4753 Mark Scheme June M1 ±1/3 cos 3x seen or A1 1 1

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Georgiana Berry
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1 4753 Mark Scheme June 4 ± /6 /6 ( sin 3 )d cos3 3 ±/3 cos 3 seen ( sin u )[d u ] i.e. condone sign err 3 cos3 ( u cos u) condone + c 3 3 = /6 /3 cao o.e., must be eact isw after crect answer seen y = ln( cos), let u = cos dy/d = dy/du. du/d /( cos) soi = (/u). sin d/d ( cos ) = ±sin 3 3 = 4 sin cao cos d y sin( / 3) substituting /6 3 into their deriv must be in at least two places When = /6, d cos( / 3) = 3 cao isw after crect answer seen 3 = 4, = ½ 3 = 4, = ½ not 3/( ) If 3 me final answers offered, f each increct additional answer f final ans written as an inequality (3 ) = 6 squaring both sides (3 ) = 4 is M + 9 [= ] crect quadratic o.e. but with single term = ½, ½ 5

2 4753 Mark Scheme June 4 4 (i) a =, b = ½ BB [] 4 (ii) y = + cos ½ y = + cos ½ y (may be seen later) = cos ½ y subtracting [their] a from both sides need not substitute f a, b (first) arccos( ) = ½ y arccos( [their] a) = [their] b y with y, need not subst f a, b y = f () = arccos( ) cao cos ( ) may be implied by flow diagram Domain 3 domain to 3, range to if not stated, assume first is domain Range y crectly specified: must be, f domain, y f f () f range allow [, 3], [, ] not 36 (not f) 5 dv/dr = 4r B r /3, condone dr/dv, dv/dr dv/dt = B Condone use of other letters f t dv/dt = (dv/dr)(dr/dt) a crect chain rule soi o.e. e.g. dr/dt = (dr/dv)(dv/dt) = 4.64.dr/dt dr/dt =.4 cm s o.e. (soi) must be crect. better /56 5/8 mark final answer 6 (i) V = e.t when t =, V = B (soi) art 64 so car loses ( )36 B condone no, must be to nearest B f crect answer 6 (ii) When t =, V = 3 3 = 5 e k k [ln e] = ln(3/5) taking lns crectly oe e.g. ln 3 = ln5 k [lne] k =.43 =.43 (3sf) * cao NB AG must show some wking if 4 th d.p. not shown [] If k =.43 verified,e.g. 5 e.43 = 3[.3 ], SCB need not have substituted f V and A e.g. k = ln(3/5) =.43 6

3 4753 Mark Scheme June 4 6 (iii) 5e.43t = e.t * must be crect, but could use a me If M, SCB f 5 5. years from accurate value f k crect calculations f each car, rot e.g. t = 5, 7358 (Brian), 7338(Kate) ( 7334 with me accurate k ) (5/) = e (.43.)t dep dep * o.e. e.g. ln5.43t = ln.t t = ln.75 /.57 = 5.5 years so after 5 years 7 (i) False e.g. neither 5 and 7 are prime as 5 is div by 5 and 7 by 3 7 (ii) True: one has fact of, the other 4, so product must have fact of 8. 8 (i) 8 (ii) f( ) = ( ) cao accept answers in the range 5 5. B B [] crect counter-eample identified justified crectly B algebraic proofs: e.g. n(n+) = 4n(n+) = 4 even odd no so div by 8 [] substituting f in f() Need not eplicitly say false B f stating with justification div by 4 e.g. both even, from 4(n + n) 4pq,, ( ) ( ) st line must be shown, must have f( ) = = f() f() oe somewhere MA Rotational symmetry of der about O B must have rotate and O and der oe e.g. reflections in both - and y-aes 8 ½ turn f() = ( ) /.. ( ). ( ) ( ) 3/ 3/ * quotient product rule used ½ u / ½ v 3/ soi crect epression QR: condone udv ± vdu, but u, v and denom must be crect 3/ / ( / )( ). ( ). 3/ ( ) ( ) NB AG When =, f () = / 3/ = / B oe e.g. /, /, / /, but not / 3/ allow isw on these seen A 7

4 4753 Mark Scheme June 4 8 (iii) A = [d ] B crect integral and limits limits may be inferred from subsequent wking, condone no d let u = +, du = d v = ( + ), dv = ( + ) / d 8 (iv) (A) 3 d u [d u ] [d v] u k( + ) / u condone no du dv, but not / 3 u [u / ] o.e. (but not /u / ) [v] k = = 3 cao must be eact isw approimations y squaring (crectly) must show [ (+ )] = + (o.e.) d u /y = ( + )/ = / + * equivalent algebra NB AG If argued backwards from given result without err, SCB [] 8 (iv) (B) y 3 dy/d = 4 3 BB LHS, RHS condone dy/d y 3 unless pursued dy/d = 4 3 / y 3 = y 3 / 3 * B NB AG Not possible to substitute = and y = into this epression B soi (e.g. mention of /) 9 (i) e = m may be implied from nd line e = m = ln m = ½ ln m * NB AG If = ½ ln m, y = ½ ln m e ln m substituting crectly so P lies on y = m = ½ ln m m Condone can t substitute = o.e. (i.e. need not mention y = ). Condone also division by is infinite dividing by, subtracting ln o.e. e.g. [ln ] = ln m + [ln ] factising: (e m) = 8

5 4753 Mark Scheme June 4 9 (ii) let u =, u =, v = e, v = e * product rule consistent with their derivs dy/d = e e o.e. crect epression.( ln m).( ln m) dep subst = ½ ln m into their deriv dep = e.( ln m) e * = e lnm + e lnm cao condone e ln m [= m + m ln m] lnm not simplified but not ( ½ ln m), but mark final ans 9 (iii) m + mln m = m ln m = m = e * y + ½mlnm = m( + lnm)( + ½lnm) = lnm, y= ½mlnm=m(+lnm)( ½lnm) + ln m =, ln m =, m = e their gradient from (ii) = m NB AG B f fully crect methods finding - intercept of equation of tangent and equating to ln m At P, = B y = e B isw approimations not e 9 (iv) Area under curve = e d u =, u =, v = e, v = ½ e parts, condone v = k e, provided it is used consistently in their parts fmula e e d ft ft their v igne limits until 3 rd e e 4 e e o.e 4 = ( ½ e ¼ e ) ( ¼ e ) crect epression need not be simplified [= ¼ ¾ e ] Area of triangle = ½ base height ft their, e [e /] o.e. using isosceles triangle = ½ e may be implied from.67 So area enclosed = ¼ 5e /4 cao o.e. must be eact, two terms only isw [7] 9

PMT. GCE Mathematics (MEI) Unit 4753: Methods for Advanced Mathematics. Advanced GCE. Mark Scheme for June Oxford Cambridge and RSA Examinations

PMT. GCE Mathematics (MEI) Unit 4753: Methods for Advanced Mathematics. Advanced GCE. Mark Scheme for June Oxford Cambridge and RSA Examinations GCE Mathematics (MEI) Unit 4753: Methods for Advanced Mathematics Advanced GCE Mark Scheme for June 014 Oxford Cambridge and RSA Examinations 4753 Mark Scheme June 014 1. Annotations and abbreviations