Mark Scheme (Results) June AEA Mathematics (9801)
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1 Mark Scheme (Results) June 0 AEA Mathematics (980)
2 Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Epert service helpful. Ask The Epert can be accessed online at the following link: June 0 Publications Code UA080 All the material in this publication is copyright Edecel Ltd 0
3 June Advanced Etension Award Mathematics Mark Scheme Question Scheme Marks Notes. sin( θ + 35) cos( θ 53) M Use of correct defns for tan = and cot cos( θ + 35) sin( θ 53) 0 = cos( θ 53)cos( θ + 35) sin( θ + 35)sin( θ 53) Use of cos(a+b) rule to M reach single trig function 0 = cos(θ ) AA A for 5 and A for θ 8 = 90, 70 so = 5, () Use of tan (A + B) doesn't score until tanθ = tan( ) ALT tan( θ + 35) = tan[90 ( θ 53)] M Use of cot = + tan(90+) θ + 35 = 90 ( θ 53) or θ + 35 = 90 ( θ 53) + 80 M either. ( + tan ) = + tan ( ) + tan ( M Attempt to multiply ) 3 terms at least correct = sec ( ) + tan ( ) ( ( ) ( )) ( ) ( ) sec + tan d = tan + ln sec M M A Use of sec α = + tan α M for attempt to integrate (ktanθ or k lnsecθ) A for all correct 0 ( )... d = tan + ln sec (0) M Use of limits seen (provided some int. attempt) 3. (a) k, kp, kpq; = + ln = + ln 3 kp q, kp q, kp q [Need one line clearly showing factorisation or split ] Identify: k + kpq+ kp q is GP with a = k, r = pq Identify: kp kp q kp( pq) is GP with a = kp, r = pq A A (7) M A//0 (3) MA MA a = b = (Accept ln) AA dep. on th M only M for st 3 terms A//0 (- eeoo) for net 3 M for splitting into series A for st a and r M for identifying nd GP A for nd a and r (c) S n AEA Mathematics (980) Summer 0 n n ( ( ) ) ( ( ) ) k pq kp pq = + pq pq n ( + )( ( ) ) k p pq = pq 3 3 = i.e. k = 6, p=, q= r = pq = ( r < S formula can be used) 5 7 k( + p) S = =, = = 70 pq 3 5 M Acso (6) B M A,A () (3) Use of Sn formula twice. One correct ft their a & r Identify link with above and values for k, p and q Attempt to find r. (S+ for noting r < etc) A for an epression can be in k, p or q. ft their values A for 70
4 Question Scheme Marks Notes. M Use of sint (a) y = sint cost = sint M Use of cost = cos t = cos t = cos t M Successfully eliminating t and eqn. for circle + y = (c) ( ) ( ) ( ) ( ) so centre (,0), + y = r = Area of R = 3 cos α sinαcosα = cos αsinα da 3 cosα cos α 3cos αsin α dα = da 0 cos α( cos α 3sin α) 0 dα = = cos α = 0 α = or tan α α = = (or 30 o ) 3 6 AA B() MA M A A A for centre A for radius Some evidence of y leading to given result M for use of product rule M for setting derivative =0 and attempting to solve A for trig =..A for α =.. Can ignore α = but consider for S+ A = sinα cos α(3 8cos α) and show <0 for α = 6 or argument based on α = gives min so this is ma M Some check that this value of α gives a ma Maimum area is (o.e.) B (7) (3) Single fraction with rational denom ALT (a) y + y = cos t or = sin t y + y = or + = ( ) + y = Then as in scheme M M M Epression in and y for cos t or sin t Equation in just or y An attempt to complete the square Marks for Style Clarity and Presentation (up to a ma of 7 marks) S For a fully correct and succinct solution to questions or or (n ) and some S+ in questions 3-7 Or for succinct solution of full marks on questions 3 7 but no S+ points seen S For a fully correct and succinct solution to questions 3 to 7 with some S+ points evident T For a good attempt at the whole paper (>50% on each question) Pick the best 3 S/S scores to form the total AEA Mathematics (980) Summer 0
