Mark Scheme (Results) Summer Pearson Edexcel GCE in Core Mathematics 3 (6665/01)

Transcription

1 Mark Scheme (Results) Summer 06 Pearson Edexcel GCE in Core Mathematics (6665/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding bo. 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 50 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 06 Publications Code 6665_0_606_MS All the material in this publication is copyright Pearson Education Ltd 06

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. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. 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 75. 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

5 . 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. 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 Scheme Marks (a) 8 fg( x) x 8 Sets fg( x) x x x 8 ( x)( x ) x x0 0 ( x6)( x5) 0 x6, x 5 a 6 Alt (a) fg( x) x g( x) f ( x) x x 7 x x0 0 ( x6)( x5) 0 x6, x 5 0 x x S. Case Uses gf ( x ) instead fg( x ) Makes an error on fg( x ) 7 Sets fg( x) x x x 7 ( x ) M0 7x d A () B ft () 5 marks d A marks 7x x 0 x x6 0 (7x)( x) 0 ( x)( x) 0 x, 7 x x, x d A0 out of marks (a) 8 Sets or implies that fg( x) Eg accept fg( x) 7 7 x x followed by fg( x) x x Alternatively sets g( x) f ( x) where f ( x) 7 8 Note that fg( x) 7 x is M0 7( x) Sets up a TQ (= 0) from an attempt at fg( x) x or g( x) f ( x) d Method of solving TQ (= 0) to find at least one value for x. See ''General Priciples for Core Mathematics'' on page for the award of the mark for solving quadratic equations This is dependent upon the previous M. You may just see the answers following the TQ. A Both x 6 and x 5 Bft For a 6 but you may follow through on the largest solution from part (a) provided more than one answer was found in (a). Accept 6, a 6 and even x 6 Do not award marks for part (a) for work in part.

9 (a) Question Scheme Marks (a) A f'( ) x x 5 xx y x 5 x 5 0 x x 5 0 x x 5 A A 0 0 x Critical values of 5 x 5, x 5 or equivalent da () 7 marks vu ' uv ' Attempt to use the quotient rule with u x and v x 5. If the rule is quoted it must be v vu ' uv ' correct. It may be implied by their u x, u A, v x 5, v Bx followed by their v If the rule is neither quoted nor implied only accept expressions of the form A x 5 xbx, AB, 0 You may condone missing (invisible) brackets x 5 u / v x Alternatively uses the product rule with and () v / u x 5. If the rule is quoted it must be correct. It may be implied by their u x, u A, v x 5, v Bx x 5 followed by their vu ' uv '. If the rule is neither quoted nor implied only accept expressions of the form 5 5 x correct (unsimplified). For the product rule look for versions of x 5 xx x 5 A x x Bx x Simplifies to the form f'( x) A Bx x 5 When the product rule has been used the A of A x 5 5 x f'( x) x 5 vu ' uv ' oe. This is not dependent so could be scored from v must be adapted. A CAO. Accept exact equivalents such as x 0 x 5, or x 5 x 0x 5 Remember to isw after a correct answer Sets their numerator either 0, 0, 00 0, 0 0 and proceeds to at least one value for x For example 0 x..0 x.. 5 will be dm0 A0. It cannot be scored from a numerator such as or indeed 0 x d Achieves two critical values for their numerator 0 and chooses the outside region Look for x smaller root, x bigger root. Allow decimals for the roots. Condone x 5, x 5and expressions like 5 x 5 If they have x 0 0 following an incorrect derivative they should be choosing the inside region A Allow x 5, x 5 x 5orx 5 x: x 5 5 x x 5 Do not allow for the A x 5an 5. 5 x 5 or x: x 5 5 x but you may isw following a correct answer.

