PhysicsAndMathsTutor.com. Mark Scheme (Results) Summer GCE Core Mathematics C3 (6665) Paper 1

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1 Mark (Results) Summer 0 GCE Core Mathematics C (6665) Paper

2 Edecel and BTEC Qualifications Edecel 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 Epert 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 UA0956 All the material in this publication is copyright Pearson Education Ltd 0

3 Summer Core Mathematics C Mark General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. When eaminers 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 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel 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 and can be used if you are using the annotation facility on epen. 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 4. All A marks are correct answer only (cao.), unless shown, for eample, as 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 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 ( + b + c) = ( + p)( + q), where pq = c, leading to =... ( a + b + c) = ( m + p)( n + q), where pq = c and mn = a, leading to =. Formula Attempt to use correct formula (with values for a, b and c), leading to =. Completing the square Solving + b + c = 0 b : ( ) ± ± q ± c, q 0, leading to = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( n n ). Integration Power of at least one term increased by. ( n n+ ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent eaminers 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. Eact answers Eaminers reports have emphasised that where, for eample, an eact 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 . = + At any stage B 9 4 ( )( ) Eliminating the common factor of (+) at any stage ( + ) = ( )( + ) B Use of a common denominator + + ( )( ) (9 4) (9 4)( + ) (9 4)( + ) or ( + ) ( ) ( )( + ) ( + )( ) 6 6 or ( )( + ) 9 (4 marks) Notes B For factorising 9 4 = ( )( + ) using difference of two squares. It can be awarded at any stage of the answer but it must be scored on E pen as the first mark B For eliminating/cancelling out a factor of (+) at any stage of the answer. For combining two fractions to form a single fraction with a common denominator. Allow slips on the numerator but at least one must have been adapted. Condone invisible brackets. Accept two separate fractions with the same denominator as shown in the mark scheme. Amongst possible (incorrect) options scoring method marks are ( + ) (9 4) Only one numerator adapted, separate fractions (9 4)( + ) (9 4)( + ) + Invisible brackets, single fraction ( )( + ) 6 ( )( + ) This is not a given answer so you can allow recovery from invisible brackets. Alternative method ( + ) ( + )( + ) (9 4) 8 + has scored 0,0,,0 so far = = (9 4) ( + ) (9 4)( + ) (9 4)( + ) 6( + ) = is now,,,0 ( + )( )( + ) 6 = and now,,, ( )( + )

7 . (a) = 0 + = 4 (b) awrt ( + ) = 4 4 4( ) = = ( + ) ( + ) d* =.4, =.0 =., (c ) Choosing (.75,.75) or tighter containing root.7998 () () Notes (a) f (.75) = ( + ) f (.75) = Change of sign α=.7 Moves from f()=0, which may be implied by subsequent working, to ( ± ) = ± ± 4 by separating terms and factorising in either order. No need to factorise rhs for this mark. d Divides by (+) term to make the subject, then takes square root. No need for rhs to be factorised at this stage * CSO. This is a given solution. Do not allow sloppy algebra or notation with root on just numerator for instance. The -4 needs to have been factorised. () (9 marks) (b) Note that this appears B,B,B on EPEN An attempt to substitute 0 = into the iterative formula to calculate. This can be awarded for the sight of 4( ) 8,, and even.4 (+ ) 4 =.4. The subscript is not important. Mark as the first value found, is A0 = awrt.0 = awrt.. Mark as the second and third values found. Condone. for (c ) Note that this appears on EPEN Choosing the interval (.75,.75) or tighter containing the root Continued iteration is not allowed for this question and is M0 Calculates f(.75) and f(.75), or the tighter interval with at least correct to sig fig rounded or truncated. Accept f(.75) = sf rounded or truncated. Also accept f(.75) = -0.0 dp Accept f(.75) = (+)0.008 sf rounded or truncated. Also accept f(.75) = (+)0.0 dp Both values correct (see above), A valid reason; Accept change of sign, or >0 <0, or f(.75) f(.75)<0 And a (minimal) conclusion; Accept hence root or α=.7 or QED or

