Mark Scheme (Results) Summer 01 Pearson Edexcel GCE in Further Pure Mathematics FP1 (6667/01)
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General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
EDEXCEL GCE MATHEMATICS 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) Marks should not be subdivided.. 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. All A marks are correct answer only (cao.), unless shown, for example, as A1 ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. 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. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.
General Principles for Further Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic: 1. 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 mn a, leading to x =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square b Solving x bx c 0 : x q c 0, q 0, leading to x = Method marks for differentiation and integration: 1. Differentiation Power of at least one term decreased by 1. ( x n x n1 ). Integration Power of at least one term increased by 1. ( x n x n1 )
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 small errors 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 may 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.
1.(a) (b) z p 1 i 1. i Multiplying top and bottom by z 1 i 1 i conjugate p pi i 5 p p, i 5 5 At least correct terms in the numerator, evidence that i = -1 and denominator real. Real + imaginary with i factored out. Accept single denominator with numerator in correct form. Accept a= and b=. z1 p p z 5 5 ` Accept their answers to part (a). Any erroneous i or i award M0 p p 5 5 1 p p or 1 5 5 p 8p16 p 8p 169 1 or 5 5 5p 0 5 p 81 p 9 z z 1 z1 A1, A1 1 or 1 d z d:attempt to solve their quadratic in p, dependent on both previous Ms. A1:both 9 and -9 da1 () OR z z 1 p 5 1 5 p 169 1 or oe 5 p Finding moduli Any erroneous i or i award M0 p oe Equating to 1 d 81 p 9 d:attempt to solve their quadratic in p, dependent on both previous Ms A1:both 9 and -9 da1 () Total 8
. (a) (b) f (1.1) 1.65960, f (1.5).0117 f( x) x x 5 x Sign change (and f( x) is continuous) therefore a root / α is between x 1.1 and x 1.5 5 f( x) x x x 5 15 f'( x) x x 15 (c) f (1.1) 1.1 1.1 8.585 5 Attempts to evaluate both f(1.1) and f(1.5) and evaluates at least one of them correctly to awrt (or trunc.) sf. Both values correct to awrt (or trunc.) sf, sign change (or a statement which implies this e.g. -1.6.. < 0 <.01..) and conclusion. : x term n n 1 x for at least one A1:Correct derivative oe Attempt to find f (1.1). Accept f (1.1) seen and their value. " 1.65960" 1.1 Correct application of N-R "8.585" 1.91 cao A1 A1 A1 () () () Total 7
. x px 0xq 0 (a) 1 5i (b) OR ( x 15i x 15 i ) x x 6 (( x ) x 15 i ) x ( 5 i) x(1 5 i) x x6 x x px 0x q : 1. Attempt to expand or. Use sum and product of the complex roots. A1: Correct expression Uses their third factor with their quadratic with at least terms in the expansion A1 p, q 5 May be seen in cubic A1, A1 f(1+5i)=0 or f(1-5i)=0 Substitute one complex root to achieve equations in p and / or q q p 0 and 10 p 0 0 Both equations correct oe A1 (1) Solving for p and q p, q 5 May be seen in cubic A1, A1 (5) 5 1 5i : Conjugate pair correctly positioned and labelled with 1+5i, 1-5i or (1,5),(1,-5) or axes labelled 1 and 5. (c) O 1 : The correctly positioned relative to conjugate pair and labelled. 5 1 5i () Total 8
. 1 1 A = 1, B = 5 1 1 (i)(a) (b) 1 5 6 1 1 5 6 11 1 1 5 1 11 1 1 1 15 5 BA 1 1 1 1 5 (ii) (det kk-- -1 C = : x matrix with a number or numerical expression for each element A:cao (-1 each error) Only 1 error award A1A0 Allow any convincing argument. E.g.s BA is a x matrix (so AB BA) or dimensionally different. Attempt to evaluate product not required. NB Not commutative only is B0 C=) Correct attempt at determinant 1 k k k 6 : 1 k their det C k A1:cao oe A A1 () () Total 7
5.(a) (r1 )r r 1 Proof by induction will usually score no more marks without use of standard results n n r 1 r r 1 r1 r1 r r 1 1 1. nn1n1. nn1, n 6 : An attempt to use at least one of the standard results correctly in summing at least terms of their expansion of 1 r A1 A1: Correct underlined expression oe : 1 n 1 n n n n 6 6 6 Attempt to factor out 1 n before given answer 1 1 n n cso A1 (6) n (b) rn1 r1 f( n) f( n)orf(n1) Require some use of the result in part (a) for method. 1 1 n.n 1.n.n 1 Correct expression A1 18 16 1 n n n 11 1 n n Accept a, b 11 A1 () Total 9