5 Question Scheme Marks Notes 5. (a) U is ( 0, ) B () For y coordinate M for attempt to diff. (Two M, parts and one correct) dy ( ) ( ) Wrong formula used is M0 =, = A A when num. simplified d ( ) ( ) ( a Gradient of normal at P = ) a ( a ) ( ) a Equation of normal: y = a a a ( ) a a = 0 gives y = (*) a M M M Acso (6) Use of perpendicular gradient rule and = a Attempt at eqn of normal can ft their changed grad M clear use of = 0 in norm A for no incorrect working seen (c)(i) No use of circle is 0/5 for (i) Centre is at (0, k) [where k is y-coord from part ] Radius = y coord of their centre 0.5 B B May be implied by a sketch radius touches at U (ii) (iii) ( a ) a Radius to P = a + k a or a + 6 From and k - 0.5: ( a a ) ( a ) = a + a 6 a ( a ) ( a ) = a + ( a ) (*) 6 a a ( a ) ( a ) ( a ) + = a + ( a ) 6 6 a a + a = + = ( a ) a ( ) a = (*) =± so a=± 3 or ± 5 M M Acso M Acso A Epression for radius from centre to P For attempt at a suitable equation in a NB r = LHS implies BB [When cancel a and consider a = 0 for S+] ( a Remove ) 6 and cancel For a = 5 or better, 3 can be ignored and + Dependent on 3 rd M a k = 5 = so centre is 9 (0, ) rad is Allow them to start at (ii) but 3 rd M is critical AA (7) [S+ for reason to reject 3 ] A for centre, A for radius (Dependent on 3 rd M) [May imply some Bs] AEA Mathematics (980) Summer 0
6 Question Scheme Marks Notes t 7 0 5t (a) uuur M Attempt vector PR PR = 3+ 3t = 5+ 3t A 8+ t 7 5+ t 5 uuur M Attempt suitable scalar PR 3 = t 5 + 9t t = 0 product 7 A 50t = 75 t = uuur uuur uuur M Strategy using known If X is midpoint of PP then OP = OP + PX vectors 5 7 NB X is uuur ( 9, 5, 6 OP ) = + = 3 A 7 5 (6) Let t = then can see A lies on L B () Showing t = works 0 5 M uuur (c) uuur uuur uuur Attempt suitable vectors(+) AP = 7, AP = AP AP = = M Attempt suitable scalar So PA P = 0 o product (+) Acso (3) No incorrect working seen (d) = 50 = 5 6 Area = AB PP = 50 3 AB = 0 or 50 o.e. uuur uuur AX = 50 so AB = AX or when t = in equation of L 3 7 uuur OB = 3 ignore t = P P = ( ) B MA M A Attempt PP' (oe) or use sin60 M for attempt at equation giving length of AB Strategy for finding B (e) 0 0 uuur uuur uuur uuur o AP = 7, PB = AP PB = 0 so angle is 90 (*) 7 M Acso () Full method to find angle (f) ALT (c)(d) Since APB is right angle AB is a diameter 7 3 So centre is at midpoint = 8 0 uuur uuur Finding AP and AP uuur uuur AP = AP = 50 or (-,6,) M A () (9) M M Using angle in semicircle theorem (S+ for mentioning) May show PAB = 60 AEA Mathematics (980) Summer 0 B B for PP' from (d)
7 7. (a) (i) (ii) uuur PP = ; sin( PAX ) = = PAP = 0 50 ( 3 ) + ( 5) dy = or y = 3 + y = + d 3 y = 0 = or 5 o Acso ( 3 ) ( 3 ) A is (, -) and B is (5, -0) Horizontal translation 3 to left so p = q = -(q - 0), so q = 6 D is (, ) Then as in main scheme M A M AA B MA BB M for an attempt to differentiate A any correct ver. Find stat points Full coords M for a correct identifiable strategy for b e.g. eqn for q (B, B) (c) (i) 5 y = 3y y = 5 3 y y 3y+ 5= + y ( + ) = 3y+ 5+ y y + y+ 0 + =± o.e. (Accept +, or +) y y + y so m ( ) + + = = M M A A () Set y = and st step Isolate s or set up as 3TQ and attempt to solve for [S+ for reason for choosing ] Must choose (ii) (iii) Domain is range of m() i.e. ( ) If, m( t) = m ( t) then m() intersects with y = B () M Suitable strategy leading to an eqn for t. ft their m t 5 = t 3 t t 3t 5( = 0) ( t )( t+ ) = 5 0 t = - (or.5) A M A A A correct quadratic equation Solving correct 3TQ correct factors (A) (- only) [S+ for reason] Can t be.5 since not in domain for m() (0) AEA Mathematics (980) Summer 0
8 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 FN Telephone Fa Order Code UA080 June 0 For more information on Edecel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE
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