GCE Further Mathematics A Y540/01: Pure Core 1 Advanced GCE Mark Scheme for Autumn 2021

Document Content and Description Below

Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y540/01: Pure Core 1 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 2021Y540/01 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.Y540/01 Mark scheme October 2021 3 Question Answer Marks AO Guidance 1 (a) (i) Circle Centre 1−2i, Radius 3 B1 B1 1.1 2.2a Be generous over circles drawn freehand If the axes are scaled then a mark at (1, -2) will do. For radius, an indication that the radius is 3 will do (e.g. passing through (4, -2) etc if marked will do.) [2] (ii) Straight vertical line B1 1.1 12 x = B1 2.2a Can be seen by x = ½ being labelled on the axis and vertical line through it [2] (b) Inside circle B1 1.1 And to the left of 1 2 x = B1 2.2a Or their line if it is vertical. [2]Y540/01 Mark scheme October 2021 4 Question Answer Marks AO Guidance 2 (a) (i) f (0) π 4 = B1 1.1 Not for 450 [1] (ii) ( )2 1 1 f '( ) f '(0) 1 1 2 x x = ⇒ = + + M1 A1 2.1 1.1 Diffn – Must be seen 2 1 f '( ) 1 x x = + is M0 [2] (iii) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 f '( ) 1 1 2 2 1 f ''( ) 1 2 2 2 2 2 2 2 2 2 1 f ''(0) 4 2 x x x x x x x x x x x = = + + + + ⇒ = × − × + + + − + = + +   − ⇒ = = −     M1 A1 A1 2.1 2.1 2.1 Diffn their f’(x) oe, e.g. ( ) ( ) ( ) ( )2 2 2 1 f 1 1 x x x + ′′ = − + + f’’(0) must be seen. The substitution must be seen (implied by 2 4 − ) AG [3] (b) 2 f ( ) f (0) f '(0) f ''(0) x2 x x = + + M1 1.1 Using the formula and substituting their value for f;(0) 2 2 1 1 4 2 2 2 4 2 4 x x x x π π = + − × = + − A1 2.2a ft their values from (a) [2]Y540/01 Mark scheme October 2021 5 Question Answer Marks AO Guidance 3 (a) e.g. α β γ 2 2 2 + + = −5 means that at least one root is complex But complex roots come in complex pairs so there are 2 complex roots. Given that there are 3 roots and 2 are complex one is real. B1 B1 B1 2.4 2.4 2.4 [3] (b) α β γ + + = 3 B1 1.1 ( ) ( ) ( ) 2 2 2 2 2 9 5 2 But 7 k k α β γ α β γ αβ βγ γα αβ βγ γα αβ βγ γα + + = + + + + + = − + + + = + + ⇒ = M1 A1 3.1a 1.1 Attempt to obtain identity and substitute Condone missing 2 and sign errors [3] (c) 1 1 1 3 2 3 7 5 0 u u u             − + − =       M1 1.1 For the substitution “=0” not necessary here but needed for A1 ⇒ − + − + = 5 7 3 1 0 u u u 3 2 oe A1 1.1 Allow in terms of z Allow ft from their k in (b) Alternate method 1 1 1 7 , α β γ 5 + + = 1 1 1 1 1 1 3 α β β γ γ α 5 + + = , 1 1 1 1 α β γ 5 = Answer as above M1 A1 For calculating the sum, product and sum of product of pairs of reciprocals of α, β, γ [2]Y540/01 Mark scheme October 2021 6 Question Answer Marks AO Guidance 4 (a) 3 3 3 4 3 Equation of is 2 3 0 3 AB x AB y z λ   −   =             −             = +             oe oe uuur B1 M1 1.1 1.1 soi 3 3 3 their   −         soi r = or x y z       =     is not required for M1 ⇒ − = − = 4 2 x y z A1 1.1 Allow equivalent equations e.g. ⇒ − = − = − 1 5 3 x y z from using B [3]Y540/01 Mark scheme October 2021 7 Question Answer Marks AO Guidance 4 (b) 3 4 3 4 3 3 , is 2 3 2 3 3 0 3 3 4 3 1 3 3 2 3 4 3 2 3 2 3 2 3 3 3 Perpendicular to 3 2 . 