2EDEXCEL June 2008 Core Mathematics C2 Mark SchemeQuestion Scheme Marks number 1. (a) Attempt to find f(–4) or f(4). ()20)4(39)4(3)4(2)4(f23+−−−−−=− M1 ()2015648128++−−= = 0, so (x + 4) is a factor. A1 (2) (b) )5112)(4(203932223+−+=+−−xxxxxx M1 A1 )5)(12.....(−−xx(The 3 brackets need not be written together)M1 A1cso (4) or )102(21.....−⎟⎠⎞⎜⎝⎛−xxor equivalent 6 (a) Long division scores no marks in part (a). The factor theorem is required. However, the first two marks in (b) can be earned from division seen in (a)... ... but if a different long division result is seen in (b), the work seen in (b) takes precedence for marks in (b). A1 requires zero and a simple conclusion (even just a tick, or Q.E.D.), or may be scored by a preamble, e.g. 'If f(–4) = 0, (x + 4) is a factor.....' (b) First M requires use of (x + 4) to obtain 0,0),2(2≠≠++babaxx, even with a remainder. Working need not be seen... this could be done 'by inspection'. Second M for the attempt to factorise their three-term quadratic. Usual rule: kpqbcddqxcpxbaxkx==++=++andwhere),)(()(2. If 'solutions' appear before or after factorisation, ignore... ... but factors must be seen to score the second M mark. Alternative (first 2 marks):04)4()8(2)2)(4(232=+++++=+++bxbaxaxbaxxx, then compare coefficients to find values of a and b. [M1] a = –11, b = 5 [A1] Alternative: Factor theorem: Finding that ∴=⎟⎠⎞⎜⎝⎛021ffactor is, (2x – 1) [M1, A1] Finding that ()∴=05f factor is, )5(−x [M1, A1] “Combining” all 3 factors is not required. If just one of these is found, score the first 2 marks M1 A1 M0 A0. Losing a factor of 2:)5(21)4(−⎟⎠⎞⎜⎝⎛−+xxx scores M1 A1 M1 A0. Answer only, one sign wrong: e.g.)5)(12)(4(+−+xxx scores M1 A1 M1 A0 www.mymathscloud.com
3EDEXCEL June 2008 Core Mathematics C2 Mark Scheme Question Scheme Marks number 2. (a) 1.732, 2.058, 5.196 awrt (One or two correctB1 B0, All correct B1 B1)B1 B1 (2) (b)......5.021× B1 ()(){}630.3646.2058.22196.5732.1......++++ M1 A1ft = 5.899 (awrt 5.9, allowed even after minor slips in values) A1 (4) 6 (a) Accept awrt (but less accuracy loses these marks). Also accept exact answers, e.g. 3 at x = 0, 33or 27 at x = 2. (b) For the M mark, the first bracket must contain the 'first and last' values, and the second bracket must have no additional values. If the only mistake is to omit one of the values from the second bracket, this can be considered as a slip and the M mark can be allowed. Bracketing mistake: i.e. )630.3646.2058.2(2)196.5732.1(5.021++++× scores B1 M1 A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). x values: M0 if the values used in the brackets are x values instead of y values. Alternative: Separate trapezia may be used, and this can be marked equivalently. ⎥⎦⎤⎢⎣⎡+++++++)196.5630.3(41)630.3646.2(41)646.2058.2(41)058.2732.1(41www.mymathscloud.com
4EDEXCEL June 2008 Core Mathematics C2 Mark Scheme Question Scheme Marks number 3. (a) ......101)1(10axax+=+(Not unsimplified versions)B132)(68910)(2910axax××+×+ Evidence from one of these terms is sufficientM1 332232120,45or)(120,)(45xaxaaxax++++ A1, A1 (4) (b) 23452120aa×=⎟⎠⎞⎜⎝⎛=75.0,12090e.g.equiv.or 43a Ignore a = 0, if seen M1 A1 (2) 6 (a) The terms can be ‘listed’ rather than added. M1: Requires correct structure: ‘binomial coefficient’ (perhaps from Pascal’s triangle) and the correct power of x. (The M mark can also be given for an expansion in descending powers of x). Allow ‘slips’ such as: 22910ax×, ()323910ax××, 22910x×, 3323789xa××× However, 332212045xaxa+++ or similar is M0. ⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛310and210 or equivalent such as 210C and 310Care acceptable, and even ⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛310and210 are acceptable for the method mark. 1st A1: Correct 2x term. 2nd A1: Correct 3x term (These must be simplified). If simplification is not seen in (a), but correct simplified