Mark Scheme (Results) Summer GCE Core Mathematics 2 (6664/01)

Transcription

1 Mark (Results) Summer 0 GCE Core Mathematics (666/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere 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 0 Publications Code UA0566 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 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) 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) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper 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. 8. In some instances, the mark distributions (e.g., B and A) printed on the candidate s response may differ from the final mark scheme.

5 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 correct formula (with values for a, b and c).. Completing the square b Solving 0 :, 0, leading to x =... x + bx+ c= x± ± q± c q Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( x x ). Integration Power of at least one term increased by. ( n n x x + ) 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 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. Answers without working The rubric says that these may not gain full 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.

6 . (a) { r = } B { } 8 (c) { } (a) (c) Alternative method for (c) () p = B cao () 5 8( ( ) ) S5 = { } S5 = S5 = awrt A () [] Notes for Question B: Accept 8, 0.6 or 0.6 recurring, or even (sf) but not 0.6 or 0.67 B: accept 8 only 5 8( (their r) ) : Applies this formula S5 =, can be implied by their answer. For this mark (their r) they may use any value for r except r = or r = 0 (even / or -6 may be used) A: Answers which round to : (Adding terms is an unlikely method for this question) Need to see 5 terms listed as or can be implied by correct answer A: awrt Answer only : 5.9 is M0A0 with no working, but with no working is A

7 . (a) Alternative method (a) (a) + x - Mark (a) and together ( ) + C ( x) + C ( x) + C ( x) + ( x) First term of 6 B C... x + C... x + C... x + C... x ( ) ( ) ( ) ( ) = (6 + ) 96x + 6x + 6x + 8x Must use Binomial otherwise A0, A A A0 () ( x) = 6 96x + 6x 6x + 8x Bft () 5 ( ) + x = x ( + ) ( + C ( ) + C ( ) + C ( ) + ( ) ) x x x x is applied exactly as before Notes for Question B: The constant term should be 6 in their expansion : Two binomial coefficients must be correct and must be with the correct power of x. Accept C or or as a coefficient, and C or or 6 as another.. Pascal s triangle may be used to establish coefficients. A: Any two of the final four terms correct (i.e. two of 96x + 6x + 6x + 8x ) in expansion following Binomial Method. A: All four of the final four terms correct in expansion. (Accept answers without + signs, can be listed with commas or appear on separate lines) Bft: Award for correct answer as printed above or ft their previous answer provided it has five terms ft and must be subtracting the x and x terms Allow terms in to be in descending order and allow +-96x and +-6x in the series. (Accept answers without + signs, can be listed with commas or appear on separate lines) e.g. The common error + C x + C x + Cx + x = (6) + 96x + 7x + x + x would earn B,, A0, A0, and if followed by = (6) 96x + 7x x + x gets Bft so /5 Fully correct answer with no working can score B in part (a) and B in part. The question stated use the Binomial theorem and if there is no evidence of its use then M mark and hence A marks cannot be earned. Squaring the bracket and squaring again may also earn B M0 A0 A0 B if correct Omitting the final term but otherwise correct is B A A0 B0ft so /5 If the series is divided through by or a power of at the final stage after an error or omission resulting in all even coefficients then apply scheme to series before this division and ignore subsequent work (isw)

8 . (a) Either (Way ) : Attempt f () or f ( ) Or (Way ): Assume a = -9 and attempt f () or f ( ) f() = 5 5+ a + 8= 0 a = 7 a = 9* f() = 0 so (x -) is factor A * cso () Or (Way ): ( x 5x + ax+ 8) ( x ) = x + px+ q where p is a number and q is an expression in terms of a Sets the remainder 8+ a +9 = 0 and solves to give a = 9 A* cso () Either (Way ): f( x) = ( x )(x + x 6) A = ( x )(x )( x + ) A () Or (Way ) Uses trial or factor theorem to obtain x = - or x = / Uses trial or factor theorem to obtain both x = - and x = / A Puts three factors together (see notes below) Correct factorisation : ( x )(x )( x + ) or ( x)( x)( x + ) or A ( x )( x )( x + ) oe () Or (Way ) No working three factors ( x )(x )( x + ) otherwise need working AA (c) y { } y y y { } ( ) = = or g() = 0 B =.5 log = log.5 or y = log.5 { y } = y= awrt 0.7 A () [9] Notes for Question (a) for attempting either f () or f ( ) with numbers substituted into expression A for applying f () correctly, setting the result equal to 0, and manipulating this correctly to give the result given on the paper i.e. a = 9. (Do not accept x = -9) Note that the answer is given in part (a). If they assume a = -9 and verify by factor theorem or division they must state (x ) is a factor for A (or equivalent such as QED or a tick). st : attempting to divide by ( x ) leading to a TQ beginning with the correct term, usually x. (Could divide by ( x), in which case the quadratic would begin - x.) This may be done by a variety of methods including long division, comparison of coefficients, inspection etc. st A: usually for x + x 6 Credit when seen and use isw if miscopied nd : for a valid* attempt to factorise their quadratic (* see notes on page 6 - General Principles for Core Mathematics Marking section ) nd A is cao and needs all three factors together. Ignore subsequent work (such as a solution to a quadratic equation.) NB: ( x )( x )( x + ) is AM0A0, ( x )( x )(x + ) is AA0, but ( x )( x )( x + ) is AA. (c) B: y = seen as a solution may be spotted as answer no working needed. Allow also for g() = 0. y ky : Attempt to take logs to solve = α or even = α, but not 6 y = α where α > 0 and α & was a root of f(x) = 0 (ft their factorization) A: for an answer that rounds to 0.7. If a third answer is included (and not rejected ) such as ln(-) lose final A mark

