Mark Scheme January 2009

Transcription

1 Mark January 009 GCE GCE Mathematics (87/87,97/97) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on , our GCSE team on , or visit our website at If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Expert service helpful. Ask The Expert can be accessed online at the following link: January 009 Publications Code UA0094 All the material in this publication is copyright Edexcel Ltd 009

3 Contents. 666 Core Mathematics Core Mathematics Core Mathematics Core Mathematics Further Pure Mathematics (legacy) Further Pure Mathematics (new) Mechanics Mechanics Mechanics Statistics Statistics Decision Mathematics 97 87_974 GCE Mathematics January 009

4 87_974 GCE Mathematics January 009 4

5 January Core Mathematics C Mark Question (a) 5 ( ± 5 is B0) ( their 5) or their 5 = 5 or 0.04 ( ± is A0) 5 B A () () [] follow through their value of 5. Must have reciprocal and square. 5 is not sufficient to score this mark, unless follows this. 5 A negative introduced at any stage can score the but not the A, e.g. 5 = = scores A = = scores A Correct answer with no working scores both marks. Alternative: 5 or = 565 ( 5 ) = A 5 (reciprocal and the correct number squared) 87_974 GCE Mathematics January 009 5

6 6 8 4 ( I = ) x x + x+ c = x x + x+ c AAA [4] n for an attempt to integrate x x 6 4 (i.e. ax or ax or ax, where a is any non-zero constant). Also, this M mark can be scored for just the + c (seen at some stage), even if no other terms are correct. st 6 A for x nd 4 A for x rd A for x + c (or x + k, etc., any appropriate letter can be used as the constant) n+ x Allow x + c, but not + c. Note that the A marks can be awarded at separate stages, e.g. 6 x x 4 + x scores nd A 6 6 x x + x + c 6 scores rd A 6 4 x x + x scores st A (even though the c has now been lost). Remember that all the A marks are dependent on the M mark. If applicable, isw (ignore subsequent working) after a correct answer is seen. 6 4 Ignore wrong notation if the intention is clear, e.g. Answer x x + x + c dx. 87_974 GCE Mathematics January 009 6

7 , or 7 4 or an exact equivalent such as = 49 A [] for an expanded expression. At worst, there can be one wrong term and one wrong sign, or two wrong signs. e.g is (one wrong term ) is (two wrong signs + 7 and + 4 ) is (one wrong term +, one wrong sign + 7 ) is (one wrong term 7, one wrong sign + 4 ) is M0 (two wrong terms 7 and ) is M0 (two wrong terms 4 and 4 ) If only terms are given, they must be correct, i.e. (7 4) or an equivalent unsimplified version to score. The terms can be seen separately for the. Correct answer with no working scores both marks. 87_974 GCE Mathematics January 009 7

8 4 x x ( f( x) = ) 7x( + c) = x x 7 x ( + c) f(4) = = c c = st for an attempt to integrate ( x or x seen). The x term is insufficient for this mark and similarly the + c is insufficient. st A for x x or (An unsimplified or simplified correct form) nd A for all three x terms correct and simplified (the simplification may be seen later). The + c is not required for this mark. Allow 7x, but not 7x. nd for an attempt to use x = 4 and y = in a changed function (even if differentiated) to form an equation in c. rd A for c = with no earlier incorrect work (a final expression for f(x) is not required). AA Acso (5) [5] 87_974 GCE Mathematics January 009 8

9 5 (a) y Shape maximum., touching the x-axis at its - x Through (0,0) & marked on x-axis, or (,0) seen. Allow ( 0, ) if marked on the x-axis. Marked in the correct place, but, is A0. A (-,-) Min at (, ) A () y Correct shape (top left - bottom right) B - x Through and max at (0, 0). Marked in the correct place, but, is B0. Min at (, ) B B () (-,-) [6] (a) as described above. Be generous, even when the curve seems to be composed of straight line segments, but there must be a discernible 'curve' at the max. and min. st A for curve passing through and the origin. Max at (,0) nd A for minimum at (, ). Can simply be indicated on sketch. st B for the correct shape. A negative cubic passing from top left to bottom right. Shape: Be generous, even when the curve seems to be composed of straight line segments, but there must be a discernible 'curve' at the max. and min. nd B for curve passing through (,0) having a max at (0, 0) and no other max. rd B for minimum at (, ) and no other minimum. If in correct quadrant but labelled, e.g. (,), this is B0. In each part the (0, 0) does not need to be written to score the second mark having the curve pass through the origin is sufficient. The last mark (for the minimum) in each part is dependent on a sketch being attempted, and the sketch must show the minimum in approximately the correct place (not, for example, (, ) marked in the wrong quadrant). The mark for the minimum is not given for the coordinates just marked on the axes unless these are clearly linked to the minimum by vertical and horizontal lines. 87_974 GCE Mathematics January 009 9

10 6 (a) (a) x or p = (Not x x ) x or x or q = d y = 0x + x dx = 0x + x st B for p =.5 or exact equivalent nd B for q = n n x for an attempt to differentiate x (for any of the 4 terms) st A for 0x (the must 'disappear') nd Aft for x or x. Follow through their p but they must be differentiating p x, where p is a fraction, and the coefficient must be simplified if necessary. rd 0 Aft for (not the unsimplified x ), or follow through for correct q q differentiation of their x (i.e. coefficient of x is ). If ft is applied, the coefficient must be simplified if necessary. 'Simplified' coefficient means b a where a and b are integers with no common factors. Only a single + or sign is allowed (e.g. must be replaced by +). If there is a 'restart' in part it can be marked independently of part (a), but marks for part (a) cannot be scored for work seen in. Multiplying by 4 x : (assuming this is a restart) e.g. y = 5x x x + x x dy 45 7 = x x + 4x x dx scores A0 A0 (p not a fraction) Aft. Extra term included: This invalidates the final mark. 4 e.g. y = 5x + x x x dy = x x x x scores A A0 (p not a fraction) A0. dx Numerator and denominator differentiated separately: For this, neither of the last two (ft) marks should be awarded. Quotient/product rule: Last two terms must be correct to score the last marks. (If the M mark has not already been earned, it can be given for the quotient/product rule attempt.) B B () AAftAft (4) [6] 87_974 GCE Mathematics January 009 0

