Mark Scheme (Results) June AEA Mathematics (9801)

Transcription

1 Mark Scheme (Results) June 0 AEA Mathematics (980)

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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

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