Edexcel GCE Final Further Pure Maths Unit no. 667/0 June 006 (Results) Edexcel GCE Further Pure Maths 667/0
General Instructions. The total number of marks for the paper is 75.. Method (M) marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated.. Accuracy (A) marks can only be awarded if the relevant method (M) marks have been earned.. (B) marks are independent of method marks. 5. Method marks should not be subdivided. 6. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. Indicate this action by MR in the body of the script (but see also note 0). 7. If a candidate makes more than one attempt at any question: (a) If all but one attempt is crossed out, mark the attempt which is NOT crossed out. (b) If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 8. for each question, or part of a question, must appear in the right-hand margin and, in addition, total marks for each question, even where zero, must be ringed and appear in the right-hand margin and on the grid on the front of the answer book. It is important that a check is made to ensure that the totals in the right-hand margin of the ringed marks and of the unringed marks are equal. The total mark for the paper must be put on the top right-hand corner of the front cover of the answer book. 9. For methods of solution not in the mark scheme, allocate the available M and A marks in as closely equivalent a way as possible, and indicate this by the letters OS (outside scheme) put alongside in the body of the script. 0. All A marks are correct answer only (c.a.o.) unless shown, for example, as A f.t. to indicate that previous wrong working is to be followed through. In the body of the script the symbol should be used for correct f.t. and for incorrect f.t. After a misread, however, the subsequent A marks affected are treated as A f.t., but manifestly absurd answers should never be awarded A marks.. Ignore wrong working or incorrect statements following a correct answer.
June 006. (a) z+ iw= iz i w= i Adding z+ iz= + i Eliminating either variable + i z = + i A + i i z = + i i 8+ + i+ 6i = 5 = + i A () (b) arg z = π arctan.0 cao A () [6]. Use of r dθ π π Limits are 8 and ( θ ) 6a cos θ 8a cos = + sin θ ( + cosθ) dθ = θ + A A = a θ + sin θ π π 8 π π = a + ( 0 ) 8 π = a = a ( π ) cso A (7) [7]
June 006. (a) y = sinx+ 6xcosx y = cos x xsin x A Substituting cos x xsin x+ xsin x= kcos x k = A () (b) General solution is y = Acos x+ Bsin x+ xsin x 0, A = ( ) π π π π π, = B+ B = π y = cosx sinx+ xsinx Needs y = A () [8]. (a) + i is a solution x i x + i = x 6x+ ( )( ) ( x) = ( x x+ )( x + ax+ b) f 6 b = 6 Coefficients of x a 6= 6 or equivalent a = 0 A x + 6= 0 x= 6i, 6i A (7) (b) I O R Conjugate complex pair on imaginary axis Conjugate complex () pair in correct quadrants [9]
June 006 5. (a) (r+ ) = 8r + r + 6r+ (r ) = 8r r + 6r + = + ( A, B ) (r ) (r ) r = = A () Accept r = 0 B = and r = A+ B = 6 A= for both (b) (c) 5 ( n ) ( n ) n r = = + = + M + = n + r 0 0 r= r= ( ) n n + = r + n ft their B A Aft r = 8n + n + n = = n( n + n+ ) = n( n+ )( n+ ) 6 6 cso A (5) (r ) = (9r 6r+ ) = 9 0 8 6 0 + 0 6 = 980 A () [0]
June 006 6. (a) f ( 0.) 0.058, f ( 0.8) = 0.089 accept sf Change of sign (and continuity) α ( 0., 0.8) A () ( ) (b) f ( 0.6) 0.07 α ( 0., 0.6) f ( 0.5) 0.00 ( α ( 0.5, 0.6) ) f ( 0.55) 0.00 α ( 0.55, 0.6) accept sf A () (c) f() 0.05 at least sf cos x f ( x) = + A x f 0.8 at least sf A ( ) 0.05 β +.6 cao A (6) 0.8 If f ( ) 0.8 accepted for three marks A A. is produced without working, this is to be []
June 006 7. (a) Leading to ( x ) x + x 6= 6 x x + x 6= 0 + = 7 x = ± 7 surds required A x x x Leading to x x= 0 x= 0, A, A (6) + 6= 6 (b) Accept if parts (a) and (b) done in reverse order y Curved shape Line At least intersections () O x (c) Using all CVs and getting all into inequalities x > 7, x < 7 both Aft ft their greatest positive and their least negative CVs 0< x < A () []
June 006 0 t ln( 0 t) e = 0 t A 8. (a) dt = ln ( 0 t) ( ) ds S + = ( 0 t) dt ( 0 t) ( 0 t) d S = dt ( 0 t) ( 0 t) or integral equivalent S = + C 0 t 0 t A ( ) ( ) ( ) ( 0, 6) 6 0 0 C C = + = 600 ( 0 t) 0 t S = accept C = awrt 0.007 A (8) 600 (b) Substituting ( t) ds 0 = + dt 600 ds = 0 t = 5 A dt 9 8 S = (kg) A () []
June 006 8.Contd. Alternative forms for S are t t t+ 0 0 t S = 6 + = 0 600 600 ( )( ) ( t ) 600 + 90t t 565 = = 5 600 600 Alternative for part (b) S can be found without finding t Using d S = 0 in the original differential equation dt S 0 t = Substituting for t into the answer to part (a) 6S S = S A 600 Solving to 9 8 S = (kg) A ()