3 3 2 3 AB OM CM AB λ λ λ λ λ λ λ λ λ λ λ λ λ         − − −         = + = +                               − −       = + − = −                   − +    − −   ⇒ −       + uuur uuur uuur uuur 0 9 9 9 6 9 6 0 1 27 9 . 3 Coordinates of are (3, 3, 1) M λ λ λ λ λ      =     ⇒ − + + − + + = ⇒ = ⇒ = ⇒ M1 A1 M1 A1 3.1a 1.1 1.1 1.1 Attempt to find general point on AB to get vector CM. Can use (1, 5, 3) Allow working throughout that uses e.g. 1 − 1 1           ft their vector from (i). Use of dot product to solve Do not accept a vector answer Alternative method for last two marks 2 2 2 2 2 3 3 Minimise 3 2 3 2 (3 3 ) (3 2) (3 2) 1 Coordinates of are (3, 3, 1) 3 CM M λ λ λ λ λ λ λ   −   = −       + = − + − + + ⇒ = ⇒ uuur M1 A1 Express as a function of λ and minimise the quadratic in λ [4]Y540/01 Mark scheme October 2021 8 Question Answer Marks AO Guidance 4 (c) 2 2 2 2 2 2 2 2 2 1 3 14 3 3 3 27 CM AB = + + = = + + = B1 B1 1.1 1.1 B1 for each distance ft their M ft their AB 1 Area = AB . CM 2 3 42 2 ⇒ = uuur uuur M1 A1 3.1a 1.1 Formula for area ft their M Alternative method 1 2 2 2 3 0 12 1 1 1 Area = 3 1 15 2 2 2 3 5 3 1 1 3 12 15 3 378 42 2 2 2 AB BC       −       × = − × = −                   − = + + = = uuur uuur M1 M1 A1 A1 Formula for area Cross product Alternative method 2 1 Area = sin where . cos 2 12 4 cos 27 26 78 8 31 sin 1 39 39 1 31 3 Area = 27 26 42 2 2 39 AB BC θ θ AB BC AB BC θ θ = − − ⇒ = = ⇒ = − = ⇒ = uuur uuur uuur uuur uuur uuur M1 A1 M1 A1 For use of dot product, formula for area Pythagoras to find sinθ [4]Y540/01 Mark scheme October 2021 9 Question Answer Marks AO Guidance 5 1 5 5 5 3 3 5 5 5 3 5 3 5 Take cos isin cos isin 1 2isin 1 10 5 1 32isin 5 10 1 1 1 32isin 5 10 2isin 5 10isin 3 20isin 5 5 1 sin sin sin 3 sin 5 . 8 16 16 i.e. z z z z z z z z z z z z z z z z z z A θ θ θ θ θ θ θ θ θ θ θ θ θ θ − = + ⇒ = − ⇒ − =     − = = − + − + −         ⇒ = − − − + −             = − + ⇒ = − + = 5 5 1 , , 8 16 16 B C = − = M1 A1 M1 A1 2.1 1.1 1.1 2.2a Use of z and de Moivre Sight of i is necessary Both sides Attempt conversion into sin soi All three stated Alternative method 1 ( ) ( ) ( ) (( ) ( ) ( )) ( ) 5 i i 5 5i 3i i i 3i 5i 5 5i 5i 3i 3i i i 5 e e sin 2i 1 e 5e 10e 10e 5e e 2i 1 e e 5 e e 10 e e 32i 1 sin 5 5sin 3 10sin 16 5 5 1 sin sin sin 3 sin 5 8 16 16 5 5 1 , , 8 16 16 A B C θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ − − − − − − −   − =     = − + − + − = − − − + − = − + ⇒ = − + ⇒ = = − = M1 A1 M1 A1 Sight of i in the denominator is necessary Collection to convert back All three statedY540/01 Mark scheme October 2021 10 Alternative method 2 ( ) ( ) ( ) 5 5 5 4 3 2 2 3 4 5 4 2 3 5 2 2 2 3 5 3 5 cos sin cos5 sin 5 cos sin cos 5 cos sin 10cos sin 10 cos sin 5cos sin sin sin 5 5cos sin 10cos sin sin 5 1 sin sin 10 1 sin sin sin 5sin 10sin 5sin 10si z i z i i i i i θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = + ⇒ = + = + = + − − + + ⇒ = − + = − − − + = − + − ( ) 3 5 5 3 5 3 5 3 2 2 3 2 3 3 5 5 n 10sin sin 5sin 20sin 16sin cos3 sin 3 cos sin cos 3 cos sin 3cos sin sin sin 3 3cos sin sin 3sin 4sin sin 5 5sin 3 10sin 16sin 16sin 10sin 5sin 3 sin 5 10 z i i i i A θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ + + = − + = + = + = + − − ⇒ = − = − ⇒ − = − + ⇒ = − + ⇒ = 5 5 1 , , 16 8 8 16 = = − = B C M1 A1 M1 A1 De Moivre for both Eliminate sin3θ All three stated [4]Y540/01 Mark scheme October 2021 11 Question Answer Marks AO Guidance 6 ( ) ( ( ) ) ( ) 2 2 9 2 4 3 For , 1 4 12.566... For , dy 37 10 d 356.05.... Total 356.05... 