January 2015 (IAL) PhysicsAndMathsTutor.com. Mark Scheme (Results) January Pearson Edexcel International A Level Core Mathematics 12 (WMA01_01)

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1 January 05 (IAL) Mark (Results) January 05 Pearson Edexcel International A Level Core Mathematics (WMA0_0)

2 January 05 (IAL) 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: January 05 Publications Code IA008 All the material in this publication is copyright Pearson Education Ltd 05

3 January 05 (IAL) 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 January 05 (IAL) EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5.. 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.. 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) 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. 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 January 05 (IAL) 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 January 05 (IAL) General Principles for Core 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 =. Formula Attempt to use correct formula (with values for a, b and c). a, leading to x =. Completing the square Solving x + bx + c = 0 : b ( x ) q c, 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 small 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.

7 January 05 (IAL) January 05 International A Level WMA0/0 Core Mathematics C Mark Number. (a) (b) x x or x or 0.5x B [] B, B [] marks (a) B: This answer only (b) B: For k x as final answer, k can even be 0. Also accept for B but is not simplified and is B0 B: for x to power (independent mark) so kx with k a constant (could even be ) as final answer n.b. Can score B0B or BB0 or B0B0 or BB Mark the final answer on this question 5 Also note : Candidates who misread question as x should get x This is awarded BB0 x Special case: The answer x is awarded B0 B as x may be in a bracket with power outside.

8 January 05 (IAL) Number. x 5 8 y (a) State h =, or use of B aef { (.50.5 )} For structure of { }... A {7.67} ( = 6.555) = awrt 6.6 A [] (b) (b) Adds 9+ half of their answer from (a) seen ( allow use of half of 6.555) So required estimate = 9 +. =. Way : Begins again with trapezium rule A [] 7 marks x 5 8 y Uses { (.5+.76) } =. A [] (a) B: for using or.5 or equivalent or just states h = : requires the correct{...} bracket structure. It needs the first bracket to contain first y value plus last y value and the second bracket to be multiplied by and to be the summation of the remaining y values in the table with no additional values. If the only mistake is a copying error or is to omit one value from nd bracket this may be regarded as a slip and the M mark can be allowed (An extra repeated term forfeits the M mark however). M0 if values used in brackets are x values instead of y values A: for the completely correct bracket {...} A: for answer which rounds to 6.6 after attempt at trapezium rule NB: Separate trapezia may be used: B for.5, for / h(a + b) used or times (and A if it is all correct ) Then A for 6.6. Special case: Bracketing mistake.5 ( ) + ( ) scores B A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). (b) Way : : Adds Area of Rectangle [ ] : Half answer to part (a) seen A: Accept awrt. = 9 or dx = x Or Way : (If they begin again with a trapezium rule) : for correct table : for correct use of trapezium rule A: awrt. to their. or to their 6.6 or to their 5.9

9 January 05 (IAL) Number. (a) Shape- similar to before but with indication of stretch in y direction by at least one correct from the three traits: y intercept, (0, 7) maximum point (, ) or asymptote indicated at 9 Intercept (0,7), max (,) and x intercept (.5,0) all three of these seen B B B [] Shape (reflection in y axis) B (b) (,), (0,9) and (.5,0) seen B y = (must be equation) B [] 6 marks

10 January 05 (IAL) (a) B: Correct shape with curve crossing x axis and one label correct from the three listed (i.e. a correct new y value). Condone slight imperfections in the curvature of the sketches. B: All three specified labels given to indicate the three new point positions. Do not need coordinates if clearly labelled on the axes. Accept 7 and accept.5 and even allow (7, 0) and (0,.5) on y and x axes respectively. B: Equation of asymptote correct (asymptote on figure takes precedence) Asymptote does not need to be drawn dotted. (b) B: Correct shape (maximum in nd quadrant, intercept on negative x - axis and approaches asymptote for large positive x) Condone slight imperfections in the curvature of the sketches. B: All three specified labels given to indicate the three new point positions. Accept 9 and accept -.5 and even allow (9, 0) and (0, -.5) on y and x axes respectively. B: Equation of asymptote correct (asymptote on figure takes precedence) Do not award this mark if they merely copy the original graph. If there is no sketch the maximum mark in part (a) is B0BB and in part (b) is B0BB0 so /6 Special case: Stretch in y direction of scale factor /. If there is a graph of the correct shape with (0,), (, /), (.5,0) and asymptote y = then award B0B0B

