2015 Mathematics. New Higher Paper 1. Finalised Marking Instructions

Transcription

1 National Qualifications 0 0 Mathematics New Higher Paper Finalised Marking Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Assessment team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 General Comments These marking instructions are for use with the 0 Higher Mathematics Examination. For each question the marking instructions are in two sections, namely Illustrative Scheme and Generic Scheme. The Illustrative Scheme covers methods which are commonly seen throughout the marking. The Generic Scheme indicates the rationale for which each mark is awarded. In general, markers should use the Illustrative Scheme and only use the Generic Scheme where a candidate has used a method not covered in the Illustrative Scheme. All markers should apply the following general marking principles throughout their marking: Marks must be assigned in accordance with these marking instructions. In principle, marks are awarded for what is correct, rather than deducted for what is wrong. One mark is available for each. There are no half marks. Working subsequent to an error must be followed through, with possible full marks for the subsequent working, provided that the level of difficulty involved is approximately similar. Where, subsequent to an error, the working for a follow through mark has been eased, the follow through mark cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Throughout this paper, unless specifically mentioned in the marking instructions, a correct answer with no working receives no credit. In general, as a consequence of an error perceived to be trivial, casual or insignificant, e.g. 6 6 =, candidates lose the opportunity of gaining a mark. But note the second example in comment 7. 6 Where a transcription error (paper to script or within script) occurs, the candidate should be penalised, eg This is a transcription error and so the mark is not awarded. Eased as no longer a solution of a quadratic equation. x + x+ 7 = 9x+ x x+ = 0 x = Exceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt. x x x = 9 + ~~ x x+ = 0 ( x )( x ) = 0 x = or Page

3 7 Vertical/horizontal marking Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed marking instructions for the question. Example: Point of intersection of line with curve 6 Illustrative Scheme: x =, x = x = y = 6 6 y =, y = 7 x = y = 7 Markers should choose whichever method benefits the candidate, but not a combination of both. 8 In final answers, numerical values should be simplified as far as possible, unless specifically mentioned in the detailed marking instructions. Examples: should be simplified to or 0 should be simplified to 0 6 must be simplified to 8 should be simplified to should be simplified to The square root of perfect squares up to and including 00 must be known. 9 Commonly Observed Responses (COR) are shown in the marking instructions to help mark common and/or non-routine solutions. CORs may also be used as a guide when marking similar non-routine candidate responses. 0 Unless specifically mentioned in the marking instructions, the following should not be penalised: Working subsequent to a correct answer; Correct working in the wrong part of a question; Legitimate variations in numerical answers, eg angles in degrees rounded to nearest degree; Omission of units; Bad form (bad form only becomes bad form if subsequent working is correct), e.g. ( x + x + x+ )(x+ ) written as ( x + x + x+ ) x+ x + x + 6x + x+ x + x + x+ x + x + 8x + 7x+ gains full credit; Repeated error within a question, but not between questions. In any Show that... question, where the candidate has to arrive at a required result, the last mark of that part is not available as a follow through from a previous error unless specifically stated otherwise in the detailed marking instructions. Page

4 All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate s response. Marks may still be available later in the question so reference must be made continually to the marking instructions. All working must be checked: the appearance of the correct answer does not necessarily indicate that the candidate has gained all the available marks. If you are in serious doubt whether a mark should or should not be awarded, consult your Team Leader (TL). Scored out working which has not been replaced should be marked where still legible. However, if the scored out working has been replaced, only the work which has not been scored out should be marked. Where a candidate has made multiple attempts using the same strategy, mark all attempts and award the lowest mark. Where a candidate has tried different strategies, apply the above ruling to attempts within each strategy and then award the highest resultant mark. For example: Strategy attempt is worth marks Strategy attempt is worth marks From the attempts using strategy, the resultant mark would be. Strategy attempt is worth mark Strategy attempt is worth marks From the attempts using strategy, the resultant mark would be. In this case, award marks. 6 In cases of difficulty, covered neither in detail nor in principle in these instructions, markers should contact their TL in the first instance. Page