12.566... 368.61... 369 cm to3 f b a AB V BC V x y y V s π π π = × × = = = − − = ⇒ = + = = ∫ ∫ M1 A1 A1 3.3 1.1 3.4 4π Split into two parts and use formulae An integral and an attempt at the volume of a cylinder must be seen Integration – ignore limits BC 340 3 π Units are not required 352 3 π [3]Y540/01 Mark scheme October 2021 12 Question Answer Marks AO Guidance 7 (a) 0 sin 3 0 0, 3 0, 3 r θ π θ θ π   = ⇒ =   ⇒ =   ⇒ = B1 1.1 Both required Don’t give if any extras within range. Ignore values outside range [1] (b) 3 sin , i.e. 1, 6 6 6     π π π         B1 B1 1.1 1.1 For r For θ [2] (c) DR 3 3 2 2 0 0 1 1 Area = d sin 3 d 2 2 r π π ∫ ∫ θ θ θ = ( ) 3 0 1 1 cos 6 d 4 π = − θ θ ∫ 3 0 1 1 sin 6 4 6 π = −       θ θ M1 M1* DepM1 1.1 3.1a 1.1 Correct use of formula – ignore limits Attempt to use double angle formula (Could be wrong way round, 2 missing or sign wrong) Integrate their integrand 1 0 4 3   π = −     = 1 12 π A1 1.1 Use correct limits, must be seen [4] (d) ( ) ( ) 3 3 4 2 3 2 2 2 2 2 3 sin 3 3sin 4sin 3 4 3 4 3 4 y y r r r r r y y x y y x y y θ θ θ = −   ⇒ = −     ⇒ = − + = + − oe e.g. 3 ( ) x y x y y 2 2 2 3 + = − 2 M1 A1 1.1 1.1 Using triple angle formula and y r = sinθ isw [2]Y540/01 Mark scheme October 2021 13 Question Answer Marks AO Guidance 8 (a) 2 2 4sinh 3cosh d 4cosh 3sinh d 0 when 4cosh 3sinh 0 e e e e 4 3 0 2 2 1 e 7 which is not possible as e 0 so no turning points x x x x x x y x x y x x x x x − − = + ⇒ = + = + =     + − ⇒ + =         ⇒ = − > M1 M1 A1 1.1 2.1 2.4 Diffn (Hyperbolics or exponentials) Set = 0 and use exponential forms – can change to exponentials before diffn. Conclusion with justification Alternative method 4sinh 3cosh d 4cosh 3sinh d 4 0 when tanh 3 But tanh <1 for all . d So there are no values of for which 0 d So no turning points y x x y x x x x x x y x x = + ⇒ = + = = − = M1 M1 A1 Differentiate Set = 0 and use formula for tanh Conclusion with justification [3]Y540/01 Mark scheme October 2021 14 (b) 2 4sinh 3cosh 5 e e e +e 4 3 5 2 2 7e e 10 7e 10e 1 0 10 100 28 5 32 5 32 e or 14 7 7 5 32 But e 0 so cannot = 7 5 32 So the only root is e 7 5 4 2 ln 7 x x x x x x x x x x x y x x x − − − = + =     − ⇒ + =         ⇒ − = ⇒ − − = ± + + − = = − > + =   + ⇒ =       M1 M1 A1 A1 A1 3.1a 3.1a 1.1 2.3 1.1 Use of exponentials equation of the form a b c e e 0 2x x + + = (for non-zero a, b and c) Two roots for ex One rejected plus reason Ignore inclusion of 2nd root Alternative method (see appendix for full working) 2 2 2 2 1 4sinh 3cosh 5 4sinh 5 3cosh 16sinh 16(cosh 1) 25 30cosh 9cosh 7 cosh 30cosh 41 0 15 16 2 cosh 1 cosh 7 15 16 2 15 16 2 4 43 30 2 cosh ln 7 7 But the negative root does not work in th x x x x x x x x x x x x x − + = ⇒ = − ∴ = − = − + + − = − + ≥ ⇒ =   ⇒ = − + − + + − = ±       e original equation since LHS would be negative while RHS would be positive (but equal when squared). 