Edexcel GCE. Mathematics. Pure Mathematics P Summer FINAL Mark Scheme. Mathematics. Edexcel GCE
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1 Edexcel GCE Mathematics Pure Mathematics P 667 Summer 00 FINAL Mark Scheme Edexcel GCE Mathematics
2 General Instructions. The total number of marks for the paper is 7.. 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.. 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. Marks 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. 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
3 June Pure P Mark Scheme Question Number Scheme Marks (a) dy dx = 6 8x M A () (b) yd x 6x x C M AA [] (a) M is for x n x n in at least one term, 6 or x is sufficient. A is fully correct answer. Ignore subsequent working. (b) M: Correct power of x in at least one term (C sufficient) First A: 6x + C Second A: + x 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
4 (a) a = B x 8x 9 ( x ) 6 ( 9), b = MA (b) x ( ), [ ] x cao c =, d = Or: x, 80 6 M A M A A [6] (a) B is for ( x ) or a M requires ( x p) p ( 9), p 0, A is for b Answer only: full marks. Note: Can score B0MA [e.g. (x + ) ] Comparing coefficients: M is for comparing x coefficients and constant term (b) M is for full method leading to x " a"... or x... (formula) First A: c = Second A: d = (or ) one of Notes: (i) If not seen anywhere withhold final A, so answer only c =, d = (or ) ( one of ) is MAA0 ( so is + ) (ii) x " a" ve number can score MA for x = +. (iii) using correct quadratic formula with a, b 8, c 9 ending with x = 8 6 score as MAA Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
5 Question Number Scheme Marks (a) r 6,. (*) MA () (b) Area of sector OAB = r. ( 7.) MA Area of triangle OCD = 0 sin. (...) Shaded area = 7.. = 97 m cao MA A [7] () (a) M is for applying correct formula or quoting and attempting to use correct formula (b) For each area M is for attempting to use correct formula or complete method in case of Δ * A is for a numerically correct statement (answer is not required just there as check) Final A is for 97 only. e.g. splitting triangle into two triangles: For guidance 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
6 (a) y ( ) ( x 9) MA x y 0 or y x 0 (condone terms equation: e.g. (b) Equation of l : y x y ) Solve l and simultaneously to find P: l x, y 6 (c) C: 0, 7 or OC = 7 (may be on diagram) x A B M AA B () Area of triangle OCP = OC xp 0 (must be exact) MA [0] (a) M for finding the equation of a straight line: If using y y ( ) m x x or equivalent, m must be ⅓. If using y m x c, m must be ⅓ and c found. First A: unsimplified form [ y 9 = ⅓(x + ) M0, y = ⅓ (x -9) MA0, y = ⅓ (x + 9) M0 ] (b) M: solving two linear equations to form linear equation in one variable A for first coordinate if correct A f.t. : For second coordinate correct for candidate after substituting in y = x Watch (-, 6) [which usually scores MA0A ] (c) B f.t. : Correct y value when x = 0 in candidate s equation in (a) M: For ½ x candidate s OC x candidate s x co-ord in (b) SC: If x found when y = 0, allow M for finding area for their configuration. 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
7 (a) arctan 6. ( ) seen anywhere 0, ( 0) 80 ( 6.), 80 (.7) (One of these) x = 7.9,., 7. B MM M AA (6) (b) sin 0 x ( sin x) or ( cos 9 x ) cos x 0 9 M sin x or 9 8 cos x or 9 tan x or sec x or cos x = 8 9 A x 9., 9. AA () [0] (a) First M: Subtracting (allow adding) 0º from Second M: Dividing that result by (order vital!) [So. gains BMM] Third M: Giving a third quadrant result First A is for correct solutions, Second A for third correct solution. B: Allow 0.98 (rads) or 6.6 (grad), and possible Ms but A0A0] (b) M for use of sin x cos x or sin x and cos x in terms of cosx For use of tan (A + B) see separate sheet Note : Max. deduction of for not correcting to dec. place. Record as 0 first time occurs but then treat as f.t. Answers outside given interval, ignore Extra answers in range, max. deduction of in each part [Final mark] (i.e. or more answers within interval in (a), from any gained A marks; or more answers within interval in (b), from any gained A marks 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
