Cambridge International Examinations Cambridge International General Certificate of Secondary Education. Published
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1 Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/ Paper October/November 06 MARK SCHEME Maximum Mark: 80 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 06 series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 6 printed pages. UCLES 06 [Turn over
2 Page Mark Scheme Syllabus Paper Cambridge IGCSE October/November Abbreviations awrt cao dep FT isw oe rot SC soi www answers which round to correct answer only dependent follow through after error ignore subsequent working or equivalent rounded or truncated Special Case seen or implied without wrong working (a) (i) 0 (iii) 4 (b) (i) Q R P Q =, or { } a=, b=, c = B for each y + 5y = 0, for 5y or 5log x, for y=, y = x=, x = x=.44, x = 9, for correct attempt at the solution of their quadratic equation for dealing with one base logarithm correctly for each 4 (i) x x + x B for each term, powers of x must be 9 simplified Coefficients needed: 80 their + their = 48 ( ) for dealing with terms Allow for 7 48x UCLES 06
3 Page Mark Scheme Syllabus Paper Cambridge IGCSE October/November (i) d y = dx ( x+ ) dy When x=, y= 0, = dx Equation of normal: y= x + for correct derivative of log function for y = 0 for attempt at a gradient of a perpendicular from differentiation and the equation of the normal Q 0, 9 or ( ) 0,0. or better R 0, ln or ( 0, 0.5 ) or better Area of PQR = ln + 9 = ft Follow through on their c from part (i) Allow (a) YX, XZ B B for both with no extras for correct with or without extras for both correct with extras B0 for anything else (b) (i) 7 8 4, for 8, for 7 4 C = A - B 7 4 = = 0, for pre-multiplication for any correct pair of elements, but must be from correct matrices UCLES 06
4 Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE October/November ( 0, ) or ( 0,.7 ) or better 7 (i) π, 6 or ( ) 0.54, or better, for each (iii) π cos x = 0 6 π x = oe or.09 or better for correct attempt to solve trigonometric equation (iv) π sin x ( +c) 6 (v) Area = π sin x 6 = + = 8 (i) 47 4 = θ θ =, so θ =.97 θ =.9 to dp CD θ sin = CD = awrt 9.6 or 9.7 π 0 or better (iii) Area of sector = awrt 8 Area of triangle AOB = awrt 67 or 68 Area of segment = awrt 70 or 7 AD AB + segment area = 45 leading to AD = awrt 8. or 8.0 Alternative method: Area of sector = awrt 8 Difference in length between BC (or AD) and OM where M is the midpoint of CD = 6.88, allow awrt 6.9 Remaining area consists of two trapezia each of width 9.85 and each of area 4.4 ( BC 6.88 ) 9.85 = 4.4 oe leading to AD = awrt 8. or 8.0 for correct use of their limits, in radians, π into ksin x. 6 for complete correct method to get θ = must have evidence of working to more than dp, allow if.96 seen (truncated) for a complete method, may use cosine rule to get CD for sector area, allow unsimplified for a correct attempt at area for segment area (their sector area their triangle area) for complete method to find AD Allow for 8 for sector area for attempt to find difference between parallel sides for area of one trapezium ( BC their 6.88 ) their 9.85 oe for attempt to find either BC or AD UCLES 06
5 Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE October/November a 9 b p = a 8 + b = 0 4 leading to 9a+ 4b + 4= 0 oe and 7a+ 4b 48 = 0 oe 9 (i) p : ( = 0) leading to a= 4, b = 5 ( ) for attempt at p for differentiation and attempt at p for solution of simultaneous equations, to get either a or b for both ( )( ) x+ x oe, for attempt at long division or factorisation (iii) ( )( ) x+ x = x + x+ = 0, x = ( x ) = leading to x=, x = Must be using ( x + ) correctly using part to get x = for solution of the quadratic equation 0 (a) (i) U 0U + U + 0 = 65 leading to U = 6 D for realising that area under the graph is needed and attempt to find an area for equating their area to 65 and attempt to solve Gradient of line: 0., for use of the gradient, must be negative (b) (i) 7 t = 8ln4 t =. or better for a correct attempt to solve t e 8 = 4 (iii) acceleration = t t t e 8 e 8 4 8, for a correct attempt to differentiate using the chain rule When t =, a = 6.98, for use of t = in their acceleration UCLES 06
6 Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (i) ln y = ln A+ xln b 0. Gradient: ln b =, = b = Intercept: ln A = 0.6 A =.0 Alternative ln y = ln A+ xln b 0. = 4ln b+ ln A 0.08 = ln b+ ln A A =.0 and b = Alternative 4. = Ab.08 = Ab A =.0 and b = When x= 6, ln y = 0.7 y =.9 (iii) When y =., ln y = x = D D,, may be implied, if equation not seen specifically, by correct values for A and b for use of gradient to obtain ln b Allow for e 0.05 for use of one of the given points correctly Allow for 0.6 e or. for one correct equation for attempt to obtain either lna or lnb from simultaneous equations 0.05 Allow for b = e 0.6 and a = e or. for correct attempt to obtain b or A, must already have B 0.05 Allow for b = e 0.6 and a = e or. for ln y = their ln A + 6 their ln b or y = their A ( their b) 6 allow awrt.8 to.0 for ln. = their ln A + x their ln b or ( theirb). = theira allow 0.5 to.5 x UCLES 06
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