Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 1 (6663_01)

Transcription

1 Mark Scheme (Results) Summer 0 Pearson Edexcel GCE in Core Mathematics (666_0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 0 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 0 Publications Code UA089 All the material in this publication is copyright Pearson Education Ltd 0

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 7. The Edexcel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

5 . For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

6 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. Factorisation ( x + bx + c) = ( x + p)( x + q), where pq = c, leading to x = ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn = a, leading to x =. Formula Attempt to use the correct formula (with values for a, b and c).. Completing the square b Solving x + bx + c = 0 : x ± ± q ± c = 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+ )

7 Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are small errors in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.

8 Question Number Scheme Marks. (8x + ) dx = 8x + x M, A = x + x+ c A ( marks) Notes n n M x x + so x x or x or x A This is for either term with coefficient unsimplified (power must be simplified) so (accept x ) 8 x or x A Fully correct simplified solution with c i.e. x 0 + x + c [ allow x + x + cx ] If the answer is given as x + x+ c, with an integral sign having never been seen as the fully correct simplified answer without an integral sign then give MAA0 but allow anything before the = sign e.g. y = x + x + c, f(x) = x + x + c, = x + x+ c, etc. If this answer is followed by (for example) x + x+ kthen treat this as isw (ignore subsequent work) If they follow it by finding a value for c, also isw, provided correct answer with c has been seen and credited

9 Question Number Scheme Marks. (a) = or or (b) For or 0. as coefficient of k x, for any value of k including k = 0 A Correct index for x so A x or o.e. for any value of A x B B M () = x or 0. x A cao Marks () Notes (a) B Answer must be in part (a) for this mark (b) Look at their final answer M For or or or 0. in their answer as coefficient of k x for numerical value of k (including k = 0) so final answer is M B0 A0 0 A B Ax or or equivalent e.g. Ax 0 or Ax i.e. correct power of x seen in final answer x A May have a bracket provided it is ( Ax) or x A or x x or 0. x oe but must be correct power and coefficient combined correctly and must not be followed by a different wrong answer. Poor bracketing: x earns M0 B A0 as correct power of x is seen in this solution (They can recover if they follow this with x etc ) Special case ( x) as a final answer and can have M0 B A0 if the correct expanded answer is not seen x The correct answer etc. followed by x x or ( x), treat as final answer so M B A isw x But the correct answer x etc clearly followed by the wrong x or x, gets M B A0 do not ignore subsequent wrong work here

10 Question Number Scheme Marks. (a) x 7> x x > 0 M x>., x>, < x o.e. A (b) Obtain x 9x 6 and attempt to solve x 9x 6 = 0 e.g. ( x )( x+ ) = 0 so x =, 9 ± 8+ or x = M, A x MA (c ). < x Acso (7 marks) () () () Notes (a) M Reaching px > q with one or both of p or q correct. Also give for -x < -0 A Cao x >. o.e. Accept alternatives to. like and even allow 0 and allow < x o.e. This answer must occur and be credited as part (a) A correct answer implies MA Mark parts (b) and (c) together. (b) M Rearrange TQ 0 or TQ = 0 or even TQ > 0 Do not worry about the inequality at this stage AND attempt to solve by factorising, formula or completion of the square with the usual rules (see notes) A and seen as critical values M Inside region for their critical values must be stated not just a table or a graph A x Accept x and x or [-, ] For the A mark: Do not accept x or x nor - < x < nor (-, ) nor x, x However allow recovery if they follow these statements by a correct statement, either in (b) or as they start part (c) N.B. 0 and x, x are poor notation and get MA0 here. (c) A cso. < x Accept x. and x Do not accept x>. or x > Allow 0 Accept (., ] A graph or table is not sufficient. Must follow correct earlier work except for special case Special case (c) x>., x ;. < 0 are poor notation but if this poor notation has been penalised in (b) then allow A here. Any other errors are penalised in both (b) and (c).