8 6 (a) S = a + (a + d) + (a + d) + + [a + (n )d] S = [a + (n )d] + [a + (n )d] + + a or equiv. B M Add: S = n[a + (n )d] S = n[a + (n )d] cso (*) (b), 8, (c) a = d = n (*) Sum = n nn M A B B M A () () (d) 00 Finding e.g. 00 r r (= 0000) M r or stated Sum of first terms: 00 r r r S(00) S() ALT: Working with, 8,,. B M A () [] a = B; Finding n and d M Applying S = n[a + (n )d] with candidate s, n= 9 or 96, d = M (a) B: requires min of terms, including the last. (b) First M generous; second M hard. Note: Result is given so check working carefully. (c) For B f.t. terms must be in AP, But allow M for candidate s a and d in given result in (a) (d) First M for substitution of 00 in result from (c) S.C. Allow second M for S(00) S() 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
9 7 (a) dy dx x 6 M A (b) x 6 0, x x = (*) x x 6x 0 dx x 0x x x 0x 0 ( = = 9.8 ) Finding area of trapezium = 6 ( ) 6 0 [A = (, 6), B = (, )] M A () M A A M A M A Or by integration: x x Area of R = 9.8 =. (a) First M for decrease of in power of x of at least one term (disappearance of 0 sufficient) A (8) [] dy Second M for putting 0 dx and finding x =. (b) First M: Power of at least one term increased by / x First A: For Second A: For x 0x Second M for limits requires (allow candidate s ) and some processing of integral, y is M0 A requires and substituted in candidate s -termed integrand (unsimplified) x y A B B or d x A correct unsimplified Area of trapezium: M attempt at ½ y ( x ) See separate sheet for " line curve" 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
10 8 (a) f (some working needed) = 0, so (x ) is a factor of (b) x x 0x x (c) and x x (d) f ( x) x 6x f = fx M A M A A B M A A () () (e) x 6x = M x 6x 0 (x 7)( x ) 0 x = dep M A x-coordinate of S is 7 (a) M is for substituting x = in f(x) A requires f = 0 and statement or or.6 or.67 6 (b) M for quadratic factor x ax SC: M for one other linear factor found (d) M for differentiation: at least one term has power of x reduced by (e) First M : equating their gradient function to their answer to (d) Second M: Solving by factors requires ax bx c ( mx p)( nx q) where pq c and mn a, leading to x... Solving by quadratic formula requires attempt to use correct formula with candidate s values of a, b and c used. A f.t. only for correct (real) answers to their quadratic Final A for a correct exact form. A () [] 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
11 EXTRAS (a) Using expansion of tan( x 0 ) Getting as far as tan x = number (0.78..) tan x = 6.º, 6.º,.7º 6.º First M B x =. º, 7. º, 7.9 º Third quad result Divide by Answers as scheme Third M Second M AA 6(c) r B 7(b) n( n ) = n n n n(n ) = * (cso) Attempting integral ( equation of line equation of curve)! Performing integration: 8 = ( x x ) dx M A Third M Fourth A First M 8 7 ( x x ) ( x ) Limits MA Answer A ( 8 7 ( x x ) allow as follow through in this case. x ) First A Second A Second M Third A Fifth A 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
12 GENERAL PRINCIPLES FOR P MARKING Method mark for solving term quadratic:. Factorisation ( x bx c) ( x p)( x q), where pq c, leading to x = ( ax bx c) ( mx p)( nx q), where pq c and. Formula Attempt to use correct formula (with values for a, b and c). mn a, leading to x =. Completing the square Solving x bx c 0 : ( x p) q c, p 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but will be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may gain no credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice. Misreads (See the next sheet for a simple example). A misread must be consistent for the whole question to be interpreted as such. These are not common. In clear cases, please deduct the first A (or B) marks which would have been lost by following the scheme. (Note that marks is the maximum misread penalty, but that misreads which alter the nature or difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written). Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written. 667 Pure Maths P Final MS June 00 Advanced Subsidiary/Advanced Level in GCE Mathematics
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