8 Alternative to (a) working backwards (a) 4( ) 4( ) ( + ) ( + ) = = + = ( ) 4( ) + = = 0 d States that this is f()=0 * Alternative starting with the given result and working backwards Square (both sides) and multiply by (+) d Epand brackets and collect terms on one side of the equation =0 A statement to the effect that this is f()=0 () An acceptable answer to (c ) with an eample of a tighter interval Choosing the interval (.75,.70). This contains the root.79(98) Calculates f(.75) and f(.70), with at least correct to sig fig rounded or truncated. Accept f(.75) = sf rounded or truncated f(.75) = -0.0 dp Accept f(.70) = (+) sf rounded or f(.70) = (+) truncated sf Both values correct (see above), A valid reason; Accept change of sign, or >0 <0, or f(.75) f(.70)<0 And a (minimal) conclusion; Accept hence root or α=.7 or QED or f() ( ) An acceptable answer to (c ) using g() where g()= ( + ) nd Calculates g(.75) and g(.75), or the tighter interval with at least correct to sig fig rounded or truncated. g(.75) = Accept g(.75) = awrt (+) sf rounded or awrt truncated. g(.75)= Accept g(.75) = awrt sf rounded or awrt truncated.

9 . (a) e sin e cos d 0 d e ( sin + cos ) = 0 tan = π π = = 9 (6) (b) At =0 d = B y 0 Equation of normal is = or any equivalent y = 0 () (9 marks) (a) Applies the product rule vu +uv to e sin. If the rule is quoted it must be correct and there must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, ie. terms are written out u=,u =.,v=.,v =.followed by their vu +uv ) only accept answers of the form Ae sin e B cos d Correct epression for e sin e cos d Sets their d y 0 d e producing an equation in sin and cos Achieves either tan = or tan = Correct order of arctan, followed by. 5π 5π π π Accept = = or = = but not = arctan( ) 9 9 π CS0 = Ignore etra solutions outside the range. Withhold mark for etra inside the range. 9 (b) B Sight of for the gradient A full method for finding an equation of the normal. Their tangent gradient m must be modified to y 0 Eg = or equivalent is acceptable their ' m' 0 y 0 y = or any correct equivalent including =. 0 and used together with (0, 0). m

10 Alternative in part (a) using the form Rsin( + α ) JUST LAST MARKS. (a) e sin e cos d 0 d e ( sin + cos ) = 0 π ( )sin( + ) = 0 π π = = 9 (6) π π Achieves either ( )sin( + ) = 0 or ( ) cos( ) = 0 6 Correct order of arcsin or arcos, etc to produce a value of π Eg accept + = 0 or π or π =... π Cao = Ignore etra solutions outside the range. Withhold mark for etra inside the range. 9 Alternative to part (a) squaring both sides JUST LAST MARKS. (a) e sin e cos d 0 d e ( sin + cos ) = 0 sin = cos cos ( ) = or sin ( ) = 4 4 = arcos( ± ) 4 oe = π 9

11 4.(a) Shape including cusp B (-.5, 0) and (0, 5) B () (b) Shape B (0,5) B () ( c) Shape B (0,0) B (-0.5, 0) B () (7 marks) (a) Note that this appears as on EPEN B Shape (inc cusp) with graph in just quadrants and. Do not be overly concerned about relative gradients, but the left hand section of the curve should not bend back beyond the cusp B This is independent, and for the curve touching the -ais at (-.5, 0) and crossing the y-ais at (0,5) (b) Note that this appears as on EPEN B For a U shaped curve symmetrical about the y- ais B (0,5) lies on the curve (c ) Note that this appears as BB on EPEN B Correct shape- do not be overly concerned about relative gradients. Look for a similar shape to f() B Curve crosses the y ais at (0, 0). The curve must appear in both quadrants and B Curve crosses the ais at (-0.5, 0). The curve must appear in quadrants and. In all parts accept the following for any co-ordinate. Using (0,) as an eample, accept both (,0) or written on the y ais (as long as the curve passes through the point) Special case with (a) and (b) completely correct but the wrong way around mark - SC(a) 0, SC(b) 0, Otherwise follow scheme

