GCE Edexcel GCE Core Mathematics C (666) January 006 Mark Scheme (Results) Edexcel GCE Core Mathematics C (666)
January 006 666 Core Mathematics C Mark Scheme. x( x 4x + ) Factor of x. (Allow ( x 0) ) M = x( x )( x ) Factorise term quadratic M A Total marks () Alternative: ( x x)( x ) or ( x x)( x ) scores the second M (allow ± for each sign), then x( x )( x ) scores the first M, and A if correct. Alternative: Finding factor ( x ) or ( x ) by the factor theorem scores the second M, then completing, using factor x, scores the first M, and A if correct. Factors split : e.g. x( x 4x + ) ( x )( x ). Allow full marks. Factor x not seen: e.g. Dividing by x ( x )( x ). M0 M A0. If an equation is solved, i.s.w.
. (a) u = ( ) 4 B = =, u4 4 u = For u, ft ( u Bft, B ) (b) u = 0 4 Bft () () Total 4 marks (b) ft only if sequence is oscillating. Do not give marks if answers have clearly been obtained from wrong working, e.g. u = ( ) 0 = u = (4 ) = u4 = (5 ) = 4
. (a) y = 5 ( ) = (or equivalent verification) (*) B (b) Gradient of L is B () y ( ) = ( x ) (ft from a changed gradient) M Aft x y 5 = 0 (or equiv. with integer coefficients) A (4) Total 5 marks (a) y ( ) = ( x ) y = 5 x is fine for B. Just a table of values including x =, y = is insufficient. (b) M: eqn of a line through (, ), with any numerical gradient (except 0 or ). y ( ) For the M Aft, the equation may be in any form, e.g. =. x Alternatively, the M may be scored by using y = mx + c with a numerical gradient and substituting (, ) to find the value of c, with Aft if the value of c follows through correctly from a changed gradient. Allow x y = 5 or equiv., but must be integer coefficients. The = 0 can be implied if correct working precedes.
4. (a) dy dx = 4x + 8x 4 4 M: x x or x x M A () (b) x 6x + C M: x x or x x or + C M A A () = x + x + C First A: x + C Second A: 6x Total 5 marks In both parts, accept any correct version, simplified or not. Accept 4x for 4x. + C in part (a) instead of part (b): Penalise only once, so if otherwise correct scores M A0, M A A.
5. (a) 5 (or a = ) B () (b) ( + 5) ( + 5) ( 5) ( + 5) ( 5)( + 5) = 9 5 ( = 4) M (Used as or intended as denominator) B ( + 5)( p ± q 5) =... 4 terms ( p 0, q 0) or ( 6 + 5)( p ± q 5) =... 4 terms ( p 0, q 0) (Independent) M [Correct version: ( + 5)( + 5) = 9 + 5 + 5 + 5, or double this.] (4 + 6 5) 4 = 7 + 5 st A: b = 7, nd A: c = A A (5) Total 6 marks (b) nd M mark for attempting ( + 5)( p + q 5) is generous. Condone errors.
6. (a) (See below) M Clearly through origin (or (0, 0) seen) A labelled (or (, 0) seen) A () (b) 6 Stretch parallel to y-axis M and 4 labelled (or (, 0) and (4, 0) seen) A 4 6 labelled (or (0, 6) seen) A () (c) Stretch parallel to x-axis M and 8 labelled (or (, 0) and (8, 0) seen) A 8 labelled (or (0, ) seen) A () Total 9 marks (a) M: (b) M: with at least two of: (, 0) unchanged (4, 0) unchanged (0, ) changed (c) M: with at least two of: (, 0) changed (4, 0) changed (0, ) unchanged Beware: Candidates may sometimes re-label the parts of their solution.