11 January 05 (IAL) Number. (a) 0 x x 0 8 x 0 7 x + = = 0, + 80x+ 70x + 0 x... B, A A [] (b) State or Use x = 0. B Estimate = 0 + "80" (0.) + 0 (0.)... = 59. or 59.0 or 59 or 59. (after correct working) A [] 7 marks (a) : The method mark is awarded for an attempt at Binomial to get one or more of the terms in x need correct binomial coefficient multiplied by the correct power of x. Ignore bracket errors or errors (or omissions) in powers of or or bracket errors. Accept any notation for 0 C, 0 C and 0 C, e.g. 0, 0 and 0 (unsimplified) or 0. 5 and 0 from Pascal s triangle This mark may be given if no working is shown, but any of the terms including x is correct. 0 B: must be simplified to 0 ( writing just is B0 ). If miscopied later then isw A: is cao and is for two correct from 80 x, A: is c.a.o and is for all of 80 x, 70x and 70x and 0x 0x correct (ignore extra terms) if divided by or then isw Allow terms given separately without + signs and with commas. Ignore extra terms. Ignore subsequent work once correct answer is seen in simplified form. N.B. If the series is given in Descending Order the first M mark may be awarded and if the whole expansion is given (all terms) then full marks is possible. (b) B: States or Uses x = 0. x : Uses their solution of = 0.05 substituted in to their series expansion If no equation stated could see evidence of use of 0. or 0.0 (not 0.05) substituted consistently for example A: This is cao and must follow. NB 59.5 or 59.5 is A0 (used 0.05 ) But correct working followed by an answer 59 or 59. can be awarded A

12 January 05 (IAL) Number 5. (a) S = a + ( a+ d) + ( a+ d) ( a+ ( n ) d) n S = ( a+ ( n ) d) + ( a+ ( n ) d) ( a+ d) + a n (b) Sn = ( a+ ( n ) d) + ( a+ ( n ) d) ( a+ ( n ) d) n Sn = [ a+ ( n ) d] * See notes below for those who use triangle numbers in their A* proof Uses either n or ( 97) 7 ( a+ ( n )7) i i= 7 i.e ( ) or ( ) or (7 + 97) or 7 (7) A [] = 789 A [] 7 marks (a) : List terms including at least first two and a last term which may be a + nd or a + (n )d or L : List terms in reverse including at least their last term ( or correct last term) and finally their first term : The LHS should be S. The RHS must follow from at least two terms correctly matching in the addition and should include at least two terms which are each correctly{ a+ ( n ) d} or (a + L) or should be n{ a+ ( n ) d} or n(a + L) A: Need some indication of at least three terms being added (i.e at least three terms and their pairs listed with terms correctly matching or three additions seen) and also need to achieve final answer with no errors and if L was used need to state that L = a + (n )d NB: Some candidates use a variation of n n n n ( n ) ( a + ( r ) d = a + d ( r ) = na + d ( n + ) dn or na + d ( n) r= r= r= n And conclude that Sn = [ a+ ( n ) d]. This gains the full marks A, but must be completely correct. (b) :Uses correct formula (with their a and n) with d =7 or with last term correct A: Uses consistent and correct a and n A: Correct answer

13 January 05 (IAL) Number 6. (a) Use or state log ( ) log ( ) Use or state x+ = x+ log or = = (x + ) Use or state log x+ log (x ) = log x(x ) or log (x+ ) log x= log etc x (x+ ) = x(x ) or equivalent including correct rational equations A Then 9 8 x + x+ = x x and so x 6x 9= 0 * A* (b) (x + )(x - 9) = 0 so x = (or use other method e.g formula or completion of square) 9 x = ( or ) A [] 7 marks (a) : Uses power law for logs : Connects with correctly : Uses addition ( or subtraction) law correctly (x + ) e.g. log x+ log (x ) = log x(x ) or log (x+ ) log x= log or x (x + ) log (x+ ) log x log (x ) = log or even log x+ log = logxor x(x ) log (x ) + log = log(x ) or log (x ) + log + log x= log x(x ) etc (x + ) A: Correct equation (unsimplified) after correct work. e.g. = x(x ) A: Obtains printed answer correctly (This is a given answer so needs previous A mark to have been awarded and needs correct expansion) Special case : log ( x + ) x + x + 9 log ( x + ) = + log x( x ) so = so = log x( x ) x x This can have,,, A0, A0 so /5 losing accuracy because of the error in the second step. (b) Some candidates who did not achieve marks in part (a) begin the log work again and make more progress here. Mark the better work. So credit for (a) may be given in (b). Credit for (b) should not be given in (a) : Uses solution of their quadratic or of printed quadratic(see notes). This must be in part (b) 9 A: x =.5 and discards x = (any equivalent form) Giving x =, is A0 This must be in part (b) [5]

