6663 Core Mathematics C1 January 2008 Advanced Subsidiary/Advanced Level in GCE Mathematics2January 2008 6663 Core Mathematics C1 Mark SchemeQuestion Scheme Marks number 1. 323kxx→ or 654kxx→ or kx→−7(k a non-zero constant) M1 333x or 646x(Either of these, simplified or unsimplified) A1 xxx73263−+ or equivalent unsimplified, such as 16376433xxx−+ A1 + C (or any other constant, e.g. + K) B1 (4) 4M: Given for increasing by one the power of x in one of the three terms. A marks: ‘Ignore subsequent working’ after a correct unsimplified version of a term is seen. B: Allow the mark (independently) for an integration constant appearing at any stage (even if it appears, then disappears from the final answer). This B mark can be allowed even when no other marks are scored. www.mymathscloud.com
6663 Core Mathematics C1 January 2008 Advanced Subsidiary/Advanced Level in GCE Mathematics3Question Scheme Marks number 2. (a) 2 B1 (1) (b) 9x seen, or ()3(a)answer to seen, or 33)2(x seen. M1 98x A1 (2) 3 (b) M: Look for 9x first... if seen, this is M1. If not seen, look for ()3(a)answer to, e.g. 32... this would score M1 even if it does not subsequently become 8. (Similarly for other answers to (a)). In 33)2(x, the 32 is implied, so this scores the M mark. Negative answers: (a)Allow 2−. Allow 2±. Allow ‘2 or 2−’. (b) Allow 98x±. Allow ‘98x or 98x−’. N.B. If part (a) is wrong, it is possible to ‘restart’ in part (b) and to score full marks in part (b). www.mymathscloud.com
6663 Core Mathematics C1 January 2008 Advanced Subsidiary/Advanced Level in GCE Mathematics4Question Scheme Marks number 3. ()()()()32323235−−×+− M1 ()⎟⎟⎠⎞⎜⎜⎝⎛+−=+−−=...33710...33532102 M1 ()3713−=⎟⎟⎠⎞⎜⎜⎝⎛−13713Allow 13 (a = 13) A1 37− (b = −7) A1 (4) 4 1st M: Multiplying top and bottom by ()32−. (As shown above is sufficient). 2nd M: Attempt to multiply out numerator ()35−()32−. Must have at least 3 terms correct. Final answer: Although ‘denominator = 1’ may be implied, the 3713− must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M1 A1 A1 is not an option). The A marks cannot be scored unless the 1st M mark has been scored, but this 1st M mark could be implied by correct expansions of both numerator anddenominator. It is possible to score M1 M0 A1 A0 or M1 M0 A0 A1 (after 2 correct terms in the numerator). Special case: If numerator is multiplied by ()32+ instead of ()32−, the 2nd M can still be scored for at least 3 of these terms correct: ()23353210−+−. The maximum score in the special case is 1 mark: M0 M1 A0 A0. Answer only: Scores no marks. Alternative method: )32)(3(35++=−ba33232)32)(3(+++=++baaba M1: At least 3 terms correct. ba325+=ba21+=−a =... or b =... M1: Form and attempt to solve simultaneous equations. a = 13, b = −7 A1, A1www.mymathscloud.com
6663 Core Mathematics C1 January 2008 Advanced Subsidiary/Advanced Level in GCE Mathematics5Question Scheme Marks number 4. (a) ⎟⎠⎞⎜⎝⎛−=−−=−−−−−−−−=21147or 147,)6(843or 86)3(4m M1, A1 Equation: ))6((214−−−=−xy or )8(21)3(−−=−−xy M1 022=−+yx(or equiv. with integer coefficients... must have ‘= 0’) A1 (4) (e.g. 014714=−+xy and 014714=−−yx are acceptable) (b) 22))3(4()86(−−+−− M1 22714+ or 227)14(+− or 22)7(14−+ (M1 A1 may be implied by 245) A1 22714+=AB or )12(7222+ or 245 57 A1cso (3) 7 (a) 1st M: Attempt to use 1212xxyym−−= (may be implicit in an equation of L). 2nd M: Attempting straight line equation in any form, e.g ()11xxmyy−=−, mxxyy=−−11, with any value of m (except 0 or ∞) and either (–6, 4) or (8, –3). N.B. It is also possible to use a different point which lies on the line, such as the midpoint of AB (1, 0.5). Alternatively, the 2nd M may be scored by using cmxy+= with a numerical gradient and substituting (–6, 4) or (8, –3) to find the value of c. Having coords the wrong way round, e.g. )4(21)6(−−=−−xy, loses the 2nd M mark unless a correct general formula is seen, e.g. )(11xxmyy−=−. (b) M: Attempting to use 212212)()(yyxx−+−. Missing bracket, e.g. 22714+− implies M1 if no earlier version is seen. 22714+− with no further work would be M1 A0. 22714+− followed by ‘recovery’ can score full marks. www.mymathscloud.com
6663 Core Mathematics C1 January 2008 Advanced Subsidiary/Advanced Level in GCE Mathematics6Question Scheme Marks number 5. (a) ⎟⎟⎠⎞⎜⎜⎝⎛+−−12132xx1,21−=−=qp B1, B1 (2) (b) ⎟⎟⎠⎞⎜⎜⎝⎛++−=−−1213275xxxy5dd⎟⎠⎞⎜⎝⎛=xy()05or x(75−xcorrectly differentiated)B1 Attempt to differentiate either px2 with a fractional p, giving )0(1≠−kkxp, (the fraction p could be in decimal form) or qx3 with a negative q, giving )0(1≠−kkxq. M12232233,31221−−−−−−⎟⎟⎠⎞⎜⎜⎝⎛=×−×−xxxx A1ft, A1ft (4) 6 (b): N.B. It is possible to ‘start again’ in (b), so the p and q may be different from those seen in (a), but note that the M mark is for the attempt to differentiate px2 or qx3. However, marks for part (a) cannot be earned in part (b). 1st A1ft: ft their px2, but p must be a fraction and coefficient must be simplified (the fraction p could be in decimal form). 2nd A1ft: ft their qx3, but q must be negative and coefficient must be simplified. 'Simplified' coefficient means ba where a and b are integers with no common factors. Only a single + or − sign is allowed (e.g. −− must be replaced by +). Having +C loses the B mark.www.mymathscloud.com
