4037 ADDITIONAL MATHEMATICS
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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge Ordinary Level MARK SCHEME f the October/November 05 series 07 ADDITIONAL MATHEMATICS 07/, maximum raw mark 80 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 befe 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 Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the October/November 05 series f 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.
2 Page Cambridge O Level October/November Abbreviations Awrt Cao dep FT isw oe rot SC soi www answers which round to crect answer only dependent follow through after err igne subsequent wking equivalent rounded truncated Special Case seen implied without wrong wking i A B C ii A B C iii A B C cos x = ± oe x division by and square root =,, + oe x = +, +, x = 0 and x= 0 and D crect der of operations in der to obtain a solution A//0 A f solutions and no extras in the range f solutions A0 f one solution no solutions Cambridge International Examinations 05
3 Page 07 a B,,0 B f 6 elements crect, f 5 elements crect b = m + n B,,0 B f crect elements in X f crect elements in X = 6m 8 = m giving m = F m = using crect I 8 = m + n 76 = 8m + n n = complete method to obtain n c Cambridge O Level October/November 05 i a 6 = 0 so a = ± 6 B,,0 7 + BC = 7 BC = + BC = B f a = ± 6 a = ±.5, with no increct statements seen f a = ± 6 a = ±.5 seen f a = 6 and no increct wking crect use of the area crect rationalisation Dependent on all method being seen Alternative method 7 + BC = + a + b = 7 ii Leading to a + b = 7 and a + b = 0 Solution of simultaneous equations BC = + + Dependent on all method seen including solution of simultaneous equations = FT 6 crect FT terms seen AC = 98 AC = 7 p = 7 cao 98 and 7 98 and p = 7 Cambridge International Examinations 05
4 Page Cambridge O Level October/November 05 When x = 5, y= = 5sec x = 0 When x =, Equation of nmal y = 0 y + x 0 6 x 0 = 0 0 y + x 0.8 = 0 oe 07 y= 5 sec x 0 from differentiation y their = allow unsimplified shape intercepts on x-axis intercept on y-axis f a curve with a maximum and two arms x their0 i ii iii, 6 k =0 k > 6, ±6 seen, k where k > 0, 6 x = and y = 6 only Cambridge International Examinations 05
5 Page 5 Cambridge O Level October/November 05 = sin x +c = +c 7 = sin x + y = cosx + x +d = cos + + d 9 y = cos x + x 8 a sin x finding constant using = k sin x + c making use of = and x = 9 Allow with c = FT FT integration of their k sin x finding constant d f k cos x + cx + d Allow y = cos x x better + kx 8 = kx + 79k x + 79k x k= p = q = 8 b 07 FT FT C x x 8 8 x 6 6 leading to x FT 79 multiplied by their k FT 79 multiplied by their k D crect term seen Term selected and and 9C crectly evaluated Cambridge International Examinations 05
6 Page 6 9 Cambridge O Level October/November 05 a i Number of arrangements with Maths books as one item =!! ii 5! 8 6! 7 b i 00 ii M W = 70 M W = 60 M W = 756 5M = 6 86!! oe!!! oe Maths books can be arranged! ways and Histy! ways =!!!!! = 8 07 f 8 5! their answer to i f 6 their answer to i 6 6 C 9 complete crect method using cases, may be implied by wking. Must have at least one crect any crect 5 subtracted Cambridge International Examinations 05
7 Page 7 0 i Cambridge O Level October/November 05 0 = cos ABC 07 crect cosine rule statement crect ABC equating areas statement f sin oe ABC = better XY =.970 Arc length 6 oe f XY may be implied by later wk, allow on diagram crect arc length unsimplified their + 6 their angle C ABC 5 sin = 6 ABC = sin ii Perimeter = + 6 = 9.0 iii =.50.5 better sect area using their C area of ABM where M is the midpoint of AC, s ABY and BXY ABC Answers to sf better Cambridge International Examinations 05
8 Page 8 Cambridge O Level October/November 05 x x = 0 y 6 y + 5 = 0 07 substitution and simplification to obtain a three term quadratic equation in one variable leading to, 5 and,, f each pair from a crect quadratic equation, crectly obtained. Midpoint, cao midpoint Gradient Perpendicular bisect y = x Meets the curve again if x + 0 x 5 = 0 y 8 y + = 0 leading to x = 5 ± 0, y = 9 m 0, CD = 0 CD = perpendicular bisect, must be using their perpendicular gradient and their midpoint substitution and simplification to obtain a three term quadratic equation in one variable. f each pair Pythagas using their codinates from solution of second quadratic. x x + y y must be seen if not using crect codinates. f 8 5 from so far. Cambridge International Examinations 05 0 and all crect
9 Page 9 a Cambridge O Level October/November 05 y x = y x x+ y = 7 and 07 expressing x+y, 8 as powers of and 9y x, 7y as powers of x + x + y = 7 oe y x = y oe leading to x =, y = Crect equation from crect wking Crect equation from crect wking f both Example of Alternative method Method mark as above x + x + y = 7 As befe One of the crect equations in x and y Crect, unsimplified, equation in x y only Both answers solution of quadratic crect solution leading to y = 8 x y x = y 8 x 8 x Leading to x = Leading to x = and y = Crectly substituted in b 5 5 z z + = 0 z leading to.5 = 5 z = 5 z = 0.5 z= D log 0.5 z = 0. better log5 crect attempt to solve.5 z = k, where k is positive must have one solution only Cambridge International Examinations 05
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