c 6. xy c ct (a) (b) (c) c dy c y c x c x x dx x 1 dy xy c x y 0 dx at,. dy dy dt c 1.. their dx dt dx t c t dy kx dx Correct use of product rule. The sum of two terms, one of which is correct and rhs = 0 dy 1 dt dx their dt dy dy dy c 1 c x or x y 0 or. dx dx dx t c Correct differentiation A1 or equivalent expressions c y their mt x ct or t c y mxcwith their mt and ( ct, )in c 1 y x ct ( t ) t d t t an attempt to find ' c '. Their m T must have come from calculus and should be a function of t or c or both c and t. t y x ct (Allow x t y ct) Correct solution only. A1* 1 (a) Candidates who derive x t y ct, by stating that m T, with no t justification score no marks in (a). ct c ct c ct c y 0 x A, 0 or equivalent form t t t y 0 x ct B( ct,0). ct ct c AB " ct " " " or PA= ct t t 1 and PB= ct t 1 Area APB = 1 c their AB t 1 ct c c c t 1 = ct t t t 1 8 t 1 t 1 8 8 8 81 or or or 8 t t t t Attempt to subtract their x-coordinates either way around. Valid complete method for the area of the triangle in terms of t or c and t. Use of c = and completes to one of the given forms oe simplest form. Final answer should be positive for A mark. A1 () () () Total 9
7.(i)(a) (b) (c) 0 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 : Multiplies their (b) x their (a) in the correct order A1: Correct matrix Correct matrix seen A1 A1 1 11 k k 6 det 6 1 1 1 6 6 k 6 "1" 6 1 k (ii) Area triangle T = Area triangle T = : Correct method for area for T A1: k : Correct method for determinant A1: 1 Uses 6 and their determinant correctly to form an equation in k. Accept 1 k or A1 A1 1 k A1 () (6) Total 10
8.(a) k 8k m Valid attempt to find gradient in terms of k k k k : Correct straight line method with their gradient in terms of k. y8k x k or k If using y mx cthen award M provided they attempt to find c yk x k or k A1: Correct equation. A1 8k If using y mx c, awardwhen they y x k 8k obtain c oe ky k x 16k ky x 8 k * or Correct completion to printed answer with ky 1k x k ky x 8 k * at least one intermediate step. A1* () (b) (Focus) (, 0) Seen or implied as a number (Directrix) x = - Seen or implied as a number k Attempt negative reciprocal of grad l Gradient of l is 1 as a function of k k Use of their changed gradient and y0 x numerical Focus in either formula, as printed oe, A1 k Substitute numerical directrix into their xy line ( y )6k oe A1 (7) Total 11
9. n n f( n) 8 is divisible by 6. Way Way 1 1 f(1) 8 6, Shows that f(1) 6 Assume that for n k, k k f( k) 8 is divisible by 6. f( 1) f( ) 8k 1 k 1 8k k k k k 8 k 81 1 7 8 k k Attempt f(k + 1) f(k) 6 8 k 8 k k 68 f k : Attempt f(k + 1) f(k) as a multiple of 6 A1 A1: rhs a correct multiple of 6 f( k 1) 6 8 k f k Completes to f(k + 1) = a multiple of 6 A1 If the result is true for n k, then it is now true for n k 1. As the result has been shown to be true for n 1, then the result is true for all n (.) A1cso Do not award final A if n defined incorrctly e.g. n is an integer award A0 (6) Total 6 1 1 f(1) 8 6, Shows that f(1) 6 Assume that for n k, k k f( k) 8 is divisible by 6. f( k 1) 8 k 1 k 1 88 k k k. k Attempts f(k + 1) in terms of k and 8 k f( k 1) 8 k 1 k 1 8f( k) k. k :Attempts f(k + 1) in terms of f(k) A1: rhs correct and a multiple of 6 f( k1) 8 f( k) 6. k Completes to f(k + 1) = a multiple of 6 If the result is true for n k, then it is now true for n k 1. As the result has been shown to be true for n 1, then the result is true for all n (.) 1 1 f(1) 8 6, Shows that f(1) 6 Assume that for n k, k k f( k) 8 is divisible by 6. A1 A1 A1cso f( k1) 8f( k) 8 k 1 1 8.8 k 8. Attempt f(k + 1) 8f(k) Any multiple m replacing 8 award f( k1) 8f( k) 8 k 18 k 1 8.. 6. : Attempt f(k+1) f(k) as a multiple of 6 A1 A1: rhs a correct multiple of 6 f( k1) 8 f( k) 6. k Completes to f(k + 1) = a multiple of 6 A1 General Form for multiple m f( k1) 6.8 k ( m)(8 k ) If the result is true for n k, then it is now true for n k 1. As the result has been shown to be true for n 1, then the result is true for all n (.) A1cso
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