17January 2008 6664 Core Mathematics C2 Mark Scheme Question Number Scheme Marks 1.a)i) ii) (b) Notes: (a) (b) f(3) = 33 - 2 x 32 - 4 x 3 + 8 ; = 5 M1; A1 f(–2) = (– 8 – 8 + 8 + 8) = 0 ( B1 on Epen, but A1 in fact) B1 (3) M1 is for attempt at either f(3) or f(–3) in (i) or f(–2) or f(2) in (ii). [(x + 2)]( x2 – 4 x + 4) (= 0 not required) [must be seen or used in (b)] M1 A1 (x + 2) (x – 2)2 (= 0) ( can imply previous 2 marks) M1 Solutions: x = 2 or – 2 (both) or (–2, 2, 2) [no wrong working seen] A1 (4) [7] No working seen: Both answers correct scores full marks One correct ;M1 then A1B0 or A0B1, whichever appropriate. Alternative (Long division)Divide by (x – 3) OR (x + 2) to get ,2baxx!!a may be zero [M1] 5and12!"!xx seen i.s.w. (or “remainder = 5”) [A1] 0and442!"xx seen (or “no remainder”) [B1] First M1 requires division by a found factor ; e.g )2(),2("!xxor what candidate thinks is a factor to get zerobemay),(2abaxx!!. First A1 for [ (x + 2)] ( x2 – 4 x + 4) or (x – 2)( x2 – 4) Second M1:attempt to factorise their found quadratic. (or use formula correctly) [Usual rule: .||||),)((2bcdwheredxcxbaxx#!!#!!] N.B. Second A1 is for solutions, not factors SC: (i) Answers only: Both correct, and no wrong, award M0A1M0A1 (as if B1,B1) One correct, (even if 3 different answers) award M0A1M0A0 (as if B1) (ii) Factor theorem used to find two correct factors, award M1A1, then M0, A1 if both correct solutions given. ( –2,2,2 would earn all marks) (iii) If in (a) candidate has )4)(2(2"!xx B0, but then repeats in (b), can score M1A0M1(if goes on to factorise)A0 (answers fortuitous) Alternative (first two marks)02)2()2())(2(232#!!!!!#!!!cxcbxbxcbxxx and then compare with 084223#!""xxx to find b and c. [M1] 4,4#"#cb [A1] Method of grouping)2(4,)2(842223$"#!""xxxxxx M1; = )2(4)2(2"""xxx A1 [= 22)2)(2(])2)(4("!#""xxxx M1 Solutions: 2,2"##xx both A1 www.mymathscloud.com
18Question Number 2. (a) (b) (c) : Notes 3. (a) (b) Notes: Scheme Marks Complete method, using terms of form ark, to find r M1 [e.g. Dividing ar6 = 80 by ar3 = 10 to find r; r6 – r3 = 8 is M0] r = 2 A1 (2) Complete method for finding a M1 [e.g. Substituting value for r into equation of form ark = 10 or 80 and finding a value for a. ] (8a = 10 ) a = 41145# (equivalent single fraction or 1.25) A1 (2) Substituting their values of a and r into correct formula for sum. M1 %&%&12451120"#""#rraSn (= 1310718.75) 1 310 719 (only this) A1 (2) [6] (a) M1: Condone errors in powers, e.g. ar4 = 10 and/or ar7= 80, A1: For r = 2, allow even if ar4 = 10 and ar7 = 80 used (just these) (M mark can be implied from numerical work, if used correctly) (b) M1: Allow for numerical approach: e.g. 310cr'210cr'cr10'10 In (a) and (b)correct answer, with no working, allow both marks. (c) Attempt 20 terms of series and add is M1 (correct last term 655360) If formula not quoted, errors in applying their a and/or r is M0 Allow full marks for correct answer with no working seen. 10211()*+,-!x= 1 + (()*++,-11032213102121021()*+,-(()*++,-!()*+,-(()*++,-!()*+,-xxx M1 A1 = 1 + 5x ; + 445(or 11.25)x2 + 315x( coeffs need to be these, i.e, simplified) A1; A1 (4) [Allow A1A0, if totally correct with unsimplified, single fraction coefficients) (1 + 01021..)10 = 1 + 5(0.01) + %&201.0)25.11445(or + 15(0.01)3 M1 A1! = 1 + 0.05 + 0.001125 + 0.000015 = 1.05114 cao A1 (3) [7] (a) For M1 first A1: Consider underlined expression only. M1 Requires correct structure for at least two of the three terms: (i)Must be attempt at binomial coefficients. [Be generous :allow all notations e.g. 210C, even ()*+,-210; allow “slips”.] (ii) Must have increasing powers of x, (iii) May be listed, need not be added; this applies for all marks. First A1: Requires all three correct terms but need not be simplified, allow 110 etc, 210Cetc,and condone omission of brackets around powers of ½xSecond A1: Consider as B1: 1 + 5x(b) For M1: Substituting their (0.01) into their (a) result [0.1, 0.001, 0.25, 0.025,0.0025 acceptable but not 0.005 or 1.005] First A1 (f.t.): Substitution of (0.01) into their 4 termed expression in (a) Answer with no working scores no marks (calculator gives this answer) www.mymathscloud.com
19Question Number 4. (a) (b) Notes Scheme Marks 3 sin2! – 2cos2! = 1 3 sin2! – 2(1 – sin2!) = 1 (M1: Use of )1cossin22#!// M1 3 sin2! – 2 + 2 sin2! = 1 5 sin2! = 3 cso AG A1 (2) 53sin2#/, so sin/ = (±)0 0.6 M1 Attempt to solve both sin/ = +.. and sin/ = –...(may be implied by later work) M1 / = 50.7685o awrt #/ 50.8° (dependent on first M1 only) A1 / (= 180º – 50.7685co ); = 129.23...o awrt 129.2º M1; A1 ! [f.t. dependent on first M and 3rd M] sin / = –0 0.6 / = 230.785o and 309.23152o awrt 230.8º, 309.2º (both) M1A1 (7) [9] (a) N.B: AG; need to see at least one line of working after substituting cos2!. (b) First M1: Using 5 3sin2#/ to find value for sin/ or / [Allow such results as 53sin,53sin#/#/ .....for M1] econd M1: Considering the – value for sin./ (usually later) First A1: Given for awrt 50.8°. Not dependent on second M. Third M1: For (180 – candidate’s 50.8)°, need not see written down Final M1: Dependent on second M (but may be implied by answers) For (180 + candidate’s 50.8)° or (360 – candidate’s 50.8)° or equiv. Final A1: Requires both values. (no follow through) [ Findsk#/2cos(k = 2/5) and so #/cos (±)...M1, then mark equivalently]NB Candidates who only consider positive value for sin/can score max of 4 marks: M1M0A1M1A1M0A0 – Very common. Candidates who score first M1 but have wrong sin/ can score maximum M1M1A0M1A!M1A0 SC Candidates who obtain one value from each set, e.g 50.8 and 309.2 M1M1(bod)A1M0A0M1(bod)A0 Extra values out of range – no penalty www.mymathscloud.com
20Question Number 5. Scheme Marks Method 1 (Substituting a = 3b into second equation at some stage) Using a law of logs correctly (anywhere) e.g. log3ab = 2 M1 Substitution of 3b for a (or a/3 for b) e.g. log3 3b2 = 2 M1 Using base correctly on correctly derived log3 p= q e.g. 2233#b M1 First correct value b = 0 3 (allow 3½) A1 Correct method to find other value ( dep. on at least first M mark) M1 Second answer a = 3b = 3 0 3 or !27 A1 Method 2 (Working with two equations in log3a and log3b) “Taking logs” of first equation and “separating” ba333log3loglog!# M1 ( = 1 + b3log ) Solving simultaneous equations to find log 3a or log 3 b M1 [ log 3a = 1½, log 3b=½ ] Using base correctly to find a or b M1 Correct value for a or ba = 3 0 3 or b = 0 3 A1 Correct method for second answer, dep. on first M; correct second answer M1;A1[6][Ignore negative values] Notes: Answers must be exact; decimal answers lose both A marks There are several variations on Method 1, depending on the stage at which a = 3b is used, but they should all mark as in scheme. In this method, the first three method marks on Epen are for (i)First M1: correct use of log law, (ii) Second M1: substitution of a = 3b, (iii) Third M1: requires using base correctly on correctly derived log3 p= q Three examples of applying first 4 marks in Method 1:(i) 2log3log33#!bb gains second M1 2loglog3log333#!!bb gains first M1 (2##bb33log,1log½) no mark yet 3#b½ gains third M1, and if correct A1 (ii) 2)(log3#ab gains first M1 23#ab gains third M1 2233#b gains second M1 (iii) 23log23#b has gained first 2 M marks 23log23#1bor similar type of error 13log3#1b33#1b does not gain third M1, as 13log3#b not derived correctly www.mymathscloud.com
21Question Number 6. (a) (b) Notes: Scheme Marks N C / B º 500m 15! 700m ° A BC2 = 7002 +5002 – 2 . 500 . 700 cos 15° M1 A1 ( = 63851.92... ) BC = 253 awrt A1 (3) BCBscandidate'15sin700sin# M1 sin B = sin 15 . 700 /253c = 0.716.. and giving an obtuseB ( 134.2°) dep on 1st M M1 / = 180º – candidate’s angle B (Dep. on first M only, B can be acute) M1 /= 180 – 134.2 = (0)45.8 (allow 46 or awrt 45.7, 45.8, 45.9) A1 (4) [7] [46 needs to be from correct working] (a) If use cos 15º = ....., then A1 not scored until written as BC2 = ... correctlySplitting into 2 triangles BAX and CAX, where X is foot of perp. from B to AC Finding value for BX and CX and using Pythagoras M1 BC2 = 22)15cos500700()15sin500(!!"! A1 BC = 253 awrt A1 (b) Several alternative methods: (Showing the M marks, 3rd M dep. on first M)) (i) cosBCBCBs'xcandidate500x2700scandidate'500222"!# or ccxBCxBC5002500700222"!# M1 Finding angle B M1 dep., then M1 as above (ii) 2 triangle approach, as defined in notes for (a) tan CBX= valueforBXvalueforAX"700 M1 Finding value for 2CBX ( 593°) dep M1 )]'75(180[CBXscandidate2!"#/!! M1 (iii) Using sine rule (or cos rule) to find C first: Correct use of sine or cos rule for CM1, Finding value for C M1 Either B =180° – (15º + candidate’s C) or #/(15º + candidate’s C) M1 (iv) 700/cos50015cosBC!#! M2 {first two Ms earned in this case} Solving for /; #/45.8 (allow 46 or5.7, 45.8, 45.9) M1;A1 Note: S.C. In main scheme, if / used in place of B, third M gained immediately; Other two marks likely to be earned, too, for correct value of / stated. www.mymathscloud.com
22Question Number 7 (a) (b) (c) Scheme Marks Either solving 0 = x(6 – x) and showing x = 6 (and x = 0) B1 (1) or showing (6,0) (and x = 0) satisfies y = 6x – x2[allow for showing x = 6] Solving 2x = 6x – x2 (x 2 = 4x) to x = .... M1 x = 4 ( and x = 0) A1 Conclusion: when x = 4, y = 8 and when x = 0, y = 0 , A1 (3) (Area =) 4")4()0(2)6(xxdx Limits not required M1 Correct integration 3332xx" (+ c) A1 Correct use of correct limits on their result above (see notes on limits) M1 [“3332xx"”]4 – [“3332xx"”]0 with limits substituted [= 48 – 32263121# ] Area of triangle = 2 . 8 =16 (Can be awarded even if no M scored, i.e. B1) A1 Shaded area = $(area under curve – area of triangle ) applied correctly M1 ( = 3210)163226#" (awrt 10.7) A1 (6)[10] Notes (b) In scheme first A1: need only give x = 4 If verifying approach used: Verifying (4,8) satisfies both the line and the curve M1(attempt at both), Both shown successfully A1 For final A1, (0,0) needs to be mentioned ; accept “clear from diagram” (c) Alternative Using Area = $}2;)6{()4()0(2xxx""4dxapproach (i) If candidate integrates separately can be marked as main scheme If combine to work with = $)4()4()0(24"xxdx, first M mark and third M mark = (±) [3232xx" (+ c) ] A1, Correct use of correct limits on their result second M1, Totally correct, unsimplified ± expression (may be implied by correct ans.) A1 10" A1 [Allow this if, having given - 10", they correct it] M1 for correct use of correct limits: Must substitute correct limits for their strategy into a changed expression and subtract, either way round, e.g 5 6 5 6}{04"$If a long method is used, e,g, finding three areas, this mark only gained for correct strategy and all limits need to be correct for this strategy. Final M1: limits for area under curve and triangle must be the same. S.C.(1) 44#""606022)6(xdxdxxx56....336026032#"789:;<"xxxaward M1A1 MO(limits)AO(triangle)M1(bod)A0 (2) If, having found ± correct answer, thinks this is not complete strategy and does more, do not award final 2 A marks Use of trapezium rule: M0A0MA0possibleA1for triangle M1(if correct application of trap. rule from x = 0 to x = 4) A0 www.mymathscloud.com