11 7 (a) ( ) b 4ac> 0 6 4k 5 k > 0 or equiv., e.g. 6 > 4k(5 k) So k 5k+ 4> 0 (Allow any order of terms, e.g. 4 5k + k 0 ) (*) > A Acso () (a) Critical Values ( k )( k ) 4 = 0 k =. k = or 4 k < or k > 4 Choosing outside region For this question, ignore (a) and labels and award marks wherever correct work is seen. A A (4) [7] for attempting to use the discriminant of the initial equation (> 0 not required, but substitution of a, b and c in the correct formula is required). If the formula b 4ac is seen, at least of a, b and c must be correct. If the formula b 4ac is not seen, all (a, b and c) must be correct. This mark can still be scored if substitution in b 4ac is within the quadratic formula. This mark can also be scored by comparing b and 4 ac (with substitution). However, use of b + 4ac is M0. st A for fully correct expression, possibly unsimplified, with > symbol. NB must appear before the last line, even if this is simply in a statement such as b 4ac > 0 or discriminant positive. Condone a bracketing slip, e.g. 6 4 k 5 k if subsequent work is correct and convincing. nd A for a fully correct derivation with no incorrect working seen. Condone a bracketing slip if otherwise correct and convincing. Using b 4ac > 0 : Only available mark is the first (unless recovery is seen). st for attempt to solve an appropriate TQ st A for both k = and 4 (only the critical values are required, so accept, e.g. k > and k > 4). ** nd for choosing the outside region. A diagram or table alone is not sufficient. Follow through their values of k. The set of values must be 'narrowed down' to score this M mark listing everything k <, < k < 4, k > 4 is M0. nd A for correct answer only, condone "k <, k > 4" and even "k < and k > 4", but " > k > 4 " is A0. ** Often the statement k > and k > 4 is followed by the correct final answer. Allow full marks. Seeing and 4 used as critical values gives the first A by implication. In part, condone working with x's except for the final mark, where the set of values must be a set of values of k (i.e. marks out of 4). Use of (or ) in the final answer loses the final mark. 87_974 GCE Mathematics January 009

12 8 (a) ( ) ( ) ( ) a = + =4 (, 4) or y = 4 is also acceptable y (i) Shape or anywhere B () B - x Min at (,0) can be on x-axis. Allow ( 0, ) if marked on the x-axis. Marked in the correct place, but, is B0. (, 0) and (0, ) can be on axes B B (ii) Top branch in st quadrant with intersections Bottom branch in rd quadrant (ignore any intersections) B B (5) (c) ( intersections therefore) (roots) Bft () [7] st B for shape or Can be anywhere, but there must be one max. and one min. and no further max. and min. turning points. Shape: Be generous, even when the curve seems to be composed of straight line segments, but there must be a discernible 'curve' at the max. and min. nd B for minimum at (,0) (even if there is an additional minimum point shown) rd B for the sketch meeting axes at (, 0) and (0, ). They can simply mark on the axes. The marks for minimum and intersections are dependent upon having a sketch. Answers on the diagram for min. and intersections take precedence over answers seen elsewhere. 4 th B for the branch fully within st quadrant having intersections with (not just touching ) the other curve. The curve can touch the axes. A curve of (roughly) the correct shape is required, but be very generous, even when the arc appears to turn 'inwards' rather than approaching the axes, and when the curve looks like two straight lines with a small curve at the join. Allow, for example, shapes like these: 5 th B for a branch fully in the rd quadrant (ignore any intersections with the other curve for this branch). The curve can touch the axes. A curve of (roughly) the correct shape is required, but be very generous, even when the arc appears to turn 'inwards' rather than approaching the axes. (c) Bft for a statement about the number of roots - compatible with their sketch. No sketch is B0. The answer incompatible with the sketch is B0 (ignore any algebra seen). If the sketch shows the correct intersections and, for example, one other intersection, the answer here should be, not, to score the mark. 87_974 GCE Mathematics January 009

13 9 (a) (c) a + 7 d = 5 or equiv. (for st B), a + 0 d =. 5 or equiv. (for nd B), Solving (Subtract) d = 7.5 so d =.5 a = so a = -7.5 (*) n = 5 + ( n ) { = n(5n 75) } = n n 5 n ( ) 5n= (*) B, B () Acso () Aft Acso (4) (d) (a) (c) (d) n 5n = 0 or n 5n 00 = 0 ( n 55)( n+ 40) = 0 n = n = 55 (ignore - 40) A () [] Mark parts (a) and as one part, ignoring labelling. Alternative: st.5 5 B: d =.5 or equiv. or d =. No method required, but a = 7. 5 must not be assumed. nd B: Either a + 7 d = 5 or a + 0 d =. 5 seen, or used with a value of d or for listing terms or similar methods, counting back 7 (or 0) terms. : In main scheme: for a full method (allow numerical or sign slips) leading to solution for d or a without assuming a = 7. 5 In alternative scheme: for using a d value to find a value for a. 5 A: Finding correct values for both a and d (allowing equiv. fractions such as d = ), with no 6 incorrect working seen. In the main scheme, if the given a is used to find d from one of the equations, then allow A if both values are checked in the nd equation. st for attempt to form equation with correct S n formula and 750, with values of a and d. st Aft for a correct equation following through their d. nd for expanding and simplifying to a term quadratic. nd A for correct working leading to printed result (no incorrect working seen). st forming the correct TQ = 0. Can condone missing = 0 but all terms must be on one side. First can be implied (perhaps seen in (c), but there must be an attempt at (d) for it to be scored). nd for attempt to solve TQ, by factorisation, formula or completing the square (see general marking principles at end of scheme). If this mark is earned for the completing the square method or if the factors are written down directly, the st is given by implication. A for n = 55 dependent on both Ms. Ignore 40 if seen. No working or trial and improvement methods in (d) score all marks for the answer 55, otherwise no marks. 87_974 GCE Mathematics January 009

14 0 (a) (c) (d) (a) (c) (d) y 5 y 5= ( x ) or equivalent, e.g. =, y = x+ x x= y = ( ) + 6 = 7 (therefore B lies on the line) (or equivalent verification methods) AB = ( ) + (7 5), = = 0, AB = 0 = 5 ( ) ( p ) C is (p, p + 6 ), so AC = p Therefore 5 = p 4 p+ 4 + p p =.5p 5p + 5 or 00 = 5p 0 p+ 0 (or better, RHS simplified to terms) Leading to: 0= p 4p 6 (*) A The version in the scheme above can be written down directly (for marks), and A0 can be allowed if there is just one slip (sign or number). If the 5 and are the wrong way round the M mark can still be given if a correct formula (e.g. y y = m( x x) ) is seen, otherwise M0. If (, 5) is substituted into y = mx + c to find c, the M mark is for attempting this and the st A mark is for c = 6. Correct answer without working or from a sketch scores full marks. A conclusion/comment is not required, except when the method used is to establish that the line through (,7) with gradient has the same eqn. as found in part (a), or to establish that the line through (,7) and (,5) has gradient. In these cases a comment 'same equation' or 'same gradient' or 'therefore on same line' is sufficient. for attempting AB or AB. Allow one slip (sign or number) inside a bracket, i.e. do not allow ( ) (7 5). st A for 0 (condone bracketing slips such as = 4 ) nd A for 5 or k = (Ignore ± here). st for ( p ) + (linear function of p). The linear function may be unsimplified but must be equivalent to ap + b, a 0, b 0. nd (dependent on st M) for forming an equation in p (using 5 or 5) and attempting (perhaps not very well) to multiply out both brackets. st A for collecting like p terms and having a correct expression. nd A for correct work leading to printed answer. Alternative, using the result: p = ± 5 and use one or both of the two solutions to find the Solve the quadratic ( ) length of AC or C C : e.g. AC = ( + 5 ) + ( 5 5 5) scores st, and st A if fully correct. Finding the length of AC or AC for both values of p, or finding C C with some evidence of halving (or intending to halve) scores the nd. Getting AC = 5 for both values of p, or showing C C = 5 scores the nd A (cso). A, Acao () B (), A, A () A Acso (4) [] 87_974 GCE Mathematics January 009 4