11 Question Number Scheme Marks. (a) - accept (, 0) B (b) () Shape Touches at (0,0) Crosses at (,0) only B B B () (c ) solutions as curves cross twice B ft () ( marks) Notes N.B. Check original diagram as answer may appear there. (a) B The x coordinate of A is. Accept - or (,0) on the diagram or stated with or without diagram Allow (0, -) on the diagram if it is on the correct axis. (b) If no graph is drawn then no marks are available in part (b) B Correct shape. The position is not important for this mark but the curve must have two clear turning points and be a +ve x curve ( with a maximum and minimum) B The graph touches the origin. Accept touching as a maximum or minimum. There must be a sketch for this mark but sketch may be wrong and this mark is independent of previous mark. Origin is where axes cross and may not be labelled. This may be a quadratic or quartic curve for this mark. B The graph crosses the x-axis at the point (,0) only. If it crosses at (,0) and (0,0) this is B0. Accept (0,) or marked on the correct axis. Accept (, 0) in the text of the answer provided that the curve crosses the positive x axis. There must be a sketch for this mark. Do not give credit if (,0) appears only in a table with no indication that this is the intersection point. (If in doubt send to review ) Graph takes precedence over text for third B mark. (c) Bft Two (solutions) as there are two intersections (of the curves) N.B. Just states with no reason is B0 If the answer states roots and two intersections or crosses twice this is enough for B BUT B0 If there is any wrong reason given e.g. crosses x axis twice, or crosses asymptote twice Isw is not used for this mark so any wrong statement listed to follow a correct statement will result in B0 Allow ft so if their graph crosses the hyperbola once allow one solution as there is one intersection And if it crosses three times allow three solutions as there are three intersections or four etc.. If it does not cross at all (e.g.negative cubic) allow no solutions as there are no intersections However in (c) if they have sketched a curve (even a fully correct one) but not extended it to intersect the hyperbola and they put "no points of intersection so no solutions" then this scores B0. Accept lines or curves cross over twice, or touch twice, or meet twice etc as explanation, but need some form of words)

12 Question Number Scheme Marks. (a) 7 = a a =.. M a = A (b) a = "" and a = "7" M () r= r= a = a + a + a + a r = dm = 98 A ( marks) () Notes (a) M Writes 7= a and attempts to solve leading to an answer for a. If they rearrange wrongly before any substitution this is M0 A Cao a = Special case: Substitutes n = into n and obtains answer. This is fortuitous and gets M0A0 but full marks are available on (b). (b) M Attempts to find either their a or their a using a = n a + n, a = 7 Needs clear attempt to use formula or is implied by correct answers or correct follow through of their a dm (Depends on previous M mark) Sum of their four adjacent terms from the given sequence. n.b May be given for 9 + a + a as they may add + 7 to give 9 (dm0 for sum of an Arithmetic series) A cao 98 Special case (a) a = is M0 A0 (b) Adds for example = or is M M A0 Total mark possible is / (This is not treated as a misread as it changes the question)

13 Question Number Scheme Marks 6. (a) 80 = 6 80 = B Method Method (b) 80 c or + + (p+q )( +)= 80 = or + + p +q +p+q = = 0 0 p + q = 0 or p + q = = p =, q = Bft M A Acao () () ( marks) Notes (a) B Accept or c = no working necessary (b) (Method ) Bft Only ft on c See 80 + c + M State intention to multiply by or in the numerator and the denominator A Obtain denominator of ( for ) or - (for ) or correct simplified numerator of 0 or ( ) or 0 or ( ) So either numerator or denominator must be correct A Correct answer only. Both numerator and denominator must have been correct and division of numerator and denominator by has been performed. Accept p=, q = or accept or + Also accept (Method ) Bft Only ft on c (p+q )( +)= 80 or c M Multiply out the lhs and replace 80 by c A Compare rational and irrational parts to give p + q =, and p + q = 0 A Solve equations to give p =, q = Common error: 80 + product) then A0 = 0 = gets B M A (for correct numerator denominator is wrong for their Correct answer with no working send to review have they used a calculator? Correct answer after trial and improvement with evidence that (- )( +)= 80 could earn all four marks

