9231 FURTHER MATHEMATICS

Transcription

1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mar This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have considered the acceptability of alternative answers. Mar schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademar of Cambridge International Eaminations.

2 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Mar Scheme Notes Mars are of the following three types: M A B Method mar, awarded for a valid method applied to the problem. Method mars are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mar and in some cases an M mar can be implied from a correct answer. Accuracy mar, awarded for a correct answer or intermediate step correctly obtained. Accuracy mars cannot be given unless the associated method mar is earned (or implied). Mar for a correct result or statement independent of method mars. When a part of a question has two or more "method" steps, the M mars are generally independent unless the scheme specifically says otherwise; and similarly when there are several B mars allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mar is dependent on an earlier M or B (asterised) mar in the scheme. When two or more steps are run together by the candidate, the earlier mars are implied and full credit is given. The symbol implies that the A or B mar indicated is allowed for wor correctly following on from previously incorrect results. Otherwise, A or B mars are given for correct wor only. A and B mars are not given for fortuitously "correct" answers or results obtained from incorrect woring. Note: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. The mars indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mar, allow the candidate the benefit of the doubt. Unless otherwise indicated, mars once gained cannot subsequently be lost, e.g. wrong woring following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mar unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mar if a value is obtained which is correct to s.f., or which would be correct to s.f. if rounded ( d.p. in the case of an angle). As stated above, an A or B mar is not given if a correct numerical answer arises fortuitously from incorrect woring. For Mechanics questions, allow A or B mars for correct answers which arise from taing g equal to 9.8 or 9.8 instead of. Cambridge International Eaminations

3 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 The following abbreviations may be used in a mar scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra checing is needed to ensure that the detailed woring leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) Correct Woring Only often written by a fortuitous' answer Ignore Subsequent Woring Misread Premature Approimation (resulting in basically correct wor that is insufficiently accurate) See Other Solution (the candidate maes a better attempt at the same question) Special Ruling (detailing the mar to be given for a specific wrong solution, or a case where some standard maring practice is to be varied in the light of a particular circumstance) Penalties MR PA A penalty of MR is deducted from A or B mars when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B mars then become "follow through " mars. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B mars in the case of premature approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations

4 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars & 6cos t sin t, y& 6sin t cost dy tant (OE) d d y sec t sec tcosect AG d 6cos t sin t 6 m m ( m ) m t t CF: Ae Bte soi PI : pt qt r & pt q & p p 8pt q pt qt r 7 t p, q, r t t GS: Ae Bte t t n in formula gives a e ae e ae d a a a a a ( e ) e. ae e ae H is true oe d a a a a Assume H is true, i.e. ( ) d d a ( e ) a e a a a d e a d a e ( ) a e a a a a e e a a e a a e H is true, hence by PMI H n is true for all positive integers n.. B B MA [] M A M M A A [6] 6 B B B M A A [6] 6 (i) K 6. 8 MA A [] n 6 (ii) 6 >.9.n.79n.79( > ) n n n >.K so terms required. MA dma [] 7 Cambridge International Eaminations

5 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & α β γ p p Solution Mars B 6 ( α β ) q ( αβ βγ γα) ( α β γ ) γ q ( 8) 7 6 α ( [ β γ ]) α a α 6 α, α, e.g. since > 8 or other reason βγ 7 6 r αβγ (etra answer penalised) i j 6 λ : 8 ~ oe i j λ : ~ oe M A [] M MA B A [] 8 MA A [] 7 7 λ 8 MA [] D, P (or other multiples or permutations). B B Det P (or depending on permutation). B Adj P P ( or other permutations). MA [] Cambridge International Eaminations

6 Page 6 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Cambridge International Eaminations Qn & Solution Mars 7 ~ ~ r(m) t z y t z y E.g. Set z λ and t µ µ λ y and µ λ 9 9, is Basis 9 b a µ λ Solving: λ and µ a, b. MA A M M A [6] M M A A [] 8 ( ) ) )( ( y ) ( < AC B no stationary points 8 6 < > for no stationary points. When : Vertical asymptote: Oblique asymptote: y y Aes and asymptotes Each branch. M A MA A [] B MA B BB [6]

