Mark Scheme (Results) Summer GCE Further Pure Mathematics 3 (6669/01R)

Transcription

1 Mark (Results) Summer 03 GCE Further Pure Mathematics 3 (6669/0R)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 03 Publications Code UA All the material in this publication is copyright Pearson Education Ltd 03

3 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.

4 EDEXCEL GCE MATHEMATICS 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. 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 4. All A marks are correct answer only (cao.), unless shown, for example, as A 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. 8. In some instances, the mark distributions (e.g. M, B and A) printed on the candidate s response may differ from the final mark scheme

5 General Principles for Pure Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving 3 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 mn = a, leading to x =. Formula Attempt to use correct formula (with values for a, b and c). 3. Completing the square b Solving 0 :, 0, leading to x =... x + bx+ c= x± ± q± c q 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 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. Answers without working The rubric says that these may not gain full 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.

6 Foci ( ±5, 0), Directrices x = ±. a 9 ( ± ) ae = ( ± )5 and( ± ) =( ± ) Correct equations (ignore ± s) B e 5 M: Solves using an 5 a 9 so e = = a = 9 appropriate method to find a 5 5 a or a MA or a = e a 3 e e = 5 = 3 = A: a = 9or a = ( ± )3 b = a e a b = 5 9 so b = 6 ( b= 4) b = a e b = 9 9 = 6 ( b= 4) 5 or ( ) b So x y = 9 6 M: Use of b a ( e ) = to obtain a numerical value for b or b A: : b = 6 or b= ( ± )4 x y M:Use of = with a b their a and b A: Correct hyperbola in any form. (7)

7 . i j k (a) n = l : (i j + k) + λ(4i + 3j + k) l : (3i + 7j + k) + λ(-4i + 6j + k) = 9i j + 36k M: Correct attempt at a vector product between 4i + 3j + k and -4i + 6j + k (if the method is unclear then components must be correct) allowing for the sign error in the y component. A: Any multiple of (3i + 4j - k) MA (b) Way a-a =±(i+8j+k) So p = 9 8 i ± 78 ± 78 p = = = 5 39 M: Attempt to subtract position vectors A: Correct vector ±(i + 8j + k) (Allow as coordinates) Correct formula for the distance using their vectors: " ±(i+8j +k) "" i n" " n" M: Correctly forms a scalar product in the numerator and Pythagoras in the denominator. (Dependent on the previous method mark) A: (not -) M d () (b) Way (i j + k).(3i + 4j - k) = -3 (d ) (3i + 7j + k).(3i + 4j - k) = 3 (d ) M: Attempt scalar product between their n and either position vector A: Both scalar products correct MA (5) p = ± d d " " " " n n or ( = ) d " n " Divides either of their scalar products by the magnitude of their normal vector. dord " n" M: Correct attempt to find the required distance i.e. subtracts their d d and " n" " n" or doubles their d " n" if d = d. (Dependent on the previous method mark) A: (not -) M d (5) Total 7

8 3. (a) y O P N x = 8 x A closed curve approximately symmetrical about both axes. A vertical line to the right of the curve. A horizontal line from any point on the ellipse to the vertical line with both P and N clearly marked. B () 3. (b) (c) x + 8 M is, y = (X, Y) 6cos 8 or θ +,3sin θ = ( X, Y ) (X 8) Y + = 36 9 M: Finds the mid-point of PN A: Correct mid-point M: Attempt cartesian equation A: Correct equation The next 3 marks are dependent on having the equation of a circle. Convincing argument allow Circle because equation may be follow through provided they do written ( x 4) + y = 3 have a circle! Can be implied by their centre and radius. M: Use their circle equation to The centre is (4, 0) and the radius is 3 find centre and radius A: Correct centre and radius Special Case: In (b) they assume the locus is a circle and find the intercepts on the x-axis as (, 0) and (7, 0) and hence deduce the centre (4, 0) and radius 3. This approach scores no marks in (b) but allow recovery in (c). MA Bft MA (4) (3) Total 8

9 s+ t 0 + s+ t 0 t M: Writes Π as a single vector A: Correct statement MA s+ t 0 + s+ t 0 t + s+ t+ 6 6t = + s+ 4t + t + s+ t+ 4 4t MA M: Correct attempt to multiply A: Correct vector in any form 8+ s 4t = 4+ s+ 6t 3+ s t Correct simplified vector B i j k n = 4 6 (8i 4j + 3k).(i k) = 8-6 r. (i - k) = 8 4 r = 4 + s + t 6 3 M: Attempts cross product of = 0i +0k their direction vectors A: Any multiple of 0i +0k Attempt scalar product of their normal vector with their position vector Correct equation (accept any correct equivalent e.g. r. (-0i + 0k) = -0) MA M A (9)