14 Question Number Notes Scheme 7. (a) ( x) = x+ x M d d ( ) ( ) 8 d = x dx + x = + x o.e. MA Alternative method using chain rule: Answer of - ( x) (b) x + 6 x x x x x x Marks MMA = + 6, = x + x M,A Attempts to differentiate x to give k M 9 = x x o.e. A Quotient Rule ( May rarely appear) See note below () (7 marks) (a) M Attempt to multiply out bracket. Must be or term quadratic and must have constant term n n M x x. Follow through on any term in an incorrect expression. Accept a constant 0 A + 8x Accept - ( x ) or equivalent. This is not cso and may follow error in the constant term Following correct answer by - + x apply isw Correct answer with no working assume chain rule and give MMA i.e. / Common errors: ( x) = x+ x is M0, then allow MA for - + 8x ( x) = x is M0 then -8x earns MA0 or ( x) = x is M0 then -x earns MA0 Use of Chain Rule: MM: first M for complete method so ± ( )( x) second M for ( x) (as power reduced) Then A for - ( x ) or for - + 8x So (i) ( x) gets M0 MA0 for reducing power and (ii) ( x) gets M MA0 (b) M x An attempt to divide by x first. This can be implied by the sight of the following x x Some correct working e.g. + 6 or ( x 6 x)( x ) p q + leading to ax + bx in either case x x p or can be implied by x + x (after no working) i.e. both coefficients correct and power correct Common error: ( x + 6 x)x is M0 (may earn next M mark for the differentiation x A M Writing the given expression as x 6 + or x 0.x + x 6 or 0.x + x or etc x x ) 9 x A Cao x x 9 o.e. e.g. x then isw. Allow factorised form. Do not x x penalise Use of Quotient Rule : Send to review if doubtful 9 x + used instead of 9 x x(x + x ) xx ( + 6 x) 6x 8x M,A:Reaching, = x x 6 MA: Simplifying (e.g.dividing numerator and denominator by ) to reach x 9x 6 x () () o.e.

15 Question Scheme Marks Number 8. th (a) Use n term = a+ ( n ) d with d = 0; a = 0 and n = 8, or a = 60 and n = 7, or a = 70 and n = 6 : = or or M = 0* (Or gives clear list see note) A* () Or If answer 0 is assumed and 0 + (n ) 0 =0 or variation is solved for n= M Then n = 8, so 007 is the year (must conclude the year) A* () n n (b) Use Sn = { a+ ( n )0} Or S { } n = a+ l and l= a + (n )0 M = 7(00+ 0) or 7(0 + 80) A = 7 0 = 00 A () (c) Cost in year n = 900+(n ) -0 M Sales in year n = 0+(n ) 0 Cost = Sales 900+(n ) -0 = (0+(n ) 0) M 900 0n+0 = 0+0n 0 00 = 0n n =0 M Year is 009 A As n is not defined they may work correctly from another base year to get () the answer 009 and their n may not equal 0. If doubtful send to review. (9 marks) Notes th (a) M Attempt to use n term = a+ ( n ) d with d = 0, and correct combination of a and n i.e. a = 0 and n = 8 or a = 60 and n = 7, or a = 70 and n = 6 A * Shows that 0 computers are sold in 007 with no errors Note that this is a given solution, so needed or or or equivalent. Accept a correct list showing all values and years for both marks Just 0,60,70,80,90,00,0,0 is MA0 Need some reference to years as well as the list of numbers of computers for A. n (b) M Attempts to use Sn = { a+ ( n ) d} with d = 0, and correct combination of a and n i.e. a = 0 and n =, or a = 60 and n =, or a = 70 and n = n n A Uses Sn = { a+ ( n ) d} with a = 0, d = 0 and n = [N.B. S { } n = a+ l needs l= a + (n - )d as well NB A0 for a = 60 and n = or a = 70 and n = unless they then add the first, or first two terms respectively. A Cao 00. This answer (with no working) implies correct method MAA. Special case: If a complete list is seen, then there is an error finding the sum then score MAA0, but incomplete or wrong lists score M0A0A0 (c ) M Writes down an expression for the cost = 900+(n ) 0 or writes (n ) d and states d = -0 Allow n 0. Allow recovery from invisible brackets. M Attempts to write down an equation in n for statement cost = sales 900+(n-) -0 = (0+(n-) 0). Accept the on the wrong side and allow use of 0 instead of -0 and allow n (consistently) instead of n for this mark. Ignore signs in equation. M Solves the correct linear equation in n to achieve n = 0 (for those using n ) or n = 9 (for those using n). Ignore signs. A Cso Year 009 (A0 for the answer Year 0 if 009 is not given ) Special case. Just answer or trial and improvement with no equation leading to answer scores SC 0,0,, Equations satisfying the method mark descriptors followed by trial and improvement could get all four marks