7 Page 7 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars e 9 ln d ln I n (ln ) n.ln d e n e n [(ln ) ( ln ) ] ( n )(ln ). ( ln )d e [ I I ] e n ( )(ln ) (ln )d ( n ) n n n (AG) Alternative for obtaining reduction formula: I n [ (ln ) ] e e e n n n (ln ) d n(ln ) d I n e nin I n e ( n ) In I n nin In ( n ) In I n ( n ) [ In In] (AG) Similarly I I e [] [ ln e ] e (e ) e I I ( [e ]) 6 I I ( ) e B M MA MA [6] MA A B M A [6] B B M A MV I 6 e e e M A [6] Cambridge International Eaminations

8 Page 8 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars (cosθ isinθ) cosθ isinθ ( c is) c c si c s ic s i cs s i c s c s s tan θ c c s cs Divide numerator and denominator by c (stated or shown): tanθ tan θ tan θ tan θ (AG) tan θ tan θ tan θ θ π, π, π, π, π t t t has roots tan π, tan π, tan π, tan π, tanπ t t has roots tan π, tan π, tan π, tan π. t tan π t tan π since tan π tan π and tan π tan π. has roots tan π and tan π. (AG) sec α tan α y y ( y ) ( y ) y y 6 B MA M A [] B B M A [] M M A [] Cambridge International Eaminations

9 Page 9 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Cambridge International Eaminations Qn & Solution Mars E E.g. ~ 8 j i. (AG) ) ( 7 j i j i p Line AP: t r For Q q t t E.g. AB, BQ j i cos cos 8 cos.. cos (AG) MA MA [] B MA MA [] B MA MA []

10 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & O Solution Closed curve through pole with correct orientation. Completely correct. π π a ( cosθ cos θ )dθ a π cosθ cosθ dθ π a θ sinθ sin θ a π π π Mars B B [] MM MA A [] ds dθ a ( cosθ cos θ sin θ) a ( cosθ ) a.sin θ a sin θ (AG) B MA s π a sin θ π d π a cos θ a π θ M A MA [7] Cambridge International Eaminations

11 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mar This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have considered the acceptability of alternative answers. Mar schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademar of Cambridge International Eaminations.

12 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Mar Scheme Notes Mars are of the following three types: M A B Method mar, awarded for a valid method applied to the problem. Method mars are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mar and in some cases an M mar can be implied from a correct answer. Accuracy mar, awarded for a correct answer or intermediate step correctly obtained. Accuracy mars cannot be given unless the associated method mar is earned (or implied). Mar for a correct result or statement independent of method mars. When a part of a question has two or more "method" steps, the M mars are generally independent unless the scheme specifically says otherwise; and similarly when there are several B mars allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mar is dependent on an earlier M or B (asterised) mar in the scheme. When two or more steps are run together by the candidate, the earlier mars are implied and full credit is given. The symbol implies that the A or B mar indicated is allowed for wor correctly following on from previously incorrect results. Otherwise, A or B mars are given for correct wor only. A and B mars are not given for fortuitously "correct" answers or results obtained from incorrect woring. Note: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. The mars indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mar, allow the candidate the benefit of the doubt. Unless otherwise indicated, mars once gained cannot subsequently be lost, e.g. wrong woring following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mar unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mar if a value is obtained which is correct to s.f., or which would be correct to s.f. if rounded ( d.p. in the case of an angle). As stated above, an A or B mar is not given if a correct numerical answer arises fortuitously from incorrect woring. For Mechanics questions, allow A or B mars for correct answers which arise from taing g equal to 9.8 or 9.8 instead of. Cambridge International Eaminations

