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GCE Core Mathematics C2 (6664) January 2010 3 January 2010 Core Mathematics C2 6664 Mark SchemeQuestion Number Scheme Marks Q1 Notes ()6654263336()3()2xxx⎛⎞⎡⎤− = + × ×− + × ×−⎜⎟⎣⎦⎝⎠ = 729, 1458x−, 21215x+M1 for either the x term or the 2x term. Requires correct binomial coefficient in any form with the correct power of x – condone lack of negative sign and wrong power of 3. This mark may be given if no working is shown, but one of the terms including x is correct. Allow 66,or12 (must have a power of 3, even if only power 1) First term must be 729 for B1, ( writing just 63 is B0 ) can isw if numbers added to this constant later. Can allow 729(1... Term must be simplified to –1458x for A1cao. The x is required for this mark. Final A1is c.a.o and needs to be 21215x+ (can follow omission of negative sign in working) Descending powers of x would be 6524636( ) 3( ) ..4xxx⎛⎞+××− + × ×− +⎜⎟⎝⎠i.e. 65 418135..xx x−+ + This is M1B1A0A0 if completely “correct” or M1 B0A0A0 for correct binomial coefficient in any form with the correct power of x as before M1 B1,A1, A1[4]AlternativeNB Alternative method: ()66233633 (1 6 ()()..)2xxx⎛⎞−= +×−+×−+⎜⎟⎝⎠is M1B0A0A0 – answers must be simplified to 729, 1458x−, 21215x+ for full marks (awarded as before) The mistake ()66333(1)xx−=−=23363(16()()..)2xx⎛⎞+×− +× ×− +⎜⎟⎝⎠may also be awarded M1B0A0A0Another mistake 623 (1 615 ...) 729...xx−+ = would be M1B1A0A0 www.mymathscloud.com
GCE Core Mathematics C2 (6664) January 2010 4 Question Number Scheme Marks Q2 (a) (b) (a) (b)()25 sin1 2 1 sinxx=+ −22sin5sin3 0xx+−=(*) ()()21s3 0s−+= giving s = []sin3 has no solutionx=− so 12sinx=30, 150x∴=M1 for a correct method to change 22cos into sinxx (must use 22cos1 sinxx=−) A1 need 3 term quadratic printed in any order with =0 included M1 for attempt to solve given quadratic (usual rules for solving quadratics) (can use any variable here, s, y, x, or sinx ) A1 requires no incorrect work seen and is for 12sinx=or x = 112sin−12y= is A0 (unless followed by x = 30) B1 for 30 (α ) not dependent on method2nd B1 for 180 - αprovided in required range (otherwise 540 - α ) Extra solutions outside required range: Ignore Extra solutions inside required range: Lose final B1 Answers in radians: Lose final B1 S.C. Merely writes down two correct answers is M0A0B1B1 Or 12sinx=30, 150x∴= is M1A1B1B1Just gives one answer : 30 only is M0A0B1B0 or 150 only is M0A0B0B1 NB Common error is to factorise wrongly giving()()2 sin1 sin30xx+−=[]sin3 gives no solutionx=12sin210, 330xx=− ⇒ =This earns M1 A0 B0 B1ft Another common error is to factorise correctly ()()03sin1sin2=+−xxand follow this with 21sin=x, 3sin=x then oo051,30=xThis would be M1 A0 B1 B1 M1 A1cso(2)M1 A1 B1, B1ft (4)[6] www.mymathscloud.com
GCE Core Mathematics C2 (6664) January 2010 5 Question Number Scheme Marks Q3 (a) (b) (a) (b) ()12111f26842ab=×+×+×−()31112424f5 ab=− ⇒ + = or 23ab+= f ( 2)16 426ab−=−+ − −f ( 2) 0 4222ab−= ⇒ − =Eliminating one variable from 2 linear simultaneous equations in a and ba = 5 and b = - 1 ()()322256 223xxx x xx+−−=+ +− = ()()()22 31xxx++−NB ()()()32222xx x++ − is A0 But ()()()32221xxx++−is A1 1st M1 for attempting f(12±) Treat the omission of the –5 here as a slip and allow the M mark. 1st A1 for first correct equation in a and b simplified to three non zero terms (needs –5 used) s.c. If it is not simplified to three terms but is correct and is then used correctly with second equation to give correct answers- this mark can be awarded later. 