16 Question Number Scheme Marks 9. (a) x+ y = 6 y = 6 ± xand attempt to find m from y = mx + c M 6 ( y = x ) so gradient = A Gradient of perpendicular = their gradient (= ) M Line goes through (0,0) so y = x A (b) Solves their y = x with their x+ y = 6 to form equation in x or in y M Solves their equation in x or in y to obtain x = or y = dm ( ) x= or any equivalent e.g. 6/9 or y = 6 o.a.e A B= (0, 6 ) used or stated in (b) B Method ( see other methods in notes below) 6 x= Area = "" "6" dm = (oe with integer numerator and denominator) A (6) (0 marks) Notes (a) M Complete method for finding gradient. (This may be implied by later correct answers.) e.g. Rearranges x + y = 6 y = mx + c so m = 8 0 Or finds coordinates of two points on line and finds gradient e.g. (, 0) and (,8) so m = - condone x if they continue correctly. Ignore errors in constant A States or implies that gradient = term in straight line equation M Uses m m = to find the gradient of l. This can be implied by the use of A y = x or y x = 0 Allow isw y = x+ 0 Also accept y=x, y=9/6x or even their gradient y 0 = ( x 0) and

17 Notes Continued (b) M Eliminates variable between their y = x and their (possibly rearranged) x+ y = 6 to form an equation in x or y. (They may have made errors in their rearrangement) dm (Depends on previous M mark) Attempts to solve their equation to find the value of x or y A x = or equivalent or y = 6 or equivalent B y coordinate of B is 6 (stated or implied) - isw if written as ( 6, 0). Must be used or stated in (b) dm (Depends on previous M mark) Complete method to find area of triangle OBC (using their values of x and/or y at point C and their 6/) A Cao 0 or 6 or 78 o.e Method : Uses the area of a triangle formula ½ OB (x coordinate of C) Alternative methods: Several Methods are shown below. The only mark which differs from Method is the last M mark and its use in each case is described below: Method in 9(b) using BC OC dm Uses the area of a triangle formula ½ BC OC Also finds OC (= ) and BC= ( ) Method in 9(b) using dm States the area of a triangle formula or equivalent with their values Method in 9(b) using area of triangle OBX area of triangle OCX where X is point (, 0) dm Uses the correct subtraction " 6 " "6" Method in 9(b) using area = ½ (6 ) + ½ ( 8/) drawing a line from C parallel to the x axis and dividing triangle into two right angled triangles dm for correct method area = ½ ( 6 ) + ½ ( [ 6/ - 6 ]) Method 6 Uses calculus dm 6 x x " " d x = 0 6 x x x 0

18 Question Number 0. (a) Scheme f ( x) = x 0x + dx 8 Marks n n+ x x x x f ( x) = x( + c) M, A, A Substitute x =, y = = c c = M x f ( x) = 0x + x+ A 8 ( ) (b) Sub x= into f ( x) = x 0x + 8 M f () = f () = A Gradient of tangent = Gradient of normal is / dm () Substitute x =, y = into line equation with their changed gradient e.g. y = ( x ) ± k(y+ x ) = 0 o.e. (but must have integer coefficients) dm Acso () (0 Marks) Notes (a) M n n Attempt to integrate x x + A Term in x or term in x correct, coefficient need not be simplified, no need for +x nor +c A ALL three terms correct, coefficients need not be simplified, no need for +c M For using x =, y = in their f(x) to form a linear equation in c and attempt to find c A x = 0x + x+ cao (all coefficients and powers must be simplified to give this answer- do 8 not need a left hand side and if there is one it may be f(x) or y). Need full expression with These marks need to be scored in part (a) (b) M Attempt to substitute x = into f( x) must be in part (b) A f( x) = at x = dm (Dependent on first method mark in part (b)) Using m m = to find the gradient of the normal from their tangent gradient (Give mark if gradient of becomes - as they will lose accuracy) dm (Dependent on first method mark in part (b)) Attempt to find the equation of the normal (not tangent). Eg use x=, y= in y= / x+c to find a value of c or use y ' ' = with their adapted gradient. x A cso ± k(y+ x ) = 0 (where k is any integer)