15 (a) (c) (a) (c) dy = 4+ 8x (4 or 8x for sign can be wrong) dx x = m = 4 + = 8 The first 4 marks could be earned in part y = 9 8 = Equation of tangent is: y+ = ( x ) y = x (*) Gradient of normal = Equation is: y x + = ( A :), ( B :) 8 or better equivalent, e.g. y = x 4 A Area of triangle is: ( x B ± x A ) yp with values for all of xb, x A and yp 45 8 = or.5 A (4) 4 [] st for 4 or 8x (ignore the signs). st A for both terms correct (including signs). nd for substituting x = into their d y (must be different from their y) dx B for y P =, but not if clearly found from the given equation of the tangent. rd for attempt to find the equation of tangent at P, follow through their m and y P. Apply general principles for straight line equations (see end of scheme). NO DIFFERENTIATION ATTEMPTED: Just assuming m = at this stage is M0 nd Acso for correct work leading to printed answer (allow equivalents with x, y, and terms such as x + y = 0 ). Bft for correct use of the perpendicular gradient rule. Follow through their m, but if m there must be clear evidence that the m is thought to be the gradient of the tangent. for an attempt to find normal at P using their changed gradient and their y P. Apply general principles for straight line equations (see end of scheme). A for any correct form as specified above (correct answer only). st B for and nd B for 8. B Acso (6) Bft A B, B () for a full method for the area of triangle ABP. Follow through their xa, xb and their y P, but the mark is to be awarded generously, condoning sign errors.. The final answer must be positive for A, with negatives in the working condoned. x y Determinant: Area = x y = =... (Attempt to multiply out required for ) x y 8 0 Alternative: AP = ( 0.5) + ( ), BP = ( 8) + ( ), Area = AP BP =... Intersections with y-axis instead of x-axis: Only the M mark is available B0 B0 A0. 87_974 GCE Mathematics January 009 5

16 87_974 GCE Mathematics January 009 6

17 January Core Mathematics C Mark Question ( ) 5 4 x = 4, 5 ( ) ( x) = 80x () ( x ) = + 080x + B, B A (4) [4] Notes Special cases First term must be 4 for B, writing just second line of special cases below). Term must be simplified to 80x for B The x is required for this mark. 5 is B0 (Mark their final answers except in The method mark () is generous and is awarded for an attempt at Binomial to get the third term. There must be an x (or no x- i.e. not wrong power) and attempt at Binomial Coefficient and at dealing with powers of and. The power of should not be one, but the power of may be one (regarded as bracketing slip). 5 So allow or 5 or 5 C or 5 5 C or even or 5 or use of 0 (maybe from Pascal s triangle) May see 5 C() ( x) or 5 C() ( x ) or 5 5 C() ( x ) or 0() ( x) which would each score the Ais c.a.o and needs 080x (if 080x is written with no working this is awarded both marks i.e. A.) x + 080x is BB0A (condone no negative signs) Follows correct answer with 7 90x + 0x can isw here (sp case) full marks for correct answer 5 4 Misreads ascending and gives x + 40x 70x is marked as BB0A0 special case and must be completely correct. (If any slips could get B0B0A0) 5 Ignores and expands (± x) is 0/4 4, -80x, 080x is full marks but 4, -80, 080 is B,B0,,A NB Alternative method ( ( ) ( ) x) = 5 x + x +.. is B0B0A0 answers must be simplified to 4 80x + 080x for full marks (awarded as before) 5 5 Special case ( ( ) ( ) x) = 5 x + x +.. is B0, B0,, A0 5 Or ( x) is B0B0M0A0 87_974 GCE Mathematics January 009 7

18 y = ( + x)(4 x) = 4 + x x M: Expand, giving (or 4) terms ( 4 + x x ) x x dx = 4x + M: Attempt to integrate A 64 4 [...] = = = 0 = A (5) [5] Notes needs expansion, there may be a slip involving a sign or simple arithmetical error e.g. 4= 5, but there needs to be a constant an x term and an x term. The x terms do not need to be collected. (Need not be seen if next line correct) n n Attempt to integrate means that x x + for at least one of the terms, then is awarded ( even 4 becoming 4x is sufficient) one correct power sufficient. A is for correct answer only, not follow through. But allow x x or any correct equivalent. Allow + c, and even allow an evaluated extra constant term. : Substitute limit 4 and limit into a changed function (must be ) and indicate subtraction (either way round). A must be exact, not 0.8 or similar. If recurring indicated can have the mark. Negative area, even if subsequently positive loses the A mark. Special cases (i) Uses calculator method: for expansion (if seen) for limits if answer correct, so 0, or marks out of 5 is possible (Most likely M0 M0 A0 A0 ) (ii) Uses trapezium rule : not exact, no calculus 0/5 unless expansion mark gained. (iii) Using original method, but then change all signs after expansion is likely to lead to: A0, A0 i.e. /5 87_974 GCE Mathematics January 009 8

19 (a).84, 4.4, 4.58 (Any one correct B B0. All correct B B) B B () 0.4, { ( ) } B, Aft = 7.85 (awrt 7.9) A (4) ( ) [6] Notes (a) B for one answer correct Second B for all three correct Accept awrt ones given or exact answers so, 49, score the marks. 5 B is for using 0. or 0.4 as h or 4 5, and requires first bracket to contain first plus last values and second bracket to include no additional values from those in the table. If the only mistake is to omit one value from nd bracket this may be regarded as a slip an can be allowed ( An extra repeated term forfeits the M mark however) x values: M0 if values used in brackets are x values instead of y values. Separate trapezia may be used : B for 0., for ha ( + b) used 4 or 5 times ( and Aft all e.g.. 0.( +.47) + 0.( ) + 0.( ) + 0.( ) is A0 equivalent to missing one term in { } in main scheme Aft follows their answers to part (a) and is for {correct expression} Final A must be correct. (No follow through) Special cases Bracketing mistake: i.e. 0.4( ) + ( ) scores B A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). Need to see trapezium rule answer only (with no working) is 0/ or 87_974 GCE Mathematics January 009 9