5 Detailed Marking Instructions for each question Question Generic Scheme Illustrative Scheme Max Mark. equate scalar product to zero + t + 6 = 0 state value of t t = 9 Candidate A + t + 6 = t = 7 or 8. know to and differentiate 6x evaluate dy dx evaluate y-coordinate state equation of tangent y = x+. is only available if an attempt has been made to find the gradient from differentiation.. At mark accept y+ = ( x+ ), y x= or any other rearrangement of the equation. Page

6 . Method know to use x = interpret result and state conclusion ( ) ( ) ( ) 0 + = 0 ( x + ) is a factor. Method remainder = 0 ( x + ) is a factor. Method x x+ x x x+ 0 x + x = 0 ( x + ) is a factor. x 6x+ 8stated or implied by state quadratic factor ( x+ )( x )( x ) factorise completely. Communication at must be consistent with working at that stage ie a candidate s working must arrive legitimately at 0 before is awarded.. Accept any of the following for : f( ) = 0 so ( x+ ) is a factor since remainder is 0, it is a factor the 0 from the table linked to the word factor by eg so, hence,,,. Do not accept any of the following for : double underlining the zero or boxing the zero without comment x = is a factor, ( x ) is a factor, x = is a root, ( x ) is a root, ( x + ) is a root the word factor only, with no link. At the expression may be written in any order.. An incorrect quadratic correctly factorised may gain 6. Where the quadratic factor obtained is irreducible, candidates must clearly demonstrate that b ac < 0 to gain 7. = 0 must appear at or for to be awarded. 8. For candidates who do not arrive at 0 at the stage not available. 9. Do not penalise candidates who attempt to solve a cubic equation. However, within this working there may be evidence of the correct factorisation of the cubic. Page 6

7 Candidate A 0 0 x is a factor ( x )( x x ) ( x )( x )( x+ ) x+ is a factor. state the value of p state the value of q Candidate B 0 0 x is a factor p = q = state the value of r r =. These are the only acceptable responses for p, q and r. (a). let y = 6 x and rearrange. state expression. 6 y 6 x x= or y = 6 x x x 6 g ( x) = or or Method Method equates composite function to x gg ( ( x)) x = this gains start to rearrange. state expression. g x = x 6 x x x 6 g ( x) = or or 6 ( ). At accept any equivalent expression with any distinct variables. (b). state expression x. Candidates using method may be awarded at line one. g g ( x ), accept. For candidates who attempt to find the composite function ( ) 6 x 6 for.. In this case may be awarded as follow through where an incorrect g, provided it includes the variable x. ( x) is found at Page 7

8 6. use laws of logs log 6 7 use laws of logs evaluate log log 6 7 Candidate A log6 + log69 log 6( 9) log write in differentiable form Candidate B log 6 ( 7) log 6 log6 Award out of ^,^ x x differentiate first term 9... x + differentiate second term... + x evaluate derivative at x = 9 8. must involve a fractional index.. must involve a negative index.. is only available as a consequence of substituting into a derivative containing a fractional or negative index.. If no attempt has been made to expand the bracket at then & are not available. is still available as follow through. Candidate A f( x) = x x f '( x) = x + x = + x x f '() = + = + 8 Page 8

9 8. interpret information xx ( ) < express in standard quadratic form x x < 0 factorise state range ( x )( x+ ) < 0 < x < Candidate A xx ( ) = x x = 0 x =, ^ Candidate C x x< x > Award / Candidate E x x< x > and x < Award / Inequalities linked by and Candidate B - Mistaking perimeter for area x < 9 x < Award / Candidate D x x< Inequalities not x > linked by and x < Award / Page 9

10 9. find gradient of AB m = AB calculate gradient of BC m = BC interpret results and state conclusion mab mbc points are not collinear. m = AB Method AB makes 0 with positive direction of the x axis. 0 0 so points are not collinear.. The statement made at must be consistent with the gradients or angles found for and. 0(a). state value of cos x Candidate A Candidate B cos x = cos x = cos x = cos x =... cos x = cos x = cos x = invalid answer 0(b). use double angle formula cos x =... evaluate cos x 0. Ignore the inclusion of. 0. At the double angle formula must be equated to the candidates answer to part (a). Page 0

11 (a). state coordinates of centre ( 8, ) find gradient of radius state perpendicular gradient determine equation of tangent y = x. is only available as a consequence of trying to find and use a perpendicular gradient.. At mark accept y+ = ( x+ ), y x=, y x+ = 0or any other rearrangement of the equation. Page

