Mark (Results) January 011 GCE GCE Core Mathematics C (6666) Paper 1 Edexcel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WC1V 7BH
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General Instructions for Marking 1. 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. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. 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 011 Core Mathematics C 6666 Mark 1. x cos x cos x xsin xdx= + dx sin x =... + π π [... ] = A1 0 A1 A1 [6]. ( ) At 3 di t 16ln 0.5 0.5 dt A1 di 16ln 0.5 0.5 dt = ln0.5= ln A1 t = ( ) 3 [5] GCE Core Mathematics C (6666) January 011 1
3. (a) (b) A B = + x 1 3x+ x 1 3x+ 5 ( )( ) ( ) ( ) 5= A 3x+ + B x 1 x 1 5= 5A A= 1 A1 5 x 5= 3 3 3 B B = A1 (3) dx ( )( ) 5 1 3 = dx x 1 3x+ x 1 3x+ = ln x 1 ln 3x+ + C ft constants A1ft A1ft ( ) ( ) ( ) (3) (c) 5 1 dx dy ( x 1)( 3x ) = + y ln ( x 1) ln ( 3x ) ln y ( C) K( x 1) Using ( ) + = + A1 y = depends on first two Ms in (c) dep 3x + K, 8 8 = depends on first two Ms in (c) dep 8 6( x 1) y = A1 (6) 3x + [1] GCE Core Mathematics C (6666) January 011
. (a) ( ) AB = i + j k i 3j + k = 3i+ 5j 3k A1 () (b) = 3 + + λ ( 3 + 5 3 ) or r = i + j k+ λ ( 3i+ 5j 3k ) r i j k i j k A1ft () (c) AC = i + pj k ( i 3j + k ) ( p ) = i + + 3 j 6k or CA B1 1 3 AC. AB = p + 3. 5 = 0 6 3 3+ 5p + 15 + 18 = 0 Leading to p = 6 A1 () (d) ( 1) ( 6 3) ( ) ( 6) AC = + + + = AC = 6 accept awrt 6.8 A1 () [10] GCE Core Mathematics C (6666) January 011 3
5. 3 = 1 x (a) ( x) 3 B1 3 3 3. 3 3. 3. 3 A1 1 x = 1+ ( ) x + x + x +... 1. 1..3 7 7 3 = 1+ 3 x+ x + x +... ( ) 1 3 7 7 3 3 x = + x+ x + x +... A1 (5) 16 8 f 1 3 7 7 x = a+ bx + x+ x + x +... 16 8 3a b Coefficient of x; + = 0 ( 3a+ b= 0) 7a 3b 9 Coefficient of x ; + = ( 9a+ b= 3) A1 either correct A1 16 16 Leading to a= 1, b= 3 A1 (5) (b) ( ) ( ) 3 (c) Coefficient of 7a 7b 7 7 + = 1 + 3 8 16 8 16 A1ft 7 = 16 cao A1 (3) 3 x is ( ) [13] GCE Core Mathematics C (6666) January 011
6. (a) dx 1 =, dt t dy t dt = dy t dx = A1 Using mm = 1, at t = 3 1 m = 18 A1 1 y 7= ( x ln3) 18 A1 (6) (b) x= ln t t = e x B1 x y = e A1 (3) V = π e dx x (c) ( ) ( x ) ( x e d e e x x = + ) dx x x e e = + x A1 ln x x e e π + x = π ( 6 3 + ln ) ( 8 + ln ) ln ( 36 ln ) = π + A1 Alternative to (c) using parameters ( ) d V = π t x dt dt ( ) 1 3 t dt = t t+ dt t t t = t + lnt A1 The limits are t = and t = t ln t π 6 3 ln 8 ln t π + = ( + ) ( + ) ( 36 ln ) = π + A1 (6) [15] (6) GCE Core Mathematics C (6666) January 011 5
7. (a) x= 3 y = 0.187 awrt B1 x= 5 y = 0.1667 awrt or 1 B1 6 () 1 0. 0.1667 0.187 0.175 B1 A1ft (b) I + + ( + ) 0.53 0.5 or 0.53 A1 () dx du B1 1 1 dx = ( u ) du + ( x 1) u 8 = du u A1 = u 8lnu A1 x= u = 5, x= 5 u = 6 B1 (c) ( u ) [ u 8ln u] 6 ( 1 8ln 6) ( 10 8ln 5) = 5 5 = + 8ln A1 6 (8) [1] GCE Core Mathematics C (6666) January 011 6
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