terms are seen in (b), these marks can be awarded. However, if wrong simplification is seen in (a), this takes precedence. Special case: If 32)(and)(axax are seen within the working, but then lost... ... A1 A0 can be given if 32120and45axax are both achieved. (b) M: Equating their coefficent of 3xto twice their coefficient of 2x... ... or equating their coefficent of 2xto twice their coefficient of 3x. (... or coefficients can be correct coefficients rather than their coefficients). Allow this mark even if the equation is trivial, e.g. 120a = 90a. An equation in a alone is required for this M mark, although... ...condone, e.g.⇒=223390120xaxa()⇒=2390120aaa = 43. Beware: a = 43following 120a = 90a, which is A0. www.mymathscloud.com
5EDEXCEL June 2008 Core Mathematics C2 Mark Scheme Question Scheme Marks number 4. (a) 7logor5log7log5==xx (i.e. correct method up to x =...) M1 1.21 Must be this answer (3 s.f.) A1 (2) (b) ()()5575−−xx Or another variable, e.g. ()()57−−yy, even ()( )57−−xx M1 A1 ()55or 75==xxx = 1.2 (awrt) ft from the answer to (a), if used A1ft x = 1 (allow 1.0 or 1.00 or 1.000) B1 (4) 6 (a)1.21 with no working: M1 A1 (even if it left as 21.15). Other answers which round to 1.2 with no working: M1 A0. (b) M: Using the correct quadratic equation, attempt to factorise ()()5575±±xx, or attempt quadratic formula. Allow 5log7logor7log5 instead of 1.2 for A1ft. No marks for simply substituting a decimal answer from (a) into the given equation (perhaps showing that it gives approximately zero). However, note the following special case: Showing that 75=x satisfies the given equation, therefore 1.21 is a solution scores 0, 0, 1, 0 (and could score full marks if the x = 1 were also found). e.g. If 75=x, then4952=x, and 035844935)5(1252=+−=+−xx, so one solution is x = 1.21 (‘conclusion’ must be seen). To score this special case mark, values substituted into the equation must be exact. Also, the mark would not be scored in the following case: e.g. If 75=x, 21.149503584522=⇒=⇒=+−xxx (Showing no appreciation that 22)5(5xx=) B1: Do not award this mark if x = 1 clearly follows from wrong working. www.mymathscloud.com
6EDEXCEL June 2008 Core Mathematics C2 Mark Scheme Question Scheme Marks number 5. (a) 22)13()38(−+− or 22)13()38(−+− M1 A1 kyx=±+±22)1()3( or kyx=±+±22)3()1( (k a positive value) M1 29)1()3(22=−+−yx (Not()229 or 239.5) A1 (4) (b) Gradient of radius = 52(or exact equiv.)Must be seen or used in (b) B1 Gradient of tangent =25−(Using perpendicular gradient method)M1 )8(253−−=−xy(ft gradient of radius, dependent upon both M marks) M1 A1ft 04625=−+yx (Or equiv., equated to zero, e.g. 010492=−−xy) A1 (5) (Must have integer coefficients) 9 (a) For the M mark, condone one slip inside a bracket, e.g. 22)13()38(++−, 22)31()18(−+− The first two marks may be gained implicitly from the circle equation. (b) 2nd M: Eqn. of line through (8, 3), in any form, with any grad.(except 0 or ∞). If the 8 and 3 are the ‘wrong way round’, this M mark is only given if a correct general formula, e.g. )(11xxmyy−=−, is quoted. Alternative: 2nd M: Using (8, 3) and an m value in y = mx + c to find a value of c. A1ft: as in main scheme. (Correct substitution of 8 and 3, then a wrong c value will still score the A1ft) (b) Alternatives for the first 2 marks: (but in these 2 cases the 1st A mark is not ft) (i) Finding gradient of tangent by implicit differentiation 0dd)1(2)3(2=−+−xyyx (or equivalent) B1 Subs. x = 8 andy = 3 into a 'derived' expression to find a value for xyd/d M1 (ii) Finding gradient of tangent by differentiation of 26201xxy−++=())26(62021dd212xxxxy−−+=− (or equivalent) B1 Subs. x = 8 into a 'derived' expression to find a value for xyd/d M1 Another alternative: Using 0)()(1111=++++++cyyfxxgyyxx0192622=−−−+yxyx B1 019)3()8(3,38=−+−+−+yxyx M1, M1 A1ft (ft from circle eqn.) 04625=−+yxA1www.mymathscloud.com