9 . (a) x y { x } At =.5, y =.58 (only) B cao [] 0.5 ; B oe { ( their )} For structure of {...} ; Aft + + ( )} { } 0.5 { (5 0.5).5 their = (.956) = 6.9 = awrt 6. A (c) Adds Area of Rectangle or first integral = or [ x] to previous answer 0 So required estimate = {"6.9" + = "8.9"} = "awrt 8." (or + previous answer). Aft N.B. 7 + previous answer is M0A0 (added seven times because 7 numbers in table) [] 7 Notes for Question (a) B:.58 B: for using 0.5 or or equivalent. : requires the correct{...} bracket structure. It needs the first bracket to contain first y value plus last y value and the second bracket to be multiplied by and to be the summation of the remaining y values in the table with no additional values. If the only mistake is a copying error or is to omit one value from nd bracket this may be regarded as a slip and the M mark can be allowed ( An extra repeated term forfeits the M mark however). M0 if values used in brackets are x values instead of y values Aft: for the correct bracket {...} following through candidate s y value found in part (a). A: for answer which rounds to 6.. NB: Separate trapezia may be used : B for 0.5, for / h(a + b) used 5 or 6 times (and Aft if it is all correct ) Then A as before. Special case: Bracketing mistake 0.5 (5 0.5) (.5 their ) scores B A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). An answer of 0.8 usually indicates this error. (c) : Relates previous answer ( not integral of previous answer) to this question by integrating between limits, and adding, or by using geometry to find rectangle and adding. Aft: for + answer to Alternative method (c) Those who do a trapezium rule for part - using the table from (a) with added to each cell of the table " their " their = (structure must be correct allow Get: for { ( )} one copying error only) And Aft: for awrt 8. (or + previous answer). []

10 5. (a) Mark (a) and together. Usually answered in radians: Uses either sin(angle) ab or () (angle) or both Area = ()()sin 0.6 or () ( π 0.6) { = or } A Area = ()()sin () ( π 0.6) { = } A { Area = } = awrt 6.5 (m ) or 6.(m ) or 6.6 (m ) A () CDE = ( angle), = ( π 0.6) CDE = , A { } (a) { } AE = + ()()cos(0.6) AE = or AE = AE = Perimeter = = = awrt 80. (m) A Notes for Question 5 : uses either area of triangle formula as given in mark scheme, or area of sector or both (may be implied by answer) A: one correct area expression (with correct angle used) ()()sin 0.6 or () ( π 0.6) see awrt 8. or awrt 80 (80 may be split as 6.(semicircle) minus 6.(small sector)) A: two correct area expressions (with correct angles) added together (allow.5 as implying π 0.6 ) or see awrt 8. + awrt 80 ( or 6-6 ) A: answers which round to 6.5 or 6. or 6.6 st for attempt to use s = r θ (any angle) st A for π 0.6 in the formula (or.5) nd : Uses correct cosine rule to obtain AE or AE (this may appear in part (a)) rd (independent): Perimeter = + + their AE + their CDE nd A: awrt 80. (ignore units even incorrect units) (5) [9] or Degrees (a) Uses either sin(angle) ab or anglein degrees π () or both for 60 (80-6.7) Area = ()()sin6.7 or π () { = 60 awrt 8... or 80} A (80-6.7) Area = ()()sin6.7 + π () 60 { = awrt } A Final mark as before Angle in degrees CDE = π, = π{ CDE = }, A Final three marks as before

