4753 Mark Scheme June 2015 [6] M1 M1. substituting u = 2x 1 in integral ½ o.e. 3 M1 integral of u 1/3 = u 4/3 /(4/3) (oe) soi not x 1/3

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1 75 Mark Scheme June 05 y = e cos product rule used consistent with their derivs dy/d = e cos e sin A cao mark final ans e.g. e e tan is A0 dy/d = 0 e (cos sin ) = 0 their derivative = 0 cos = sin = sin /cos = tan sin / cos = tan used or sin + cos = used =. A. or 0.5 or better, or arctan, not 6. but condone ans given in both degrees and radians here y =.09 Acao art. no choice let u =, du = d d d u u [6] substituting u = in integral ½ o.e..0787, 0.56, penalise incorrect rounding / i.e. u or integral of u / = u / /(/) (oe) soi not / 8 u c ( ) c 8 or d= ½ ( ) / / ( ) c 8 Acao Acao [] o.e., but must have + c and single fraction mark final answer ( ) / seen / (oe) soi ½ o.e., but must have + c and single fraction mark final ans u seen in integral condone no du, or d instead of du so ( ) c is M0A0 e.g. correct power of ( ) e.g. ¾ ( ) / seen so ( ) is A0 8 7

2 75 Mark Scheme June 05 let u = ln, dv/d =, du/d = /, v = ¼ u, u, v, v all correct ln d ln. d A ln. [d ] ignore limits ln d dep simplifying / = in second term (soi) dep st ln 6 Acao ln o.e. 6 = ln 5/6 Acao o.e. must be eact, but can isw must evaluate ln = 0 and combine + /6 h = r so V = h / B o.e. e.g h tan 5 / [5] dv/dt = 5 B soi (can be implied from V = 5t) e.g. from a correct chain rule dv/dh = h Bft must be dv/dh soi, ft their h / but must have substituted for r dv/dt = (dv/dh). dh/dt any correct chain rule in V, h and t (soi) e.g. dh/dt = dh/dv dv/dt, 5 = 00 dh/dt dh/dt = 5/00 = 0.06 cm s A 0.06 or better; accept /(0) o.e., but mark final answer or V = 5t so h / = 5t h dh/dt = 5 or 5 dt/dh = h o.e. B dh/dt = 5/h = 5/00 = 0.06 cm s A 0.06 or better; accept /(0) o.e., but mark final answer [5] penalise incorrect rounding penalise incorrect rounding 8

3 75 Mark Scheme June 05 5 y + ln y = + ln = so (, ) lies on the curve. dy dy y ln y.. d y d dy ln y [ ] d y / y when =, y =, = ½ dy ln d B clear evidence of verification needed at least + 0 = A cao Acao 6 (i) arcsin = /6 = sin /6 6 (ii) sin / = cos / = / 7 (i) arcsin (/ ) = arccos (/ ) [6] d/d (y ) = ydy/d d/d ( ln y) = ln y + /y dy/d substituting both = and y = into their dy/d or their equation in, y and dy/d not from wrong working = ½ A allow unsupported answers [] = / B o.e. e.g. /, must be eact; SCB ff( ) f( ) substituting ( )/(+) for in f() * [] A correctly simplified to NB AG f () = f() = ( )/(+) B or just f () = f() must be correct must be correct condone dy/d = unless pursued dy dy ln d d 9

4 75 Mark Scheme June 05 7 (ii) ( ) g( ) ( ) substituting for in g() condone use of f for g g( ) [] if brackets are omitted or misplaced allow A0 A must indicate that g( ) = g() somewhere condone use of f for g Graph is symmetrical about the y-ais. B allow reflected, reflection for symmetrical must state ais (y-ais or = 0) 8 (i).( ) ( ) f'( ) [] A quotient (or product) rule, condone sign errors only correct ep, condone missing brackets here.( ) ( ) e.g. PR: ( - ).( ) + (/).( ) = ( )/ = / * A simplified correctly NB AG with correct use of brackets or f() = ( + )/ = + / A epanding bracket and dividing each term by correctly f () = / * A not from wrong working NB AG f "( ) 8 / B o.e. e.g. 8 or 8/ must be terms: ( )/ is M0 e.g. + / is A0 f() = 0 when =, = ± = ± found from / = 0 allow for = unsupported so at Q, =, y = 8. A (, 8) f ( ) [= ] < 0 so maimum Bdep [7] dep first B. Can omit, but if shown must be correct. Must state < 0 or negative. must use nd derivative test 0