10 5(a) 5 5 n I n ( ) ( ) d n = x x nx x x n 5 5 n n 3 ( )( ) d I = nx x x x I = 5 3 ni + ni n n n n n (n+ ) I = ni * n n M: Parts in the correct direction including a valid attempt to integrate (x ) A: Fully correct application may be un-simplified. (Ignore limits) Obtains a correct (possibly un-simplified) expression using the limits 5 and and writes (x ) as(x )(x ) Replaces and I n- n x (x ) dxwith I n n x (x ) dx with Correct completion to printed answer with no errors seen B dm Acso (b) 5 0 ( ) d 5 = = (x ) = I 0 = B 5I = I+ 74and 3I = I0 + 4 M: Correctly applies the given reduction formula twice A: Correct equations for I and I (may be implied) 6 So I = and I =... or 3 Completes to obtains a I0 + 4 numerical expression for I 5I = + 74 and I =... 3 dm 54 I = 5 B (5) Total 0 (5)

11 6. (a) M: Multiplies the given matrix by the given eigenvector M: Equates the x value to λ b 0 =..., = λ, λ = 8 M, M, A a A: λ = 8 (3) (b) M: Their + b = λ or 8 their a + = 0 + b = "8" So a = - and b = 7 A: b = 7 or a = - a + 0 A: b = 7 and a = - A (3) (c) 4 λ 3 7 λ 0 = 0 8 λ M (4 λ)(7 λ)(8 λ) (8 λ) + 3( + (7 λ)) = 0 Correct attempt to establish the Characteristic Equation. = 0 is required but may be implied by later work Allow this mark if the equation is in terms of a, b, c Attempts to factorise i.e. (8 λ)(30 λ+ λ ) or (6 λ)(40 3 λ+ λ ) 3 or (5 λ)(48 4 λ+ λ ) (NB 40 8λ + 9λ λ = 0) M: Attempt to factorise their cubic an attempt to identify a linear factor and processes to obtain a simplified quadratic factor A: Correct factorisation into one linear and one quadratic factor M: Solves their equation to Eigenvalues are 5 and 6 obtain the other eigenvalues A: 5 and 6 (5) Total 8

12 7(a) Put 6cosh x = 9 - sinh x M 6 ( e + e ) = 9 ( e e ) x x x x Replaces coshx and sinhx by the correct exponential forms M x x 4e + e 9= 0 + = x x 4e 9e 0 M: Multiplies by e x A: Correct quadratic in e x in any form with terms collected (b) x So or e = and x = ln or ln 4 Area is (9 sinh x 6 cosh x) dx ± (9x cosh x 6sinh x) or x x ± (9x 4e + e ) 4 M: Solves their quadratic in e x A: Correct values of x (Any correct equivalent form) (9 sinh x 6cosh x) dx or (6cosh x (9 sinh x))dx or the equivalent in exponential form M: Attempt to integrate A: Correct integration ± ([ 9ln cosh ln 6sinh ln ] -[ ] M 9ln coshln 6sinhln ) dm Complete substitution of their limits from part (a). Depends on both previous M s ln ln ln ln ln ln ln ln = ± (9ln( ) (e + e ) 3(e e ) + (e + e ) + 3(e e )) M 4 Combines logs correctly and uses cosh and sinh of ln correctly at least once (6) 5 8 9ln =9ln8 4 or 7ln 4 4 Any correct equivalent Subtracting the wrong way round could score 5/6 max Note x x If they use 4e 9e + in (b) to find the area no marks Acao (6) Total

13 8(a) dy = x dx Correct derivative (may be unsimplified) B s = 8 + ( x ) dx = ( + )dx x A correct formula quoted or implied. There must be some working before the printed answer. () (b) dx x = sinh u = sinh ucosh u Correct derivative B du cosh u ( + ) coth ( + x) = + cosech u = coth u x = u or ( + x) = sinh u B (may be implied by later work) s = coth u.sinh ucosh udu= M: Complete substitution cosh udu A: cosh udu = M: Uses cosh u = ± cosh u± or changes to exponentials in an attempt u+ sinh u or u u 4e + u 4e to integrate an expression of the form kcosh u A: Correct integration x= 8 u = arsinh 8 = ln(3 + ), x= u = arsinh= ln(+ ) or [ u+ sinh u] arsinh 8 arsinh arsinh 8 u u 4e + u 4e arsinh = arsinh 8 + sinh(arsinh 8) (arsinh+ sinh(arsinh)) arsinh 8 -arsinh = e + arsinh 8 e or arsinh x sinh(arsinh x) + = arsinh 8 + sinh(arsinh 8) (arsinh+ sinh(arsinh)) M: The limits arsinh 8 and arsinhor their ln(3 + ) and ln( + ) used correctly in their f(u) or the limits 8 and used correctly if they revert to x Dependent on both previous M s A: A completely correct expression ln( + ) + 5 A B d ddma (9) Total

14 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA Summer 03 For more information on Edexcel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE