Mark Scheme (Results) Summer Pearson Edexcel International A Level in Further Pure Mathematics F1 (WFM01/01)

Transcription

1 Mark Scheme (Results) Summer 07 Pearson Edexcel International A Level in Further Pure Mathematics F (WFM0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. 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 07 Publications Code WFM0_0_706_MS All the material in this publication is copyright Pearson Education Ltd 07

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 PEARSON EDEXCEL IAL 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) 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 o.e. or equivalent (and appropriate) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark

5 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.

6 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:. 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:. 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 )

7 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.

8 Question Number May 07 WFM0 Further Pure Mathematics F Mark Scheme Scheme s Marks. x -5x += 0 has roots a, b a + b = 5, ab = a b + b a = a + b ab =... a + b = ( a + b) - ab =... a b + b 5 a = ( ) - ( ) ( ) = 9 9 = 9 Both a + b = 5 and ab =, seen or implied B Attempts to substitute at least one of their (a + b ) or their ab into a + b ab Use of a correct identity for a + b (May be implied by their work) dependent on ALL previous marks being awarded 9 or 57 9 or 6 or 6. o.e. from correct working Question s. Finding a + b = 5, ab = 5 + by writing down a, b =, 5 - or by applying 6 6 a + b = = and ab = = scores B0. M M A cso (4) 4 Those candidates who then apply a + b = 5, ab = having written down/applied a, b = 5 +, Give M0M0A0 for a b + b a = in part (a) can only score the M marks. ( ) 5 - ( 6 ) + Give M0M0A0 for a b + b a = a + b ab = 5 - ( 6 ) Give M0M0A0 for a b + b ( a = a + b ) - ab ab 5 + ( 6 ) = ( 6 ) ( ) = 5 + ( 6 )( 5-6 ) (( ) ( )) ( 5 + = )( 5-6 ) 5 + ( 6 )( 5-6 ) Allow B for both S = 5 and P = or for å = 5 and Õ = = 9 Give final A0 for 6. or 6. without reference to 9 or 57 9 or 6

9 Question Number. (a) AB = = = Scheme s Marks 4 - k k k k k + k - 4 Obtains a matrix consisting of 4 elements with at least two correct elements which can M be simplified or un-simplified Correct un-simplified matrix for AB A () (b) { det(ab) = 0 Þ } (-k)(-4) - ( + k) = 0 Þ 4k -56-6k = 0 Þ -k = 56 Þ k = - 56 or - 78 or -7 Applies "ad - bc" = 0 on their matrix for AB and solves the resulting equation to give k =... k = or - or -7 Accept any exact equivalent form for k Condone () 4 Question s. (a) Give A (ignore subsequent working) for a correct un-simplified answer which is later followed by an incorrect simplified answer. (b) Give MA for sight of the correct answer in part (b). Condone the sign error in applying... - ( + k) = 0 to give k = 0 (o.e.) E.g. Allow M for - k + k -4 = 0 Þ 4k k = 0 Þ k =... Give final A0 for -7.0 or - 7. or without reference to - 56 or - 78 or -7 M A

10 Question Number. Required to prove by induction the result Scheme s Marks n r(r + )(r + ) = - å, n λ + (n + )(n + ) r = Shows or states LHS = Way n = : LHS =, RHS = - ()() = (Assume the result is true for n = k ) k + r(r +)(r + ) = - å + (k +)(k + ) (k +)(k ++)(k ++ ) r = = - (k +)(k + ) + (k +)(k + )(k + ) = - (k + ) (k +)(k + )(k + ) + (k +)(k + )(k + ) or = - (k + ) - (k +)(k + )(k + ) and shows either RHS = - (+)( +) = or RHS = - ()() = or RHS = - 6 = Adds the (k +) th term to the sum of k terms dependent on the previous M mark Makes (k +)(k + )(k + )a common denominator for their second and third fractions = - Obtains - or (k + )(k + ) - (k ++)(k ++ ) (k + )(k + ) by correct solution only If the result is true for n = k, then it is true for n = k +. As the result has been shown to be true for n =, then the result is true for all n (λ + ) Final A is dependent on all previous marks being scored in that part. It is gained by candidates conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution. Way The MdMA marks for Alternative Way k + r(r +)(r + ) = - å + (k +)(k + ) (k +)(k ++)(k ++ ) r = = = (k +)(k + )(k + ) - (k + ) + () (k +)(k + )(k + ) Adds the (k +) th term to the sum of k terms dependent on the previous M mark Makes (k +)(k + )(k + )a common denominator for their three fractions k + 6k + 9k + 4 (k +)(k + )(k + ) = (k +)(k + 5k + 4) (k +)(k + )(k + ) = k + 5k + 4 (k + )(k + ) - = (k + )(k + ) (k + )(k + ) = - (k + )(k + ) Obtains - (k + )(k + ) or - (k ++)(k ++ ) by correct solution only B M dm A A cso M dm A (5) 5

11 Question s. LHS = RHS by itself or LHS = RHS = is not sufficient for the st B mark. Way The st A can be obtained by e.g. using algebra to show that å gives r(r +)(r + ) (k + 5k + 4) (k + )(k + ) and by using algebra to show that - (k + )(k + ) also gives (k + 5k + 4) (k + )(k + ) Moving from - (k +)(k + ) + (k +)(k + )(k + ) to - (k + )(k + ) with no intermediate working is nd M0 st A0 nd A0. k + r = Way The MdMA marks for Alternative Way k + r(r +)(r + ) = - å + (k +)(k + ) (k +)(k ++)(k ++ ) r = = - (k +)(k + ) + (k +)(k + ) - (k + )(k + ) = - (k + )(k + ) Obtains - (k + )(k + ) Adds the (k +) th term to the sum of k terms dependent on the previous M mark This step must be seen in Way or - (k ++)(k ++ ) by correct solution only M dm A

12 Question Number 4. (a) Way (a) Way (b) ì íx = 4t, y = 4 î t Scheme s Marks ü Þý 4 þ t - (4t) = 0 Substitutes x = 4t and y = 4 into the printed t equation to obtain an equation in t only 8t + 0t - = 0 or 4t + 5t - 6 = 0 (can be implied) (8t - 6)(t + ) = 0 Þ t =... or (4t - )(t + 4) = 0 Þ t =... or (4t - )(t + ) = 0 Þ t =... x = 4 4 = and y = 4 x = 4 - A, 6 x 0 + x ( 4 ) = 6 ( ) = - 8 and y = ( 4 -) = - A correct term quadratic : E.g. - 8t = 0t, 8t + 0t - = 0 { } or 8t + 0t = are acceptable for this mark dependent on the previous M mark Correct method (e.g. factorising, completing the square or applying the quadratic formula) of solving a TQ to find t =... dependent on both the previous M marks Correct substitution at least one of their values for t into the given parametric equations and obtains two sets of corresponding values for x =... and y =..., B( -8, -) or A: x =, y = 6 and B: x = -8, y = - Identifies the correct coordinates for A and B = 6 y -0 6 x - x = 0 y - 6 y = 0 x + 0x - 48 = 0 or x + 5x - 4 = 0 or x + 0 x - 6 = 0 or y - 5y - 6 = 0 or y - 0y - = 0 (can be implied) e.g. (x +6)(x - ) = 0 Þ x =... or (x + 8)(x - ) = 0 Þ x =... or (y -6)( y + ) = 0 Þ y =... E.g. x = Þ y = 6 x = - 8 Þ y = 6-8 = - A, 6 + (-8), 6 y = 6 Either substitutes their rearranged y - x = 0 into xy = k or substitutes either y = k x or x = k, k ¹ 0, into y - x = 0 y to form an equation in either x only or y only A correct term quadratic : 0x + x = 48, y - 0y = or x + 5x - 4 = 0 { } are acceptable for this mark dependent on the previous M mark Correct method (e.g. factorising, completing the square or applying the quadratic formula) of solving a TQ to find either x =... or y =... dependent on both the previous M marks. Correct substitution of at least one of their values for x or y into either y - x = 0or their rearranged y - x = 0 or y = k x or x = k, k ¹ 0, and y obtains two sets of corresponding values for x =... and y =..., B( -8, -) or A: x =, y = 6 and B: x = -8, y = - Identifies the correct coordinates for A and B + (- ) ; = - 5, 5 Uses their (x, y ) and (x, y ) from part (a) to apply x + x, y + y M A dm ddm A cao M A dm ddm A cao o.e. M; Correct answer A (5) (5) () 7