13 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 The following abbreviations may be used in a mar scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra checing is needed to ensure that the detailed woring leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) Correct Woring Only often written by a fortuitous' answer Ignore Subsequent Woring Misread Premature Approimation (resulting in basically correct wor that is insufficiently accurate) See Other Solution (the candidate maes a better attempt at the same question) Special Ruling (detailing the mar to be given for a specific wrong solution, or a case where some standard maring practice is to be varied in the light of a particular circumstance) Penalties MR PA A penalty of MR is deducted from A or B mars when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B mars then become "follow through " mars. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B mars in the case of premature approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations

14 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars & 6cos t sin t, y& 6sin t cost dy tant (OE) d d y sec t sec tcosect AG d 6cos t sin t 6 m m ( m ) m t t CF: Ae Bte soi PI : pt qt r & pt q & p p 8pt q pt qt r 7 t p, q, r t t GS: Ae Bte t t n in formula gives a e ae e ae d a a a a a ( e ) e. ae e ae H is true oe d a a a a Assume H is true, i.e. ( ) d d a ( e ) a e a a a d e a d a e ( ) a e a a a a e e a a e a a e H is true, hence by PMI H n is true for all positive integers n.. B B MA [] M A M M A A [6] 6 B B B M A A [6] 6 (i) K 6. 8 MA A [] n 6 (ii) 6 >.9.n.79n.79( > ) n n n >.K so terms required. MA dma [] 7 Cambridge International Eaminations

15 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & α β γ p p Solution Mars B 6 ( α β ) q ( αβ βγ γα) ( α β γ ) γ q ( 8) 7 6 α ( [ β γ ]) α a α 6 α, α, e.g. since > 8 or other reason βγ 7 6 r αβγ (etra answer penalised) i j 6 λ : 8 ~ oe i j λ : ~ oe M A [] M MA B A [] 8 MA A [] 7 7 λ 8 MA [] D, P (or other multiples or permutations). B B Det P (or depending on permutation). B Adj P P ( or other permutations). MA [] Cambridge International Eaminations

16 Page 6 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Cambridge International Eaminations Qn & Solution Mars 7 ~ ~ r(m) t z y t z y E.g. Set z λ and t µ µ λ y and µ λ 9 9, is Basis 9 b a µ λ Solving: λ and µ a, b. MA A M M A [6] M M A A [] 8 ( ) ) )( ( y ) ( < AC B no stationary points 8 6 < > for no stationary points. When : Vertical asymptote: Oblique asymptote: y y Aes and asymptotes Each branch. M A MA A [] B MA B BB [6]

17 Page 7 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars e 9 ln d ln I n (ln ) n.ln d e n e n [(ln ) ( ln ) ] ( n )(ln ). ( ln )d e [ I I ] e n ( )(ln ) (ln )d ( n ) n n n (AG) Alternative for obtaining reduction formula: I n [ (ln ) ] e e e n n n (ln ) d n(ln ) d I n e nin I n e ( n ) In I n nin In ( n ) In I n ( n ) [ In In] (AG) Similarly I I e [] [ ln e ] e (e ) e I I ( [e ]) 6 I I ( ) e B M MA MA [6] MA A B M A [6] B B M A MV I 6 e e e M A [6] Cambridge International Eaminations

18 Page 8 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars (cosθ isinθ) cosθ isinθ ( c is) c c si c s ic s i cs s i c s c s s tan θ c c s cs Divide numerator and denominator by c (stated or shown): tanθ tan θ tan θ tan θ (AG) tan θ tan θ tan θ θ π, π, π, π, π t t t has roots tan π, tan π, tan π, tan π, tanπ t t has roots tan π, tan π, tan π, tan π. t tan π t tan π since tan π tan π and tan π tan π. has roots tan π and tan π. (AG) sec α tan α y y ( y ) ( y ) y y 6 B MA M A [] B B M A [] M M A [] Cambridge International Eaminations