12 (b). Method arrange equation of tangent in appropriate form and equate y tangent to y parabola Method x = x + px + p 6 6 rearrange and equate to 0 x + ( px ) + p = know to use discriminant and identify a, b,and c ( p) ( p ) 7 8 p p+ 0 = 0 8 simplify and equate to 0 9 start to solve 9 ( p 0)( p ) = 0 0 state value of p 0 p = 0 Method Method arrange equation of tangent in x = x + px + p appropriate form and equate y tangent to y parabola 6 find dy dy for parabola 6 x p dx dx = + 7 equate to gradient of line and = x+ p 7 rearrange for p p = + x 8 substitute and arrange in standard form 9 factorise and solve for x 0 state value of p 0= x x 8 0 = xx ( ) 9 x= 0, x= 0 p = 0. At 6 accept x + x px + p = 0.. At 7 accept a=, b= ( p), and c= ( p ). Just using the parabola a = b= p c= p ^ 6 ^ b ac = p ( )( p) 7 = p 8p+ 8= 0 8 p = ± p= + 8 as p> Page

13 . interpret integral below (accept area below x axis = ) x axis evaluate. For candidates who calculate the area as award out of. (a) calculate b (b)(i) reflecting in the line y = x. If the reflected graph cuts the y axis, is not awarded. Page

14 (b)(ii) calculate y intercept state coordinates of image of Q (, 0) see note state coordinates of image of P (, ). can only be awarded if (,0) is clearly identified either by their labelling or by their diagram.. is awarded for the appearance of, or (,0) or (0, ).. is awarded for the appearance of (,). Ignore any labelling attached to this point. Candidate A Candidate B y = f( x) reflected in x-axis y = f( x) reflected in y-axis (0,-) (0,) (,-) (-,) (c) 6 state x coordinate of R 6 x = 7 state y coordinate of R 7 y = 7. identify length of radius determine value of k tangent to circle Circle passes through origin Page

15 . know to integrate 6 integrate a term complete integration... or... 0 t kt... kt or t... 0 find constant of integration find value of k c = 00 k = 6 T = t t state expression for T 0. Accept unsimplified expressions at and stage.., and 6 are not available for candidates who have not considered the constant of integration.. may be implied by. [END OF MARKING INSTRUCTIONS] Page

16 National Qualifications 0 0 Mathematics New Higher Paper Finalised Marking Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Assessment team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

17 General Comments These marking instructions are for use with the 0 Higher Mathematics Examination. For each question the marking instructions are in two sections, namely Illustrative Scheme and Generic Scheme. The Illustrative Scheme covers methods which are commonly seen throughout the marking. The Generic Scheme indicates the rationale for which each mark is awarded. In general, markers should use the Illustrative Scheme and only use the Generic Scheme where a candidate has used a method not covered in the Illustrative Scheme. All markers should apply the following general marking principles throughout their marking: Marks must be assigned in accordance with these marking instructions. In principle, marks are awarded for what is correct, rather than deducted for what is wrong. One mark is available for each. There are no half marks. Working subsequent to an error must be followed through, with possible full marks for the subsequent working, provided that the level of difficulty involved is approximately similar. Where, subsequent to an error, the working for a follow through mark has been eased, the follow through mark cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Throughout this paper, unless specifically mentioned in the marking instructions, a correct answer with no working receives no credit. In general, as a consequence of an error perceived to be trivial, casual or insignificant, e.g. candidates lose the opportunity of gaining a mark. But note the second example in comment 7. 6 Where a transcription error (paper to script or within script) occurs, the candidate should be penalised, e.g. This is a transcription error and so the mark is not awarded. Eased as no longer a solution of a quadratic equation. Exceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt. x x x = 9 + ~~ x x+ = 0 ( x )( x ) = 0 x = or Page