12 5. (a) 4 4cosec θ cosec θ = sin θ sin θ = 4 (sinθ cos θ ) sin θ B B 4 4 (sinθ cos θ ) sin θ 4sin θ cos θ sin θ (b) = Using cos θ = sin θ (c ) Note (a) and (b) can be scored together (a) B One term correct. Eg. writes cos θ = sin θ cos θ sin θ cos θ sin θ sin θ cos θ = = sec θ cos θ * sec θ = 4 secθ = ± cosθ = ± π π θ =,, 4 (sinθ cos θ ) 4cosec θ as or cosec θ as (9 marks) () (4) (). Accept terms like sin θ cos θ cosec θ = + cot θ = +. The question merely asks for an epression in sin θ and cos θ sin θ 4 B A fully correct epression in sin θ and cos θ. Eg. Accept equivalents (sinθ cos θ ) sin θ Allow a different variable say s instead of θ s but do not allow mied units. b) Attempts to combine their epression in sinθ and cosθ using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted Attempts to form a single term on the numerator by using the identity cos θ = sin θ Cancels correctly by sin θ terms and replaces with sec θ cos θ * Cso. This is a given answer. All aspects must be correct IF IN ANY DOUBT SEND TO REVIEW OR CONSULT YOUR TEAM LEADER c ) For sec θ = 4 leading to a solution of cosθ by taking the root and inverting in either order. Similarly accept tan θ =, sin θ = leading to solutions of tan θ, sin θ. Also accept cosθ = 4 π Obtains one correct answer usuallyθ = Do not accept decimal answers or degrees π π Obtains both correct answers. θ =, Do not award if there are etra solutions inside the range. Ignore solutions outside the range.

13 6. (a) f()> B () (b) ln fg( ) e, = + = +, () (c ) e + = 6 e = 4 + = ln 4 ln 4 = or ln- + + (d) Let y = e + y = e ln( y ) = (4) f ( ) ln( ),. = >, Bft () (e) Shape for f() B (0, ) B Shape for f - () B (, 0) B (4) (4 marks) (a) B Range of f()>. Accept y>, (, ), f>, as well as range is the set of numbers bigger than but don t accept > (b) For applying the correct order of operations. Look for ln e +. Note that ln e + is M0 ln Simplifies e + to +. Just the answer is acceptable for both marks (c ) Starts with e + + = 6 and proceeds to e + =... e + = 4 Takes ln s both sides, + = ln.. and proceeds to =. ln 4 = oe. eg ln - Remember to isw any incorrect working after a correct answer

14 (d) Note that this is marked on EPEN y Starts with y = e + or = e + and attempts to change the subject. All ln work must be correct. The must be dealt with first. Eg. y = e + ln y = + ln = ln y ln is M0 = There must be some form of bracket f ( ) ln( ) or y= ln( ) or y= ln Bft Either >, or follow through on their answer to part (a), provided that it wasn t y R Do not accept y> or f - ()>. (e) B B B B Shape for y=e. The graph should only lie in quadrants and. It should start out with a gradient that is appro. 0 above the ais in quadrant and increase in gradient as it moves into quadrant. You should not see a minimum point on the graph. (0, ) lies on the curve. Accept written on the y ais as long as the point lies on the curve Shape for y=ln. The graph should only lie in quadrants 4 and. It should start out with gradient that is appro. infinite to the right of the y ais in quadrant 4 and decrease in gradient as it moves into quadrant. You should not see a maimum point. Also with hold this mark if it intersects y=e (, 0) lies on the curve. Accept written on the ais as long as the point lies on the curve Condone lack of labels in this part Eamples Scores,0,,0. Both shapes are fine, do not be concerned about asymptotes appearing at =, y=. (See notes) Both co-ordinates are incorrect Scores 0,,, Shape for y = e is incorrect, there is a minimum point on the graph. All other marks an be awarded

15 7. (a)(i) (ii) d (ln( )) = d d ( ln( )) = ln( ) + d y = 0 d ( ) 5 4 d ( ) 0 ( 0 ) 5( ) () (b) 80 = d ( ) tan y 6sec y Uses sec y tan 6 d = = d = 6sec y = + y and uses tan y = = = ( ) d 6( + ( ) ) 8 + Note that this is marked B on EPEN (a)(i) Attempts to differentiate ln() to B. Note that is fine. () (5) ( marks) Attempts the product rule for ln( ). If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted nor implied from their stating of u, u, v, v and their subsequent epression, only accept answers of the form B ln( ) A +, A, B > 0 Any correct (un simplified) form of the answer. Remember to isw any incorrect further work d ln( ) ( ln( )) = ln( ) + = ( + ) = ( ln + ) d Note that this part does not require the answer to be in its simplest form (ii) Applies the quotient rule, a version of which appears in the formula booklet. If the formula is quoted it must be correct. There must have been an attempt to differentiate both terms. If the formula is not quoted nor implied from their stating of u, u, v, v and their subsequent epression, only accept answers of the form