19 Question Number Scheme Marks. (a) Discriminant = b ac = 8, = 0 M, A () (b) x + 8x+ = ( x +...) or p= B = (( x + ) ±...) or q = M = ( x + ) or p =, q = and r = - A () (c ) Method A: Sets derivative"x+ 8" = x=, x = - M, A Substitute x = - in y = x + 8x+ ( y = ) dm Substitute x = - and y= - in y = x + c or into (y + )=(x + ) and expand dm c = or writing y = x + Acso () Method B: Sets derivative"x+ 8" = x=, x = - M, A Substitute x = - in x + 8x+ = x+ c dm Attempts to find value of c dm c = or writing y = x + Acso () Method : Sets x + 8x+ = x+ c and collects x terms together M Obtains x + x+ c= 0 or equivalent A States that b ac = 0 dm ( c) = 0 and so c = dm c = Acso () Method : Sets x + 8x+ = x+ c and collects x terms together M Obtains x + x+ c= 0 or equivalent A Uses ( x+ ) + c= 0 or equivalent dm Writes - + c = 0 dm So c = Acso () Also see special case for using a perpendicular gradient (overleaf) (0 marks) Notes (a) M Attempts to calculate A Cao 0 b ac using 8 - must be correct not just part of a quadratic formula (b) B See (.) or p = M..(( x + ) ±...) is sufficient evidence or obtaining q = A Fully correct values. ( x + ) or p =, q =, r = cso. Ignore inclusion of =0. [In many respects these marks are similar to three B marks. p = is B; q = is B and p =, q = and r = - is final B but they must be entered on epen as B M A] Special case: Obtains + + = ( x + ) This may have first B, for p = then M0A0 x 8x

20 (c ) Method A (Differentiates and puts gradient equal to. Needs both x and y to find c) M Attempts to solve their d y =. They must reach x =... (Just differentiating is M0 A0) dx A x = (If this follows d y dx x + 8, then give M A by implication) dm (Depends on previous M mark) Substitutes their x = - into f(x) or into their f(x) from (b) to find y dm A (Depends on both previous M marks) Substitutes their x = - and their y = - values into y = x + c to find c or uses equation of line is (y + )=(x + ) and rearranges to y = mx +c c = or allow for y = x + cso (c ) Method B (Differentiates and puts gradient equal to. Also equates equations and uses x to find c) MA Exactly as in Method A above dm (Depends on previous M mark) Substitutes their x = - into x + 8x+ = x+ c dm Attempts to find value of c then A as before (c) Method ( uses repeated root to find c by discriminant) M Sets x + 8x+ = x+ c and tries to collect x terms together A Collects terms e.g. x + x+ c= 0 or x x + c= 0or x + x+ = cor even x + x= c Allow =0 to be missing on RHS. dm (If the line is a tangent it meets the curve at just one point so repeated root and b ac = 0) Stating that b ac = 0is enough dm Using b ac = 0 to obtain equation in terms of c (Eg. ( c) = 0 ) AND leading to a solution for c A c = or allow for y = x + cso (c) Method ( Similar to method but uses completion of the square on the quadratic to find repeated root ) M Sets x + 8x+ = x+ c and tries to collect x terms together. May be implied by x + 8x+ x± con one side A Collects terms e.g. x + x+ c= 0 or x x + c= 0or x + x+ = c or even x + x= c Allow =0 to be missing on RHS. dm Then use completion of square ( x+ ) + c= 0 (Allow ( x+ ) k+ c= 0 ) where k is non zero. It is enough to give the correct or almost correct (with k) completion of the square dm c = 0 AND leading to a solution for c (Allow c = 0) (x = - has been used) A c = cso In Method they may use part (b) and differentiate their f(x) and put it equal to They can earn M, but do not follow through errors. In Methods and they may use part (b) to write their ( x+ ) = x+ c. They need to expand and collect x terms together for M Then expanding gives x + x+ c= 0 for A do not follow through errors Then the scheme is as before If they just state c = with little or no working please send to review, PTO for special case

21 Special case uses perpendicular gradient (maximum of /) Sets x+ 8= x=, x = 6 M A0 Substitute 69 x = in y = x + 8x+ ( y = ) 6 8 M0 69 Substitute x = and y = into y = x + c or into (y )=(x + ) and expand 6 M A0

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