20 4 ( ) 4 x log 5 x = log 5 x, log5 (4 x) log5 ( x ) = log5 B, x 4 x = + = x log log 5 5x x 4 0 or 5x x 4 + = o.e. A 4 d A ( 5x 4)( x + ) = 0 x = (x = ) 5 (6) [6] Notes B is awarded for logx = logx anywhere. for correct use of log A log B = log A B for replacing by log k k. A for correct quadratic ( log(4 x) log x = log 5 4 x x = 5 is BM0A0 M0A0) d for attempt to solve quadratic with usual conventions. (Only award if previous two M marks have been awarded) A for 4/5 or 0.8 or equivalent (Ignore extra answer). Alternative log 5(4 x) = log5 x so log 5(4 x) log5 5 = log5 x 4 x log5 = log5 x 5 then could complete solution with log 5 x = log 5 ( x ) B Special cases 4 x 5 = x x + x = Then as in first method ( 5x 4)( x + ) = 0 x = (x = ) 5 Complete trial and error yielding 0.8 is M and B for 0.8 A, A awarded for each of two tries evaluated. i.e. 6/6 Incomplete trial and error with wrong or no solution is 0/6 Just answer 0.8 with no working is B If log base 0 or base e used throughout - can score BA0A0 A d A (6) [6] 87_974 GCE Mathematics January 009 0

21 5 (a) 0 PQ: m = ( = ) 9 ( ) and QR: 0 4 m = 9 a 8 6 A m m = : = a = (*) () 9 a Alt for (a) Alternative method (Pythagoras) Finds all three of the following (a) 9 + (0 ), ( ie..08), 9 a + (0 4), a + (4 ) ( ( )) ( ) ( ( )) Using Pythagoras (correct way around) form equation Solve (or verify) for a, a = (*) Centre is at (5, ) e.g. a 6a 9 40 a 8a = + + to A B () Alt for ( r = ) ( 0 ) + (9 5) or equiv., or ( d = ) ( ) ( x 5) + ( y ) = 65 or x y x y ( ) + (4 ) A = 0 A (5) Uses ( x a) + ( y b) = r or x + y + gx+ fy+ c= 0 and substitutes (-, ), (9, 0) and (, 4) then eliminates one unknown Eliminates second unknown Notes Obtains g = 5, f =, c= or a = 5, b =, r = 65 A, A, Bcao (5) [8] (a) -considers gradients of PQ and QR -must be y difference / x difference (or considers three lengths as in alternative method) Substitutes gradients into product = - (or lengths into Pythagoras Theorem the correct way round ) A Obtains a = with no errors by solution or verification. Verification can score /. Geometrical method: B for coordinates of centre can be implied by use in part for attempt to find r, d, r or d( allow one slip in a bracket). A cao. These two marks may be gained implicitly from circle equation for ( x ± 5) + ( y± ) = k or ( x ) ( y 5) k numerical. ± + ± = ft their (5,) Allow k non A cao for whole equation and rhs must be 65 or ( 65 ), (similarly B must be 65 or ( 65 ), in alternative method for ) 87_974 GCE Mathematics January 009

22 Further alternatives (i) A number of methods find gradient of PQ = / then give perpendicular gradient is / This is They then proceed using equations of lines through point Q or by using gradient QR to obtain equation such as 4 0 = (may still have a 9 x in this equation rather than a and there may be a small slip) They then complete to give (a )= A (ii) A long involved method has been seen finding the coordinates of the centre of the circle first. This can be done by a variety of methods Giving centre as (c, ) and using an equation such as ( c 9) + 7 = ( c+ ) + (equal radii) or 6 = (perpendicular from centre to chord bisects chord) c Then using c ( = 5) to find a is Finally a = A (iii) Vector Method: States PQ. QR = 0, with vectors stated i +8j and (9 a)i + 6j is Evaluates scalar product so 08 a + 48 = 0 () solves to give a = (A) A A A 87_974 GCE Mathematics January 009

23 6 (a) f () = a + b or f( ) = 5 a+ b A Alternative for (a) Alternative for Finds nd remainder and equates to st a+ b= 5 a+ b A a = 0 Acso (5) 4 f ( ) = ( ) + 5( ) a + b = 0 Aft b = 0 gives b = -6 A cso (a) Uses long division, to get remainders as b + a + 56 or b a - 4 or correct equivalent A Uses second long division as far as remainder term, to get A b + a + 56 = b a - 4 or correct equivalent a = 0 Acso (5) 4 Uses long division of x + 5x 0x+ b by (x + ) to obtain x + x 6x+ a+ 8( with their value for a ) Giving remainder b + 6 = 0 and so b = -6 Aft A cso Notes (a) : Attempts f( ± ) or f( ± ) A is for the answer shown (or simplified with terms collected ) for one remainder : Attempts other remainder and puts one equal to the other A: for correct equation in a (and b) then A for a = 0 cso : Puts f ( ± ) = 0 A is for f( -) = 0, (where f is original function), with no sign or substitution errors (follow through on a and could still be in terms of a ) A: b = -6 is cso. Alternatives 4 (a) : Uses long division of x + 5x + ax+ b by (x ± ) or by (x ± ) as far as three term quotient A: Obtains at least one correct remainder : Obtains second remainder and puts two remainders (no x terms) equal A: correct equation A: correct answer a = -0 following correct work. : complete long division as far as constant (ignore remainder) Aft: needs correct answer for their a A: correct answer Beware: It is possible to get correct answers with wrong working. If remainders are equated to 0 in part (a) both correct answers are obtained fortuitously. This could score AM0A0A0AA0 () [8] () [8] 87_974 GCE Mathematics January 009

24 7 (a) (a) (c) r θ = 6. = 9.6 ( cm ) π. = π. =.04 (rad) (c) DAC = 6 4sin.04 ( 0.7) Total area = sector + triangles = 6 ( ) A () A () Aft cm A (4) [8] : Needs θ in radians for this formula. Could convert to degrees and use degrees formula. A: Does not need units. Answer should be 9.6 exactly. Answer with no working is A. This A can only be awarded in part (a). : Needs full method to give angle in radians A: Allow answers which round to.04 (Just writes.04 no working is /) : Use 6 4sinA (if any other triangle formula e.g. b h is used the method must be complete for this mark) (No value needed for A, but should not be using.) A: ft the value obtained in part need not be evaluated- could be in degrees : Uses Total area = sector + triangles or other complete method A: Allow answers which round to 6. (Do not need units) Special case degrees: Could get M0A0, M0A0, AA0 Special case: Use BDC BAC Both areas needed for first Total area = sector + area found is second NB Just finding lengths BD, DC, and angle BDC then assuming area BDC is a sector to find area BDC is 0/4 87_974 GCE Mathematics January 009 4