2nd M1 for attempting f( 2m) 2nd A1 for the second correct equation in a and b. simplified to three terms (needs 0 used) s.c. If it is not simplified to three terms but is correct and is then used correctly with first equation to give correct answers - this mark can be awarded later. 3rd M1 for an attempt to eliminate one variable from 2 linear simultaneous equations in a and b3rd A1 for both a = 5 and b = -1 (Correct answers here imply previous two A marks) 1st M1 for attempt to divide by (x+2) leading to a 3TQ beginning with correct term usually 22x2nd M1 for attempt to factorize their quadratic provided no remainder A1 is cao and needs all three factors Ignore following work (such as a solution to a quadratic equation). M1 A1 M1 A1 M1 A1 (6) M1 M1A1 (3) [9] (a) (b) Alternative; M1 for dividing by (21)x−, to get 212()constantaxx+++with remainder as a function of a and b, and A1 as before for equations stated in scheme . M1 for dividing by (2)x+, to get 22(4) ...xax+− (No need to see remainder as it is zero and comparison of coefficients may be used) with A1 as before Alternative; M1 for finding second factor correctly by factor theorem, usually (x – 1) M1 for using two known factors to find third factor , usually (23)x±Then A1 for correct factorisation written as product ()()()22 31xxx++−www.mymathscloud.com
GCE Core Mathematics C2 (6664) January 2010 6 Question Number Scheme Marks Q4 (a) (b) []38.176.1ˆ=−=πDBC Sector area = ()[]21241.7611.0 ~ 11.1π××− =3.794212×× is M0 Area of ()[]125 4 sin 1.769.8ABC∆=××× = or 101sin4521××× Required area = awrt 20.8 or 20.9 or 21.0 or gives 21 (2sf) after correct work. M1 M1 M1, A1 (4)M1 M1 A1(3)[7](a) (b)1st M1 for correct use of sine rule to find ACB or cosine rule to find b (M0 for ABC here or for use of sin x where xcould be ABC) 2nd M1 for a correct expression for angle ACB (This mark may be implied by .7835 or by arcsin (.7058)) and needs accuracy. In second method this M1 is for correct expression for b – may be implied by 6.96. [Note 3.86.0cos10≈] (do not need two answers) 3rd M1 for a correct method to get angle ABC in method (i) or sinABC or cosABC , in method (ii) (If sin B >1, can have M1A0) A1cso for correct work leading to 1.76 3sf . Do not need to see angle 0.1835 considered and rejected. 1st M1 for a correct expression for sector area or a value in the range 11.0 – 11.1 2nd M1 for a correct expression for the area of the triangle or a value of 9.8 Ignore 0.31 (working in degrees) as subsequent work. A1 for answers which round to 20.8 or 20.9 or 21.0. No need to see units. (a) Special case If answer 1.76 is assumed then usual mark is M0 M0 M0 A0. A Fully checked method may be worth M1 M1 M0 A0. A maximum of 2 marks. The mark is either 2 or 0. Either M1 for BCAˆis found to be 0,7816 (angles of triangle) then M1 for checking ˆsin()sin 0.654ACB= with conclusion giving numerical answers This gives a maximum mark of 2/4OR M1 for b is found to be 6.97 (cosine rule) M1 for checking sin()sin 0.64ABCb= with conclusion giving numerical answers This gives a maximum mark of 2/4Candidates making this assumption need a complete method. They cannot earn M1M0. So the score will be 0 or 2 for part (a). Circular arguments earn 0/4. .Either ˆsin()sin 0.654ACB=ˆ arcsin(0.7058...)ACB∴=[0.7835.. or 2.358]=Use angles of triangle ˆˆ0.6ABCACBπ=− −(But as AC is the longest side so) ˆ1.76ABC= (*)(3sf) ]76.17.100[Allow→oIn degrees oo44.9BCˆA,377.346.0==or2224525cos0.6bb=+−××210 cos 0.6(100 cos 0.6 36)2b±−∴= = [6.96 or 1.29] Use sine / cosine rule with value for b sin 0.6sin4Bb=× or 225 16cos40bB+−=(But as AC is the longest side so) ˆ1.76ABC= (*)(3sf)www.mymathscloud.com