18 7 Vertical/horizontal marking Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed marking instructions for the question. Example: Point of intersection of line with curve 6 Illustrative Scheme: x = y = 6 x = y = 7 Markers should choose whichever method benefits the candidate, but not a combination of both. 8 In final answers, numerical values should be simplified as far as possible, unless specifically mentioned in the detailed marking instructions. Examples: should be simplified to or should be simplified to should be simplified to 0 must be simplified to 8 should be simplified to The square root of perfect squares up to and including 00 must be known. 9 Commonly Observed Responses (COR) are shown in the marking instructions to help mark common and/or non-routine solutions. CORs may also be used as a guide when marking similar non-routine candidate responses. 0 Unless specifically mentioned in the marking instructions, the following should not be penalised: Working subsequent to a correct answer; Correct working in the wrong part of a question; Legitimate variations in numerical answers, eg angles in degrees rounded to nearest degree; Omission of units; Bad form (bad form only becomes bad form if subsequent working is correct), e.g. ( x + x + x+ )(x+ ) written as ( x + x + x+ ) x+ x + x + 6x + x+ x + x + x+ x + x + 8x + 7x+ gains full credit; Repeated error within a question, but not between questions. In any Show that... question, where the candidate has to arrive at a required result, the last mark of that part is not available as a follow through from a previous error unless specifically stated otherwise in the detailed marking instructions. Page

19 All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate s response. Marks may still be available later in the question so reference must be made continually to the marking instructions. All working must be checked: the appearance of the correct answer does not necessarily indicate that the candidate has gained all the available marks. If you are in serious doubt whether a mark should or should not be awarded, consult your Team Leader (TL). Scored out working which has not been replaced should be marked where still legible. However, if the scored out working has been replaced, only the work which has not been scored out should be marked. Where a candidate has made multiple attempts using the same strategy, mark all attempts and award the lowest mark. Where a candidate has tried different strategies, apply the above ruling to attempts within each strategy and then award the highest resultant mark. For example: Strategy attempt is worth marks Strategy attempt is worth marks From the attempts using strategy, the resultant mark would be. Strategy attempt is worth mark Strategy attempt is worth marks From the attempts using strategy, the resultant mark would be. In this case, award marks. 6 In cases of difficulty, covered neither in detail nor in principle in these instructions, markers should contact their TL in the first instance. Page

20 Paper Question Generic Scheme Illustrative Scheme Max Mark (a) calculate gradient of AB m = AB use property of perpendicular lines m alt = substitute into general equation of a line y = ( x ) demonstrate result... x y =. is only available as a consequence of trying to find and use a perpendicular gradient.. is only available if there is/are appropriate intermediate lines of working between and.. The ONLY acceptable variations for the final equation for the line in are = x y, y+ x=, = y+ x. Candidate A Candidate B ( ) For mab = = = 7 y- = x- malt = y = ( x ) y = x- is not available y = x- - not acceptable y- x=- - not acceptable x- y = Page

21 (b) calculate midpoint of AC M AC = (,) 6 calculate gradient of median 6 m BM = 7 determine equation of median 7 y = x. 6 and 7 are not available to candidates who do not use a midpoint.. 7 is only available as a consequence of using a non-perpendicular gradient and a midpoint. 6. Candidates who find either the median through A or the median through C or a side of the triangle gain mark out of. 7. At 7 accept y ( ) = ( x ( )), y = ( x ), y x+ = 0or any other rearrangement of the equation. Median through A Median through C M BC = (6, ) M AB = (,) 8 m AM = m CM = y+ = ( x 6) or y 7 = ( x+ ) y = ( x ) or y = ( x+ ) 8 8 Award / Award / (c) 8 calculate x or y coordinate 8 x= or y = 9 calculate remaining coordinate of the point 9 y = or x= of intersection 8. If the candidate s median is either a vertical or horizontal line then award out of if both coordinates are correct, otherwise award 0. For candidates who find the altitude through Candidate A B in part (b) y- = ( x-) 7 (b) x = 7 y = 8 9 y = x- -error x y = (c) y = x Leading to x= 7 and y = 8 9 Page 6

22 (a) interpret notation state a correct expression f(( + x)( x) + ) stated or implied by 0 + ( + x)( x) + stated or implied by. is not available for g( f ( x) ) = g(0 + x) but may be awarded for ( x)( x ) Candidate A f( g( x)) = ' g( f( x))' (a) = ( x)( (0 + x)) + (b) (c) = x x x x = + = ( x + 9) = x or 8 7 ( x 8x ( 9) = ( x 9) 6 0 x = or Candidate B f( gx ( )) = 0(( + x) ( x)) + Candidate C f( gx ( )) = 0(( + x)( x) + ) (0 + ) +. ^ ^ (b) write f( gx ( )) in quadratic form Method identify common factor complete the square + x x or x + x+ Method ( x x stated or implied by ( x ) + 6 Method expand completed square form and equate coefficients Method px + pqx + pq + r and p =, process for form q and r and write in required q = and r = 6 Note if p = is not available Page 7