25 8 (a) 4( cos x ) + 9cos x 6 = 0 4cos x 9cos x + = 0 (*) A ( 4cos x )(cos x ) = 0 cos x =..., 4 x = 75.5 ( α) B A () 60 α, 60 + α or 70 - α, 84.5, 45.5, A (6) [8] Special cases (a) : Uses sin x = cos x (may omit bracket) not sin x = cos x A: Obtains the printed answer without error must have = 0 : Solves the quadratic with usual conventions A: Obtains ¼ accurately- ignore extra answer but penalise e.g. -. B: allow answers which round to 75.5 : 60 α ft their value, : 60 + α ft their value or 70 - α ft A: Three and only three correct exact answers in the range achieves the mark In part Error in solving quadratic (4cosx-)(cosx+) Could yield, A0BA losing one mark for the error Works in radians: Complete work in radians :Obtains. B0. Then allow for π α, π + α or 4π α Then gets 5.0, 7.6,. A0 so /4 Mixed answer., 60., 60 +., 70. still gets B0A0 87_974 GCE Mathematics January 009 5

26 9 (a) Initial step: Two of: a = k + 4, ar = k, ar = k Or one of: k, k, k r = r = r =, k+ 4 k k+ 4 Or k = ( k+ 4)(k 5) or even k = ( k+ 4) k(k 5) k = ( k + 4)(k 5), so k = k + 8k 5k 60, A Proceed to k 7k 60 = 0 (*) A (4) ( k )( k + 5) = 0 k = (*) A () (c) (d) Common ratio: a r 6 = 4 ( ) = 64 k k-5 or = = or 0.75 k+ 4 k 6 4 A () A () [0] (a) : The initial step, scoring the first M mark, may be implied by next line of proof : Eliminates a and r to give valid equation in k only. Can be awarded for equation involving fractions. A : need some correct expansion and working and answer equivalent to required quadratic but with uncollected terms. Equations involving fractions do not get this mark. (No fractions, no brackets could be a cubic equation) A: as answer is printed this mark is for cso (Needs = 0) All four marks must be scored in part (a) : Attempt to solve quadratic A: This is for correct factorisation or solution and k =. Ignore the extra solution (k = 5 or even k = 5), if seen. Substitute and verify is A0 must be scored in part (c) : Complete method to find r Could have answer in terms of k A: 0.75 or any correct equivalent Both must be scored in (c) (d) a : Tries to use, (even with r>). Could have an answer still in terms of k. r A: This answer is 64 cao. 87_974 GCE Mathematics January 009 6

27 0 π rh + π r = 800 B (a) 400 π r 400 π r, A h =, V = π r = 400r π r π r π r (*) (4) dv A = 400 π r dr π r = 0 r =..., r = ( = 6.5 ( s.f.) ) A π (c) V = r π r = 77 cm π (accept awrt 77 or exact answer) d V = 6π r, Negative, maximum dr (Parts and (c) should be considered together when marking) = ( ) A A (6) () [] Other methods for part (c): Either:M: Find value of d V on each side of 400 r = and consider sign. dr π A: Indicate sign change of positive to negative for d V, and conclude max. dr 400 Or: M: Find value of V on each side of r = and compare with 77. π A: Indicate that both values are less than 77 or 77.5, and conclude max. Notes (a) B: For any correct form of this equation (may be unsimplified, may be implied by st ) : Making h the subject of their three or four term formula : Substituting expression for h into π rh(independent mark) Must now be expression in r only. A: cso : At least one power of r decreased by A: cao : Setting d V =0 and finding a value for correct power of r for candidate dr A : This mark may be credited if the value of V is correct. Otherwise answers should round to 6.5 (allow ± 6.5 )or be exact answer : Substitute a positive value of r to give V A: 77 or or exact answer 87_974 GCE Mathematics January 009 7

28 (c) Alternative for (a) : needs complete method e.g.attempts differentiation (power reduced) of their first derivative and considers its sign A(first method) should be 6π r (do not need to substitute r and can condone wrong r if found in ) Need to conclude maximum or indicate by a tick that it is maximum. Throughout allow confused notation such as dy/dx for dv/dr A π r π rh = +, = + is Equate to 400r B A r r r h π π Then V = 400r π r is A 87_974 GCE Mathematics January 009 8

29 January Core Mathematics C Mark Question d dx (a) ( x ) d dx ( ) ( 5 ) = ( 5x ) ( ) = 5 5x A dy 5 ( ) ( ) = x 5x + x 5x Aft dx At x =, dy 0 0 = 4 9+ = + dx 9 46 = Accept awrt 5. A (6) d sin x x cos x xsin x = 4 dx x x A+A A (4) [0] Alternative to d ( ) ( x x = x x + x ) x dx A + A cosx sinx = x cosx x sinx = x x A (4) 87_974 GCE Mathematics January 009 9

30 (a) x + x+ x+ x+ = + ( )( ) x+ ( x+ )( x+ ) ( x )( x+ ) ( x + )( x) ( x )( x+ ) x x x x x x = = x = Accept x x, x x x A A (4) ( x )( ) ( x) ( ) d x = dx x x x+ + x = = ( x ) ( x ) A cso A () [7] Alternative to (a) x + ( x + ) = = x x ( x )( x+ ) x A x + ( x + ) = x x x x = x A (4) Alternatives to ( ) x f x = = = ( x ) x x f x = x A ( ) ( )( )( ) = ( x ) ( ) ( )( ) f x = x x cso A () ( ) ( )( ) x x ( x)( )( x ) x ( x ) ( x) = = x ( x ) ( x ) f = + = ( x ) A () A 87_974 GCE Mathematics January 009 0

31 (a) y O (, 6 ) ( 7, 0 ) x Shape B, 6 B ( ) ( 7, 0 ) B () y (, 5 ) ( 7, ) Shape B, 5 B ( ) ( 7, ) B () [6] O x 87_974 GCE Mathematics January 009

32 4 x= cos( y+ π ) At π y =, 4 dx = sin( y + π ) dy A dy = Follow through their d x dx sin y+ π dy Aft ( ) dy = = dx π sin before or after substitution B π y = x 4 π y = x+ 4 A (6) [6] 87_974 GCE Mathematics January 009

33 5 (a) g( x) B () ( ) ( ) x x x fg x = f e = e + lne = x + e x A () ( fg : xa x + e ) x (c) fg ( x) B () d x (d) ( ) x e 0, dx x x + e = x+ 6xe A x x x + 6xe = x e + x x ( x x ) e 6 = 0 6x x 0 = A x = 0, 6 A A (6) [0] 87_974 GCE Mathematics January 009