11 6. (a) Seeing and. B () x( x + )( x ) = x + x 8x or x x + x 8x ( without simplifying) B 8 8 ( x x x x x x x x + x 8 x )d x = + { + c } or + { + c } Aft 0 x x 8x 8 + = (0) 6 6 or One integral x x 8x 6 + = + 6 (0) =± (.6 or awrt.7 ) or other integral = ± 6 (6.6 or awrt 6.7) 0 d A Hence Area = " their " + " their6 " or = 9 or 9. or 8 (An answer of (NOT Area = " their " " their6 " d 8 ) A = 9 may not get the final two marks check solution carefully) (7) (a) [8] Notes for Question 6 B: Need both and. May see (-,0) and (,0) (correct) but allow (0,-) and (0, ) or A = -, B = or indeed any indication of - and check graph also B: Multiplies out cubic correctly (terms may not be collected, but if they are, mark collected terms here) n n : Tries to integrate their expansion with x x + for at least one of the terms Aft: completely correct integral following through from their CUBIC expansion (if only quadratic or quartic this is A0) d: (dependent on previous M) substituting EITHER -a and 0 and subtracting either way round OR similarly for 0 and b. If their limits a and b are used in ONE integral, apply the Special Case below. A: Obtain either ± (or.6 or awrt.7)from the integral from - to 0 or ± 6 (6.6 or awrt 6.7) from the integral from 0 to ; NO follow through on their cubic (allow decimal or improper equivalents 8 or 0 ) isw such as subtracting from rectangles. This will be penalized in the next two marks, which will be M0A0. d (depends on first method mark) Correct method to obtain shaded area so adds two positive numbers (areas) together or uses their positive value minus their negative value, obtained from two separate definite integrals. A: Allow 9., 9., 9. etc. Must follow correct logical work with no errors seen. For full marks on this question there must be two definite integrals, from - to 0 and from 0 to, though the evaluations for 0 may not be seen. (Trapezium rule gets no marks after first two B marks) Special Case: one integral only from a to b: BA available as before, then x x 8x = ( + 6) 6 6 = 6 + =. d for correct use of their limits a and b and subtracting either way round. A for 6: NO follow through. Final M and A marks not available. Max 5/7 for part

12 7. (i) Method 7(i) x log = 5x + x = or 5x + 5x + 5x + or log =, or log = (see special case ) x x 5x + = x 5x + or = x x 5x + 8 or log = log 6x = 5x + x = (depends on previous Ms and must be this equation or equivalent) d x = or exact recurring decimal 0.6 after correct work log ( x) + = log (5x + ) A cso () Method So log() x + log(8) = log(5 x + ) ( replaced by log 8 ) nd Then log (6 x) = log (5x + ) (addition law of logs) st Then final A as before da (ii) loga y + loga = 5 loga 8y = 5 Applies product law of logarithms. d 5 5 y = a y = a Acao 8 8 () [7] Notes for Question 7 (i) st : Applying the subtraction or addition law of logarithms correctly to make two log terms in x into one log term in x nd : For RHS of either,, or log 8, log 8 or log 6 i.e. using connection between log base and to a power. This may follow an earlier error. Use of is M0 rd d: Obtains correct linear equation in x. usually the one in the scheme and attempts x = A: cso Answer of / with no suspect log work preceding this. (ii) (i) : Applies power law of logarithms to replace log by log or log 8 a a a d: (should not be following M0) Uses addition law of logs to give loga y = 5 or loga 8y = 5 log ( x) x Special case : log ( x) = log (5x + ) = = x = or log (5x + ) 5x + log ( x) x x log ( x) = log (5x + ) = log = = x = each log (5x + ) 5x + 5x+ attempt scores M0A0 special case Special case : log ( x) = log (5x + ) log + log x= log (5x + ), is M0 until the two log terms are 5x + combined to give log = + log. This earns x 5x + 5x + Then = or log = log gets second. Then scheme as before. x x

13 8. (i) ( α = ) x { α } (ii)(a) = + = = awrt 96. B x 0 = 80 + "56.099"... or x 0 = π + "0.98"... x = { = } = awrt -8.7 A sinθ sinθ = cosθ + cosθ cos θ = cosθ + cosθ cos θ cos θ cosθ 0 cos θ cosθ = + = + * A cso * ± ()( ) cosθ = 8 or (cos θ ± ) ± q± = 0, or ( cos θ ± ) ± q± = 0, q 0 so cos θ =... (i) (ii) (a) o One solution is 7 or o, o Two solutions are 7 and o A, A { 7,, 6, 88} θ = A (5) [] Notes for Question 8 B: 96. by any or no method : Takes 80 degrees from arctan (.5) or from their 96. May be implied by A. (Could be obtained by adding 80, then subtracting 60). A: awrt 8.7 Extra answers: ignore extra answers outside range. Any extra answers in range lose final A mark (if earned) Working in radians could earn for x 0 = π + "0.98"... so B0A0 sinθ sin : uses tanθ = or equivalent in equation (not just tan = cosθ cos, with no argument) d: uses sin θ = cos θ (quoted correctly) in equation A: completes proof correctly, with no errors to give printed answer*. Need at least three steps in proof and need to achieve the correct quadratic with all terms on one side and =0 : Attempts to solve quadratic by correct quadratic formula, or completion of the square. Factorisation attempts score M0. st A: Either 7 or, nd A: both 7 and (allow 7.0 etc.) : 60 "a previous solution" (provided that cos was being used) (not dependent on previous M) A: All four solutions correct (Extra solutions in range lose this A mark, but outside range - ignore) (Premature approximation: e.g. 7.9,., 88. and 5.9 lose first A then ft other angles) Do not require degrees symbol for the marks Special case: Working in radians : as before, A for either θ = π or θ = π or decimal equivalents, and nd A: both 5 5 : π α or π α then A0 so /5 d () ()