19 Page 9 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Cambridge International Eaminations Qn & Solution Mars E E.g. ~ 8 j i. (AG) ) ( 7 j i j i p Line AP: t r For Q q t t E.g. AB, BQ j i cos cos 8 cos.. cos (AG) MA MA [] B MA MA [] B MA MA []

20 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & O Solution Closed curve through pole with correct orientation. Completely correct. π π a ( cosθ cos θ )dθ a π cosθ cosθ dθ π a θ sinθ sin θ a π π π Mars B B [] MM MA A [] ds dθ a ( cosθ cos θ sin θ) a ( cosθ ) a.sin θ a sin θ (AG) B MA s π a sin θ π d π a cos θ a π θ M A MA [7] Cambridge International Eaminations

21 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mar This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have considered the acceptability of alternative answers. Mar schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademar of Cambridge International Eaminations.

22 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Mar Scheme Notes Mars are of the following three types: M A B Method mar, awarded for a valid method applied to the problem. Method mars are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mar and in some cases an M mar can be implied from a correct answer. Accuracy mar, awarded for a correct answer or intermediate step correctly obtained. Accuracy mars cannot be given unless the associated method mar is earned (or implied). Mar for a correct result or statement independent of method mars. When a part of a question has two or more "method" steps, the M mars are generally independent unless the scheme specifically says otherwise; and similarly when there are several B mars allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mar is dependent on an earlier M or B (asterised) mar in the scheme. When two or more steps are run together by the candidate, the earlier mars are implied and full credit is given. The symbol implies that the A or B mar indicated is allowed for wor correctly following on from previously incorrect results. Otherwise, A or B mars are given for correct wor only. A and B mars are not given for fortuitously "correct" answers or results obtained from incorrect woring. Note: B or A means that the candidate can earn or. B// means that the candidate can earn anything from to. The mars indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mar, allow the candidate the benefit of the doubt. Unless otherwise indicated, mars once gained cannot subsequently be lost, e.g. wrong woring following a correct form of answer is ignored. Wrong or missing units in an answer should not lead to the loss of a mar unless the scheme specifically indicates otherwise. For a numerical answer, allow the A or B mar if a value is obtained which is correct to s.f., or which would be correct to s.f. if rounded ( d.p. in the case of an angle). As stated above, an A or B mar is not given if a correct numerical answer arises fortuitously from incorrect woring. For Mechanics questions, allow A or B mars for correct answers which arise from taing g equal to 9.8 or 9.8 instead of. Cambridge International Eaminations

23 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 The following abbreviations may be used in a mar scheme or used on the scripts: AEF AG BOD CAO CWO ISW MR PA SOS SR Any Equivalent Form (of answer is equally acceptable) Answer Given on the question paper (so etra checing is needed to ensure that the detailed woring leading to the result is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) Correct Answer Only (emphasising that no "follow through" from a previous error is allowed) Correct Woring Only often written by a fortuitous' answer Ignore Subsequent Woring Misread Premature Approimation (resulting in basically correct wor that is insufficiently accurate) See Other Solution (the candidate maes a better attempt at the same question) Special Ruling (detailing the mar to be given for a specific wrong solution, or a case where some standard maring practice is to be varied in the light of a particular circumstance) Penalties MR PA A penalty of MR is deducted from A or B mars when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B mars then become "follow through " mars. MR is not applied when the candidate misreads his own figures this is regarded as an error in accuracy. An MR penalty may be applied in particular cases if agreed at the coordination meeting. This is deducted from A or B mars in the case of premature approimation. The PA penalty is usually discussed at the meeting. Cambridge International Eaminations

24 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars & 6cos t sin t, y& 6sin t cost dy tant (OE) d d y sec t sec tcosect AG d 6cos t sin t 6 m m ( m ) m t t CF: Ae Bte soi PI : pt qt r & pt q & p p 8pt q pt qt r 7 t p, q, r t t GS: Ae Bte t t n in formula gives a e ae e ae d a a a a a ( e ) e. ae e ae H is true oe d a a a a Assume H is true, i.e. ( ) d d a ( e ) a e a a a d e a d a e ( ) a e a a a a e e a a e a a e H is true, hence by PMI H n is true for all positive integers n.. B B MA [] M A M M A A [6] 6 B B B M A A [6] 6 (i) K 6. 8 MA A [] n 6 (ii) 6 >.9.n.79n.79( > ) n n n >.K so terms required. MA dma [] 7 Cambridge International Eaminations