34 6 (a)(i) sin θ = sin ( θ + θ) = sin θ cosθ + cos θsin θ = sinθ cos θ.cosθ + sin θ sinθ A ( ) ( ) = sinθ sin θ + sinθ sin θ = sinθ 4sin θ cso A (4) (ii) 8sin θ 6sinθ + = 0 sinθ + = 0 A sin θ = π 5π θ =, 6 6 π 5π θ =, 8 8 A A (5) sin5 = sin ( ) = sin 60 cos 45 cos 60 sin 45 = A = 6 = ( 6 ) cso A (4) [] Alternatives to sin5 = sin 45 0 = sin 45 cos0 cos 45 sin 0 ( ) Using cos θ sin = θ, = A = 6 = ( 6 ) cso A (4) cos 0 = sin 5 sin 5 = cos0 = sin 5 = A 4 ( 6 ) = ( 6+ ) = Hence sin5 = ( 6 ) 4 cso A (4) 87_974 GCE Mathematics January 009 4

35 7 (a) ( x) x x f = e + x e A ( x) x x x e + xe = e + = 0 x = A f = e B (5) ( ) x = B x = 0.57 B x = B () (c) Choosing ( , ) or an appropriate tighter interval. f ( ) = f ( ) = A Change of sign (and continuity) root ( , ) cso A ( x = 0.576, is correct to 4 decimal places) () [] Note: x = is accurate 87_974 GCE Mathematics January 009 5

36 8 (a) R = + 4 R = 5 A 4 tanα = α = 5... awrt 5 A (4) Maximum value is 5 ft their R B ft At the maximum, ( θ α) cos = or θ α = 0 θ = α = 5... ft their α A ft () (c) f () t = 0+ 5cos( 5t α ) Minimum occurs when cos( 5t α ) = The minimum temperature is ( 0 5) = 5 A ft () (d) 5t α = 80 t = 5.5 awrt 5.5 A () [] 87_974 GCE Mathematics January 009 6

37 January Core Mathematics C4 Mark (a) C: y y = x + 8 dy dy dy = y = x dx dx dx Differentiates implicitly to include either ky dy dy dy ± or ±. (Ignore = d x d x dx.) Correct equation. A dy dx ( y ) = x A correct (condoning sign error) attempt to dy dy combine or factorise their y. d x d x Can be implied. dy x = dx y x y A oe (4) y = 9 () = x + 8 Substitutes y = into C. x = 8 x= Only x = A d (4) d (,) y y 4 dx = 6 dx = dy 4 dx = from correct working. Also can be ft using their x value and y = in the dy x correct part (a) of = dx y final A. Note if the candidate inserts their x value and y = into then an answer of dy = their x, may indicate a correct follow through. dx dy x = dx y, A () [7] 87_974 GCE Mathematics January 009 7

38 (a) Area(R) = dx = (+ 4 x) dx ( + 4 x) 0 0 x Integrating ( + 4 ) to give ( + 4 x) ± k( + 4 x). =.4 0 Correct integration. Ignore limits. A = ( + 4 x) 0 ( ) 9 ( () ) = Substitutes limits of and 0 into a changed function and subtracts the correct way round. 9 = = A (units) (Answer of with no working scores M0A0M0A0.) (4) Volume = π ( + 4 x) 0 dx Use of V = π y dx. Can be implied. Ignore limits and d x. B = ( ) π 0 9 dx + 4x ± kln + 4x 9 = ( π ) ln + 4x ln + 4x A 4 ( ) ( 9 ln 9) ( 9 ln) = π Substitutes limits of and Note that ln can be implied as equal to 0. and subtracts the correct way round. d 9 So Volume π ln 9 4 Note the answer must be a one term exact value. Note, also you can ignore subsequent working here. = 9 π ln 9 4 or 9 π ln or 8 π ln 4 A oe isw 9 Note that = π ln 9 + c (oe.) would be (5) 4 awarded the final A0. [9] 87_974 GCE Mathematics January 009 8

39 (a) 7x + x+ 6 A(x+ )( x) + B( x) + C(x+ ) Forming this identity x = ( ) ( ), + 6 = B = B B= 4 x =, = 5C 75 = 5C C = Substitutes either x = or x = into their identity or equates terms or substitutes in values to write down three simultaneous equations. Both B = 4 and C = (Note the A is dependent on both method marks in this part.) A Equate x : 7 = A+ 9C 7 = A = A A = 0 x= 0, 6 = A + B + 4C 6 = A = A A= 0 Compares coefficients or substitutes in a third x-value or uses simultaneous equations to show A = 0. B (4) 4 f( x) = + (x + ) ( x) = 4(x + ) + ( x) Moving powers to top on any one of the two expressions ( ) x x = ( ) ( ) x x = + + ( ) x Either ± ( )( ) or x ( )( ) x = + ( )( ); + ( ) +... ± ( )( x) from either first d;! or second expansions respectively Ignoring and, any one ( )( ) A + ( )( x); ( x)... correct {...} expansion ! Both {...} correct. A 7 { x x...} { x x... 4 } = = 4+ 0 x ; x 4 (0 ); 9 + x x 4 A; A (6) 87_974 GCE Mathematics January 009 9

40 (c) Actual = f (0.) = (6.76)(0.8) = = = Or Actual = f (0.) = ((0.) ) ( 0.) 4 95 = +.75 = = Attempt to find the actual value of f(0.) or seeing awrt 4. and believing it is candidate s actual f(0.). Candidates can also attempt to find the actual value by using A + B + C (x + ) (x+ ) ( x) with their A, B and C. 9 Estimate = f (0.) = 4 + (0.) 4 = = 4.9 Attempt to find an estimate for f(0.) using their answer to %age error = their estimate - actual actual 00 = =.%(sf ).% A cao (4) [4] 87_974 GCE Mathematics January

41 4 (a) d = i + j 4k, d = qi + j+ k As d d q = = ( q) + ( ) + ( 4 ) 4 Apply dot product calculation between two direction vectors, ie. ( q) + ( ) + ( 4 ) d d = 0 q + 8 = 0 Sets d d = 0 q = 6 q = AG and solves to find q = A cso () Lines meet where: 5 q + λ = + µ 7 4 p First two of i : λ = 5 + qµ () j: + λ = + µ () k :7 4λ = p + µ () Need to see equations () and (). Condone one slip. (Note that q =.) () + () gives: 5 = 7 + µ µ = () gives: + λ = 4 λ = 5 Attempts to solve () and () to d find one of either λ or µ Any one of λ = 5orµ = A Both λ = 5andµ = A () 7 4(5) = p + ( ) Attempt to substitute their λ and µ into their k component to give an equation in p alone. dd p = p = p = A cso (6) (c) 5 r = + 5 or r = 7 4 Substitutes their value of λ or µ into the correct line l or l. Intersect at r = 7 or (, 7, ) 7 or (, 7, ) A () 87_974 GCE Mathematics January 009 4