S = (*) B () (b) Using a = 500, d = 00 with n = 7, 8 or 9 a + ( n ) d or listing M 7. (a) 500 + (500 + 00) = 00 or = { 000 + 00} 00 500 + (7 00) = ( )900 A () + or n { a + l}, or listing and summing terms M S 8 = 8{ 500 + 7 00} or S 8 = 8{ 500 + 900}, or all terms in list correct A = ( ) 9600 A () (d) n { 500 + ( n ) 00} = 000 M: General S n, equated to 000 M A (c) Using n { a ( n ) d} n + 4n 0 = 0 (or equiv.) M: Simplify to term quadratic M A ( n + 0)( n 6) = 0 n =... M: Attempt to solve t.q. M n = 6, Age is 6 Acso,Acso (7) (b) Correct answer with no working: Allow both marks. (c) Some working must be seen to score marks: Minimum working: 500 + 700 + 900 + (+ 900) = scores M (A). (d) Allow or > throughout, apart from Age 6. A common misread here is 00. This gives n = 4 and age 4, and can score M A0 M A0 M A A with the usual misread rule. Alternative: (Listing sums) (500, 00, 00, 00, 4500, 6000, 7700, 9600,) 700, 4000, 6500, 900, 00, 500, 8500, 000. List at least up to 000 M All values correct A n = 6 (perhaps implied by age) Acso Age 6 Acso If there is a mistake in the list, e.g. 6 th sum = 00, possible marks are: M A0 A0 A0 Alternative: (Trial and improvement) Use of S n formula with n = 6 (and perhaps other values) M Accurately achieving 000 for n = 6 A Age 6 A Total marks
5x + 8. = 5x + x x M: One term correct. M A A: Both terms correct, and no extra terms. 5 5x x f ( x ) = x + + ( + C) (+ C not required here) M Aft 5 6 = + + 4 + C Use of x = and y =6 to form eqn. in C M C = Acso 5 x + x + 4x (simplified version required) A (ft C) (7) 5 [or: x + x + 4 x or equiv.] Total 7 marks For the integration: M requires evidence from just one term (e.g. x), but not just +C. Aft requires correct integration of at least terms, with at least one of these terms having a fractional power. For the final A, follow through on C only.
9. (a) (P), (Q) (± scores B B) B, B (b) y = x x 4x + 4 (May be seen earlier) Multiply out, giving 4 terms M y dx d = x x 4 (*) M Acso (c) At x = : d y = ( ) ( ) 4 = dx Eqn. of tangent: y 6 = ( x ( )), y = x + 7 (*) M Acso () () () (d) x x 4 = (Equating to gradient of tangent ) M x x 5 = 0 ( x 5)( x + ) = 0 x = M 5 x = or equiv. A 5 5 y = 4, = = or equiv. 9 9 7 M, A (5) Total marks (b) Alternative: Attempt to differentiate by product rule scores the second M: y = dx d {( x 4) } + {( x ) x} Then multiplying out scores the first M, with A if correct (cso). dy (c) M requires full method: Evaluate and use in eqn. of line through (,6), dx (n.b. the gradient need not be for this M). Alternative: Gradient of y = x + 7 is, so solve x x 4 =, as in (d) M to get x =. Acso (d) nd and rd M marks are dependent on starting with k is a constant. x x 4 = k, where
0. (a) x + x + = ( x + ), + (a =, b = ) B, B () 6 y (b) U -shaped parabola M 5 4 Vertex in correct quadrant (ft from ( a, b) Aft (0, ) (or on y-axis) B () 4 x (c) b 4ac = 4 = 8 B Negative, so curve does not cross x-axis B () (d) b 4ac = k (May be within the quadratic formula) M k < 0 (Correct inequality expression in any form) < k < (or < k < ) M A (4) A Total marks (b) The B mark can be scored independently of the sketch. (, 0) shown on the y-axis scores the B, but if not shown on the axis, it is B0. (c). no real roots is insufficient for the nd B mark.. curve does not touch x-axis is insufficient for the nd B mark. (d) nd M: correct solution method for their quadratic inequality, e.g. k < 0 gives k between the critical values α < k < β, whereas k > 0 gives k < α, k > β. k > and k < scores the final M A, but k > or k < scores M A0, k >, k < scores M A0. N.B. k < ± does not score the nd M mark. k < does not score the nd M mark. instead of <: Penalise only once, on first occurrence.
GENERAL PRINCIPLES FOR C MARKING Method mark for solving term quadratic:. Factorisation ( x + bx + c) = ( x + p)( x + q), where pq = c, leading to x = ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn =. Formula Attempt to use correct formula (with values for a, b and c). a, leading to x =. Completing the square Solving x + bx + c = 0 : ( x ± p) ± q ± c, p 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n+ ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but will be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may gain no credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice. Misreads (See the next sheet for a simple example). A misread must be consistent for the whole question to be interpreted as such. These are not common. In clear cases, please deduct the first A (or B) marks which would have been lost by following the scheme. (Note that marks is the maximum misread penalty, but that misreads which alter the nature or difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written). Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written.
MISREADS Question 8. 5x misread as 5x 5x + 8. = 5x + x x 5 M A0 7 5x x f ( x ) = x + + ( + C) M Aft 7 0 6 = + + 4 + C M 7 7 C = 7 0 7, f ( x ) = x + x + 4x 7 7 7 A0, A