14 January 05 (IAL) number 7 (a) Obtain (b) (c) ( x ± 5) and ( y ± ) Centre is ( 5, ). See r = ( x 5) ( y ) 6 ± + ± = (= r ) or A ( r = )"5" + "9" 8 Use x = in either form of equation of circle to obtain simplified quadratic in y e.g x ( ) ( y ) ( y ) = = 6 = or y y y y ( ) ( ) = 0 6 = 0 solve resulting quadratic to give y = A [] [] Alternatives (a) OR (b) (c) y = ± A, A [] 8 marks Method : From x + y + gx + fy + c = 0 centre is ( ± g, ± f) Centre is ( g, f), and so centre is ( 5, ). A Method : Use any value of y to give two points (L and M) on circle. x coordinate of mid point of LM is -5 and Use any value of x to give two points (P and Q) on circle. y co-ordinate of mid point of PQ is (Centre chord theorem). (-5, ) is A Method : Using g + f c or A () ( r = )"5" + "9" 8 r = A () Method : Divide triangle PTQ and use Pythagoras with r ( " 5") = h, then evaluate " ± h" - then get ± A A ()

15 January 05 (IAL) Mark (a) and (b) together (a) as in scheme and can be implied by ( ± 5, ± ) A: for correct centre and (-5, ) (without working) implies A (b) for a complete and correct method leading to r = "5" + "9" 8 or r = "5" + "9" 8 or for using equation of circle in N.B. r = k or r ( x 5) ( y ) k ± + ± = form to identify r=k = k is M0 Also is M0 and r = "5" + "9" (without the 8)is M0 A r = (only and not with r = -) Again correct answer with no working implies A Special case: if centre is given as (5, ) or (5, ) or (-5, -) allow A for r = worked correctly as ( r = )"5" + "9" 8 i.e if they obtain r = after sign error give final A (So A0A) (c) For substituting x = - into an equation for the circle and attempt to simplify to term quadratic or to ( y a) = b For attempting to solve their quadratic (following usual rules see notes) A, A Answers must be given as surds A for each correct answer. To earn both A marks, answers must be simplified.

16 January 05 (IAL) Number 8. u = k, u = ( u ) u = k, u = 9k 8 A (a) u = (9k 8) = 7k 56 (ft their u ). Aft [] (b) 7k 56 = 5 so k = (c) ui = or i= 9 k = 6 or or 6. (sf) A [] ui = k+ k + 9k 8 + 7k 56 i= = 0k 6, = 7 or Aft, Acao [] 9 marks (a) : Attempt to use formula twice to find uand u A: two correct simplified answers : Attempt again to find u Aft: th term correct and simplified - follow through their u (b) : Put their th term ( not 5 th ) equal to 5 and attempt to find k = A: accept any correct fraction or decimal answer (allow 6. or better here) (c) : Uses st term and their following terms with plus signs (either numerical or in terms of k). Must be using terms from iteration and not formula for an AP or GP. May make a copying slip. Aft: for 0k 6 or follow through on their k so check 0k -6 for their k A: obtains 7 (must be exact) if exact answer given, then isw Those who use 6. will obtain 6 They should have AftA0 should have used exact k to give exact answer here. Those who use 6. will obtain 7. This should have AftA0 should have used exact k to give exact answer here. Those who use 6. will obtain 7. This should have AftA0 should have used exact k to give exact answer here. 6. will obtain 7. This should have AftA0 should have used exact k to give exact answer here. 6. will obtain 7. etc All these answers should have AftA0 should have used exact k to give exact answer here. Etc Special case: Those who use k = 6 will obtain = This is A0 A0 in part (c) as over simplified