42 uuur (d) Let OX = i + 7j k be point of intersection uuur 9 8 uuur uuur uuur Finding vector AX by finding the uuur uuur AX = OX OA = 7 = 4 difference between OX and OA. ± 6 uuur Can be ft using candidate sox. uuur uuur uuur uuur uuur OB = OA + AB = OA + AX uuur OB 9 8 = uuur + their AX d Hence, uuur OB 7 = 9 uuur or OB = 7i + j 9k 7 9 or 7i + j 9k or ( 7,, 9) A () [] 87_974 GCE Mathematics January 009 4

43 5 (a) Similar triangles r 6 h = r = h 4 Uses similar triangles, ratios or trigonometry to find either one of these two expressions oe. h 4π h V π r h π h = = = 7 AG Substitutes r = h into the formula for the volume of water V. A () From the question, d V = 8 dt dv dt = 8 B dv π h 4π h = = dh 7 9 d V π h 4π = or h B dh 7 9 dh dv dv dt dt dh 4π h π h = = = π h 8 7 Candidate s d V dt or 8 or d V ; ; dh 9 8 4π h π h oe A When dh 8 h =, = = dt 44π 8π 8 44π or 8π A oe isw Note the answer must be a one term exact value. Note, also you can ignore subsequent working 8 after 44π. (5) [7] 87_974 GCE Mathematics January 009 4

44 6 (a) tan x dx NB : sec A = + tan A gives tan A = sec A The correct underlined identity. oe = sec x dx = tan x x ( + c) Correct integration with/without + c A () ln x dx x du u = ln x = dx x = x v = = dv x dx x = ln x. dx x x x Use of integration by parts formula in the correct direction. Correct direction means that u = ln x. Correct expression. A = ln x + dx x x = ln x + + x x ( c) An attempt to multiply through k n x, n, n by and an x attempt to... integrate (process the result); correct solution with/without + c A oe (4) 87_974 GCE Mathematics January

45 (c) e x + e x dx x du x dx dx u = + e = e, =, = x dx du e du u Differentiating to find any one of the three underlined B x x x e.e ( u ).e = d x =. du x x + e u e ( u ) or =. du u ( u ) x Attempt to substitute for e = f( u), their d x = and u = + du e x ex x or e = f( u), their d x = and du u u = + e x. * ( u ) = u du ( u ) u du A u = u+ d u u = u + du u u = u + lnu + c ( ) An attempt to multiply out their numerator to give at least three terms and divide through each term by u Correct integration with/without +c d* A x ( + e ) x x = (+ e ) + ln(+ e ) + c Substitutes u = + e x back into their integrated expression with at least two terms. d* = c e x e x e x ln( e x ) = c e x e x e x ln( e x ) = e e + ln(+ e ) + c x x = k AG e x e x ln( e x ) must use a k e x e x ln( e x ) + c and " " combined. A cso (7) [] 87_974 GCE Mathematics January

46 7 (a) At A, x= + 8= 7 & y = ( ) = A(7,) A(7,) x t 8 t, y t = =, B () dx d y d t 8 t =, d t = t dy t = dx t 8 Their d y divided by their d dt d x t Correct d y dx A ( ) At A, m( T ) = = = = Substitutes for t to give any of the ( ) four underlined oe: ( ) T ( ( )) T : y their = m x their 7 = 7 + c c = = 4 9 or ( ) Hence T : y = x 5 5 Finding an equation of a tangent with their point and their tangent gradient or finds c and uses y = (their gradient) x + " c". d gives T :x 5y 9= 0 AG x 5y 9= 0 A cso (5) (c) ( t 8 t) 5t 9 = 0 Substitution of both x = t 8t and y = t into T t 5t 6t 9= 0 { } ( t+ ) (t 7t 9) = 0 { } ( t+ ) ( t+ )(t 9) = 0 9 { = } = t (at A) t atb A realisation that t + is a factor. ( ) 9 t = d A ( ) ( ) ( ) x = 8 = 6 = = 55.5 or awrt Candidate uses their value of t to find either the x or y coordinate dd 9 8 y = = = 0.5 or awrt 0. 4 One of either x or y correct. A Both x and y correct. A B 44 8, awrt (6) Hence ( ) 8 4 [] 87_974 GCE Mathematics January

47 January Further Pure Mathematics FP (legacy) Mark (a) Alt Notes n n n r r = n( n + )(n + ) n( n + ) n r= r= r= 6, A Simplifying this expression = n ( n 4) A (4) ( ) cso 0 9 ( r r ) ( r r ) = 0 (0 4) 9 (9 4) r= r= = 409 A () 0 r= 9 ( r r ) ( r r ) = r= = 409 A [6] (a) st M: Separating, substituting set results, at least two correct. nd M: Either eliminate brackets totally or factor x [...] where any product of brackets inside [...] has been reduced to a single bracket nd A: ANSWER GIVEN. No wrong working seen; must have been an intermediate step, e.g. n (n + n + n 6) M : Must be (...) (...) applied. r= r= If list terms and add, allow if terms with at most two wrong: [89, 09,, 55, 8, 09, 9, 7, 05, 4, 79] 87_974 GCE Mathematics January

48 i is a root (seen anywhere) B Notes Attempt to multiply out [x ( + i)][x ( i)] {= x 6x + 0 } f(x) = (x 6x + 0)(x x + ), A ± 4 8 ± i x =, x = 4 *, A st M: Using the two roots to form a quadratic factor. nd M: Complete method to find second quadratic factor x + ax (+ b). rd *M: Correct method, as far as x =, for solving candidate s second quadratic, DEPENDENT on both previous M marks [6] Alt (i)f(x)/{x ( + i)} = x + (-8 + i)x + (7 i)x + i {=g(x)} g(x)/{x ( - i)} = (x x + ) Attempt at complete process M; A (ii)()(x a+ib)(x a ib)( x 6x + 0 ) = f(x) and compare coeff. Either a 6= 7, or two of 0(b + a ) = 5 or 6(a + b ) 0a =, 0+ (b + a ) +4a = A; Complete method for a and b, ; AnswerA Lines and Lines and 4 87_974 GCE Mathematics January

49 Identifying as critical value e.g. used in soln Identifying 0 as critical value e.g. used in soln B B x + 5x 4( x ) x x( x + ) > 0 x > 0 or ( x + 5x )( x ) > 4( x ) o.e. or ( x )( x + x) > 0 A Notes Using their critical values to obtain inequalities. x < 0 or x > st M must be a valid opening strategy. Sketching y = x x x( x + ) or y = should mark as scheme. x A cso Alt The result 0 > x > (poor notation) can gain final M but not A. y 4 O x Identifying as critical value e.g. x = seen as asymp. Identifying 0 as critical value e.g. pt of intersection on y-axis of x + 5x y = and y = 4 x x + 5x x + 5x y = sketched for x < or y = sketched for x > x x A All correct including y = 4 drawn B B, A Using the graph values to obtain one or more inequalities x < 0 or x > A 87_974 GCE Mathematics January

