GCE Further Mathematics A Y541/01: Pure Core 2 Advanced GCE Mark Scheme for Autumn 2021

Document Content and Description Below

Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y541/01: Pure Core 2 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 2021Y541/01 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 NBOD Benefit of doubt not given 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.Y541/01 Mark Scheme October 2021 3 Question Answer Marks AO Guidance 1 12 1 5 1 20 3 7 6 3 2 4 5 a a a   − −     − −     − − − − or 4 1 5 11 4 20 3 8 7 17 a a a   − −     − −     − − − seen M1 1.1 Either product AB or BA calculated (but not if assigned incorrectly). Alternatively: equivalent correct useful entries calculated for both Condone 3 errors or omissions This mark can be implied by sight of a correct equation –12 = –4a or –1 = 11 – 4a or 7 – 6a = –8 – a or 3a – 2 = 7 or –4a – 5 = –17 M1 1.1 Finding matrix products both ways and equating entries usefully This mark can be implied by sight of a correct equation even if other entries or equations are wrong. a = 3 A1 2.2a Cannot be awarded if either AB or BA has more than 3 errors [3]Y541/01 Mark Scheme October 2021 4 Question Answer Marks AO Guidance 2 (a) (i) DR 3z1 + 4z2 = 3(3 – 7i) + 4(2 + 4i) = 17 – 5i B1 1.1 [1] (ii) DR z1z2 = (3 – 7i)(2 + 4i) = 6 + 12i – 14i – 28(–1) M1 1.1 Attempted expansion with i2 = –1 used and at least 3 correctly expanded terms – 28(–1) can be simply +28 = 34 – 2i A1 1.1 [2] (iii) DR 1 2 3 7i 3 7i 2 4i 2 4i 2 4i 2 4i zz − − − = = × + + − M1 1.1 Multiplying top and bottom by (real multiple of) conjugate of bottom 6 12i 14i 28 22 26i 11 13 i 4 16 20 10 10 − − − − − = = = − − + A1 1.1 Must see some evidence of expansion Allow − − + 11 13i 11 13i 10 10 or − [2] (b) DR 3 ( 7) 2 2 + − or tan 1 7 3 −   −     M1 1.1 Explicit working must be seen Other trig calculations could be sufficient for M1 provided that these are being used to find the argument. z1 = 58 or awrt 7.62 or argz1 = awrt –1.17 or 5.12 rads A1 1.1 z1 = − 58cis( 1.17) or z1 = 58e−1.17i or z1 = − + − 58(cos( 1.17) isin( 1.17)) or     58, 1.17 − A1 2.5 Must be in correct form with √58 exact and could be awrt 5.12 instead of –1.17. Do not condone degrees Condone round brackets [3]Y541/01 Mark Scheme October 2021 5 Question Answer Marks AO Guidance 3 (a) 1 3 2 3 2 5 4 3 2 3 λ                         − + − =                 − − . M1 1.1 Substituting the expression for a point on the line into the equation of the plane 2 + 15 – 9 + λ(6 – 10 + 6) = 4 8 + 2λ = 4 => 2λ = –4 =>λ = –2 so 1 3 5 3 2 2 7 3 2 7       −       = − + − = −                   − r M1 A1 1.1 1.1 Dotting out to form and solve equation in λ Condone coordinates [3] (b) 2 3 5 2 3 2 4 25 9 9 4 4 6 10 6 2 0.07868... 38 17 646             −         − − + + + + − + = = = . soi M1 1.1 BC. Using cosθ = a.b a b May see sinφ = a.b a b Or use of cross product θ = awrt 85.5…° soi A1 1.1 Can be implied by correct final answer or 1.49… rads (φ = 90° – 85.48…° = ) awrt 4.51° A1 1.1 or 0.0788 rads [3]Y541/01 Mark Scheme October 2021 6 Question Answer Marks AO Guidance 3 (c) 4 1 1 1 λ     = ⇒ = −       r B1 3.1a 3 2 16 2 5 5 2 3 19       −       = − =                   − − − b × M1 2.2a Method shown or at least two terms correctly evaluated So equation of l2 is 4 16 1 5 1 19 µ     −     = − +             − r oe A1 1.1 Must be r =. Allow parameter λ. [3] 4 DR 100 100 100 100 2 2 1 1 1 1 (2 3) 4 12 9 1 r r r r r r r = = = = ∑ ∑ ∑ ∑ + = + + B1 3.1a Expanding and separating 100 2 1 1 100(100 1)(2 100 1) 6 r r = ∑ = × + × + M1 1.1a Use of formula for 100 2 r 1 r ∑= 4×338350 + 12×½×100×101 + 900 = 1414900 A1 1.1 [3]Y541/01 Mark Scheme October 2021 7 Question Answer Marks AO Guidance 5 (a) DR RHS = 2cosh2x – 1 = 2 e e 2 1 2   x x + −   −   M1 2.1 Uses correct exponential form in an attempt at proof 2 2 2 2 2 2 e 2 e e 2 e 2 1 4 2 e e cosh 2 LHS 2 x x x x x x x − − −   + + + + =   − =   + = = = A1 2.1 AG Proof must be complete [2] (b) DR 2cosh2x – 1= 3coshx + 1 => 2cosh2x – 3coshx – 2 = 0 M1 3.1a Use of identity in (a) to leave a three term quadratic equation in just coshx (2coshx + 1)(coshx – 2) = 0 M1 1.1 Attempt to solve eg ( 3) 3 4 2 ( 2) ( )2 2 2 − − ± − − × × − × or 2 3 9 2 cosh 2 0 4 8 x     − − − =   coshx = 2 or –½ A1 1.1 Or solves quadratic BC coshx ≥ 1 so ≠ –½ A1 2.3 Justification must be seen and must contain no incorrect statements x = = + cosh 2 ln 2 3 −1 ( ) A1 1.1 For either correct answer seen x = − ln 2 3 ( ) A1 1.1 Both correct values for x Or x = − + ln 2 3 ( ) Mark final answer [6]Y541/01 Mark Scheme October 2021 8 Answer Marks AO Guidance 6 (a) DR A shear which leaves the x-axis invariant and which transforms the point (0, 1) to the point (2, 1). B1 2.2a Or any useful point transformed to its image not “scale factor” or sf [1] (b) DR detA = 1×1 – 0×2 = 1 and this is the area scale factor B1 2.4 Both Detailed calculation must be shown [1] (c) DR 1 0 3 1       seen B1 3.1a 1 2 1 0 7 2 0 1 3 1 3 1           =      B1 1.1 BC 1 0 7 2 0 3 1 p          M1 1.1 Correct form for stretch multiplied into their matrix in either order 7 2 7 3 p p p   =   ⇒ =   A1 1.1 Correct multiplication [4]Y541/01 Mark Scheme October 2021 9 Question Answer Marks AO Guidance 7 (a) DR 3 2 3 2 3 2 3 2 3 2 9 1 4 4 5 5 4 4 4 4 5 5 1 4 4 x x x x x x x x x x x x x x x x x + + − + + + + − ≡ + + + + + + − = + + + + So A = 1, B = 5 and C = –5 B1 3.1a Attempt to divide out improper fraction. Could be by symbolic division or other valid method (eg comparing coefficients or substitution of values for x) Allow embedded answers [1] (b) DR x x x x x 3 2 + + + = + + 4 4 1 4 ( )( ) 2 B1 3.1a Correct factorisation of cubic seen in working Could be from improper fraction 3 2 2 5 5 4 4 1 4 x D Ex F x x x x x − + = + + + + + + D x x Ex F x ( ) 2 + + + + = − 4 1 5 5 ( )( ) M1 1.2 Correct form for partial fractions equated to their remainder rational fraction from (a). Follow through their division and factorisation. x D D = − ⇒ = − ⇒ = − 1 5 10 2 A1 1.1 Or equivalent to find D correctly. x F F = ⇒ − − = − ⇒ = 0 2 5 3 A1 1.1 Allow ft A1 for second and third coefficients found. x D E E 2 : 0 2 + = ⇒ = 2 2 2 3 1 1 4 x x x + − + + + A1 1.1 Or 2 2 2 2 3 1 1 4 4 x x x x − + + + + + [5]Y541/01 Mark Scheme October 2021 10 Question Answer Marks AO Guidance 7 (c) DR 3 2 2 2 3 2 2 2 0 0 9 1 2 2 3 d 1 d 4 4 1 4 4 x x x x x x x x x x x x + + − = − + + ∫ ∫ + + + + + + *M1 3.1a Split term with denominator andx2ax + 4 in + b in numerator ( ) ( ) 2 2 1 0 3 2ln 1 ln 4 tan 2 2 x x x x −     = − + + + +         dep*M1 1.1 Correctly integrate their expression (ignore limits) 3 2 2ln 3 ln8 ln 4 π 8     − + + −   M1 1.1 Correctly substitute limits to produce exact values and evaluate their tan-1 term 2 3 2 ln 9 8 π   + +     A1 1.1 a = 2, b = 92 , c = 83 [4]Y541/01 Mark Scheme October 2021 11 8 (a) d 2 2 4e d v t F ma kv t − = = = − M1 3.3 Use of NII with m and a replaced and with 2 forces, the given force and kv F=ma can be implicit here t = ln2, v = 0.5, F = 0 => 0 = 1 – 0.5k M1 2.2a Use of given conditions to derive an equation in k Can be done first k = 2 => 