* * MATHEMATICS (MEI) 4753/01 Methods for Advanced Mathematics (C3) ADVANCED GCE. Friday 5 June 2009 Afternoon. Duration: 1 hour 30 minutes.

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1 ADVANCED GCE MATHEMATICS (MEI) 4753/0 Methods for Advanced Mathematics (C3) Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet Graph paper MEI Examination Formulae and Tables (MF) Other Materials Required: None Frida 5 June 009 Afternoon Duration: hour 30 minutes * * * * INSTRUCTIONS TO CANDIDATES Write our name clearl in capital letters, our Centre Number and Candidate Number in the spaces provided on the Answer Booklet. Use black ink. Pencil ma be used for graphs and diagrams onl. Read each question carefull and make sure that ou know what ou have to do before starting our answer. Answer all the questions. Do not write in the bar codes. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accurac appropriate to the context. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. You are advised that an answer ma receive no marks unless ou show sufficient detail of the working to indicate that a correct method is being used. The total number of marks for this paper is 7. This document consists of 4 pages. An blank pages are indicated. OCR 009 [M/0/65] OCR is an exempt Charit R 9A07 Turn over

2 Section A (36 marks) Evaluate 6 π 0 sin 3x. A radioactive substance decas exponentiall, so that its mass M grams can be modelled b the equation M Ae kt, where t is the time in ears, and A and k are positive constants. (i) An initial mass of 00 grams of the substance decas to 50 grams in 500 ears. Find A and k. [5] (ii) The substance becomes safe when 99% of its initial mass has decaed. Find how long it will take before the substance becomes safe. 3 Sketch the curve arccos x for x. 4 Fig. 4 shows a sketch of the graph of x. It meets the x- and -axes at (a, 0) and (0, b) respectivel. b O a Fig. 4 x Find the values of a and b. 5 The equation of a curve is given b e + sin x. (i) B differentiating implicitl, find d in terms of x and. (ii) Find an expression for in terms of x, and differentiate it to verif the result in part (i). [4] 6 Given that f(x) x +, show that ff(x) x. x Hence write down the inverse function f (x). What can ou deduce about the smmetr of the curve f(x)? [5] OCR /0 Jun09

3 3 7 (i) Show that (A) (x )(x + x + ) x 3 3, (B) (x + ) x + x +. [4] (ii) Hence prove that, for all real numbers x and, if x > then x 3 > 3. Section B (36 marks) 8 Fig. 8 shows the line x and parts of the curves f(x) and g(x), where f(x) e x, g(x) + ln x. The curves intersect the axes at the points A and B, as shown. The curves and the line x meet at the point C. x C f( x) B O A x g( x) Fig. 8 (i) Find the exact coordinates of A and B. Verif that the coordinates of C are (, ). [5] (ii) Prove algebraicall that g(x) is the inverse of f(x). [] (iii) Evaluate f(x), giving our answer in terms of e. 0 (iv) Use integration b parts to find ln x. Hence show that g(x) e e. [6] (v) Find the area of the region enclosed b the lines OA and OB, and the arcs AC and BC. [] OCR /0 Jun09 Turn over

4 4 9 Fig. 9 shows the curve x 3x. P is a turning point, and the curve has a vertical asmptote x a. P x a O x Fig. 9 (i) Write down the value of a. (ii) Show that d x(3x ) (3x ). [] (iii) Find the exact coordinates of the turning point P. Calculate the gradient of the curve when x 0.6 and x 0.8, and hence verif that P is a minimum point. [7] (iv) Using the substitution u 3x, show that x 3x 7 (u + + u ) du. Hence find the exact area of the region enclosed b the curve, the x-axis and the lines x 3 and x. [7] Copright Information OCR is committed to seeking permission to reproduce all third-part content that it uses in its assessment materials. OCR has attempted to identif and contact all copright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copright acknowledgements are reproduced in the OCR Copright Acknowledgements Booklet. This is produced for each series of examinations, is given to all schools that receive assessment material and is freel available to download from our public website ( after the live examination series. If OCR has unwittingl failed to correctl acknowledge or clear an third-part content in this assessment material, OCR will be happ to correct its mistake at the earliest possible opportunit. For queries or further information please contact the Copright Team, First Floor, 9 Hills Road, Cambridge CB PB. OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of Universit of Cambridge Local Examinations Sndicate (UCLES), which is itself a department of the Universit of Cambridge. OCR /0 Jun09

5 4753 Mark Scheme June (C3) Methods for Advanced Mathematics Section A π 6 π /6 sin 3x d x cos3x π cos + cos B Acao cos3 x or 3 cos u 3 substituting correct limits in ±k cos 0.33 or better. (i) 00 Ae 0 A A e 500k e 500k k ln 0.5 k ln 0.5 / (ii) 00e kt kt ln 0.0 t ln 0.0 /k 9966 ears A A [5] A 50 A e 500k ft their A if used taking lns correctl or better ft their A and k taking lns correctl art π Can use degrees or radians reasonable shape (condone extra range) B passes through (, π), (0, π) and (, 0) A good sketches look for curve reasonabl vertical at (, π) and (, 0), negative gradient at (0, π). Domain and range must be clearl marked and correct. 4 g(x) x b 0 or (0, ) x 0 x, so a or (, 0) B A Allow unsupported answers. www x is A0 www 9