25 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & α β γ p p Solution Mars B 6 ( α β ) q ( αβ βγ γα) ( α β γ ) γ q ( 8) 7 6 α ( [ β γ ]) α a α 6 α, α, e.g. since > 8 or other reason βγ 7 6 r αβγ (etra answer penalised) i j 6 λ : 8 ~ oe i j λ : ~ oe M A [] M MA B A [] 8 MA A [] 7 7 λ 8 MA [] D, P (or other multiples or permutations). B B Det P (or depending on permutation). B Adj P P ( or other permutations). MA [] Cambridge International Eaminations

26 Page 6 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Cambridge International Eaminations Qn & Solution Mars 7 ~ ~ r(m) t z y t z y E.g. Set z λ and t µ µ λ y and µ λ 9 9, is Basis 9 b a µ λ Solving: λ and µ a, b. MA A M M A [6] M M A A [] 8 ( ) ) )( ( y ) ( < AC B no stationary points 8 6 < > for no stationary points. When : Vertical asymptote: Oblique asymptote: y y Aes and asymptotes Each branch. M A MA A [] B MA B BB [6]

27 Page 7 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars e 9 ln d ln I n (ln ) n.ln d e n e n [(ln ) ( ln ) ] ( n )(ln ). ( ln )d e [ I I ] e n ( )(ln ) (ln )d ( n ) n n n (AG) Alternative for obtaining reduction formula: I n [ (ln ) ] e e e n n n (ln ) d n(ln ) d I n e nin I n e ( n ) In I n nin In ( n ) In I n ( n ) [ In In] (AG) Similarly I I e [] [ ln e ] e (e ) e I I ( [e ]) 6 I I ( ) e B M MA MA [6] MA A B M A [6] B B M A MV I 6 e e e M A [6] Cambridge International Eaminations

28 Page 8 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & Solution Mars (cosθ isinθ) cosθ isinθ ( c is) c c si c s ic s i cs s i c s c s s tan θ c c s cs Divide numerator and denominator by c (stated or shown): tanθ tan θ tan θ tan θ (AG) tan θ tan θ tan θ θ π, π, π, π, π t t t has roots tan π, tan π, tan π, tan π, tanπ t t has roots tan π, tan π, tan π, tan π. t tan π t tan π since tan π tan π and tan π tan π. has roots tan π and tan π. (AG) sec α tan α y y ( y ) ( y ) y y 6 B MA M A [] B B M A [] M M A [] Cambridge International Eaminations

29 Page 9 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Cambridge International Eaminations Qn & Solution Mars E E.g. ~ 8 j i. (AG) ) ( 7 j i j i p Line AP: t r For Q q t t E.g. AB, BQ j i cos cos 8 cos.. cos (AG) MA MA [] B MA MA [] B MA MA []

30 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Qn & O Solution Closed curve through pole with correct orientation. Completely correct. π π a ( cosθ cos θ )dθ a π cosθ cosθ dθ π a θ sinθ sin θ a π π π Mars B B [] MM MA A [] ds dθ a ( cosθ cos θ sin θ) a ( cosθ ) a.sin θ a sin θ (AG) B MA s π a sin θ π d π a cos θ a π θ M A MA [7] Cambridge International Eaminations

31 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mar This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have considered the acceptability of alternative answers. Mar schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademar of Cambridge International Eaminations.

