ADVANCED GCE MATHEMATICS (MEI) 475/0 Methods for Advanced Mathematics (C) Candidates answer on the Answer Booklet OCR Supplied Materials: 8 page Answer Booklet Graph paper MEI Eamination Formulae and Tables (MF) Other Materials Required: None Thursday 5 January 009 Morning Duration: hour 0 minutes * * 4 4 7 7 5 5 0 0 * * INSTRUCTIONS TO CANDIDATES Write your name clearly in capital letters, your Centre Number and Candidate Number in the spaces provided on the Answer Booklet. Use black ink. Pencil may be used for graphs and diagrams only. Read each question carefully and make sure that you know what you have to do before starting your 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 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. 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. The total number of marks for this paper is 7. This document consists of 4 pages. Any blank pages are indicated. OCR 009 [M/0/65] OCR is an eempt Charity R 8H6 Turn over
Section A (6 marks) Solve the inequality <. (i) Differentiate cos with respect to. (ii) Integrate cos with respect to. Given that f() = ln( ) and g() = + e, show that g() is the inverse of f(). 4 Find the eact value of 0 + 4, showing your working. [5] 5 (i) State the period of the function f() = + cos, where is in degrees. [] (ii) State a sequence of two geometrical transformations which maps the curve y = cos onto the curve y = f(). (iii) Sketch the graph of y = f() for 80 < < 80. 6 (i) Disprove the following statement. If p > q, then p <. [] q (ii) State a condition on p and q so that the statement is true. [] 7 The variables and y satisfy the equation + y = 5. (i) Show that dy = (y ). Both and y are functions of t. (ii) Find the value of dy dt when =, y = 8 and dt = 6. OCR 009 475/0 Jan09
Section B (6 marks) 8 Fig. 8 shows the curve y = ln. P is the point on this curve with -coordinate, and R is the 8 point (0, 7 8 ). y P O Q R 7 (0, 8 ( Fig. 8 (i) Find the gradient of PR. (ii) Find dy. Hence show that PR is a tangent to the curve. (iii) Find the eact coordinates of the turning point Q. [5] (iv) Differentiate ln. Hence, or otherwise, show that the area of the region enclosed by the curve y = ln, the 8 -ais and the lines = and = is 59 4 ln. [7] 4 [Question 9 is printed overleaf.] OCR 009 475/0 Jan09 Turn over
9 Fig. 9 shows the curve y = f(), where f() = The curve has asymptotes = 0 and = a. 4. y O Fig. 9 a (i) Find a. Hence write down the domain of the function. (ii) Show that dy = ( ). Hence find the coordinates of the turning point of the curve, and write down the range of the function. [8] The function g() is defined by g() =. (iii) (A) Show algebraically that g() is an even function. (B) Show that g( ) = f(). (C) Hence prove that the curve y = f() is symmetrical, and state its line of symmetry. [7] 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 009 475/0 Jan09
475 Mark Scheme January 009 475 (C) Methods for Advanced Mathematics Section A < < < < < 4 (i) y= cos dy sin cos = + B B or = ±, or squaring correct quadratic ( + )( 4) (condone factorising errors) or correct sketch showing y = to scale < < 4 (penalise once only) product rule d/ (cos) = sin oe cao (ii) d cos = ( sin ) = sin sin = sin + cos + c 4 Either y = ln( ) y = ln( y ) = ln (y ) e = y + e = y g() = + e or gf() = g( ½ ln ( )) ln( ) = + e = + = 4 0 + 4 letu = + 4, du = 4 9 = / u. du 4 9 = / 6 u = 7 6 = = or 4 6 6 6 or d / / / (+ 4 ) = 4. (+ 4 ) = 6(+ 4 ) ( 4 ) / + = / 0 ( + 4 ) 6 0 = 7 6 = = or 4 6 6 6 ft E E B cao cao [5] parts with u =, v = ½ sin + cos 4 cao must have + c ( )/ or y = e attempt to invert and interchanging with y o.e. (at any stage) e ln y = y or ln (e y ) = y used www or fg() = (correct way round) e ln( ) = or ln(e ) = www u = + 4 and du/ = 4 or du = 4 u. du 4 / / u u du = soi / substituting correct limits (u or ) dep attempt to integrate / k ( + 4) / / / (+ 4 ) = (+ 4 )... ¼ substituting limits (dep attempt to integrate) 9
475 Mark Scheme January 009 5(i) period 80 (ii) one-way stretch in -direction scale factor ½ translation in y-direction through 0 B [] condone 0 80 or π [either way round ] condone squeeze, contract for stretch used and s.f ½ condone move, shift, etc for translation used, + unit 0 only is A0 (iii) correct shape, touching -ais at -90, 90 B correct domain (0, ) marked or indicated (i.e. amplitude is ) 80 80 6(i) e.g p = and q = p > q but /p = > /q = ½ E [] stating values of p, q with p 0 and q 0 (but not p = q = 0) showing that /p > /q - if 0 used, must state that /0 is undefined or infinite (ii) Both p and q positive (or negative) B [] or q > 0, positive integers 7(i) / / dy + y = 0 / dy = / y / y y = = / * E Implicit differentiation (must show = 0) solving for dy/ www. Must show, or eplain, one more step. (ii) dy dy =. dt dt 8 =.6 = cao any correct form of chain rule 0
475 Mark Scheme January 009 8(i) When = y = (ln )/8 = B Gradient of PR = ( + 7/8)/ = 7 8.9 or better (ii) dy = 8 When =, dy/ = /8 = 7 8 Same as gradient of PR, so PR touches curve B Bdep E cao.9 or better dep st B dep gradients eact (iii) Turning points when dy/ = 0 = 0 8 = 8 = /6 = ¼ ( > 0) When = ¼, y = ln = + ln 4 6 8 4 6 8 So TP is (, + ln4) 4 6 8 cao [5] setting their derivative to zero multiplying through by allow verification substituting for in y o.e. but must be eact, not /4. Mark final answer. (iv) d ( ln ) =. +.ln = ln product rule ln Area = ( ln ) 8 = ( ln ) 8 = 8 ( ln + ) ( ln + ) 4 4 8 8 correct integral and limits (soi) condone no ln = ln used (or derived using integration by parts) ( ln ) bracket required 8 substituting correct limits = 7 + ln 8 4 = 59 ln * 4 4 E [7] must show at least one step
475 Mark Scheme January 009 9(i) Asymptotes when ( )( ) = 0 ( ) = 0 = 0 or so a = Domain is 0 < < Bft or by verification > 0 and <, not (ii) y = ( ) / let u =, y = u / dy/du = ½ u /, du/ = dy dy du / =. = ( ).( ) du = * / ( ) dy/ = 0 when = 0 =, y = / ( ) = Range is y B E B Bft [8] chain rule (or within correct quotient rule) ½ u / or ( ) / or ( / ) in quotient rule ( ) www penalise missing brackets here etraneous solutions M0 (iii) (A) g( ) = = = g( ) ( ) (B) g( ) = ( ) = = = f( ) + E E Epression for g( ) must have g( ) = g() seen must epand bracket (C) f() is g() translated unit to the right. But g() is symmetrical about Oy So f() is symmetrical about =. dep both s or f( ) = g( ), f( + ) = g() f( + ) = f( ) f() is symmetrical about =. E [7] or f( ) = = = ( ) + f( + ) = = = + ( + ) +