5 75 Mark Scheme June 05 8 (ii) f() = ( ) / = f() = () / = B verifying f() = and f() = ( ) d ( / )d / ln = (8 6 + ln) ( ½ + ln) = ln ½ Acao Area enclosed = rectangle curve soi epanding bracket and dividing each term by terms: / is M0 or ( ) = 5 + = 0 ( )( ) = 0, =, if u = u d u ( u )du u u A / + ln u / u + ln (u + ) = (ln ½) = 7½ ln Acao o.e. but must combine numerical terms and evaluate ln mark final ans or Area = ( ) [ ]d no need to have limits (5 / )d 5 / ln = 0 8 ln (5 ½ ln) = 7½ ln A A Acao [6] epanding bracket and dividing each term by 5 / 5 / ln o.e. but must combine numerical terms and evaluate ln mark final ans 8 (iii) [g() =] f( + ) soi [may not be stated] ( ) * A A [] correctly simplified not from wrong working NB AG must be terms in ( ) epansion

6 75 Mark Scheme June 05 8 (iv) Area is the same as that found in part (ii) award for ± ans to 8(ii) (unless zero) ln 7½ Acao need not justify the change of sign 9 (i) At P, (e ) = 0 e = [±], e = [ or] [] square rooting condone no ± or (e ) e + = 0 epanding to correct quadratic and solve by factorising or using quadratic formula (e )(e ) = 0, e = or 9 (ii) f () = (e )e condone e^^ = [0 or] ln A -coordinate of P is ln ; must be eact condone P = ln, but not y = ln = 0 when e =, = ln * [] A A chain rule correct derivative not from wrong working NB AG e.g. u their deriv of e (e ) is M0 or verified by substitution or f() = e e + epanding to term quadratic with (e ) or e condone e^^ f () = e e A correct derivative, not from wrong working = 0 when e = e, e =, = ln * A or e (e ) = 0 e =, = ln not from wrong working NB AG y = f(ln()) = B [] or verified by substitution

7 75 Mark Scheme June 05 9 (iii) ln ln [(e ) ]d [(e ) e ]d 0 0 ln [e e ]d 0 e e ln = (.5 + ln) (0.5 ) 0 epanding brackets must have terms: (e ) is M0, condone e^^ or if u = e, A e e + [d] (condone no d) = u + /u du B Aft e = ½ e [½ e e + ] = ln [so area = ln] A condone ln as final ans; mark final ans [5] 9 (iv) y = (e ) y = (e y ) + = (e y ) attempt to solve for y (might be indicated by epanding and then taking lns) ± ( + ) = e y (+ for y ln ) A condone no ± + ( + ) = e y y = ln( + ( + )) = f () A must have interchanged and y in final ans Domain is B must be and (not y) Range is y ln B or f () ln, must be (not or f()) if > and y > ln SCB f() = [ ½ u u + ln u] [ u u ] / u du or if and y not interchanged yet or adding (or subtracting) if not specified, assume first ans is domain and second range (,ln) f () A recognisable attempt to reflect curve, or any part of curve, in y = good shape, cross on y = (if shown), correct domain and range indicated. [see etra sheet for eamples] y = shown indicative but not essential e.g. and ln marked on aes (ln, ) [7]

8 75 Mark Scheme June 05 Appendi Annotation notes. All questions on practice and standardisation scripts should be fully annotated, unless they are all correct or worth no credit.. For all other scripts, each part question should be annotated according to the following guidelines: if part marks awarded, annotate fully or: If fully correct, one tick; if worth no credit, one cross, or a yellow line if one mark only is lost, you can indicate this by A0 or B0 in the appropriate place if mark only is earned, you can indicate this by or B in the appropriate place if M0, then you need not annotate dependent A marks. The following questions can be divided into sub-parts, which can be treated as above: 8(i) First marks, last marks 8(ii) First marks, last marks 9(iv) first marks, net marks, last marks NB Please annotate all blank pages with a BP mark (especially AOs), and blank answer spaces with a mark to show they have been seen (e.g. with a tick, a carat or yellow line). Don t forget to scroll down to the bottom of each page, and annotate this if necessary to show this. Also, please indicate you have seen the spare copy of the graph in 9(iv) Please put the annotations close to where they apply, rather than using the margins.

9 75 Mark Scheme June 05 Appendi : Marked eamples of 9(iv) A0 (incorrect domain) A (condone ve gradient for > ) A0 (domain not indicated) A0 A A0 5

10 75 Mark Scheme June 05 A0 boda (condone gradient at (-, ln) A0 A0 A bod A0 (evidence of reflection of part of curve here) 6

Friday 12 June 2015 Morning

Friday 12 June 2015 Morning Oxford Cambridge and RSA Friday 1 June 015 Morning A GCE MATHEMATICS (MEI) 4753/01 Methods for Advanced Mathematics (C3) QUESTION PAPER * 3 0 9 8 4 1 8 * Candidates answer on the Printed Answer Book. OCR