Mark Scheme (Results) January 2009

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1 Mark (Results) January 009 GCE GCE Mathematics (666/0) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 January Core Mathematics C Mark Question (a) 5 ( ± 5 is B0) ( their 5) or their 5 = 5 or 0.04 ( ± is A0) 5 B M A () () [] M 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 M but not the A, e.g. 5 = = scores M A = = scores M A Correct answer with no working scores both marks. Alternative: 5 or = 565 ( 5 ) = A 5 M (reciprocal and the correct number squared) 666/0 GCE Mathematics January 009

3 6 8 4 ( I = ) x x + x+ c = x x + x+ c M AAA [4] n M 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. 666/0 GCE Mathematics January 009

4 , or 7 4 or an exact equivalent such as = 49 M A [] M for an expanded expression. At worst, there can be one wrong term and one wrong sign, or two wrong signs. e.g is M (one wrong term ) is M (two wrong signs + 7 and + 4 ) is M (one wrong term +, one wrong sign + 7 ) is M (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 M. The terms can be seen separately for the M. Correct answer with no working scores both marks. 666/0 GCE Mathematics January 009 4

5 4 x x ( f( x) = ) 7x( + c) = x x 7 x ( + c) f(4) = = c c = st M 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 M 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). M AA M Acso (5) [5] 666/0 GCE Mathematics January 009 5

6 5 (a) y Shape maximum., touching the x-axis at its M - 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) M 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. 666/0 GCE Mathematics January 009 6

7 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 M 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 M 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 M 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 () M AAftAft (4) [6] 666/0 GCE Mathematics January 009 7

8 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 ) (*) > MA 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. M A M A (4) [7] M 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 M (unless recovery is seen). st M 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 M 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 M 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. 666/0 GCE Mathematics January 009 8

9 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) ( 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. 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. 666/0 GCE Mathematics January 009 9

10 9 (a) 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 () M Acso () MAft M Acso (4) (d) (a) (d) n 5n = 0 or n 5n 00 = 0 M ( n 55)( n+ 40) = 0 n = M 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. M: 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 MA if both values are checked in the nd equation. st M 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 M for expanding and simplifying to a term quadratic. nd A for correct working leading to printed result (no incorrect working seen). st M forming the correct TQ = 0. Can condone missing = 0 but all terms must be on one side. First M can be implied (perhaps seen in, but there must be an attempt at (d) for it to be scored). nd M 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 M 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. 666/0 GCE Mathematics January 009 0

11 0 (a) (d) (a) (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 (*) M A The version in the scheme above can be written down directly (for marks), and M 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. M 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 M for ( p ) + (linear function of p). The linear function may be unsimplified but must be equivalent to ap + b, a 0, b 0. nd M (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 M, 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 M. Getting AC = 5 for both values of p, or showing C C = 5 scores the nd A (cso). MA, Acao () B () M, A, A () M M A Acso (4) [] 666/0 GCE Mathematics January 009

12 (a) (a) dy = 4+ 8x (4 or 8x for M 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 MA M Area of triangle is: ( x B ± x A ) yp with values for all of xb, x A and yp M 45 8 = or.5 A (4) 4 [] st M for 4 or 8x (ignore the signs). st A for both terms correct (including signs). nd M 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 M 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. M 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 M Acso (6) Bft MA B, B () M 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 M) x y 8 0 Alternative: AP = ( 0.5) + ( ), BP = ( 8) + ( ), Area = AP BP =... M Intersections with y-axis instead of x-axis: Only the M mark is available B0 B0 M A0. 666/0 GCE Mathematics January 009

Mark Scheme (Results) January 2008

Mark Scheme (Results) January 008 GCE GCE Mathematics (666/0) Edexcel Limited. Registered in England and Wales No. 446750 Registered Office: One0 High Holborn, London WCV 7BH January 008 666 Core Mathematics