10 Question Scheme Marks.(a) R 5 B tan 6.57 A () 5 5 cos sin 5cos( 6.6 ) 7 cos( 6.6 ) awrt A awrt.0or awrt 7.9 A 6.660their '59.5 ' d awrt 7.9and awrt.0 A (5) (c) their 6.57 their awrt 86. A () (0 marks) (a) B 5 R. Condone R 5 Ignore decimals tan, tan... If their value of R is used to find the value of only accept cos OR sin... R R A awrt 6.57 Attempts to use part (a) cos( their 6.6 ) K, K A 7 cos( their 6.6 ) awrt Can be implied by ( their 6.6 ) awrt 59.5 / A One solution correct, awrt.0or awrt 7.9 Do not accept for.0. d Obtains a second solution in the range. It is dependent upon having scored the previous M. Usually for their their A Both solutions awrt.0 and awrt 7.9. Do not accept for.0. Extra solutions inside the range withhold this A. Ignore solutions outside the range 0 60 (c) their 6.57 their Alternatively their 6.6 their If the candidate has an incorrect sign for, for example they used cos( 6.57 ) in part it would be scored for their 6.57 their A awrt 86. ONLY. Allow both marks following a correct (a) and They can restart the question to achieve this result. Do not award if 86. was their smallest answer in. This occurs when they have cos( 6.57 ) instead of cos( 6.57 ) in Answers in radians: Withhold only one A mark, the first time a solution in radians appears FYI (a) 0.6 awrt 0.58 and awrt.78 (c) awrt.50. Require dp accuracy

11 Question Scheme Marks.(a) (i) B x x (ii) e 5 0 e xln x ln, x ln A, A (iii) 5 B x x e 5 x e x 7 xln x 7 x ln x 7 A* (c ) x ln. 7 awrt. awrt x.68, x.7 A (5) () () (d) Defines a suitable interval.65 and.75 x...and substitutes into a suitable function Eg e x 68, obtains correct values with both a reason and conclusion A () ( marks) In part (a) accept points marked on the graph. If they appear on the graph and in the text, the text takes precedence. If they don't mark (a) as (i) (ii) and (iii) mark in the order given. If you feel unsure then please use the review system and your team leader will advise. (a) (i) B Sight of. Accept (0,) Do not accept just 5 or,0 (a) (ii) Sets e and proceeds via e 5 or e to x.. Alternatively sets e x x 5 0 and proceeds via e 5e 5 0 to e.. A 5 ln or awrt 0.9 A 5 cao ln or ln 5 ln 5 ln,0 (a) (iii) B k 5 Accept also 5 or y 5 Do not accept just 5 or x 5 or y 5

12 x Sets e 5 x and makes e x x the subject. Look for e x 5condoning sign x slips.condone e 5 x and makes e x the subject. Condone for both marks a solution with xa/ x An acceptable alternative is to proceed to e x ln x ln( x) using ln laws A* Proceeds correctly without errors to the correct solution. This is a given answer and the bracketing x must be correct throughout. The solution must have come from e 5 x with the modulus having been taken correctly. x Allow e x 5going to x ln x 7 without explanation Allow ln 7 x appearing as log 7 e x but not as log 7 x... If a candidate attempts the solution backwards they must proceed from x x x ln x 7 e x 7 e 5 x for the For the A it must be tied up with a minimal statement that this is g( x) x... (c) Subs. into the iterative formula in an attempt to find x Score for x ln. 7 x ln 7.7 or awrt. A awrt x.68, x.7 Subscripts are not important, mark in the order given please. (d) For a suitable interval. Accept.65 and.75 (or any two values of a smaller range spanning the root=.7) Continued iteration is M0 A Substitutes both values into a suitable function, which must be defined or implied by their working calculates both values correctly to sig fig (rounded or truncated) x Suitable functions could be e x 68, x ln x7, xln x7. x Using e x 68 f (.65) = -0., f (.75) = +0.0 or +0.0 Using x e x f (.65) = -0.05/-0.06, f (.75) = +0.0 Using x ln x 7 f (.65) = or , f (.75) = or Using xln x 7 f (.65) = or -0.00, f (.75) = or and states a reason (eg change of sign) and a gives a minimal conclusion (eg root or tick) g(.65) 5.7(6) (7) It is valid to compare the two functions. Eg g(.75) (75) but the conclusion should be g( x) x in between, hence root. Similarly candidates can compare the functions x and ln 7 x

