ADVANCED GCE 475/0 MATHEMATICS (MEI) Methods for Advanced Mathematics (C) MONDAY JUNE 008 Additional materials (enclosed): None Additional materials (required): Answer Booklet (8 pages) Graph paper MEI Eamination Formulae and Tables (MF) Morning Time: hour 0 minutes INSTRUCTIONS TO CANDIDATES Write your name in capital letters, your Centre Number and Candidate Number in the spaces provided on the Answer Booklet. Read each question carefully and make sure you know what you have to do before starting your answer. Answer all the questions. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accuracy appropriate to the contet. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 7. You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used. This document consists of 4 printed pages. OCR 008 [M/0/65] OCR is an eempt Charity [Turn over
Section A (6 marks) Solve the inequality. [4] Find e d. [4] (i) State the algebraic condition for the function f() to be an even function. What geometrical property does the graph of an even function have? [] (ii) State whether the following functions are odd, even or neither. (A) f() (B) g() sin + cos (C) h() + [] 4 Show that 4 + d ln 6. [4] 5 Show that the curve y ln has a stationary point when e. [6] 6 In a chemical reaction, the mass m grams of a chemical after t minutes is modelled by the equation m 0 + 0e 0.t. (i) Find the initial mass of the chemical. What is the mass of chemical in the long term? [] (ii) Find the time when the mass is 0 grams. [] (iii) Sketch the graph of m against t. [] 7 Given that + y + y, find dy in terms of and y. [5] d OCR 008 475/0 Jun08
8 Fig. 8 shows the curve y f(), wheref() Section B (6 marks) P is the point on the curve with -coordinate π. + cos,for0 π. y P y f( ) O Fig. 8 (i) Find the y-coordinate of P. [] (ii) Find f (). Hence find the gradient of the curve at the point P. [5] sin (iii) Show that the derivative of + cos is. Hence find the eact area of the region enclosed + cos by the curve y f(),the-ais, the y-ais and the line π. [7] (iv) Show that f () arccos( ). State the domain of this inverse function, and add a sketch of y f () to a copy of Fig. 8. [5] [Question 9 is printed overleaf.] OCR 008 475/0 Jun08
9 The function f() is defined by f() 4 for. 4 (i) Show that the curve y 4 is a semicircle of radius, and eplain why it is not the whole of this circle. [] Fig. 9 shows a point P (a, b) on the semicircle. The tangent at P is shown. y P ( ab, ) O Fig. 9 (ii) (A) Use the gradient of OP to find the gradient of the tangent at P in terms of a and b. (B) Differentiate 4 anddeducethevalueoff (a). (C) Show that your answers to parts (A) and(b) are equivalent. [6] The function g() is defined by g() f( ), for0 4. (iii) Describe a sequence of two transformations that would map the curve y f() onto the curve y g(). Hence sketch the curve y g(). [6] (iv) Show that if y g() then 9 + y 6. [] Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. OCR is part of the Cambridge Assessment Group. 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 008 475/0 Jun08
475 Mark Scheme June 008 475 (C) Methods for Advanced Mathematics Section A 4 or ( ) 9 4 4 8 0 (4)( + )( ) 0 Let u, dv/d e v e / e d e e.. d e e + c 9 (i) f( ) f() Symmetrical about Oy. [4] [4] [] ( or ) ( or ) squaring and forming quadratic 0 (or ) factorising or solving to get, (www) parts with u, dv/d e v e e 9 +c (ii) (A) even (B) neither (C) odd [] 4 Let u + du d 4 8/ d du + u [ ln u ] 8 ½ (ln8 ln ) ½ ln(8/) ½ ln 6* 5 y ln dy. + ln d + ln dy/d 0 when +ln 0 ( + ln ) 0 ln ½ e ½ / e * E [4] E [6] / du or k ln ( + ) u ½ ln u or ½ ln( + ) substituting correct limits (u or ) must show working for ln 6 product rule d/d (ln ) / soi oe their deriv 0 or attempt to verify ln ½ e ½ or ln (/ e) ½ 8
475 Mark Scheme June 008 6(i) Initial mass 0 + 0 e 0 50 grams Long term mass 0 grams [] (ii) 0 0 + 0 e 0.t e 0.t / 0.t ln (/).0986 t.0 mins [] anti-logging correctly,.0, 0.99, 0.986 (not more than d.p) (iii) m 50 0 t [] correct shape through (0, 50) ignore negative values of t 0 as t 7 + y + y dy dy + y y 0 d + + d dy ( + y) y d dy + y d ( + y) [5] Implicit differentiation dy y d + correct equation collecting terms in dy/d and factorising oe cao 9
475 Mark Scheme June 008 Section B 8(i) y /(+cosπ/) /. [] or 0.67 or better (ii) f'( ) (+ cos ). sin sin (+ cos ) When π/, sin( π / ) f'( ) (+ cos( π / )) / 4 ( ) 9 9 [5] chain rule or quotient rule d/d (cos ) sin soi correct epression substituting π/ oe or 0.8 or better. (0.85, 0.849) (iii) deriv (+ cos ) cos sin.( sin ) (+ cos ) cos + cos + sin (+ cos ) cos + (+ cos ) * + cos π / Area d 0 + cos π / sin + cos 0 sin π / ( 0) + cos π / dep E cao [7] Quotient or product rule condone uv u v for correct epression cos + sin used dep www substituting limits or / - must be eact (iv) y /( + cos ) y /( + cos y) + cos y / cos y / y arccos(/ ) * Domain is ½ E attempt to invert equation www reasonable reflection in y [5] 0
475 Mark Scheme June 008 9 (i) y 4 y 4 + y 4 which is equation of a circle centre O radius Square root does not give negative values, so this is only a semi-circle. [] squaring + y 4 + comment (correct) oe, e.g. f is a function and therefore single valued (ii) (A) Grad of OP b/a grad of tangent a b f( ) (4 ).( ) 4 f'( a) a 4 a (C) b (4 a ) so a f'( a ) as before b (B) / E [6] chain rule or implicit differentiation oe substituting a into their f () (iii) Translation through followed by 0 stretch scale factor in y-direction 6 4 [6] Translation in -direction through or to right ( shift, move A0) 0 alone is SC 0 stretch in y-direction (condone y ais ) (scale) factor elliptical (or circular) shape through (0, 0) and (4, 0) and (, 6) (soi) if whole ellipse shown (iv) y f( ) (4 ( ) ) (4 + 4 4) (4 ) y 9(4 ) 9 + y 6 * E [] or substituting (4 ( ) ) oe for y in 9 + y 4 www