23 . Accept 6 ( x ) or ( x ) 6 at. Candidate A Candidate B ( x x ) ( x x+ ) ( x ) 6 Candidate D + x x x x px pqx pq r p and = q= r = 6 Candidate E eased Candidate C x + x+ + ( x ) ( x ) Candidate F + x x x x x ( ) 6 eased Eased, unitary coefficient of (lower level skill) x (c) 6 identify critical condition + x x x x x ( ) 6 + x x = x + so ( ) 6 ( x ) + 6 = 0 or f(( gx ( )) = x x ( x + )... x + + ( ) 6 7 identify critical values 7 and. Any communication indicating that the denominator cannot be zero gains 6.. Accept x= and x= or x and x at 7.. If x= and x= appear without working award /. Candidate A Candidate B + x ( x ) ^ f( gx ( )) f( gx ( )) > 0 x=, x= < x x< 6 7 (a) determine the value of the required term 9 or or 7. Do not penalise the inclusion of incorrect units.. Accept rounded and unsimplified answers following evidence of correct substitution. Page 8

24 (b) Method (Considering both limits) know how to calculate limit know how to calculate limit Method or L = L + or L = L + calculate limit calculate limit 6 interpret limits and state conclusion Method (Frog first then numerical for toad) know how to calculate limit calculate limit determine the value of the highest term less than 0 determine the value of the lowest term greater than 0 6 interpret information and state conclusion Method (Numerical method for toad only) continues numerical strategy exact value determine the value of the highest term less than 0 determine the value of the lowest term greater than 0 6 interpret information and state conclusion Method (Limit method for toad only) & know how to calculate limit 8 6 > 0 toad will escape Method or L = L > 0 toad will escape Method numerical strategy > 0 toad will escape & Method or L = L + & calculate limit 6 interpret limit and state conclusion & 6 > 0 toad will escape Page 9

25 . 6 is unavailable for candidates who do not consider the toad in their conclusion.. For candidates who only consider the frog numerically award / for the strategy. Error with frogs limit Frog Only LF = LF = > 0 6 frog will escape. Using Method - Toad Only Using Method - Toad Only Toad Conclusions Limit = This is greater than the height of the well and so the toad will escape award 6. However Limit = and so the toad escapes - 6 ^. missing ^ 0... missing ^ 6 0. > rounding error so the toad escapes > 0 so the toad escapes. Using Method - Toad Only rounding error > 0 so the toad escapes. Iterations f = f f f f f f f f = 667 = 6 = 7 07 = 7 80 = 7 9 = = 7 99 = t = t t t t t t t t t t t = 7 = 0 06 = 7 = = 7 = 09 = 6 79 = = 9 07 = 9 80 = 0 Page 0

26 (a) know to equate f( x)and gx ( ) x x+ = x x+ solve for x x =. and are not available to candidates who: (i) equate zeros, (ii) give answer only without working, (iii) arrive at x = with erroneous working. Candidate A Candidate B y = x x+ x x= y = x x+ x x= subtract to get 0= x x = x = In this case the candidate has equated zeros Candidate C f( x) = x x+ g( x) = x x+ f '( x) = x g'( x) = x.. x= x= x = Page

27 (b) know to integrate interpret limits use upper lower ( x x+ ) ( x x+ ) dx 6 7 integrate substitute limits 6 7 x + x accept 8 unsimplified integral 7 7 ( 8 ) evaluate area between f( x) and hx ( ) 9 state total area If limits x= 0 and x= appear ex nihilo award.. If a candidate differentiates at 6 then 6, 7 and 8 are not available. However, 9 is still available.. Candidates who substitute at 7, without attempting to integrate at 6, cannot gain 6, 7 or 8. However, 9 is still available. 6. Evidence for 8 may be implied by is a strategy mark and should be awarded for correctly multiplying their solution at 8, or for any other valid strategy applied to previous working. 8. For both a term containing a variable and the constant term must be dealt with correctly. 9. In cases where is not awarded, 6 may be gained for integrating an expression of equivalent difficulty ie a polynomial of at least degree two. 6 is unavailable for the integration of a linear expression must be as a consequence of substituting into a term where the power of x is not equalto or 0. Page