13 Question Scheme Marks 5 (i) x x x y e cos x cos xe e sin x A Sets x x cosx e e sinx 0 cosx sinx 0 (ii) x arctan x awrt 0.96 dp A x sin y sin y cos y A Uses sin y sin ycos y in their expression (5) (ii) Alt I (ii) Alt II (ii) Alt III siny cosecy siny A (5) (0 marks) x sin y x cos y nd siny st A cosec y siny A (5) d sin y cos y d A Uses x sin y AND sin y sin ycos y in their expression cosec y siny A (5) x sin y y invsin x x d x x A Uses x sin y, x cosy and sin y sin ycos y in their expression cosec y siny A (5) (i) x x Uses the product rule uv vu to achieve Ae cosx Be sin x A, B 0 The product rule if stated must be correct A x x Correct (unsimplified) cos x e e sin x

14 Sets/implies their d y 0 factorises/cancels)by e x to form a trig equation in just sin x and cos x Uses the identity sin x tan x, moves from tan, 0 cos x x C C using correct order of operations to x... Accept x awrt 0.6 (radians) x awrt 9. (degrees) for this mark. If a candidate elects to pursue a more difficult method using R cos( ), for example, the minimum expectation will be that they get () the identity correct, and () the values of R and α correct to dp. So for the correct equation you would only accept 5cos( x+awrt 0.9 or 5sin(x awrt 0.6) before using the correct order of operations to x... Similarly candidates who square cosx sinx 0 then use a Pythagorean identity should proceed from either sin x or cos x before using the correct order of operations 5 5 A x awrt Ignore any answers outside the domain. Withhold mark for additional answers inside the domain (ii) Uses chain rule (or product rule) to achieve P sin ycos y as a derivative. There is no need for lhs to be seen/ correct If the product rule is used look for d x Asin ycos y Bsin ycos y, A Both lhs and rhs correct (unsimplified). siny cosy sinycosy or siny cosy Uses sin y sin ycos y in their expression. You may just see a statement such as sinycosy siny which is fine. Candidates who write d x Asin xcos x can score this for d x A sin x Uses d y for their expression in y. Concentrate on the trig identity rather than the d x coefficient in awarding this. Eg d x sin d y y cosecyis condoned for the If d x a b do not allow a b A cosec y If a candidate then proceeds to write down incorrect values of p and q then do not withhold the mark. NB: See the three alternatives which may be less common but mark in exactly the same way. If you are uncertain as how to mark these please consult your team leader. In Alt I the second M is for writing x sin y x cos yfrom cos y sin y In Alt II the first M is for writing x sin y and differentiating both sides to Px Qcos y oe 0.5 In Alt the first M is for writing y invsin x oe and differentiating to N x d 0.5 x M x 0.5

15 Question Scheme Marks 6(a) x x x6 x x x 7x6 x x 6x x 7x6 x x8 x A (a) x x x 7x6 ( x) x x x 6 ( x )( x ) x A ( x ) () f'( x) x Aft ( x ) Subs x into f'( x ) ( ) Uses m f'() with (, f ()) (,6) to form eqn of normal y6 x or equivalent cso A (5) (9 marks) Divides x x x 7x 6 by x x 6 to get a quadratic quotient and a linear or constant remainder. To award this look for a minimum of the following +.+x x A x x6 x x x 7x6 x + x -6x + Cx D If they divide by ( x ) first they must then divide their by result by ( x ) before they score this method mark. Look for a cubic quotient with a constant remainder followed by a quadratic quotient and a constant remainder Note: FYI Dividing by ( x ) gives x x xand x x x x x with a remainder of. Division by ( x ) first is possible but difficult...please send to review any you feel deserves credit. A Quotient = x and Remainder = x Factorises x x 6 and writes their expression in the appropriate form. x x x 7x6 Their Linear Remainder Their Quadratic Quotient x x6 ( x)( x) It is possible to do this part by partial fractions. To score under this method the terms must be correct and it must be a full method to find both ''numerators''