16 ( ) ± 0 ( 0 ) C( ) 0 or 7 or 5 ( ) 5 4 y Any un simplified form of the answer. Eg d (( ) ) 80 Cao. It must be simplified as required in the question = 6 d ( ) 5 4 d ( ) 0 ( 0 ) 5( ) = 5 (b) Knows that tan y differentiates to C sec y. The lhs can be ignored for this mark. If they sin y writetan y as this mark is awarded for a correct attempt of the quotient rule. cos y d Writes down 6sec y = 6sec y d 6 cos y 6cosy sin y sin y Accept from the quotient rule or even cos y cos y An attempt to invert their d y = f ( y), or changes the subject of their implicit d. differential to achieve a similar result d y f ( y) d Replaces an epression for sec y in their d y with by attempting to use d sec y = + tan y. Alternatively, replaces an epression for y in d y d arctan( ) Any correct form of d y d d = or 6( + ( ) ) d 8 + 6sec (arctan( )) 7. (a)(ii) Alt using the product rule 0 5 Writes as ( 0 )( ) and applies vu +uv. 5 ( ) See (a)(i) for rules on how to apply 5 6 ( ) 0 + ( 0 ) 5( ) 6 Simplifies as main scheme to 80 ( ) or equivalent (b) Alternative using arctan. They must attempt to differentiate to score any marks. Technically this is A Rearrange = tan y to y = arctan( ) and attempt to differentiate A Differentiates to a form, = or oe + ( ) ( ( ) ), A + 6( + ( ) ) () (5)

17 8. (a) R=5 B 4 0 tan α = α = (awrt)7.7 7 () (b).5 cos( + theirα ) = their R their ' α ' = + their α = = 0 0 ' ' their 00 or their = awrt.,7. (c ) Attempts to use epression = AND sin = sin cos in the cos cos (5) = + = 7cos 4sin cos 48sin cos 7(cos ) 4sin (d) 4 cos 48sin cos = R cos( + α) + 7 () Maimum value = R + c = cao () ( marks) (a) B Accept 5, awrt 5.0, 65. Condone ± For tanα = ± tanα = ± sinα = ±, cosα = ± 7 4 their R their R 0 α = (awrt)7.7. The answer.87 (radians) is A0 (b).5 For using part (a) and dividing by their R to reach cos( + theirα ) = their R Achieving their α 60 (0) + =. This can be implied by. (0) /. (0) or 7. (0) /7. (0) or (0) / (0) /-6.9 (0) Finding a secondary value of from their principal value. A correct answer will imply this mark Look for 60 ± 'their' principal value±'their' α = awrt. /. OR 7. / = awrt. Ignore solutions outside of range. Penalise this mark for etra. AND 7. solutions inside the range

18 (c ) Attempts to use cos = cos and sin = sin cos in epression. Allow slips in sign on the cos term. So accept cos = ± cos ± Cao = 7 cos 4sin + 7. The order of terms is not important. Also accept a=7, b=-4, c=7 (d) This mark is scored for adding their R to their c cao Radian solutions- they will lose the first time it occurs (usually in a with.87 radians) Part b will then be marked as follows.5 (b) For using part (a) and dividing by their R to reach cos( + theirα ) = their R π The correct principal value of or awrt.05 radians. Accept 60 (0) This can be implied by awrt 0. radians or awrt or.97 radians or awrt.0 radians Finding a secondary value of from their principal value. A correct answer will imply this mark Look for π ± 'their' principal value±'their' α Do not allow mied units. = awrt.97 OR.0. = awrt.97 AND.0. Ignore solutions outside of range. Penalise this mark for etra solutions inside the range

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20 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code UA0956 Summer 0 For more information on Edecel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Esse CM0 JE