15 16 2 4 43 30 2 ln 7 x   ∴ =   − + + −     M1 M1 A1 A1 A1 Use Pythagoras Quadratic in cosh (or sinh) Two roots One rejected plus reason [5]Y540/01 Mark scheme October 2021 15 Question Answer Marks AO Guidance 9 (a) 2 1 2 1 0 x x kx kx x +    =           − − M1 A1 3.1a 1.1 ( ) ( ) 2 same line (2 ) for all 0 1 (2 ) 2 1 0 1 i.e. x k x kx x k k k k k y x ⇒ − = + ≠ ⇒ − = + ⇒ + + = ⇒ = − = − M1 A1 2.1 1.1 Value of k can be implied by the correct equation [4] (b) 2 1 2 1 0 x x x x x x x        −        = =        − − − − so each point maps to itself and it is a line of invariant points B1 2.4 Must have a reason e.g. it is sufficient to test one point other than (0, 0) [1]Y540/01 Mark scheme October 2021 16 Question Answer Marks AO Guidance 10 ( )( ) 1 2 1 2 1 2 1 2 1 (2 1) (2 1) 1 1, 0 1 1 , 2 2 A B r r r r A r B r A B A B A B = + − + − + ⇒ + + − = ⇒ − = + = ⇒ = = − M1 M1 A1 3.1a 1.1 1.1 partial fractions Allow any method to determine A and B Both values 1 ( )( ) 1 1 1 1 ... 1 1 1 3 3 5 2 1 2 1 2 1 1 1 1 2 3 2 1 2 1 2 1 1 1 1 2 2 1 n r r r n n n n n =             − + − + ⇒ =       − +         + − + −        − − − +      = −   ∑ +  oe M1 M1 A1 3.1a 2.1 1.1 Use of differences Deal with subtraction 1 1 1 0.49 2 2 1 0.98 0.49 0.49 24.5 0.02 n n n n   − +   ⇒ + ⇒ = ≥ ≥ ≥ M1 3.1a Use of inequality on their formula ⇒ = n 25 A1 3.2a No marks for a purely numerical solution. [8]Y540/01 Mark scheme October 2021 17 Question Answer Marks AO Guidance 11 (a) (i) For SHM λ = 0 B1 3.3 [1] (a) (ii) The door should close, but in SHM the motion continues indefinitely B1 3.5b [1] (b) Over- or critical- damping implies λ 2 −12 0 ≥ So λ ≥ 2 3 M1 A1 3.3 3.4 Consider discriminant with ≥ > or Ignore λ ≤ -2 3 [2] (c) e.g. B1 3.4 Graph of under-damped system. Start anywhere non-zero on θ-axis with zero gradient. Each peak must be lower than before At least two peaks (not including start point) The graph must look as though it is approaching the t axis [1]Y540/01 Mark scheme October 2021 18 Appendix 8(b) Alternate solution 2 2 2 2 2 1 4sinh 3cosh 5 4sinh 5 3cosh 16sinh 16(cosh 1) 25 30cosh 9cosh 7cosh 30cosh 41 0 30 30 4 7 41 30 2048 cosh 2 7 14 30 32 2 15 16 2 cosh 1 cosh 14 7 15 16 2 15 16 2 15 16 2 cosh ln 7 7 7 x x x x x x x x x x x x x x − + = ⇒ = − ∴ = − = − + + − = − ± − × × − − ± = = × − + − + ≥ ⇒ = = − + − + − +  ⇒ = = ± +   2 1 15 16 2 225 512 480 2 49 ln 7 49 49 15 16 2 688 480 2 15 16 2 4 43 30 2 ln ln 7 49 7 But the negative root does not work in the original equation since LHS would be neg                −   = ± + −   − + + −         = ± + =  − + −   ± − + + −          ( ) ( ) 2 ative while RHS would be positive (but equal when squared). 15 16 2 4 43 30 2 ln 7 NB 5 3 2 25 18 30 2 43 30 2 and 5 3 2 0 15 16 2 4 5 3 2 5 4 2 ln ln 7 7 x x   ∴ =   − + + −     − = + − = − − > ∴ =         − + + − =         +Y540/01 Mark scheme October 2021 19 Question 2(a)(ii) Alternative solution ( ) ( ) 1 2 2 2 2 tan 1 1 tan d 1 sec . d d 1 1 1 d sec 1 tan 1 1 y x x y y y x y x y y x − = + ⇒ + = ⇒ = ⇒ = = = + + +OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre Education and Learning [Show More]

Last updated: 1 year ago

Preview 1 out of 21 pages