13 Question 4 s 4. (a) SC If the two previous M marks have been gained then award Special Case ddm for finding their correct points by writing either x =, y = 6 or x = -8, y = - or, 6 A decimal answer of e.g. A(, 5.), B -8, - Writing coordinates the wrong way round or (-8, -) ( ) (without a correct exact answer) is nd A0 E.g. writing x =, y = 6 and x = -8, y = - followed by A 6,, B( -8, -) is nd A0 Imply the dm mark for writing down the correct roots for their quadratic equation. E.g. x + 0x - 48 = 0 or x + 5x - 4 = 0 or x + 0 x = 6 x =, - 8 y - 5y - 6 = 0 or y - 0y - = 0 y = 6, - 8t + 0t = or 4t + 5t - 6 = 0 t = 4, - For example, give dm0 for 8t + 0t = or 4t + 5t - 6 = 0 t =, - [incorrect solution] 4 with no intermediate working. You can also imply the st A dm marks for either x 0 + x y -0 = 6 or 6 x y = 6 or y - 6 y 4 t - (4t) = 0 x =, -8 4 t - (4t) = 0 y = 6, - - x = 0 x =, -8 = 0 y = 6, - with no intermediate working. You can imply the st A dm ddm marks for either x 0 + x = 6 or 6 x - x = 0 x =, -8 and y = 6, - 4 t - (4t) = 0 x =, -8 and y = 6, - with no intermediate working. You can then imply the final A mark if they correctly identify the correct pairs of values or coordinates which relate to the point A and the point B. Give nd A0 for a final answer of both A, 6 (b) A decimal answer of e.g. -.5,.67, B( -8, -) and A -8, - ( ) (without a correct exact answer) is A0 Allow A for - 5, 0 6 or -, - or exact equivalent. ( ), B, 6,

14 Question Number Scheme s Marks 5. Given f(x) = 0-7 x - x5, x > 0 and root of f(x) = 0 lies in the interval [,.] (a) Way (a) Way Attempts to evaluate at least one of f() or f(.) f() = or f(.) = and evaluates f(.05) f() or f(.)correct awrt (or truncated) to sf f(.05) = and f(.05) correct awrt (or truncated) to sf dependent on the previous M mark f(.05) =... Evaluates f(.05) (and not f(.075)) f(.05)correct awrt (or truncated) to sf and correct interval. Allow or.05 < x <.05 f(.05) = or or.05 < a <.05 or [.05,.05] or (.05,.05)equivalent in words. Condone so interval is (.05,.05) or (.05,.050) Allow a mixture of ends. Do not allow incorrect statements such as.05 < a <.05 or (.05,.05) or unless they are recovered. Ignore the subsequent iteration of f(.075) that some candidates only indicate the sign of f and not its value. In this case the M marks can still score as defined but not the A marks. Common approach in the form of a table (use the mark scheme above) a f(a) b f(b) a + b ( ) f a + b so interval is.05 < a <.05 would score full marks in part (a) M A dm A (4) (b) f (x) = - 7 x- - 5x 4 At least one of either - 7 x ± Ax- or - x 5 ± Bx 4 M where A and B are non-zero constants. At least one of either - 7 x- or - 5x 4 simplified or un-simplified A { a = } Þ a =.04 ( dp) Correct differentiation simplified or un-simplified A dependent on the previous M mark Valid attempt at Newton-Raphson using dm their values of f() and f () dependent on all 4 previous marks.04 on their first iteration (Ignore any subsequent iterations) A cso cao Correct differentiation followed by a correct answer of.04 scores full marks in part (b) Correct answer with no working scores no marks in part (b) (5) 9 Question 5 s 5. (a) Give nd M0 for evaluating both f(.05) and f(.075) Do not allow interval = f(.05) to f(.05) unless recovered. A method of evaluating f(.05) followed by f(.05) with no evidence of evaluating at least one of either f() or f(.) is M0A0M0A0