28 Candidate A Valid Strategy Candidates who use the strategy: Total Area = Area A + Area B A B Then mark as follows: Mark Area A for to 8 then mark Area B for to 8 and award the higher of the two. 9 is available for correctly adding two equal areas. Candidate B Invalid Strategy For example, candidates who integrate each of the four functions separately within an invalid strategy Gain if limits correct on f( x) and hx ( ) is unavailable or gx ( ) and kx ( ) Gain 6 for calculating either f( x) or gx ( ) and hx ( ) or kx ( ) Gain 7 for correctly substituting at least twice Gain 8 for evaluating at least two integrals correctly 9 is unavailable Candidate C ( x x+ 8 x x+ ) dx ~~~~~~~~~ ( x x) dx 8 x x 8 6 Candidate E... = cannot be negative so = 8 however, = so Area = 8 Candidate D 0 0 ( x x+ x x+ ) dx 9 8 ( x x + 6) dx Candidate F 0 x x 6x 9 ( x x+ 8 x x+ ) dx ~~~~ 7 ( 8 x + ) 0 x + x 7 8 x dx 6 6 Page

29 (a) state centre of C (, ) state radius of C calculate distance between centres of C and 0 C calculate radius of C. For to be awarded radius of C must be greater than the radius of C.. Beware of candidates who arrive at the correct solution by finding the point of contact by an invalid strategy.. is for Distancecc rc but only if the answer obtained is greater than r c. Page

30 (b) find ratio in which centre of C divides line joining centres of C and C 6 determine centre of C 7 calculate radius of C 8 state equation of C : 6 (6,7) 7 r = 0 (answer must be consistent with distance between centres) 8 ( x 6) + ( y 7) = 00. For accept ratios ± : ±, ± : ±, : ±, : ± (or the appearance of ).. 7 is for r c +r c. 6. Where candidates arrive at an incorrect centre or radius from working then 8 is available. However 8 is not available if either centre or radius appear ex nihilo (see note ). 7. Do not accept 0 for For candidates finding the centre by stepping out the following is the minimum evidence for and 6 : Candidate A Candidate B using the mid-point of centres: C = (, ) C (9,) r = 0 centre C = (,) 6 : radius of C = C ( x- ) + ( y- ) = 00 8 = note 6 C = (0, ) 7 x + ( y+ ) = 00 8 Candidate C touches C internally only Candidate D - touches C internally only 6 centre C = (,) 7 radius of C = radius of C= 8 ( x ) + ( y ) = Candidate E centre C collinear with C,C 6 e.g. centre C = (,7) 7 radius of C = (touch C internally only) ( x ) + ( y 7) = Correct follow through using the ratio : 6 Correct answer using the ratio : centre C = (,) 7 radius of C = radius of C= ( x ) + ( y ) = 8 6 Page

31 6 (a) Expands p.q + p.r Evaluate p.q Completes evaluation =. For p.( q + r) = pq + pr with no other working is not available. 6 (b) correct expression -q+ p+ r or equivalent 6 (c) correct substitution - q.q + q.p + q.r 6 start evaluation r cos0 o = 9 7 find expression for r 7 r = cos0. Award for -q +q.p + q.r Candidate A Candidate B 9 -q.q + q.p + q.r = 9-9 -q.q + q.p + q.r = 9-9 o r cos0 = o r cos 0 = 9-7 r = cos0 r = 6 Page 6

32 7 (a) integrate a term complete integration with constant sin x OR x+ c sin x+ c x 7 (b) substitute for cos x substitute for and complete ( x x) cos sin... or...(sin x+ cos x) ( )... sin x+ cos x = cos x sin x. Any valid substitution for cos x is acceptable for.. Candidates who show that cos x sin x= cos x+ may gain both marks.. Candidates who quote the formula for cos x in terms of A but do not use in the context of the question cannot gain. Candidate A cos x+ = ( cos x ) + ( cos x ) + ( sin x) + = cos x sin x Candidate B cos x sin x= (cos x+ ) ( cos x) = cos x + 7 (c) 6 interpret link state result... 6 sin x x+ c Candidate A sin x cos = (cos x + ) dx sin x+ x+ c x dx 6 Page 7

