GCE Mathematics (MEI) Advanced GCE Unit 7: Further Methods for Advanced Mathematics Mark Scheme for January Oxford 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 pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, 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 will not enter into any discussion or correspondence in connection with this mark scheme. OCR Any enquiries about publications should be addressed to: OCR Publications PO Box Annesley NOTTINGHAM NG DL Telephone: 87 77 Facsimile: E-mail: publications@ocr.org.uk
7 Mark Scheme January (a)(i) x = r cos θ, y = r sin θ, x + y = r M Using at least one of these r = (cos θ + sin θ) r = r(cos θ + sin θ) x + y = x + y A (ag) Working must be convincing x x + y y = (x ) + (y ) = which is a circle centre (, ) radius M Recognise as circle or appropriate algebra leading to (x a) + (y b) = r (ii) Area = (b)(i) G G M Attempt at complete circle with centre in first quadrant A circle with centre and radius indicated, or centre (, ) indicated and passing through (, ), or (, ) and (, ) indicated and passing through (, ) = cos sin d Integral expression involving r in terms of θ = cos sin cos sin d M Multiplying out = sin cosd A cos θ + sin θ = used = cos sin etc. A Correct result of integration with correct limits. Give A for one error = M Substituting limits. Dep. on both Ms = A Mark final answer 7 f( x) = M Using Chain Rule x x A Correct derivative in any form f( x) x x x... M Correctly using binomial expansion (ii) x x... 8 A Correct expansion f x x x x... c M Integrating at least two terms A But c = because arctan() = A Independent 9
7 Mark Scheme January (a)(i) z n + z n = cos nθ B z n z n = j sin nθ B (ii) (z + z ) = z + z + z + + z + z + z M Expanding (z + z ) = z + z + (z + z ) + (z + z ) + cos θ = cos θ + cos θ + cos θ + cos θ = cos cos cos cos θ = M Using z n + z n = cos nθ with n =, or. Allow M if omitted, etc. cos cos cos A (ag) (iii) (z z ) = z + z (z + z ) + (z + z ) B sin θ = cos θ cos θ + cos θ M Using (i) as in part (ii) A Correct expression in any form sin θ = cos cos cos cos θ sin θ = cos cos M Attempting to add or subtract A OR cos θ = (cos θ + ) B This used cos θ = cos θ + 8 cos θ + M Obtaining an expression for cos θ cos θ = cos θ + cos θ + 8 8 A Correct expression in any form cos θ sin θ = cos θ cos θ + cos θ = cos cos MA Attempting to add or subtract j (b)(i) z = 8e z = e j Correctly manipulating modulus and M argument j7 e A 8, 7 or. Condone r(c + js) j z 9 = 8e z = e j Correctly manipulating modulus and M argument j 9 e A, or. Condone r(c + js) 9 9 w z z (ii) z z = e = e j j7 e 7 9 j 8 j 9 = e Lies in second quadrant G G M A A Moduli approximately correct Arguments approximately correct Give GG for two points approximately correct Correctly manipulating modulus and argument Accept any equivalent form 9
7 Mark Scheme January (i) det(m λi) = ( λ)[( λ)( λ) + 8] M Obtaining det(m λi) + [( λ) ] + [8 + ( λ)] A Any correct form = ( λ)(λ λ + ) + (λ) + ( λ) = λ + λ λ + 8λ + λ = M Simplification λ λ + 8λ = A (ag) www, but condone omission of = (ii) λ λ + 8λ = M Factorising and obtaining a quadratic. If M, give B for substituting λ = (λ )(λ λ + ) = A Correct quadratic λ λ + = b ac = 8 M Considering discriminant o.e. so no other real eigenvalues A Conclusion from correct evidence www x (iii) λ = y z x y + z = x z = x + y z = M Two independent equations x = z = k, y = k M Obtaining a non-zero eigenvector eigenvector is A eigenvector with unit length is v B Magnitude of M n v is n B Must be a magnitude (iv) λ λ + 8λ = M M + 8M I = M Use of Cayley-Hamilton Theorem M M + 8I M = M M Multiplying by M = (M M + 8I) and rearranging A Must contain I
7 Mark Scheme January (i) sinh t + 7 cosh t = 8 (et e t ) + 7 (et + e t ) = 8 M Substituting correct exponential forms e t + e t = 8 e t 8e t + = M Obtaining quadratic in e t (e t )(e t ) = M Solving to obtain at least one value of e t e t = or AA Condone extra values t = ln( ) or ln( ) A These two values o.e. only. Exact form (ii) dy dx = sinh x + cosh x or 8ex + e x B sinh x + cosh x = sinh x + 7 cosh x = 8 x = ln( ) or ln( ) x = ln( ) or ln( ) M Complete method to obtain an x value A Both x co-ordinates in any exact form x = ln( ) y = ( ln( ), ) x = ln( ) y = ( ln( ), ) B Both y co-ordinates dy = sinh x + cosh x = dx tanh x = 7 or e x = etc. M Any complete method No solutions because < tanh x < or e x > etc. A (ag) www (,) G G Curve (not st. line) with correct general shape (positive gradient throughout) Curve through (, ). Dependent on last G 8 a cosh x7sinh x dx M Attempting integration (iii) 7 a sinh x cosh x 7 sinh a cosh a 7 = sinh a + 7 cosh a = 8 A Correct result of integration a = ln( ) or ln( ) a = ln( ) or ln( ) M Using both limits and a complete method to obtain a value of a a = ln( ) ( ln( ) < ) A Must reject ln( ), but reason need not be given 8
7 Mark Scheme January (i) a = y.. x a = y G x a =. G y.. G M Evidence s.o.i. of further investigation (A) Loops when a > A (B) Cusps when a = A 7 (ii) If x x, t t M Considering effect on t but y( t) = y(t) A (ag) Effect on y Curve is symmetrical in the y-axis B (iii) dy dx = asin t M Using Chain Rule acost A dy = a sin t = t = and ±π dx A Values of t t = T.P. is (, a) A t = ±π T.P. are (±π, + a) A Both, in any form (iv) a = : both t = and give the point (π, ) B (ag) Verification Gradients are a and a (or and ) Hence angle is arctan( ). radians x M A Complete method for angle Accept (or ) 8
OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB EU OCR Customer Contact Centre 9 Qualifications (General) Telephone: 998 Facsimile: 7 Email: general.qualifications@ocr.org.uk www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, CB EU Registered Company Number: 8 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR