ADVANCED GCE 4753/ MATHEMATICS (MEI) Methods for Advanced Mathematics (C3) FRIDAY JANUARY 8 Additional materials: Answer Booklet (8 pages) Graph paper MEI Eamination Formlae and Tables (MF) Morning Time: hor 3 mintes INSTRUCTIONS TO CANDIDATES Write yor name in capital letters, yor Centre Nmber and Candidate Nmber in the spaces provided on the Answer Booklet. Read each qestion careflly and make sre yo know what yo have to do before starting yor answer. Answer all the qestions. Yo are permitted to se a graphical calclator in this paper. Final answers shold be given to a degree of accracy appropriate to the contet. INFORMATION FOR CANDIDATES The nmber of marks is given in brackets [ ] at the end of each qestion or part qestion. The total nmber of marks for this paper is 7. Yo are advised that an answer may receive no marks nless yo show sfficient detail of the working to indicate that a correct method is being sed. This docment consists of 4 printed pages. OCR 8 [M//65] OCR is an eempt Charity [Trn over
Section A (36 marks) Differentiate 3 + 6. The fnctions f() and g() are defined for all real nmbers by f() =, g() =. (i) Find thecompositefnctionsfg() and gf(). [3] (ii) Sketch the crves y = f(), y = fg() and y = gf(), indicating clearly which is which. [] 3 The profit P made by a company in its nth year is modelled by the eponential fnction P = Ae bn. In the first year (when n = ), the profit was. In the second year, the profit was 6. (i) Show that e b =.6, and find b and A. [6] (ii) What does this model predict the profit to be in the th year? [] 4 When the gas in a balloon is kept at a constant temperatre, the pressre P in atmospheres and the volme V m 3 are related by the eqation P = k V, where k is a constant. [This is known as Boyle s Law.] When the volme is m 3, the pressre is 5 atmospheres, and the volme is increasing at a rate of m 3 per second. (i) Show that k = 5. [] (ii) Find dp dv in terms of V. [] (iii) Find the rate at which the pressre is decreasing when V =. 5 (i) Verify the following statement: p is a prime nmber for all prime nmbers p less than. [] (ii) Calclate 3 89, and hence disprove this statement: p is a prime nmber for all prime nmbers p. [] OCR 8 4753/ Jan8
3 6 Fig. 6 shows the crve e y = + y. y O P Fig. 6 (i) Show that dy d = e y. (ii) Hence find to 3 significant figres the coordinates of the point P, shown in Fig. 6, where the crve has infinite gradient. Section B (36 marks) 7 A crve is defined by the eqation y = ln( + ). (i) Find dy and hence verify that the origin is a stationary point of the crve. d (ii) Find d y, and se this to verify that the origin is a minimm point. [5] d (iii) Using the sbstittion = +, showthat + d = ( + ) d. Hence evalate d,givingyoranswerinaneactform. [6] + (iv) Usingintegrationbypartsandyoranswertopart(iii), evalate ln( + ) d. OCR 8 4753/ Jan8 [Trn over
8 Fig. 8 shows the crve y = f(), wheref() = + sin for 4 π 4 π. 4 y 4 O 4 Fig. 8 (i) State a seqence of two transformations that wold map part of the crve y = sin onto the crve y = f(). (ii) Find the area of the region enclosed by the crve y = f(),the-ais and the line = π. 4 (iii) Find the gradient of the crve y = f() at the point (, ). Hence write down the gradient of the crve y = f () at the point (, ). (iv) State the domain of f (). Addasketchofy = f () to a copy of Fig. 8. [3] (v) Find an epression for f (). [] Permission to reprodce items where third-party owned material protected by copyright is inclded has been soght and cleared where possible. Every reasonable effort has been made by the pblisher (OCR) to trace copyright holders, bt if any items reqiring clearance have nwittingly been inclded, the pblisher will be pleased to make amends at the earliest possible opportnity. OCR is part of the Cambridge Assessment Grop. Cambridge Assessment is the brand name of University of Cambridge Local Eaminations Syndicate (UCLES), which is itself a department of the University of Cambridge. OCR 8 4753/ Jan8
4753 Mark Scheme Janary 8 4753 (C3) Methods for Advanced Mathematics Section A /3 y= (+ 6 ) dy ( 6 ) = +. d 3 /3 = 4 (+ 6 ) /3 (i) fg() = f( ) = ( ) gf() = g( ) =. [3] chain rle sed /3 3 cao (mst resolve /3 ) Mark final answer forming a composite fnction mark final answer If fg and gf the wrong way rond, AA (ii) - gf() fg() ft ft [] fg mst have (, )labelled (or inferable from scale). Condone no y-intercept, nless wrong gf mst have (, ) labelled (or inferable from scale) Condone no -intercepts, nless wrong Allow ft only if fg and gf are correct bt wrong way rond. 3 (i) When n =, = A e b when n =, 6 = A e b b 6 Ae b = = e b Ae e b =.6 b = ln.6 =.47 A = /.6 = 65. (ii) When n =, P = 65 e.47 = 75,55, E [6] [] soi soi eliminating A (do not allow verification) SCB if initial B s are missing, and ratio of years =.6 = e b ln.6 or.47 or better (mark final answer) cao allow recovery from ineact b s sbstitting n = into their eqation with their A and b Allow answers from 75 to 76. 4 (i) 5 = k/ k = 5* E [] (ii) dp 5 = 5V = dv V [] NB answer given ( )V o.e. allow k/v (iii) dp dp dv =. dt dv dt When V =, dp/dv = 5/ =.5 dv/dt = dp/dt =.5 =.5 So P is decreasing at.5 Atm/s ft chain rle (any correct version) (soi) (soi).5 cao 7
4753 Mark Scheme Janary 8 5(i) p =, p = 3, prime p = 3, p = 7, prime p = 5, p = 3, prime p = 7, p = 7, prime E [] Testing at least one prime testing all 4 primes (correctly) Mst comment on answers being prime (allow ticks) Testing p = is E (ii) 3 89 = 47 = is prime, 47 is not So statement is false. E [] mst state or imply that is prime (p = is sfficient) 6 (i) e y = + y y dy dy e = + d d y dy (e ) = d dy = * y d e E Implicit differentiation allow one slip (bt with dy/d both sides) collecting terms (ii) Gradient is infinite when e y = e y = ½ y = ln ½ y = ½ ln ½ =.347 (3 s.f.) = e y y = ½ (.347) =.8465 =.9 mst be to 3 s.f. sbstitting their y and solving for cao mst be to 3 s.f., bt penalise accracy once only. 8
4753 Mark Scheme Janary 8 Section B 7(i) y = ln( + ) dy = + ln( + ) d + When =, dy/d = + ln = origin is a stationary point. E prodct rle d/d(ln(+)) = /(+) soi www (i.e. from correct derivative) (ii) d y ( + ).. = + d ( + ) + = + ( + ) + When =, d y/d = + = 4 > (, ) is a min point ft E [5] Qotient or prodct rle on their /( + ) correctly applied to their /(+) o.e., e.g. 4+ cao ( + ) sbstitting = into their d y/d www dep previos (iii) Let = + d = d ( ) d = d + = ( + ) d = ( + ) d * d = ( + ) d + ln = + E ( ) www (bt condone d omitted ecept in final answer) changing limits (or sbstitting back for and sing and ) + ln = 4 + ln ( ½ + ln ) = ln ½ [6] sbstitting limits (consistent with or ) cao (iv) A= ln( + ) d Parts: = ln( + ), d/d = /( + ) dv/d = v = = ln( + ) d + = ln ln + ½ = ½ soi sbstitting their ln ½ for cao d + 9
4753 Mark Scheme Janary 8 8 (i) Stretch in -direction s.f. ½ translation in y-direction nit p (in either order) allow contraction dep stretch allow move, shift, etc direction can be inferred from or dep translation. alone is A (ii) π /4 A = ( + sin d ) π /4 π /4 = cos π /4 = π/4 ½ cos π/ + π/4 + ½ cos ( π/) = π/ correct integral and limits. Condone d missing; limits may be implied from sbseqent working. sbstitting their limits (if zero lower limit sed, mst show evidence of sbstittion) or.57 or better cao (www) (iii) y = + sin dy/d = cos When =, dy/d = So gradient at (, ) on f() is gradient at (, ) on f () = ½ ft ft differentiating allow error (bt not + cos ) If, then mst show evidence of sing reciprocal, e.g. / (iv) Domain is. y Allow to, bt not < < or y instead of clear attempt to reflect in y = correct domain indicated ( to ), and reasonable shape π/4 O π/4 [3] (v) y = + sin y = + sin y sin y = y = arcsin( ) y = ½ arcsin( ) [] or sin = y cao