16 A x or A, B but don't penalise after a correct statement. ( x ) B B x A x x ( x ) B If they fail in part (a) to get a function in the form x A allow candidates to pick up this x Rx S method mark for differentiating a function of the form x PxQ x T using the quotient rule oe. Aft B B B x A x oe. FT on their numerical A, B for for x A only x ( x ) x Subs x into their f'( x) in an attempt to find a numerical gradient For the correct method of finding an equation of a normal. The gradient must be and the their f '() point must be,f (). Don't be overly concerned about how they found their f(), ie accept x= y =. Look for yf() ( x) or y f () f () ( x ) f() If the form y mx c is used they must proceed as far as c = A cso y6 x oe such as y x5 0but remember to isw after a correct answer. Alt (a) attempted by equating terms. Alt (a) x x x 7x6 ( x A)( x x6) B( x ) Compare terms (or substitute values) AND solve simultaneously ie x A6, x AB 7, const 6A+B= 6 A, B A,A st Mark Scored for multiplying by ( x x 6) and cancelling/dividing to achieve x x x 7x6 ( x A)( x x6) B( x ) rd Mark Scored for comparing two terms (or for substituting two values) AND solving simultaneously to get values of A and B. nd Mark A Either AorB. One value may be correct by substitution of say x th Mark A Both A andb Alt is attempted by the quotient (or product rule) ALT x x6x x 6x7 x x x 7x6x A f'( x) x x6 st Subs x into marks vu ' uv ' Attempt to use the quotient rule with u x x x 7x 6 and v x x 6 and v x x6.. x... x x x 7x6.. x.. achieves an expression of the form f'( x). x x6 Use a similar approach to the product rule with u x x x 7x 6 Note that this can score full marks from a partially solved part (a) where and v x x 6 x f( x) x x x 6

17 Question Scheme Marks 7(a) Correct position or curvature Correct position and curvature A () arcsin( x) 0 arcsin( x) ( x ) sin x da () (5 marks) (a) Ignore any scales that appear on the axes Accept for the method mark Either one of the two sections with correct curvature passing through (0,0), Or both sections condoning dubious curvature passing through (0,0) (but do not accept any negative gradients) Or a curve with a different range or an ''extended range'' See the next page for a useful guide for clarification of this mark. A A curve only in quadrants one and three passing through the point (0,0) with a gradient that is always positive. The gradient should appear to be approx at each end. If you are unsure use review If range and domain are given then ignore. Substitutes g( x) arcsin( x) in g( x ) 0 and attempts to make arcsin( x ) the subject d Accept arcsin( x ) or even g( x ). Condone Proceeds by evaluating sin and making x the subject. in decimal form awrt.07 Accept for this mark x. Accept decimal such as.866 Do not allow this mark if the candidate works in mixed modes (radians and degrees) You may condone invisible brackets for both M's as long as the candidate is working correctly with the function A oe with no other solutions. Remember to isw after a correct answer Be careful with single fractions. and are incorrect but is correct Note: It is possible for a candidate to change to 60and work in degrees for all marks

18 Question Scheme Marks 8 (a) cotx tanx tanx tan x B ( tan x) tan x tan x tan x tan x cot x A* () 6 cot x tan x cosec x cot x cosec x cot xcot x 0 cot x cot x A cot x tan x x.. x 0.9,.88,.77,.865 A,,0 (6) (0 marks) 8 (a)alt cosx cotx tanx tanx sin x B cos x sin x sin x sin x cos x cos x cos x sin x sin x cos sin x cos x sin xcos x x sin x cos x cos x sin x cot x A* 8 (a)alt ( tan x) cotx tanx tanx tanx B tan x tan x tan x tan x (tan x) tan x tan x Alt cot x A* tanx cos x 6 cot x tan x cosec x sin x sin x sin x sin xcos x sin x sin x cos x A tan x x.. x 0.9,.88,.77,.865 A,,0 (6)

