eliminate e x [3] dx B1dep dep correct derivative

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Christiana Mosley
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1 753/01 Mark Scheme January (i) y = e x sin x Product rule u their v + v their u dy/dx = e x.cos x + (e x )sin x B1 d/dx(sin x) = cos x Any correct expression but mark final answer [3] 1 (ii) ft their dy/dx but must dy/dx = 0 when cos x sin x = 0 eliminate e x derivative must have terms = tan x sin x / cos x = tan x used substituting ½ arctan into their deriv M0 x = arctan [or tan 1 ] (unless cos x = 1/ 5 and sin x = / 5 found) x = ½ arctan * NB AG must show previous step [3] (i) dy dy Rearranging for y and differentiating x y y d x dx explicitly is M0 correct equation Ignore superfluous dy/dx = unless used subsequently dy x dx y o.e., but mark final answer [3] (ii) dy 0 x = dx B1dep dep correct derivative + y = 8 y =, y = or B1B1, can isw, penalise inexact answers of ±1.1 or better once only [3] 1 for extra solutions found from using y = < x < 3 1 < x < 1 oe x 1 B1 B1 [or a = and b = 1] [] 5

2 753/01 Mark Scheme January 013 (i) = a be kt When t = 0, = = a b 15 = a b must have e 0 = 1 When t =, = = a B1 a = 100 b = 85 cao b = 85 When t = 1, = = e k 30 = a b e k (need not substitute for a and b) e k = 70/85 k = ln (70/85) = 0.19(156 ) Re-arranging and taking lns allow k = ln[(a 30)/b] ft on a, b k = or better, or ln (70/85) oe mark final ans [6] (ii) 80 = e 0.19t ft their values for a, b and k but must substitute values e 0.19t = 0/85 t = ln(/17) / 0.19 = 7.5 (min) art 7.5 or 7 min 30 s or better [] 5 (i) df/dv = 5 v d/dv(v 1 ) = v soi 5 v o.e mark final ans [] 5 (ii) When v = 50, df/dv = 5/50 (= 0.01) B1 5/50 d d d. dt dv dt df / dv dv dt dt = = cao o.e. e.g. 3/00 isw [3] 6

3 753/01 Mark Scheme January Let u = 1 + x ( 1/ 1/ )d 1 3 1/ 1/ 0 1 x(1 x) d x ( u 1) u du 1/ ( u1) u (d u) * condone no du, missing bracket, ignore limits u u u (u 1/ u 1/ )(du) 3/ 1/ 3 u u 3/ 1/ 3 u u o.e. e.g. 3/ 1/ u u ; ignore limits 3/ 1/ 1 = (16/3 ) (/3 ) dep upper lower dep 1 st and integration cao or.6 but must be exact 3 with correct limits e.g. 1, for u or 0, 3 for x or using w = (1+x) 1/ ( w 1) w (d w) w OR Let u = x, v = (1 + x) 1/ 3 = ( w 1)(d w) = 3 w w u = 1, v = (1 + x) 1/ upper lower with correct limits (w = 1,) / 1/ 1/ x x x x x x d x ignore limits, condone no dx 8/3 cao x(1 x) (1 x) 3 1/ 3/ 3 0 ignore limits *If ( 1) 1/ d 1 u u u done by parts: = ( 3 8/3) (0 /3) u 1/ (u 1) u 1/ du cao or.6 but must be exact 3 [5] [u 1/ (u 1) u 3/ /3] substituting correct limits 8/3 cao 7

4 753/01 Mark Scheme January (i) 3 5 Attempt to find counterexample + = 5 [which is not prime] If A0, allow for 3 n + correctly correct counter-example identified evaluated for 3 values of n [] 7 (ii) (3 0 = 1), 3 1 = 3, 3 = 9, 3 3 = 7, 3 Evaluate 3 n for n = 0 to or 1 = 81, to 5 allow just final digit written so units digits cycle through 1, 3, 9, 7, 1, 3, so cannot be a 5. OR 3 n is not divisible by 5 B1 all numbers ending in 5 are divisible by 5. B1 so its last digit cannot be a 5 must state conclusion for B [] 8 (i) If just vectors given withhold one A mark translation in the x direction allow shift, move only of / to the right oe (eg using vector) translation in y-direction allow shift, move of 1 unit up. oe (eg using vector) [] Translate π / is marks; if this is 1 followed by an additional incorrect transformation, SC A0 π / 1 only is MA0 8