2 4e 2 d 2 d v t v t − = − => d 2e 2 d v t v t − + = A1 1.1 AG Complete argument including F=ma [3] (b) IF = e e ∫1dt = t *B1 1.1 Or CF e e e e 2e t t t t t d d d d vt t + = = × v v ( ) −2 *M1 1.1 Multiplying by IF and writing LHS as an exact derivative Or subst correct PI into DE e 2e d 2e t t t v t c = = − + ∫ − − A1 1.1 “+ c” required Or GS � = ��−� − 2�−2� t = 0, v = 0 => c = 2 dep*M1 3.4 Use of initial conditions to derive a value for c Or using alternative boundary condition v = − 2e 2e − − t t 2 A1 3.4 [5] (c) As t→∞, v→ 0 M1 3.4 So speed starts at 0 and ends at 0 (and is continuous and positive between) so must reach a maximum somewhere in t > 0 A1 2.4 [2] (d) v is max when d 0 vdt = so t = ln2 M1 2.2a Deducing time when v is maximum Or by finding expression for ddvt and solving d 0 vdt = So vmax = 0.5 (given) (or ln 2 2ln 2 max 2 1 2e 2e 1 ) 4 2 v − − = − = − = A1 3.4 [2]Y541/01 Mark Scheme October 2021 12 Question Answer Marks AO Guidance 8 (e) d 2 2 2e 2e 2e e dt x t t t t v x d = = − ⇒ = − + + − − − − M1 3.3 Integrating to find expression for x t x = = ⇒ = − + + ⇒ = 0, 0 0 2 1 1 d d M1 3.3 Using initial conditions to find value of (new) constant Or definite integral with correct lower limit.. 0.9 2e e 1 = − + + − − t t 2 M1 3.5a Recognising that the model is only valid when x lies between 0 and 0.9 …and upper limit ( ) e 2e 0.1 0 e 2 10 3 10 10 10 ln 2.97 (3 sf) 10 3 10 t t t t − − − ± − + = ⇒ =   ⇒ =   =   − A1 2.3 Rejecting ln 10 0 10 3 10 t   =   <   + (can be implicit) [4] 9 (a) 2 3 4 4 12 8 36 , , 0 4 0 8 16 96 0 16     =     =       =     A A A B1 2.2a BC Conjecture: 2 3 2 1 0 2 n n n n n   × − =     A B1 2.2b Allow this mark for any conjecture which works for n = 1, 2, 3 and 4. [2]Y541/01 Mark Scheme October 2021 13 Question Answer Marks AO Guidance 9 (b) Basis case: n = 1: 1 0 1 1 2 3 1 2 2 3 0 2 0 2   × ×   =   = =       A A so true for n = 1 B1 2.1 Allow this mark even if the conjecture is wrong, provided that it works for n = 1 Assume true for n= k ie 2 3 2 1 0 2 k k k k   k × − =     A M1 2.1 Must have statement in terms of some other variable than n. Conjecture need not be correct. 1 1 2 3 2 2 3 0 2 0 2 k k k k k k − +   ×   = =       A A A 1 1 1 1 2 3 2 3 2 0 2 2 3( 1) 2 0 2 k k k k k k k k k + + + +   × + × =       + × =     So true for n = k⇒ true for n = k + 1. But true for n = 1. So true for all positive integer n M1 A1 2.2a 2.4 Uses inductive hypothesis properly & expands AG. Manipulating terms correctly and convincingly to obtain required form. Some intermediate working must be seen and a clear conclusion must be given for the induction process. A formal proof by induction is required for full marks. [4]Y541/01 Mark Scheme October 2021 14 Question Answer Marks AO Guidance 10 (a) DR 2 3 2 3 e 1 ... (1 ) ... 2! 3! 2! 3! x x x x x x x   = + + + + = + + + +     M1 1.1 Quoting and using the Maclaurin series 2 3 0 ... 0 2! 3! e 1 x x x x x > ⇒ + + > ⇒ > + A1 2.2a AG. Result with sufficient justification [2] (b) DR 1 e 1 e e e t e t t t x t t t = + ⇒ > ⇒ > ⇒ > − B1 3.1a AG [1] (c) DR 1 e t π = > since 2 < e < 3 and π > 3 B1 3.1a Some justification that t >1 is required e e e ( ) e π π > × = π M1 3.1a Substituting their choice into the inequality e ⇒ > eπ π (ie eπ is greater) A1 1.1 Answer without use of inequality in part (b) scores M0A0 Alternative method t = lnπ B1 Some justification that t >1 is required e eln lnπ > π M1 ln e e e eln ln( ) e π π π π π π > > > A1 [3]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

Last updated: 2 years ago

Preview 1 out of 16 pages