19 (a) B cosx States or uses the identity cotx or alternatively cotx tan x sin x tan x This may be implied by cotx. Note cotx is B0 tan x tanx tanx Uses the correct double angle identity tan x tan x sin x Alternatively uses sin x sin xcos x, cos x cos x sin x oe and tan x cos x Writes their two terms with a single common denominator and simplifies to a form ab cd. For this to be scored the expression must be in either sin x and cos x or just tan x. In alternative it is for splitting the complex fraction into parts and simplifying to a form ab cd. cos x You are awarding this for a correct method to proceed to terms like sin x cos x, cos x, sin xcos x tanx A* cso. For proceeding to the correct answer. This is a given answer and all aspects must be correct including the consistent use of variables. If the candidate approaches from both sides there must be a conclusion for this mark to be awarded. Occasionally you may see a candidate attempting to prove cot x tan x cot x. This is fine but again there needs to be a conclusion for the A* If you are unsure of how some items should be marked then please use review For using part (a) and writing 6cotx tanx as kcot x, k 0 in their equation (or equivalent) WITH an attempt at using cosec x cot x to produce a quadratic equation in just cot x / tan x A cot xcot x 0 The = 0 may be implied by subsequent working Alternatively accept tan xtan x 0 Solves a TQ=0 in cot x (or tan) using the formula or any suitable method for their quadratic to find at least one solution. Accept answers written down from a calculator. You may have to check these from an incorrect quadratic. FYI answers are cot x awrt.0, 0.0 Be aware that cot x tan x For tan x and using arctan producing at least one answer for x in degrees or radians. cot x You may have to check these with your calculator. A Two of x 0.9,.88,.77,.865 (awrt dp) in radians or degrees. In degrees the answers you would accept are (awrt dp) x 6.8,06.8, 7., 6. A All four of x 0.9,.88,.77,.865 (awrt dp) with no extra solutions in the range xx x See main scheme for Alt to using Double Angle formulae still entered M A M M A A in epen st For using part (a) and writing 6cotx tan x as kcot x, k 0 in their equation (or equivalent) cos x then using cot x, cosec x and sin x to form an equation sin and cos sin x sin x st A For sin x cos x or equivalent. Attached to the next M nd For using both correct double angle formula rd For moving from tan x C to x=..using the correct order of operations.

20 Question Scheme Marks 9(a) Subs D 5 and t = 0. x 5e 6.70 ( mg ) A e 5e.75( mg) A* 0.T 0. ( T5) (c) 5e 5e T 0.T 5e 5e e T 0.T 7.5 5e (+e ) 7.5 e d 5(+e ) e T 5ln 5ln e A, A () () () (8 marks) (a) 0.t Attempts to substitute both D 5 and t = in x De 0.8 It can be implied by sight of 5e 0., 5e or awrt 6.7 Condone slips on the power. Eg you may see -0.0 A CAO 6.70 (mg) Note that 6.7 (mg) is A0 Attempt to find the sum of two expressions with D =5 in both terms with t values of and Evidence would be 5e 5e or similar expressions such as 5e 5e 0. Award for the sight of the two numbers awrt.70 and awrt 0.05, followed by their total awrt.75 Alternatively finds the amount after 5 hours, 5e awrt 5.5 adds the second dose = 5 to get a 0. total of awrt 0.5 then multiplies this by e to get awrt.75. Sight of 5.5+5= is fine A* cso so both the expression 5e 5e and.75( mg ) are required Alternatively both the expression 5e 5e and.75( mg ) are required. Sight of just the numbers is not enough for the A* (c) Attempts to write down a correct equation involving T or t. Accept with or without correct bracketing 0.T 0. ( T5) 0.T Eg. accept 5e 5e 7.5 5e 5 e 7.5 d A A or similar equations 0.T Attempts to solve their equation, dependent upon the previous mark, by proceeding to e... m n m n An attempt should involve an attempt at the index law x x x and taking out a factor of 0. T e 0. T Also score for candidates who make e the subject using the same criteria 7.5 Any correct form of the answer, for example, 5ln 5 e CSO T5ln e Condone t appearing for T throughout this question.

21 Alt (c) using lns 0.T 0. ( T5) (c) 5e 5e T 0.T 5e 5e e T e (+e ) T ln (+e ) ln 0.5 d ln 0.5 ln (+e ) T, T 5ln 0. e A, A () (8 marks) You may see numerical attempts at part (c). Such an attempt can score a maximum of two marks. This can be achieved either by Method One 0.T 0.T st Mark (Method): 5e awrt 5.5e 7.5 e 0. T awrt nd Mark (Accuracy): T=-5ln awrt 0.7 or awrt 5.0or T=-5ln awrt 0.5 Method Two 0.T 7.5 st Mark (Method ):.75e 7.5 T 5ln nd Mark (Accuracy):.0+=5.0 Allow 5ln.75 or equivalent such as.0 Method Three (by trial and improvement) st Mark (Method): 5e 5e 7.55 or 5e 5e 7.0 or any value between nd Mark (Accuracy): Answer T =5.0.

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