32 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find independent equations for T, R A, R B : Resolve horizontally: R B T cos α M A Resolve vertically: R A W T sin α M A Tae moments about A: R B a sin θ W (a/) cos θ (a may be omitted from moment eqns) T a(sin α cos θ cos α sin θ ) or T a sin (α θ ) or T a cos θ sin α M A Tae moments about B: Tae moments about C: Tae moments about D: R A a cos θ W (a/) cos θ T a(sin α cos θ cos α sin θ ) or T a sin (α θ ) or T a sin θ cos α (M A) R A a cos θ W (a/) cos θ R B a sin θ (M A) R A a cos θ W (a/) cos θ R B a sin θ (M A) Solve for T, R A, R B (AEF in W and α): T W / sin α or ½W cosec α B R A W / B R B W / tan α or ½W cot α B 9 9 For A & B use conservation of momentum, e.g.:mv A mv B mu (allow v A v B u) Use Newton s law of restitution (consistent signs): v B v A eu M M Combine to find v A and v B : v A ( e) u/, v B ( e) u/ A, A Find e from v A v B with v B [ ] v B : ( e) 8 ( e), e / M A EITHER: Equate times in terms of reqd. distance : (d )/ v A d/v B /v B (AEF) M A [speeds need not be found: v A v B u/9, v B u/9] Use v A v B v B to solve for : d d, d M A OR: Find dist. moved by A when B reaches wall: d A (d/v B ) v A d (M A) Find reqd. distance : ½ (d d A ) d (M A) Cambridge International Eaminations

33 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find by equating equilibrium tensions: mg (a/)/a mg (a/ a) / a M A (vertical motion can earn M only) ½ /, 6/ or A Apply Newton s law at general point, e.g.: m d /dt mg (a/ )/a (lose A for each incorrect term) mg (a/ a ) / a or m d y/dt mg (a/ y)/a mg (a/ a y) / a M A Simplify to give standard SHM eqn, e.g.: d /dt ( /)g / a S.R.: B if no derivation (ma /) 8g/a A State or find period using π/ω with ω (8g/a): T π (a/8g) or π (a/g) ( on ω) or 8 (a/g) or a [s] B Substitute values in v ω ( ): 7 (8g/a){( a) ( a) } M A Solve to find numerical value of a: 9 (8g/) 7a, a 9 A Cambridge International Eaminations

34 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar EITHER: Find tension at top from F ma vertically: T mu /a mg B OR: Use energy at e.g. θ to upward vertical: ½ mv ½ mu mga ( cos θ ) Find tension T by using F ma radially: T mv /a mg cos θ Eliminate v : mu /a mg ( cos θ ) Find T at top by taing θ : T mu /a mg (B) Find u min by requiring T at top [or T > ]: u /a g so u min ag A.G. B Find v at bottom from conservation of energy: ½mv ½mu mg a M v ag ag, v (ag) A Find new speed V from conservation of momentum: m V mv with m m ¼m M V v/ (ag/) or (/) (ag) AEF A Find w at angle θ from conservation of energy: ½ m w ½ m V (condone m instead of m here since cancels out) m ga ( cos θ ) M A [ w ag (6/ cos θ )] S.R. Invalid energy method (ma /): ½ m w ½ mu [gives T (mg/)( cos θ ) ] mga ( cos θ ) ¼mga ( cos θ ) (B) Find tension T there by using F ma radially: T m w /a m g cos θ B Eliminate w : m V /a m g ( cos θ ) A Substitute for m and V: (mg/)(6/ cos θ ) AEF or mg/ (/) mg cos θ A Find cos θ when string becomes slac from T : cos θ ⅓ 6/ / or M A S.R. Allow if found from T mg (6/ cos θ ) Find or use sample mean 8 / 8 and estimate population variance: s / 9 (allow biased here: or 6 ) 8 or / or 677 M Find confidence interval (allow z in place of t) e.g.: 8 ± t ( 8 / ) M A Use of correct tabular value: t 9,.97 6[] A Evaluate C.I. correct to s.f.: ± 8[] or [ 8, 8] A Cambridge International Eaminations

