Mark (Results) January 0 GCE GCE Core Mathematics C (666) Paper Edexcel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WCV 7BH
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General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark
January 0 Core Mathematics C 666 Mark. (a) 6 = or or better 6 6 = or 0.5 (ignore ± ) () (b) x = x or x x = x or equivalent or 6 cao () (a) for a correct statement dealing with the or the power This may be awarded if is seen or for reciprocal of their s.c ¼ is A0, also is A0 ± is not penalised so (b) for correct use of the power on both the and the x terms for cancelling the x and simplifying to one of these two forms. Correct answers with no working get full marks 6 GCE Core Mathematics C (666) January 0
6 x x x 6. ( = ),, +,( +c) = 6 x x + x + c,, 5 for some attempt to integrate: a non zero constant st 6 x for or better 6 nd x for or better rd for x or better n n x x + i.e 6 ax or ax or ax or ax, where a is th for each term correct and simplified and the +c occurring in the final answer GCE Core Mathematics C (666) January 0
. ( + ) ( ) 5 +... = denominator of Numerator = 5 + 5 So 5 = + + ( ) = 5, and form simultaneous equations in p and q -p + q = 5 and p - q = - Solve simultaneous equations to give p = and q =. st for multiplying numerator and denominator by same correct expression st for a correct denominator as a single number (NB depends on M mark) Alternative: ( p q ) nd for an attempt to multiply the numerator by ( ± ) and get terms with at least correct. nd for the answer as written or p = and q =. Allow 0.5 and.5. (Apply isw if correct answer seen, then slip writing p =, q = ) Answer only (very unlikely) is full marks if correct no part marks GCE Core Mathematics C (666) January 0
(a) ( a ) = c 6 () (b) a = (their a) c (= 8 - c) a+ a + a = + "(6 c)" + "(8 c)" 6 5c = 0 ft So c = 5. o.a.e (b) st for attempting a. Can follow through their answer to (a) but it must be an expression in c. nd for an attempt to find the sum a+ a + a must see evidence of sum st ft for their sum put equal to 0. Follow through their values but answer must be in the form p + qc = 0 accept any correct equivalent answer () 5 GCE Core Mathematics C (666) January 0
5. (a) y y= y= Correct shape with a single crossing of each axis y = labelled or stated x= x= x x = labelled or stated () (b) (b) Horizontal translation so crosses the x-axis at (, 0) ± x ± x New equation is ( ) ( ) When x = 0 y = y = = for point (,0) identified - this may be marked on the sketch as on x axis. Accept x =. st for attempt at new equation and either numerator or denominator correct nd for setting x = 0 in their new equation and solving as far as y = for or exact equivalent. Must see y = or (0, ) or point marked on y-axis. Alternative f( ) = = scores A0 unless x =0 is seen or they write the point as (0, ) or give y = / Answers only: x =, y = / is full marks as is (,0) (0, /) Just and / is B0 A0 Special case : Translates unit to left (a) B0,, B0 (b) Mark (b) as before May score B0 A0 so /7 or may ignore sketch and start again scoring full marks for this part. () 7 GCE Core Mathematics C (666) January 0 5
6. (a) 0 S0 = [ a+ 9 d] or S0 = a+ a+ d+ a+ d+ a+ d+ a+ d+ a+ 5da+ 6d+ a+ 7d+ a+ 8d+ a+ 9d 6 = 0a + 5d * cso () u = a+ ( n ) d 7= a+ 5d (b) ( ) n () 0 ( b) gives 0a+ 50d = 70 (a) is 0a + 5d = 6 Subtract 5d = 8 so d =.6 o.e. Solving for a a = 7-5d so a = 9 (a) for use of S n with n = 0 () 7 (b) st for an attempt to eliminate a or d from their two linear equations nd for using their value of a or d to find the other value. GCE Core Mathematics C (666) January 0 6
7. ( f( x) ) x( c) x 8x = + + ( f( ) = 0 ) 0= ( ) + c c = 9 f( x) = x x + x+ 9 5 8. st n n for an attempt to integrate x x + st for at least terms in x correct - needn t be simplified, ignore +c nd for all the terms in x correct but they need not be simplified. No need for +c nd for using x = - and y =0 to form a linear equation in c. No +c gets M0A0 rd for c = 9. Final form of f(x) is not required. (a) b ac= ( k ) ( k) ( ) ( ) k 6k+ 9 k > 0 or k + 8k > 0 or better k + k > 0 * cso (b) ( k )( k )[ 0] + = Critical values are k = or - (choosing outside region) k > or k < - cao (a) st for attempt to find b acwith one of b or c correct nd for a correct inequality symbol and an attempt to expand. cso no incorrect working seen (b) st for an attempt to factorize or solve leading to k = ( values) nd for a method that leads them to choose the outside region. Can follow through their critical values. nd Allow, instead of or > loses the final < k < - scores A0 unless a correct version is seen before or after this one. () () 7 GCE Core Mathematics C (666) January 0 7
9. (a) ( 8 k 0) = so k = 5 (b) y = x+ k y = x+... and so m= o.e. (c) Perpendicular gradient = Equation of line is: y ( ) ft = x ft y+ x = 0 o.e. () () () c (d) y = 0, B(7,0) or x = 7 x = 7 or a ft () (e) ( 7 ) ( 0) AB = + AB = 5 or (b) for an attempt to rearrange to y = for clear statement that gradient is.5, can be m =.5 o.e. (c) ft for using the perpendicular gradient rule correctly on their.5 for an attempt at finding the equation of the line through A using their gradient. Allow a sign slip st ft for a correct equation of the line follow through their changed gradient () nd as printed or equivalent with integer coefficients allow y+ x= or y = x (d) for use of y = 0 to find x = in their equation c ft for x = 7 or a (e) for an attempt to find AB or AB for any correct surd form- need not be simplified GCE Core Mathematics C (666) January 0 8
0. (a) y (i) correct shape ( -ve cubic) Crossing at (-, 0) Through the origin Crossing at (,0) - O x (ii) branches in correct quadrants not crossing axes One intersection with cubic on each branch (6) (b) solutions Since only intersections (b) ft for a value that is compatible with their sketch dft This mark is dependent on the value being compatible with their sketch. For a comment relating the number of solutions to the number of intersections. ft dft () 8. (a) (b) (c) [ Only allow 0, or ] dy 7 = x x 8x dx 8 x= y = 6 9 + + 0 = - 7 + + 0 = -8 * cso x= 7 8 y = 6 7 = 7 = 7 Gradient of the normal = " " Equation of normal: y 8= ( x ) 7 7y x+ 6 = 0 ft () () (6) GCE Core Mathematics C (666) January 0 9
(a) st n n for an attempt to differentiate x x st for one correct term in x nd for terms in x correct rd for all correct x terms. No 0 term and no +c. (b) (c) st for substituting x = into y = and attempting note this is a printed answer Substitute x = into y (allow slips) Obtains.5 or equivalent nd for correct use of the perpendicular gradient rule using their gradient. (May be slip doing the division) Their gradient must have come from y rd for an attempt at equation of tangent or normal at P nd ft for correct use of their changed gradient to find normal at P. Depends on st, nd and rd Ms rd for any equivalent form with integer coefficients GCE Core Mathematics C (666) January 0 0
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