33 8 (a) (i) calculate T when x = 0 0 or 0 8 (a) (ii) calculate T when x = 0 or 0. Accept correct answers with no units.. Accept 6 or 0 09 or equivalent for T (0).. For correct substitution alone, with no calculation and are not available.. For candidates who calculate T when x = 0 at then is available as follow through for calculating T when x = 0 in part(ii). a) (i) (ii) 0 0 b) leading to 9 8seconds See note 0 See note 7 Page 8

34 8 (b) write function in differential form ( 6 + x ) +... start differentiation of first term ( ) complete differentiation of first term 6 complete differentiation and set candidate's derivative = 0 7 start to solve... x x( 6 + x ) = 0 ( x ) x= 6 + or x = ( 6 + x ) 8 know to square both sides 8 x = 6( 6 + x ) or x + x ( 6 ) = 6 9 find value of x 0 calculate minimum time 9 x = 8 0 T = 9 8 or 98 no units required 8. Incorrect expansion of (...) at stage only 6 and 0 are available as follow through. 6. Incorrect expansion of (...) at stage 7 only 0 is available as follow through. 7. Where candidates have omitted units, then 0 is only available if the implied units are consistent throughout their solution is only available as a follow through for a value which is less than the values obtained for the two extremes. Page 9

35 9. use compound angle formula compare coefficients process for k process for a ksin tcos a kcos tsin a k cos a = 6, k sin a = stated explicitly k = 9 a = rad or a 6 equates expression for h to 00 6 write in standard format and attempt to solve 7 solve equation for t 8 process solutions for t ( ) 6 ( ) sin t = 9sin t = 00 9 t = sin t = 08 and 8 t = 006 and 6 8. Treat ksin tcos a cos tsina as bad form only if the equations at the stage both contain k.. 9sin t cos a 9 cos tsina or 9( sin tcos a cos tsina) is acceptable for and.. Accept kcosa= 6 and ksina = for.. is not available for kcos t = 6 and ksin t =, however, is still available.. is only available for a single value of k, k > is only available for a single value of a. 7. The angle at must be consistent with the equations at even when this leads to an angle outwith the required range. 8. Candidates who identify and use any form of the wave equation may gain, and, however, is only available if the value of a is interpreted for the form ksin( t a). 9. Candidates who work consistently in degrees cannot gain Do not penalise additional solutions at 8.. On this occasion accept any answers which round to 0and6 ( significant figures required). Page 0

36 Response : Missing information in working. Candidate A Candidate B 9cosa = 6 cosa = 6 9sin a = ^ ^ s i n a = tan a = tan a = ^ 6 6 a a a = rad or 6 a = rad or 6 Does not satisfy equations at Candidate C ksin tcos a kcos tsina kcosa= 6, ksina= k = 9 or 9 tan a = 6 a a = rad or 6 or a a = 68...rad or 0 6 Response : Correct expansion of k sin( x + a)º and possible errors for and Candidate D Candidate E Candidate F kcosa= 6 kcosa= kcosa = 6 ks i n a = ksina = 6 ksina = 6 tan a = 6 tan a = tan a = 6 a = 76 rad or 67 a a = 76 rad or 67 a a = 888 rad or 7 a Response : Labelling incorrect, sin (A B) = sin A cos B cos A sin B from formula list. Candidate G Candidate H Candidate I ksin A cos B kcos A sin B kcosa= 6 ks i n a= tan a = 6 a a = rad or 6 Candidate J ksin A cos B kcos A sin B kcos t = 6 ks i n t = tan t = 6 t = rad or 6 Candidate K a ksin A cos B kcos A sin B k cos B = 6 k s i n B = tan B = 6 B = rad or 6 a 9sin( t 0 9) = sin( t 0 9) = 9 00 t 0 9 = sin sin( t 0 9) = 00 9 t 0 9 = sin [END OF MARKING INSTRUCTIONS] Page