35 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar 6 Find prob. p of head from mean variance: /p ( p)/p, p ⅔ A.G. M A (i) Find P(X ) (denoting p by q [ ⅓]): P(X ) q p /8 or 7 B (ii) (iii) Find or state P(X > ): P(X > ) [ ( q q q ) p ( q ) ] q /8 or M A Formulate condition for N: q N > 999, [ (⅓) N < ] M Tae logs (any base) to give bound for N: N > log / log ⅓ M Find N min : N > 6 9, N min 7 A (N < 6 9 or N 6 9 earns M A) 8 7 Find F() for : F() ( )/6 B Find G(y) from Y X for : G(y) P(Y < y) P(X < y) P(X < y / ) F(y / ) (result may be stated) (y / )/6 M A Find g(y) for corresponding range of y: g(y) y / / A.G. A (i) (ii) Find or state corresponding range of y : y 6 A.G. B Find median value m of Y: (m / )/6 ½ m / M A Find E(Y) [or equivalently E(X )]: E(Y) y g(y) dy y / dy / [y / ] 6 / / / or 9 7 M A 9 Cambridge International Eaminations

36 Page 6 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar 8 Find mean of sample data [for use in Poisson distn.]: λ / B State (at least) null hypothesis (AEF): H : Poisson distn. fits data or λ B Find epected values λ r e -λ /r! (to d.p.): (ignore incorrect final value here for M) M A Combine last two cells so that ep. value : O i : E i : 7 M* Calculate value of χ (to d.p.; A dep M*): χ (allow 7 9 if d.p. ep.values used) 7 99 M A State or use consistent tabular value (to s.f.): cells: χ, cells: χ, (correct) 7 cells: χ,.9 7 B State or imply valid method for conclusion e.g.: Accept H if χ < tabular value M Conclusion (AEF, requires both values correct): Distn fits or λ A Not combining cells [so χ 8 6] can earn B B M A M M B M (ma 7) 9 Calculate gradient b in y y b ( ) : S y /8 79 S /8 b S y / S 68 ( s.f.) M A Find regression line of y on : y /8 b ( 7/8) M A 68 ( 9) 68 Find y when 7: 7 9 or 8 Allow use of regression line of on y (since neither variable clearly independent): S yy 6 /8 6 b S y / S yy (M A) 7/8 b (y /8) (M A) y 6 8 Y 6 or 6 (A) Cambridge International Eaminations

37 Page 7 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find product moment correlation coefficient r: r 79 / ( 6) or ( 68 ) 797 M A* State both hypotheses (B for r ): H : ρ, H : ρ B State or use correct tabular two-tail r-value: r 8, % 77 B* State or imply valid method for conclusion e.g.: Reject H if r > tab. value (AEF) M Correct conclusion (AEF, dep A*, B*): There is non-zero correlation A 6 A Find MI of lamina about Q: I lamina ⅓m{(a) (a/) } m(9a/) M A [ (/ 8/) ma ma ] State or find MI of rod about Q: I rod (⅓ ) M (a/) [ Ma ] B Sum to find MI of object about Q: I ma Ma (8m M) a A.G. A Find MI of object about mid-point of PQ: I (/ ) ma ⅓ M (a/) Use eqn of circular motion to find d θ/dt for ais l : [ ]I d θ/dt (/) ma ¾ Ma ¾ (7m M) a A.G. M A mg (9a/) sin θ Mg (a/) sin θ M A [ (9m/ M/) ga sin θ ] [Approimate sin θ by θ and] find ω in SHM eqn: ω (m M)g / (8m M) a M Find period T for ais l from π/ω : (AEF) T π {(8m M) a / (m M)g} A Use eqn of circular motion to find d θ/dt for ais l : [ ]I d θ/dt mg a sin θ M [Approimate sin θ by θ and] find ω in SHM eqn: ω mg / (7m M) a M Find period T for ais l from π/ω : (AEF) T π {(7m M) a / mg} A Verify that T T when m M: (AEF) T π (8 a / g) T B [Taing m M throughout nd part can earn: M A M A M M A B (ma 6/8)] 8 Cambridge International Eaminations