17 January 05 (IAL) Number 9. (a) 5 = cos XAB, or (b) XAB = 0. or 0.π cos XAB = or or 0.95 or Area of sector is r θ = 0 θ Area of major sector is r (π "0.") or π r r "0.") = 7 A (c) area of triangle AXB = 0 sin XAB Way : Find angle XBA and hence area XBY area of kite = triangleaxb Area of kite = area of XBY + Area XAY d = awrt 9 = = 9 A Way : Finds length XY by cosine rule or elementary trigonometry (8.7) [] Uses area of kite = "8.7" d = awrt 9 A [] 8 marks (a) : Uses cosine rule must be a correct statement, allow statement 5 = cos XAB A: accept awrt 0. (answers in degrees gain A0). Also 0. is A0 θ (b) : Uses area formula with r = 0 and any angle in radians. If they use degrees they must use the formula π0 60 : Finds angle in major sector ft their angle from (a) and uses sector formula or subtracts minor area from circle (allow work in degrees) Must use ( π "0.") but r may be 5 instead of 0 for this mark A: Accept awrt 7 (may reach this using degrees) (c) Way : : Finds area of triangle AXB, using 0, and their angle XAB d: Doubles area of triangle AXB Way : : Finds angle XBA ( ) by valid method (cosine rule) (NOT 90 XAB) and hence area XBY = 5 5 sin.96 d: Adds areas of triangles XBY and XAY ( 7.98 and.76) Way : : Finds length XY by cosine rule or elementary trigonometry (8.7) d: Uses area of kite = "8.7" For each method A: awrt 9- do not need units A [] []

18 January 05 (IAL) Number 0. 6x + ax + bx 5 f (x) = (a) Attempts f ( ± ) or Attempts f( ± ) Or Use long division as far as remainder* Obtains 6( ) + a( ) + b( ) 5 = 0 or -6 + a b 5 = 0 or a b = or equivalent Obtains 6() + a() + b() 5= 5or 6 a b = 5 or a + b = - or equivalent Solve simultaneous equations to obtain a = -7 and b = -8 (b) 6x + ax + bx 5 = ( x + )(6 x +... x +...) A A A [5] 6x 7x 8x 5 ( x )(6 x x 5) = + A (6x x 5) = ( ax + b)( cx + d) where ac = "6"and bd = " ± 5" = ( x+ )(x 5)(x+ ) A [] 9 marks

19 January 05 (IAL) (a) : Using remainder theorem: As on scheme. One of these is sufficient do not need to equate to 0 and to - 5 *Using Long division: need at least 6 x + ( a 6) x+... as quotient, and get as far as remainder or for the other a + division reaches x + ( ) x+... as quotient, and get as far as remainder. A: Any equivalent form *e.g. - b + a = 0 (using remainder after division) The mark is earned for a b = even if =0 not explicitly seen b a+ A: Any equivalent form *e.g = 5 (using remainder after division) Must be accurate but may be unsimplified. NB Using 5 instead of -5 is A0 : Solves their linear equations to obtain a or b A: Both a and b correct. Correct answers without working can earn A. (b) : Recognises (x+) is factor and obtains quadratic expression with correct first term by any method. Use of (x -) is M0. NB Starting with (x + ) (x ) (ax + b) is also M0 A: Correct quadratic (6x x 5) : Attempt to factorise quadratic where ac = "6"and bd = " ± 5" 5 5 A: any correct combination e.g. = ( x+ )( x )(x+ ) or = 6( x+ )( x )( x+ ) etc (on one line) Following a correct value for a and for b: They may just write the factorised answer down. For a correct answer this is AA For = ( x+ )( x.5)( x+ ) award A0A0 For correct answer following incorrect quadratic give A0 A0 fortuitous If the correct answer follows incorrect a and b, it is fortuitous and again A0A0 should be given.

20 January 05 (IAL) Number (a) 0, (b) and (60, 0) and (0, 0) and (-0, 0) and (-00, 0) ) o 6 sin( x 60 ) = (= x 60 = 5 (or 65 or 95 or 5 ) or 0.6 or π radians So x = 75 or 5 or 5 or 85 (allow awrt) B B B A [] A A [5] 8 marks (a) B : Correct exact y intercept (not decimal) allow on the diagram or in the text. Allow y = B for correct x intercepts then third B for all correct x intercepts ( may or may not be given as coordinates may be given on graph) Must be in degrees. (Extra answers in the range lose the third B) o 6 o 6 (b) : Divides by first giving correct statement sin( x 60 ) = but ( x 60 ) = is o 6 o M0 and sin x sin 60 = is also M0 and sin( x 60 ) = is M0 if not preceded by correct statement A: Obtains 5 (or 65 or 95 or 5 ) : Adds 60 to their previous answer which should have been in degrees and obtained by using inverse sine A: Two correct answers second A: All four correct answers Extra answers outside range are ignored. Lose final A mark for extra wrong answers in the range. If they approximate too early allow awrt answers given for full marks. (e.g etc) Answers in mixture, degrees and radians: Allow first M A only so AM0A0A0 for 60.6 for example

