MARK SCHEME for the October/November 2015 series 0580 MATHEMATICS

Transcription

1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the October/November 05 series 0580 MATHEMATICS 0580/4 Paper 4 (Extended), maximum raw mark 30 This mark scheme is published as an aid to teachers cidates, 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 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 05 series for most Cambridge IGCSE, Cambridge International A AS Level components some Cambridge O Level components. bestexamhelp.com IGCSE is the registered trademark of Cambridge International Examinations.

2 Page Mark Scheme Syllabus Paper Cambridge IGCSE October/November Abbreviations cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent SC Special Case nfww not from wrong working soi seen or implied Question Answer Mark Part marks (a) (i) 5 4 M or M for (ii) (b) 0 00 M for 4 45 soi by 0080 (c) 9 M for 4 soi by 8.66 to 8.67 or 8.7 or 8 3 (d) (i) (ii) FT FT their (d)(i) (e) 98 or 98. to M for 00 oe or M for 00 or (a) ,,, (b) Fully correct smooth curve 4 B3 FT for 8 or 9 points or B FT for 6 or 7 points or B FT for 4 or 5 points (c) line y = x + ruled 0. to to.95 3 Line must be fit for purpose ie at least from x = 0 to x = B for correct line correct value or B for correct line or SC for no/wrong line correct values Cambridge International Examinations 05

3 Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (d) Tangent ruled at x =.5 B No daylight between tangent curve at point of contact. Consider point of contact as midpoint between two vertices of daylight, the midpoint must be between x =.6 x =.4. to 5 dep on B rise M for also dep on any tangent drawn or run close attempt at tangent at any point Must see correct or implied calculation from a drawn tangent 3 (a) Correct diagram 3 B for correct vertical plots B for correct horizontal plots B dep on at least B for reasonable increasing curve or polygon through their 6 points (b) (i) 3 to 34 If zero scored, SC for 5 out of 6 correct plots (ii) 0 reading at r = 50 FT BFT for reading at r = 50 seen (c) B for correct (d) 35. or 35 6 or 35.6 to 35.7 nfww 4 M for mid-values soi M FT for fx with x in the correct interval including boundaries Mdep for fx 0 dependent on second M earned (e) FT FT from (c) their 8 5 their 7 0 B3FT for any correct or BFT for first or second answer correct or B for 0.3 only 4 (a).6[0] or.60 to.60 3 M for or M for 0.6 cos 68 oe cos 68 = (b) 43.5 or 43.6 or to their.6 M for.9.3 or M for implicit statement A for [cos = ] 0.74 to AC Cambridge International Examinations 05

4 Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (c).33 or.33...nfww 4 M for ( ).3. = h or M for.3 ( ) M for. their. (their. must come from attempt at Pythag or from trig in triangle BCD) (d) 4. or 4.3 to M for sin = oe.9 or M for correct angle identified 5 (a) (i) 4 x( 3x + 3) x( 4x {3x 9} ) = 4 oe M x + 5x x 8x M Correct removal of all their brackets Dep on two areas added or subtracted 5x + 7x = 0 (ii) ( 5 3)( x + 4) x [= 0] 3 oe, 4 5 A M A with no errors or omissions seen at least one more line of working showing collection of like terms or division by M for ( x + a)( x + b) 5 where ab = or 5b + a = 7 [a, b integers] If zero scored SC for correct answers with no working or from other methods. (b) For correctly eliminating one variable M x = 3 y = 7 A A SC if no working shown, but correct answers given If zero scored SC for values satisfying one of the original equations (c) t = nfww 5 M for ( t 3)( t + 3) t M for denominator[s] ( t + 3) t ( t + 3) isw on RHS + or better seen t isw or for Mdep for t + t + 8 t = t + 3t oe dependent on both numerators denominator exping to give quadratics A for 9 t + 8 = 0 oe Cambridge International Examinations 05

5 Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (a) (i) 43 (ii) 6 Isosceles triangle or OYZ is isosceles Angle at centre is twice angle at circumference (iii) 30 [Opposite angles of a]cyclic quadrilateral [add up to 80 ] M for p + 5p = 80 oe (b) (i) : oe (ii) OQ MQ = NQ OM = ON Centre or O Not origin 7 (a) (i) Rotation [+]90 or 90 anticlockwise oe ( 0, ) (ii) Reflection y = oe (iii) Enlargement [s f ] oe Origin oe Not as column vector (b) 0 0 oe FT FT their s f from (a)(iii) k 0 SC for, k or 0 0 k (c) Image at (4, ) (6, ) (6, 5) (4, 3) ruled or good freeh SC for translation k k or or for 4 correct vertices not joined 3 (d) Reflection y = x oe Cambridge International Examinations 05

6 Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (a) ( 4, 6 ), (b) 4.47 or M for ( 8 4) + ( 5 3) or M for ( 8 4) + ( 5 3) or better or better (c) y = x oe 3 B for x or y = x + c oe 8 4 or M for [m =] oe soi by x 5 3 M for (3, 4) or (5, 8) or their midpoint substituted into their y = mx + c with m numerical (d) 3 3 M for use of gradient their m = soi by 9 (a) (i) M for r = their gradient 6 [+0] (ii) 56 M for [g(3) =] 8 or 3 or x (b) (c) x 5 oe final answer M for x = y + 5 or x = y 5 or better y 5 or = x + 9 6x final answer M for ( 7 3x ) + 5 (d), 0,, 3 Additional values count as errors B for one error /omission or B for two errors/omissions or M for < x oe seen or M for < x or x or x = x = or 4 < x 4 0 (a) B for correct (b) n + oe (c) (i) (n ) oe M for (n + k) for integer k (ii) 9 M for 88 or 9 seen (d) (i) n 3n final answer M for their (n ) their (n + ) soi (ii) 39 Cambridge International Examinations 05

7 Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (e) oe oe 4 8 oe Cambridge International Examinations 05