15 Question 5 s Continued 5. (b) Incorrect differentiation followed by their estimate of a with no evidence of applying the NR formula is final dm0a0. Final This mark can be implied by applying at least one correct value of either f() or f () dm in - f() f (). So just - f() with an incorrect answer and no other evidence f () scores final dm0a0. You can imply the MAA marks for algebraic differentiation for either f () = - 7 ()- - 5() 4 f ()applied correctly in Differentiating INCORRECTLY to give f (x) = - 7 x- - 5x 4 leads to This response should be awarded MAA0MA0

16 Question Number Scheme s Marks n n n 6. (a) r (r +) = r + år {: Let f(n) = å å n(n +)(n + )(n +) r= r = r = or their answer to part (a).} Attempts to expand r (r +) and attempts to = 4 n (n +) + 6 n(n +)(n +) substitute at least one correct standard formula into M their resulting expression. Correct expression (or equivalent) A = n(n +) é ën(n +) + (n +) ù û = n(n +) é ë n + 7n + ù û dependent on the previous M mark Attempt to factorise at least n(n +)having attempted to substitute both standard formulae. {this step does not have to be written} = n(n +)(n + )(n +) Correct completion with no errors. : a =, b= (b) ì 49 Way ï ü r ï íå (r +) ý îï r = 5 þï = (49)(50)(5)(48) - (4)(5)(6)(7) { = = 44650} ì 49 ï ü ( r ï íå (r +) + ) ý îï r =5 þï = "44650" + 5(); = (b) ì 49 ï Way íå r (r +) + îï r =5 ( ) Attempts to find either f(49) - f(4) or f(49) - f(5). This mark can be implied. Correct numerical expression for f(49) - f(4) which can be simplified or un-simplified. : This mark can be implied by seeing Adds 5() or equivalent to their r (r +) 49 å 49 å r =5 or clear evidence that = (49) - (4) or 50 ü ï ý = (49)(50)(5)(48) + (49) - (4)(5)(6)(7) + (4) þï = ( ) - ( ) = = r =5 Attempts to find either f(49) - f(4) or f(49) - f(5) Correct numerical expression for f(49) - f(4) which can be simplified or un-simplified. : This mark can be implied by ( ) - ( ) or Adds 50 or equivalent to their å r (r +) or clear evidence that = (49) - (4) or 50 r =5 : This mark can be implied by (... + (49)) - (... + (4)) or å r =5 dm A cso M A M A cao M A M A cao (4) (4) (4) 8

17 Question Number Scheme s Marks 6. (b) ì 49 Way ï ü ( r ï íå (r +) + ) ý = å r + å r + å îï r =5 þï r = 5 r =5 r =5 = 4 (49) (50) - 4 (4) (5) + 6 (49)(50)(99) - 6 (4)(5)(49) + ( 98-48) = ( ) + ( ) + 50 = = å ( ) or = r + r + r = 5 = 4 (49) (50) + (49)(50)(99) + (49) 6-4 (4) (5) + (4)(5)(49) + (4) 6 = ( ) - ( ) = = Attempts to find either f(49) - f(4) or f(49) - f(5) Correct numerical expression for f(49) - f(4) which can be simplified or un-simplified Adds 50 or equivalent to their å r (r +) or clear evidence that = (49) - (4) or 50 M r =5 å r =5 M A A cao Question 6 s 6. (a) Applying e.g. n =, n = to the printed equation without applying the standard formulae to give a =, b= is M0A0M0A0 Alt Alt Method : Using 4 n n + 4 n + 6 n º (9 + b) a n4 + n (6 + b) + n + b a a a n o.e. dm Equating coefficients to find both a =... and b =... and at least one of a =, b= A cso Finds a =, b= and demonstrates the identity works for all of its terms. (4) Alt Alt Method : 4 n (n +) + 6 n(n +)(n +) º n(n +)(n + )(n + b) a dm Substitutes n =, n =, into this identity o.e. to find both a =... and b =... and at least one of a =, b= A Finds a =, b= Allow final dma for 4 n n + 4 n + 6 n or n(n +0n + 9n + ) or (n4 +0n + 9n + n) n(n +)(n + )(n +) from no incorrect working.

18 Question 6 s Continued { } 6. (b) Give st M st A0 for applying f(49) - f(5). i.e = You cannot follow through their incorrect answer from part (a) for the st A mark. Give MA0MA0 for applying éë f(49) + (49) ù û - éë f(5) + (4) ù û { } i.e = Give MA0M0A0 for applying éë f(49) + (49) ù û - éë f(5) + (5) ù û { } i.e = Give st M0 st A0 for applying (49) (50) - (4) (5) = = Give st M0 st A0 for applying (49) (50) - (5) (6) = = 0800 Give M0A0M0A0 for listing individual terms. e.g = Give nd M0 for lack of bracketing in (49)(50)(5)(48) + (49) - (4)(5)(6)(7) + (4) unless recovered Give M0A0M0A0 for writing down 44600without any working. Applying f(49) - f(4) for n(n +)(n + )(n +) is = is st M st A0

19 Question Number Scheme s Marks 7. f(z) = z 4 + 4z + 6z + 4z + a, a is a real constant. z = + i satisfies f(z) = 0 (a) z = (b)(i) { } - i - i B z - z + 5 f(x) = (z - z +5)(z + 6z +) { z + 6z + = 0 Þ} Either -6 ± 6-4()() z = () Attempt to expand (z - (+i))(z - (-i)) or (z - (+i))(z - (their complex z )) or any valid method to establish a quadratic factor e.g. z = ± i Þ z - = ± i Þ z - z + = - 4 or sum of roots, product of roots 5 to give z ± (their sum)z + (their product) M z - z + 5 A Attempts to find the other quadratic factor. e.g. using long division to obtain either z ± kz +..., k ¹ 0 or z ± az + b, b ¹ 0, a can be 0 or factorising e.g. f(z) = (z - z +5)(z ± k z ± c), k ¹ 0 or f(z) = (z - z +5)(z ± a z ± b), b ¹ 0, a can be 0 M z + 6z + A dependent on only the previous M mark Correct method of applying the quadratic formula or completing the square for solving a TQ on their nd quadratic factor (z + ) = 0 Þ z =... { z = } - + i, - - i - + i and - - i A (ii) { a = } or a = 65 stated anywhere in (b) B Question 7 s 7. (b)(i) No working leading to x = -+ i, -- i is M0A0M0A0M0A0. You can assume x º z for solutions in this question. Give dma for z + 6z + = 0 Þ z = - + i, - - i with no intermediate working. Special Case: If their second term quadratic factor can be factorised then give Special Case dm for correct factorisation leading to z =... Otherwise, give rd dm0 for applying a method of factorising to solve their TQ. Reminder: Method Mark for solving a TQ, az + bz + c = 0 Formula: Attempt to use the correct formula (with values for a, b and c) Completing the square z ± b ± q ± c = 0, q ¹ 0, leading to z =... dm () (6) () 8

20 Question Number Scheme s Marks 8. C : y = 6x, P(9 p, 8p) lies on C, where p is a constant. (a) y = 6x Þ dy dx = (6)x- = dy x dx = ± k x- y = 6x Þ y dy dx = 6 dy dx = dy dt. dt dx = 8 8p their dy dt dy py dx = q their dx dt M So at P, m T = p Correct calculus work leading to m T = p A (b) (c) y -8p = p (x - 9p ) or y = p x + 9 p Correct straight line method for an equation of a tangent where m T ( ¹ m N ) is found by using calculus. : m T must be a function of p leading to py - x = 9p (*) Correct solution only A * (Directrix: x = -9 Þ) a = 9 Tangent goes through (- a,6) Þ a = 9 or a = 9 stated anywhere in this question 6 p + 9 = 9p Substitutes their value x = -"a" or their value x = "a" and y = 6 into either py - x = 9p or py - x = - 9p 9p - 6p - 9 = 0 or p - p - = 0 E.g. p = 6 ± 6-4(9)(-9) (9) dependent on the previous M mark Correct method of solving their TQ {as p > 0} p = + 0 p = or or etc. A 8 8 : Give A0 for giving two values for p as their answer to part (c) () Uses a real value of p, which is the result of (d) x = 9 + 0, y = substituting (± a,6)into py - x = ± 9p, and substitutes p into at least one of either x = 9 p or y = 8p Either x = + 0 ( + 0, ) or A or y = or y = 6( + 0) ( + 0, 6( + 0) ) Correct coordinates of P. Condone A x =..., y =... : Give nd A0 for two sets of coordinates for P () M B M dm M (4) ()