21 January 05 (IAL) Number.(a) 5 Uses (.) or finds or uses (.) or finds Finds both of the above and subtracts to give and concludes approx. 000* A* Or [] 5 Uses (.) (.), = awrt 060 = 000 ( sf ) *,A* n n (b) Puts (.) > or (.) = (.) n > (or or.6 or.6). Or (.) n = (or or.6 or.6) log( ) log( ) n > or n = log. log. ( n>.5 or n>.6 or n= 5 ) so the year is 00 A n n (c) 75000(. ) 75000(. ) Uses S = or uses S =.. Uses n = in formula A Awrt A Or: adds terms [] = awrt (see notes below) 0 marks [] []

22 January 05 (IAL) (a) : for correct expression for profit in 0 or in 00, by any method (including subtracting the sums S n+ - S n ) to give a term : for finding both correct expressions and subtracting A: answers wrt 900 and wrt 0600 subtracted or wrt 060 obtained then rounded to 000 (answer given) (b) : Correct inequality or allow equality. N.B (.) n or 0500 (.) n on LHS are also correct. : Division isw if initial fraction is correct. Not dependent on previous mark. It could follow wrong combination of a and n for example, which would give M0 log( k ) : Correct use of logs to give n or n > or log. k after (.) n > k Allow equality for this mark log. (.6 is truncated value of 0 and.6 is rounded value allow either of these if used in place of fraction) A: 00 is required. If inequalities are used and errors are seen, then this mark is A0 (even for 00) (Trial and improvement or listing can have full marks for the correct answer, need to see both th and 5 th term otherwise zero) Special case: If n is used instead of n and they reach 09 then mark profile is likely to be M0 A0 unless they recover to the correct answer when full marks may be earned If an equals sign is used throughout and then correct answer is obtained allow / Special case: Uses Sum formula Can earn M0 M0 A for correct work n 75000(. ) n log(5/) Uses S = > (M 0). > + (M 0) n> () n>.5... so 09 (A ) log. Using this method with errors can earn M0M0A0 for proceeding from. n > k with k >0 to (c) : Correct a and r but n may be wrong A: Correct use of formula with n = A: awrt (again this answer implies all marks) Or : adds terms ( mostly correct) A: lists correct terms A: correct answer = awrt (this implies two previous marks) log( k) n > log.

23 January 05 (IAL) Number. y = x x+ (a) dy 6x { 0} dx A At (, ) gradient of curve is and so gradient of normal is ( y ) = ( x ) and so x+ y = 0 * A* [5] (b) Eliminate x or y to give (x x+ ) + x = 0 or y = ( y) ( y) + Solve three term quadratic e.g 6x 7x+ = 0 or y 9y+ 7 = 0 to give x = or y. = 5 x= or y = 6 A 5 5 Both x = and y = i.e. (, ) or (0.7,.) { Ignore (, ) listed as well } A 6 6 [] (c) When this line meets the curve (x x + ) + kx = 0 So 6 x + ( k 8) x+ = 0 d Uses condition for equal roots " b = ac" on their three term quadratic to get expression in k dd So obtain ( k 8) = i.e. k 6k+ 0 = 0 * A * If they use gradient of tangent to do part (c) see the end of the notes below*. [] (d) Solve the given quadratic or their quadratic by formula or completion of the square to give k = 8 ± or 8 ± 6 or A 6 ± 96 []... 5 marks