6 4753 Mark Scheme June 009 5(i) e + sin x e d/ cos x d/ cos x e (ii) ln( + sin x) ½ ln( + sin x) d/ cos x + sinx cos x as before e B A B B E [4] Their e d/ e o.e. cao chain rule (can be within correct quotient rule with dv/ 0) /u or /( + sin x) soi www 6 x + + ff( x) x x + x x + + x x + x + x/ x* E correct expression without subsidiar denominators x+ + x x e.g. x x+ x+ f (x) f(x) B stated, or shown b inverting Smmetrical about x. B [5] 7(i) (A) (x )(x + x + ) x 3 + x + x x x 3 x 3 3 * (B) (x + ½ ) + ¾ x + x + ¼ + ¾ x + x + E E [4] expanding - allow tabulation www (x + ½ ) x + ½ x + ½ x + ¼ o.e. cao www (ii) x 3 3 (x )[(x + ½ ) + ¾ ] (x + ½ ) + ¾ > 0 [as squares 0] if x > 0 then x 3 3 > 0 if x > then x 3 > 3 * E substituting results of (i) 0

7 4753 Mark Scheme June 009 8(i) A: + ln x 0 ln x so A is (e, 0) x e B: x 0, e 0 e so B is (0, e ) C: f() e e 0 g() + ln (ii) Either b invertion: e.g. e x x x e ln x + ln x or b composing e.g. f g(x) f( + ln x) e + ln x e ln x x (iii) x x e e 0 0 e 0 e e d (iv) ln x ln x ( x) x ln x x. x x ln x x + c g( x) ( + ln x) e e [ x + xln x x] e [ x x ] (v) Area ln ) e ln e ln(e ) e * f( x) g( x) 0 e ( e ) e /e A B E E [5] E E [] Acao A Acao Bft D E [6] Acao SC if obtained using smmetr condone use of smmetr Penalise A e, B e, or co-ords wrong wa round, but condone labelling errors. taking lns or exps e + ln x or + ln(e x ) x e o.e or u x u e substituting correct limits for x or u o.e. not e 0, must be exact. parts: u ln x, du/ /x, v x, dv/ condone no c ft their x ln x x (provided algebraic ) substituting limits dep B www Must have correct limits 0.64 or better. or or Area OCB area under curve triangle e ½ ½ e Area OAC triangle area under curve ½ e ½ e Total area (½ e ) /e Acao [] OCA or OCB ½ e 0.64 or better

8 4753 Mark Scheme June 009 9(i) a /3 B [] or 0.33 or better (ii) d (3x ) x x.3 d x (3x ) 6x x 3x (3x ) 3x x (3x ) x(3x ) * (3x ) A E quotient rule www must show both steps; penalise missing brackets. (iii) d/ 0 when x(3x ) 0 x 0 or x /3, so at P, x /3 (/3) 4 when x, 3 3 (/3) 9 when x 0.6, d/ when x 0.8, d/ Gradient increasing minimum A Acao B B E [7] if denom 0 also then M0. o.e e.g. 0.6, but must be exact o.e e.g.. 0.4, but must be exact 3/6, or 0.9 or better 8/49 or 0.6 or better o.e. e.g. from negative to positive. Allow ft on their gradients, provided ve and +ve respectivel. Accept table with indications of signs of gradient. (iv) x u 3x du 3 3x ( u + ) 9 du u 3 ( u+ ) u + u+ du du 7 u 7 u ( u+ + ) du * 7 u B E ( u + ) 9 u o.e. /3 (du) expanding Condone missing du s Area x /33x When x /3, u, when x, u ( u / u) du ln 7 u u u + + [( ln ) ( + + ln)] 7 (3 + ln ) [ 7+ ln ] 7 54 B Acao [7] ln u + u + u substituting correct limits, dep integration o.e., but must evaluate ln 0 and collect terms.

9 Report on the Units taken in June Methods for Advanced Mathematics (C3) (Written Examination) General Comments This paper proved to be accessible to all suitabl prepared candidates, and there were plent of marks available to even the weakest candidates. However, there were some questions, such as 3, 6 and 7(ii) which tested the abler candidates. Fewer full marks than usual were scored the final part of question 7 proved the main stumbling block but, equall, virtuall all candidates scored above 0. There was no evidence of lack of time to complete the paper. With reference to recent examiner s reports, it was pleasing to note that fewer candidates were using graph paper for their sketch in question 3. In general, some candidates seem to be insufficientl aware of the significance of words such as verif (see question 8(ii)), and hence : most candidates missed the significance of this in questions 6 and 7(ii). Candidates also need to be clear what is meant b exact answers, or else the will lose marks in this paper. In general, the calculus topics continue to be well answered, albeit with some slopp notation used in integration, with modulus, proof and inverse trigonometric functions being less securel understood. The standard of presentation varied from chaotic to exemplar. Comments on Individual Questions Section A ) This should have been a routine test of trigonometric integration. However, man candidates confuse differentiation and integration results, using a multiplier of 3 rather than /3, and making sign errors. There were also significant numbers of evaluation errors, such as 0 for the lower limit. ) This question was ver well done exponential growth and deca questions are well understood b the large majorit of candidates. The main sources of error la in the use of 99 instead of in part (ii), and inaccurac in the final answer through premature rounding of the value of k. 3) Sketching this arccos graph proved to be quite testing perhaps more so than arcsin - and few candidates scored all three marks. The first was given for a reasonable attempt to reflect a cosine graph in x. Quite a few candidates scored the B for showing an graph through (, 0), (0, π) and (, π), even if it was a straight line! For the final A, we wanted to see the correct domain and range, and reasonabl correct gradients at x, 0 and. The use of degrees instead of radians was allowed, as no calculus was involved. 9