14 9. (a) dy = x 6x dx x x 6 = 0 x, x = =,or [or x 6x = 0 x= (no wrong work seen)] x x = 6 then squared then obtain x = A ( x = 8 ) x = A x =, y = + 0 = 8 (ignore y = 00 as second answer) A (6) d y dx = + 8x A d y ( > 0 ) y is a minimum ( there should be no wrong reasoning) A dx () [9] (a) Alternative Method: Gradient Test: for finding the gradient either side of their x-value from part (a). A for both gradients calculated correctly to significant figure, then using < 0 and > 0 respectively maybe by use of sketch or table. (See appendix for gradient values. This is not ft their x) A states minimum needs A to have been awarded. Notes for Question 9 st : At least one term differentiated correctly, so x x, or x 6 x, or A: This answer or equivalent e.g. x x nd : Sets their d y dx to 0, and solves to give x =, x = or x = after correct squaring or spots x = d y (NB = 0 so 8x 0 + = is M0 ) dx N.B. Common error: Putting derivative = 0 and merely obtaining x = 0 is M0A0, then M0A0 for next two marks. (The first two marks in (a) and marks for second derivative may be earned in part.) A: x = cao [ x = - is A0 and x = ± is also A0 ] rd : Substitutes their positive found x (NOT zero) into y = x x + 0, x > 0.Should follow attempting to set d y dx d y 0 dx A: -8 cao (Does not need to be written as coordinates) : Attempts differentiation of their first derivative with at least one term differentiated correctly. Should be seen or referred to (in part ) in determining the nature of the stationary point. A: Answer in scheme or equivalent A: States minimum (Second derivative should be correct- can follow incorrect positive x. Needs A to have been awarded- should not follow incorrect reasoning (need not say d y d y > 0 but should not have said dx dx = 0 for example )

15 0. (a) Equation of form ( x ± 5) + ( y ± 9) = k, k >0 Equation of form ( x a) + ( y b) = 5, with values for a and b x y ( + 5) + ( 9) = 5 P(8, 7). Let centre of circle = X ( 5, 9) = 5 A PX = ( 8 " 5" ) + ( 7 "9") or PX = ( 8 5) ( 7 9) Alternative for (a) Alternative for Alternative for (a) PX = ) PT ( PX ) ( 5 or 5 7 Equation of the form + = 5 with numerical PX d PT { = 00} = 0 (allow 0.0) A cso () [6] x y x y c + ± 0 ± 8 + = 0 Uses a + b 5 = c with their a and b or substitutes (0, 9) giving + 9 ± b 9+ c= 0 x + y + 0x 8y+ 8 = 0 A () An attempt to find the point T may result in pages of algebra, but solution needs to reach 8 (-8, 5) or, to get first (even if gradient is found first) 7 7 : Use either of the correct points with P (8, -7) and distance between two points d formula A: 0 Acso () Substitutes (8, -7) into circle equation so Square roots to give PT { = 00} = 0 PT = 8 + ( 7) ( 7) + 8 () da () Notes for Question 0 The three marks in (a) each require a circle equation (see special cases which are not circles) : Uses coordinates of centre to obtain LHS of circle equation (RHS must be r or k > 0 or a positive value) : Uses r = 5 to obtain RHS of circle equation as 5 or 5 A: correct circle equation in any equivalent form Special cases ( x ± 5) + ( x± 9) = (5 ) is not a circle equation so M0M0A0 Also ( x ± 5) + ( y 9) = (5 ) And ( x ± 5) ( y± 9) = (5 ) are not circles and gain M0M0A0 But ( x 0) + ( y 9) = 5 gains M0A0 : Attempts to find distance from their centre of circle to P (or square of this value). If this is called PT and given as answer this is M0. Solution may use letter other than X, as centre was not labelled in the question. N.B. Distance from (0, 9) to (8, -7) is incorrect method and is M0, followed by M0A0. d: Applies the subtraction form of Pythagoras to find PT or PT (depends on previous method mark for distance from centre to P) or uses appropriate complete method involving trigonometry A: 0 cso

16 Aliter 9. Way Gradient Test Method: dy = x 6x dx Helpful table! x dy dx

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18 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 FN Telephone Fax Order Code UA0566 Summer 0 For more information on Edexcel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE