≥ 1 in this part. 0 0 1 ! ! ! ( 1)! ! ! ! ! ! ! ! ... 2! ! 2 ! ( 1)... 3 ( 1)... 4 ... 1 q r r r q q q q p q r r r q q q q S q q q q q q S ∞ ∞ = = = + ∴ − = = + = + + + + + = + − × + − × + + + ∑ ∑ ∑ M1 2.1 Rewriting to a form in which it is clear that every term on both sides, except S, is an integer. p(q – 1)! and ! ! ... 1 ! 2! q q q + + + + are all integers but S is not which is a contradiction. A1 3.2a AGY435/01 Mark Scheme October 2021 12 Alternative Method: 1 !! r q q S r ∞ = + = ∑ M1 Expressing S as an infinite sum in terms of factorials. 0 0 1 1 1 ! ! ! !e ! ! ! ! ( 1)! ! ( 1)...( 1) since 1 q q r r r q r q q S q q q r r r p q q q q q r r q ∞ = = = = = − = − − = − − − − − + ≤ ≤ ∑ ∑ ∑ ∑ M1 Rewriting to a form in which it is clear that every term on both sides, except S, is an integer. p(q – 1)!, q! and q(q – 1)...(q – r + 1) are all integers but S is not which is a contradiction. A1 AG [3]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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