50 4 (a) f ( x0 ) At st. pt f (x) = 0, x = x 0 is undefined f ( x0 ) B () or at st. pt, tan. // to x-axis, or tan. does not cross x-axis, o.e. f (x) = x cos( x ) (may be seen in body of work), A f(0.6) = , f (0.6) =. (may be implied by correct answer) A f (0.6) Attempt to use (x ) = [0.6 ] f (0.6)... = 0.6 ( dp) ( ) A (5) (c) f(0.65) =.77 0 > 0, f(0.65) = < 0 Change of sign in f(x) in (0.65, 0.65) so 0.6 correct A () Notes ndm: If the N-R statement applied to 0.6 not seen, can be implied if answer correct; otherwise M0 [8] If no values for f(0.6), f (0.6) seen, they can be implied if final answer correct. (c) M: For candidates x, calculate f(x ) and f(x ) (or a tighter interval) A: Requires correct values of f(0.65) and f(0.65) (or their acceptable values) [may be rounded, e.g. 0, or truncated, e.g ], sign change stated or >0, <0 seen, and conclusion. 87_974 GCE Mathematics January

51 5 (a) 5i i 6 4i 5i 0 z = = + i i = i A () Im z O P(, ) Re z B, Bft () Q(, ) P: B, Q: Bft from (a) (c) grad. OP grad. OQ = ( ) π = POQ = ( ) A () Alt (c) (i) POX = tan, QOX = tan + Tan( POQ) = POQ = π ( ) A Notes (d) (e) z = + + ( + ) i = 5 i A () 5 r = + 6 = or exact equivalent (a) M: Multiplying num. and den. by i and attempt to simplify num. and denominator. If (c + id)( + i) = 5i used, need to find equations in c and d and then solve for c and d. Coords seen or clear from labelled axes. S.C: If only P and Q seen(no coords) or correct coords given but P and Q interchanged allow BB0 (c) If separate arguments are found and then added, allow but not A for decimals used e.g = ½ π. Alts: Appropriate transformation matrix applied to one point ; A Scalar product used correctly ; 0 and conclusion A Pythagoras theorem, congruent triangles are other methods seen. (d) M: Any complete method for finding centre. A: Must be complex number; coordinates not sufficient. (e) M: Correct method for radius, or diameter, for candidate s answer to (d) A () [0] 87_974 GCE Mathematics January 009 5

52 6 (a) r = x + y, y = r sinθ x y + y = or x + y = 6y o.e., A x + y () 6 r = 9 6( sin θ ) o.e. B () (c) (d) (e) y = r sinθ = 9 6(sinθ sin θ ) cos θ ( 6sin θ ) dθ o.e. ;A Or y = 9 6 sinθ cos θ (cos θ cosθ sinθ sin θ ) o.e. dθ dy = 0 [ cos θ ( 6sin θ ) =0 ] dθ and attempt to solve π ( 0 θ ) sin θ = 4 6 ( ) A (4) r = = 6 6 or 4.7 (awrt) A () C : tan. // to initial line is y = r sinθ = 6 6 = 6 6 B C : Circle, centre (0, ) (cartesian) or (, π ) (polar), passing through (0,0). tangent // to initial line has eqn y = 6 y = 6 is a common tangent A () [] Notes (a) : Use of r = x + y or r = x + y, and y = r sinθ (allow x = r sinθ ) to form cartesian equation. May be scored in (c) dy (c) st M: Finds y and attempts to find dθ Working with rcos θ instead of r sin θ, can score the M marks. d y dy dx d y If = / used throughout, = 0 etc. all marks may be gained dx dθ dθ dx (d) M: Using sin θ = to find r 6 (e) Alt. for C : dy dy 6 sin θ, ( = sin θcos θ ) and solve = 0 dθ dθ π A: Find θ = and conclude that y = 6, so common tangent M:Find y = 87_974 GCE Mathematics January 009 5

53 7 (a) d y = λxe x + λe x Use of the product rule dx d y = λxe x + λe x + λe x dx A λxe x + λe x + 4λxe x + 4λe x 5λxe x = 4e x * λ = A (4) ( P.I. is xe x ) Aux. eqn. m + 4m - 5 = 0 (m )(m + 5) = 0 m = or m = 5 C.F. is y = Ae x + Be 5x A Gen. soln. is (y =) xe x + Ae x + Be 5x [f.t: Candidate s C.F + P.I.] Aft (4) (c) = A + B d y = xe x + e x + Ae x 5Be 5x dx 4 = + A 5B A two correct unsimplified eqns. A = A 5B 4 = 6B 8 B =, A = 9 9 y = xe x 9 8 e x + 9 e 5x A (5) Notes (a) nd M dependent on first M. [] st M: Attempt to solve A.E. nd M: Only allow C.F. of form Ae ax + Be bx, where a and b are real. If seen in (a), award marks there. PI must be of form λ xe x ( λ 0) to gain final A f.t. (c) st M: Using x = 0, y = ⅔ in their general solution. nd M: Differentiating their general solution {C.F. + P.I. } (must have term in λ xe x ) (condone slips) and using dy 4 x = 0, = to find an equation in A and B. dx rd M: Solving simultaneous equations to find a value of A and a value of B. Can be awarded if only C.F. found. Insist on y =... in this part. 87_974 GCE Mathematics January 009 5

54 8 (a) d y dt = dx t dx o.e., A sin x dt t dx t cos x = t dt tcot x= cosec x dx ( ) A cso (4) I = e ln sin x = e = sin x or cosecx A dt cosx t = cosec x sin x dx sin x t cos ec x dx sin x = d t or = cos ec x dx sin x Af.t. t cot x sin x = ( + c) o.e. A cso (5) (c) t = cos x + csin x y = ( ), A cos x + csin x () (d) c + Notes c + = c = A π x =, y = ft on their c Aft () [4] dy dt (a) st M: Use of. dt dx (even if integrated /t) dy nd M: Substituting for, y, y to form d.e. in x and t only dx st M: For e cot x dx ( allow e ) and attempt at integrating nd* M: Multiplying by integrating factor (requires at least two terms correct for their IF.) (can be implied) rdaf.t: is only for those who have I.F. = sinx or sinx d ( t sin x) = dx equivalent integral (c) M: Substituting to find t = /y in their solution to (d) M: Using y =, x = π to find a value for c. 4 87_974 GCE Mathematics January