24 January 05 (IAL) n n (a) : Evidence of differentiation, so x x at least once A: Both terms correct : Substitutes x = into their derivative and uses perpendicular property : Correct method for Linear equation, using (,) and their changed gradient A: Should conclude with printed answer (this answer is given in the question) (b) : May make sign slips in their algebra; {e.g. substitute + y } does not need to be simplified so isw. But putting ( y) ( y) + = 0 instead of = y is M0 : Solve three term quadratic to give one of the two variables A: One Correct coordinate accept any equivalent A: Both correct any equivalent form. Allow decimals if correct awrt (0.7,.) ( ignore (,) given as well) (c) : Eliminate y (condone small copying errors) d: Collect into term quadratic in x or identifies a, b and c clearly (may be implied by later work). dd: Uses condition " b = ac" on quadratic in x (dependent on both previous M marks) NB M0 for b > ac or b ac or b < ac or b ac A: Need ( k 8) = or equivalent before stating printed answer *Alternative method for part (c) k : Use gradient of line = gradient of curve so "6x " = " " k k : Find x = and use line equation to get y = k + (these equations do not need to be simplified) k k : Find x = and use curve equation to get y = + ( these equations do not need to be 8 simplified) A: Puts two correct expressions for y equal and obtains printed answer without error. (d) : Solve by formula or completion of the square to give k = (Attempt at factorization is M0) A: Correct answer should be one of the forms given in the main scheme or equivalent exact form Answers only with no working marks (exact and correct) or 0 marks ( approximate or wrong)

25 January 05 (IAL) Number. (i) Way : Use sin x = tan x to give tan x = Way: complete method to find sinx = or cosx = cos x 7 7 tan x = or sin x = ± or cos x = ± A So x =., 9. A [] (ii) 0 cos θ + cosθ = ( cos θ) 9 Solves their three term quadratic cos θ + cosθ = 0 to give cos θ =... So (cos θ = ) or 7 A θ =.9,.7,.8 or 5.00 (allow 5 instead of 5.00) A A [6] 0 marks (i) : (Way ) Attempts to use sin x = tan x (there may be a sign error or may omit x and write tan = ) cos x (Way ) sin x= 7 cos x so 9sin x= 9 cos x and uses sin x+ cos x= to find sinx = or cosx = 7 7 A: must be tan x = (way ) or allow sin x = ± or cos x = ± (way ). Ignore cosx = 0 as extra answer. : One correct angle in degrees in range so need either. or 9. in most cases But If they had tan x =, then obtaining 56.8 or 6.8 is equivalent work and gains 7 7 If however they had tan x = +, then obtaining an answer in the range is not equivalent work so is M0 A: These two answers - accept awrt. and 9. Extra answers in range lose this mark Working in radians gives a maximum of AM0A0 (ii) : Replaces sin θ by ( cos θ) : Collects terms and solves their three term quadratic by usual methods (see notes) A: Both correct answers needed, but isw if one then rejected. Allow awrt -0. and 0.86 Uses inverse cosine to obtain at least two correct answers for their values of cosine (check with calculator if they have followed wrong values) A: Any two completely correct answers (allow awrt) A: All four correct (awrt) Allow 0.608π,.9 π, 0.08 π, or.59 π Extra answers outside range ignore Extra answers in the range lose final mark. Inaccurate answers to sf lose final A mark Answers in degrees lose final two marks So two of awrt 7, 87, 09 (or 09.5), 5 ( or 50.5) would earn A0A0

26 January 05 (IAL) Number 5. (a) y = x + 0x + kx dy x 0 x k dx = + + A [] (b) Substitutes x = and d y 0 dx A* to give () + 5() + k = 0 k = 78 * [] (c) When x =, y = -68 ( see this stated or see rectangle has height 68) B x + 0x 78 x ( + 68)d x= x + x x ( + 68 x+ c) A 5 Use limits 0 and to give ± or if 68x included to give ±0 db Rectangle area is 68 (= 67) or see 68x in integrated answer with limits (a) (b) (c) So R has area 67 or see +68 in original integrand = 0 A [7] marks : Fractional power dealt with correctly so becomes simplification to 5) A: All terms correct, may not be simplified x ( may be implied by : Substitutes x = and d y dx = 0 Must see () + 5() + k = 0 or k = 0 *A: This is a printed answer so all must be correct in the working and conclusion k = -78 is needed. B: Substitute into y = to find y (This may appear anywhere in the answer) : Attempt to integrate so at least one power increases A: Accept unsimplified correct answer and allow with or without their +68x, or even with their -68x db: Use limit to give but may be implied by later answer 0- needs to follow A for integration : Calculates rectangle area (may be by integration). Must be rectangle and not triangle area : Subtracts (either way round) numerical areas should be (+) (+) or (-) - (-) (subtraction may be in their original integral but penalize wrong sign here eg -68x instead of +68x ) (Again use of triangle is M0) A: 0 only (Can recover from -0 to 0) Common error: If 68x (instead of 68) is integrated this may only gain a maximum of B A db (for seeing calculated if integrals are separated) M0 M0 A0 /7

27 January 05 (IAL)

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