5 753/01 Mark Scheme January (ii) sin x g( x) sin x cos x (Can deal with num and denom separately) (sin x cos x)cos xsin x(cos xsin x) g( x) Quotient (or product) rule vu ' uv' ; allow one slip, missing brackets (sin x cos x) consistent with their derivs v sin xcos xcos xsin xcos xsin x Correct expanded expression uv' vu ' (could leave the as a is M0. Condone cos x, sin x (sin x cos x) v factor) cos x sin x (cos xsin x) (sin x cos x) (sin xcos x) (sin x cos x) * NB AG When x = /, g (/) = /(1/ +1/) substituting / into correct deriv = 1 f (x) = sec x o.e., e.g. 1/cos x f (0) =sec (0) = 1, [so gradient the same here] [7] must take out as a factor or state sin x + cos x = 1 9

6 753/01 Mark Scheme January (iii) π/ π/ sin x f( x)d x d x 0 0 cos x let u = cos x, du = sin x dx when x = 0, u = 1, when x = /, u = 1/ 1/ 1 du 1 u substituting to get 1/u (du) 1 1 du * 1/ u NB AG ln u 1 1/ [ln u] = ln 1 ln (1/) = ln = ln 1/ = ½ ln ln, ½ ln or ln(1/ ) mark final answer [] 8 (iv) Area = area in part (iii) translated up 1 unit. soi from / added or So = ½ ln + 1 / = ½ ln + /. cao oe (as above) ignore limits here, condone no du but not dx allow 1/u. du but for must deal correctly with the ve sign by interchanging limits π / π (1 tan( x π / ))d x xln sec( xπ / ) / π / π / π/ln π / π/ln B [] 9 (i) At P(a, a) g(a) = a so ½(e a 1) = a e a = 1 + a * B1 NB AG [1] 9 (ii) a 1 x A (e 1)d x correct integral and limits limits can be implied from subsequent work 0 1 e x a x 0 B1 integral of ex 1 is e x x = ½ (e a a e 0 ) = ½ (1 + a a 1) = ½ a * NB AG area of triangle = ½ a B1 area between curve and line = ½ a B1cao ½ a [6] mark final answer 10

7 753/01 Mark Scheme January (iii) y = ½(e x 1) swap x and y x = ½ (e y 1) x = e y 1 Attempt to invert one valid step merely swapping x and y is not one step x + 1 = e y ln(x + 1) = y * y = ln(x + 1) or g(x) = ln(x + 1) AG apply a similar scheme if they start with g(x) and invert to get f(x). g(x) = ln(x + 1) or g f(x) = g((e x 1)/) Sketch: recognisable attempt to reflect in y = x through O and (a, a) = ln(1 + e x 1) = ln(e x ) = x no obvious inflexion or TP, Good shape extends to third quadrant, similar scheme for fg without gradient becoming See appendix for examples too negative [5] 9 (iv) f (x) = ½ e x B1 g(x) = /(x + 1) 1/(x + 1) (or 1/u with u = x + 1) to get /(x + 1) g (a) = /(a + 1), f (a) = ½ e a either g(a) or f (a) correct B1 soi so g(a) =/e a or f (a) = ½ (a+1) substituting e a = 1 + a = 1/(½ e a ) = (a + 1)/ establishing f (a) = 1/ g (a) either way round [= 1/f(a)] [= 1/g(a)] tangents are reflections in y = x B1 must mention tangents [7] 11

8 753/01 Mark Scheme APPENDIX 1 Exemplar marking of 9(iii) A0 infl A0 infl, no 3 rd quadrant (slight infl condone) (slight infl) A0 (tp, just) A0 bod (slight infl) bod (condone TP) bod, (see 3 rd quad) A0 (TP, 3 rd quad) A0 (see 3 rd q) A0 (3 rd q bends back) 1

Wednesday 23 January 2013 Morning

Wednesday 23 January 2013 Morning A2 GCE MATHEMATICS (MEI) 4753/01 Methods for Advanced Mathematics (C3) QUESTION PAPER * 4 7 3 3 9 9 0 1 1 3 * Candidates answer on the Printed Answer Book. OCR supplied