21 Question Number { } ( 5) + (- 5) ; = 5 Scheme s Marks 5 5 ( ) + (- 5 ) or ( 5) + ( 5 ) 9. (a) z = or M 5 5 which can be implied. Correct exact answer A { arg z = arctan(-) = } = -. ( dp) -. cao or 5.8 cao B (b) Way w = li z = li ( 5-5 i) or w = 5li 5z = 5li (- i) Correct method of making w the subject and substituting for z li( 5 = + i) 5 5li(+ i) dependent on the previous M mark ( - i)( + i) = Multiplies numerator and denominator (- i)(+ i) of right hand side by ( + i) or (+ i) 5 5 dm = - + li 5 5-0l +5li to give an expression in terms + = of l which contains a real denominator = -l +li = -l +li -l +li or li -l A () (b) ( - i)(a + bi) = li Þ a + bi - ai + b = li Substitutes z and w into zw = li, Way (c) (d) a + b = 0 or - a + b = l a + b = 0, - a + b = l Þ a =... or b =... expands zw and attempts to equate either the real part of the imaginary part of the resulting equation. dependent on the previous M mark Obtains an equation in terms of a and b and obtains a second equation in terms of a, b and l and solves them simultaneously to give at least one of a =... or b =... { a = -l, b = l Þ} w = -l +li -l +li or li -l A 4 { (z + w) = } 4 ( ( - i 5 5 ) + (- + i 0 0 )); = - i Substitutes z, l and their w into 4 (z + w) M 5 C (, ) 5 0 Im O B (0, ) 0 Re M M dm - i or - 6 i or -0.4i o.e. 5 5 A Criteria plots (, - ) in quadrant plots (0, ) on the positive imaginary axis 0 plots (-, ) in quadrant 5 0 plots (0,- ) on the negative imaginary axis 5 Satisfies at least two of the four criteria B () () () D(0, ) 5 A(, ) 5 5 Satisfies all four criteria with some indication of scale or coordinates stated. All points (arrows) must be in the correct positions relative to each other. B () 0

22 Question 9 s 9. (a) M can be implied by awrt 0.45 or a truncated Give A0 for without reference to or or Give B0 for -. followed by a final answer of. (b) Be aware that ( - = + i i) 5 5

23 Question Number 0. (a) (b) (c) ì a ï í ï c îï = - - b d or Scheme s Marks ü ï = ý ï þï or = Correct matrix which is expressed in exact surds B Correct matrix which is expressed in exact surds B Multiplies their matrix from part (a) by their matrix from part (b) [either way round] and finds at least one element in the resulting matrix M At least correct exact elements A Correct exact matrix : Allow multiplication either way round Rotation (condone turn) (d) Rotation about (0, 0) B and about (0, 0) or about O or about the origin 05 degrees or 7p 05 degrees (anticlockwise) (anticlockwise) or 55 degrees clockwise or 7p clockwise B o.e. : Give nd B0 for 05 degrees clockwise () : Give B0B0 for combinations of transformations (e) Either sin75 = sin05 = + sin75 = sin05 = = + cos75 = -cos05 = - - or - or dependent on the st A mark in part (c) and states sin75 = sin05 = Question 0 s 0. (e) ALT -cos75 Comparing their matrix found in part (c) with a correct sin75 (representing a rotation 05 anti-clockwise about O) gives A db States cos75 = -cos05 and deduces a correct exact value for cos75 B -sin75 - cos75 sin75 = + (with the st A mark scored in part (c)) B cos75 = - - or - or 6-4 B () () () () 9 ()

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