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GCE Core Mathematics C2 (6664) Summer 2010SOME GENERAL PRINCIPLES FOR C2 MARKING(But the particular mark scheme always takes precedence) Method mark for solving 3 term quadratic: 1. Factorisationcpqqxpxcbxx=++=++where),)(()(2, leading to x = ... amncpqqnxpmxcbxax==++=++andwhere),)(()(2, leading to x = ... 2. FormulaAttempt to use correct formula (with values for a, b and c). 3. Completing the square Solving 02=++cbxx: 0,022≠=±±⎟⎠⎞⎜⎝⎛±qcqbx, leading to x = ... Use of a formulaWhere a method involves using a formula that has been learnt, the advice given in recent examiners’ reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values (but refer to the mark scheme first... the application of this principle may vary). Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but will be lost if there is any mistake in the working. Equation of a straight lineApply the following conditions to the M mark for the equation of a line through ),(ba: If the a and b are the wrong way round the M mark can still be given if a correct formula is seen, (e.g. )(11xxmyy−=−) otherwise M0. If (a, b) is substituted into cmxy+= to find c, the M mark is for attempting this. Answers without workingThe rubric says that these may gain no credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done “in your head”, detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice. MisreadsA misread must be consistent for the whole question to be interpreted as such. These are not common. In clear cases, please deduct the first 2 A (or B) marks which would have been lost by following the scheme. (Note that 2 marks is the maximum misread penalty, but that misreads which alter the nature or difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written). Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written. If in doubt, send the response to Review.www.mymathscloud.com
GCE Core Mathematics C2 (6664) Summer 2010 June 2010 Core Mathematics C2 6664 Mark SchemeQuestion Number Scheme Marks 1. (a) 2.35, 3.13, 4.01 (One or two correct B1 B0, all correct B1 B1) B1 B1 Important: If part (a) is blank, or if answers have been crossed out and no replacement answers are visible, please send to Review as ‘out of clip’. (2) (b)......2.021× (or equivalent numerical value) B1 ()(){}rqpk+++++65.1251, k constant, 0≠k (See notes below) M1 A1 = 2.828 (awrt 2.83, allowed even after minor slips in values) A1 The fractional answer 250707 (or other fraction wrt 2.83) is also acceptable. Answers with no working score no marks. (4) 6 (a) Answers must be given to 2 decimal places. No marks for answers given to only 1 decimal place. (b) The p, q and r below are positive numbers, none of which is equal to any of: 1, 5, 1.65, 0.2, 0.4, 0.6 or 0.8 M1 A1: ()(){}rqpk+++++65.1251 M1 A0: ()(){}qpk++++65.1251 or ()(){}rqpk++++251 M0 A0: ()(){})(65.1251svalueotherrqpk++++++ Note that if the only mistake is to omit a value from the second bracket, this is considered as a slip and the M mark is allowed. Bracketing mistake: i.e. ()()01.413.335.265.12512.021+++++× instead of ()(){}01.413.335.265.12512.021+++++×, so that only the (1 + 5) is multiplied by 0.1 scores B1 M1 A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). Alternative: Separate trapezia may be used, and this can be marked equivalently. www.mymathscloud.com
GCE Core Mathematics C2 (6664) Summer 2010Question Number Scheme Marks 2 (a)Attempting to find )3(f or )3(f−M1 9840174458140)358()3(5)3(3)3(f23−=+−−=+×−−=A1 (2) (b) {})8103()5(405853223−+−=+−−xxxxxxM1 A1 Attempt to factorise 3-term quadratic, or to use the quadratic formula (see general principles at beginning of scheme). This mark may be implied by the correct solutions to the quadratic. M1 ....0)4)(23(==+−xxx or69610010+±−=xA1 ft 32(or exact equiv.), 5,4−(Allow ‘implicit’ solns, e.g. f(5) = 0, etc.) A1 (5) Completely correct solutions without working: full marks. 7 (a) Alternative (long division): ‘Grid’ method Divide by (x− 3) to get 0,0),3(2≠≠++babaxx. [M1] 3 3 −5 −58 40 )4643(2−+xx, and –98 seen. [A1] 0 9 12 −138 (If continues to say ‘remainder = 98’, isw) 3 4 −46 −98 (b) 1st M requires use of )5(−x to obtain 0,0),3(2≠≠++babaxx. ‘Grid’ method (Working need not be seen... this could be done 'by inspection'.) 3 3 −5 −58 40 0 15 50 −40 )8103(2−+xx 3 10 −8 0 2nd M for the attempt to factorise their 3-term quadratic, or to solve it using the quadratic formula. Factorisation: bcddxcxbaxx=++=++where),)(3()3(2. A1ft: Correct factors for their 3-term quadratic followed by a solution (at least one value, which might be incorrect), or numerically correct expression from the quadratic formula for their 3-term quadratic. Note therefore that if the quadratic is correctly factorised but no solutions are given, the last 2 marks will be lost. Alternative (first 2 marks):05)5()15(3)3)(5(232=−−+−+=++−bxabxaxbaxxx, then compare coefficients to find values of a and b. [M1] a = 10, b = −8 [A1] Alternative 1: (factor theorem) M1: Finding that 0)4(f=−A1: Stating that )4(+xis a factor. M1: Finding third factor )23)(4)(5(±+−xxx. A1: Fully correct factors (no ft available here) followed by a solution, (which might be incorrect). A1: All solutions correct. Alternative 2: (direct factorisation) M1: Factors ))(3)(5(qxpxx++−A1:8−=pqM1: )4)(23)(5(±±−xxxFinal A marks as in Alternative 1. Throughout this scheme, allow ⎟⎠⎞⎜⎝⎛±32xas an alternative to ()23±x. www.mymathscloud.com