10 Report on the Units taken in June 009 4) Although good candidates did this effortlessl, there was a degree of confusion amongst some candidates over handling the modulus. Man think taking a modulus means ou have to multipl b -, so onl did this. The given diagram made most candidates aware that the signs of a and b were both positive, but there were man incorrect or dubious statements such as -, but is positive, so b ( coming from use of x when x 0). Another common mistake was to obtain a ½ from the incorrect x 0. Squaring is another rather dubious technique when applied to a given modulus graph, even more so if the is left un-squared, giving statements like (x ) 0 (true, but from wrong working). 5) (i) Most candidates made a reasonable attempt to differentiate implicitl, but some common errors were (a) starting d/.., (b) omitting the when differentiating e d/, (c) RHS + cos x, (d) LHS e. (ii) The most frequent mistake was in the mishandling of the inversion, with ln + ln sin x appearing frequentl. Even when the correct expression for was found, a surprising number needlessl used the product or quotient rules with u ½ and v ln( + sin x), and/or omitted the derivative of sin x in differentiating the latter. Some lost the last mark b not showing clearl that their results in (i) and (ii) were equivalent. 6) This question scored poorl. Although some confused composition with squaring, most candidates managed a correct expression for ff(x) b substituting (x+)/(x ) for x in f(x); however, man of these then failed to deal with the subsidiar denominators, and to correctl simplif the expression to x. Virtuall all candidates then tried to invert (x+)/(x ) to find f (x), rather than simpl writing down that f (x) f(x). Also, f (x) (x )/(x+) was quite a common error. There was some confusion in the final B between the smmetr of f(x) in x, and the fact that f(x) and f (x) are smmetrical in this line. Some candidates thought this last question referred to odd and even functions. 7) (i) This algebra proved to be an eas 4 marks for all candidates, give or take a few slips due to carelessness. (ii) On the other hand, the logic of this part eluded all but the ver best candidates. Man substituted values, or tried other letters, or x +, etc. Some recognised that x + x + had to be proved to be positive, but failed to see the connection between this and (i)(b). 0

11 Report on the Units taken in June 009 Section B 8) Plent of candidates scored well on this question, seeing the links between the various parts. (i) (ii) (iii) Most candidates got the correct coordinates for A and B, although man approximated for e we tried as much as possible to condone this b ignoring subsequent working. It is important that candidates knew wh verif was used in finding C quite a few tried to solve + lnx e x. Inverting functions is usuall well understood, and these two marks were obtained b all but a few candidates. Quite a few candidates substituted u x here, and others left the answer as e 0 e (which though arguabl in terms of e, was not what was intended), or evaluated e 0 as 0. Some candidates made errors in integrating e x, e.g. (e x )/(x ). (iv) The classic blunder here is to take u ln (whatever this means) and v x! Fortunatel, this occurred rarel, and most candidates succeeded in using the correct parts. Quoting this result was not allowed, however, as the were asked to use parts to derive it. We generousl followed through their answer to this to integrate g(x). A rather disconcerting number of candidates even good ones failed to simplif [x + xlnx x] as xlnx, which made the substitution and derivation of /e a bit more complicated than necessar. (v) This part was more demanding, but a few recovered to achieve the correct answer as the area of the square minus twice the given answer in part (iv). 9) This question was a routine test of calculus which offered plent of accessible marks. (i) (ii) (iii) This was an eas mark for all. As this was a routine application of the quotient rule, we withheld the final E mark if the bracket round (3x ) was omitted. This was not common, however, and most candidates scored 3 eas marks. The coordinates of the turning point were usuall obtained correctl from the given derivative, although a few burned their boats b using 3x 0 or numerator denominator. The gradients at 0.6 and 0.8 were also well done, and most then explained how this related to P being a minimum point. However, some candidates wasted time and effort in finding the second derivative, usuall incorrectl. We allowed this, if full correct.

12 Report on the Units taken in June 009 (iv) This part proved to be a bit more demanding, and few candidates scored all 7 marks. Errors such as x (u )/3 or x (u/3 + ) were found, though most achieved the for substituting du/3 for. We condoned missing du s and s in this instance, though this is not alwas the case. Leibnitz would not recognise his notation in some solutions! The integral was quite often incorrect usuall missing the ln u and quite a few got the limits for u incorrect, or used the x- limits of /3 and instead.