38 Page 8 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar B State hypotheses (B for ), e.g.: H : µ X µ Y, H : µ X µ Y B State assumption (AEF): Distributions have equal variances B Find sample means and estimate popln. variances:, y 8 s X (8 /) / 9 (allow biased here: 6 or 6 ) or 6 s Y (8 7 6 /) / (allow biased here: 8 or 668 ) 6 or / or 67 M Estimate (pooled) common variance: s (9 s X s Y ) / (AEF) (note s X and s Y not needed eplicitly) or (8 / /) / or 66 M A Calculate value of t (to s.f.): [-] t ( y ) / s (/ /) M A State or use correct tabular t value: t, [allow 9] B* (or can compare y 6 with 86) Correct conclusion (AEF, on t, dep *B): t > 9 so mean masses not same B S.R. Implicitly taing s X, s Y as popln. variances: z ( y ) / (s X / s Y /) (may also earn first B B M) 6 / ( 78 Comparison with z.97 and conclusion: > 96 (can earn at most /9) so mean masses not same (B) 9 State hypotheses (B for ), e.g.: H : µ X 8, H : µ X > 8 or H : µ X µ Z, H : µ X > µ Z B Calculate value of t using s X from above: t ( 8) / (s X / ) M A State or use correct tabular t value: t 9,.9 8 [allow 8] B* (or can compare with 67) Correct conclusion (A.E.F., on t, dep *B): t > 8, so claim is justified or mean mass of Royals > mean mass of Crowns B Cambridge International Eaminations

39 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mar This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have considered the acceptability of alternative answers. Mar schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademar of Cambridge International Eaminations.

40 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find independent equations for T, R A, R B : Resolve horizontally: R B T cos α M A Resolve vertically: R A W T sin α M A Tae moments about A: R B a sin θ W (a/) cos θ (a may be omitted from moment eqns) T a(sin α cos θ cos α sin θ ) or T a sin (α θ ) or T a cos θ sin α M A Tae moments about B: Tae moments about C: Tae moments about D: R A a cos θ W (a/) cos θ T a(sin α cos θ cos α sin θ ) or T a sin (α θ ) or T a sin θ cos α (M A) R A a cos θ W (a/) cos θ R B a sin θ (M A) R A a cos θ W (a/) cos θ R B a sin θ (M A) Solve for T, R A, R B (AEF in W and α): T W / sin α or ½W cosec α B R A W / B R B W / tan α or ½W cot α B 9 9 For A & B use conservation of momentum, e.g.:mv A mv B mu (allow v A v B u) Use Newton s law of restitution (consistent signs): v B v A eu M M Combine to find v A and v B : v A ( e) u/, v B ( e) u/ A, A Find e from v A v B with v B [ ] v B : ( e) 8 ( e), e / M A EITHER: Equate times in terms of reqd. distance : (d )/ v A d/v B /v B (AEF) M A [speeds need not be found: v A v B u/9, v B u/9] Use v A v B v B to solve for : d d, d M A OR: Find dist. moved by A when B reaches wall: d A (d/v B ) v A d (M A) Find reqd. distance : ½ (d d A ) d (M A) Cambridge International Eaminations

41 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find by equating equilibrium tensions: mg (a/)/a mg (a/ a) / a M A (vertical motion can earn M only) ½ /, 6/ or A Apply Newton s law at general point, e.g.: m d /dt mg (a/ )/a (lose A for each incorrect term) mg (a/ a ) / a or m d y/dt mg (a/ y)/a mg (a/ a y) / a M A Simplify to give standard SHM eqn, e.g.: d /dt ( /)g / a S.R.: B if no derivation (ma /) 8g/a A State or find period using π/ω with ω (8g/a): T π (a/8g) or π (a/g) ( on ω) or 8 (a/g) or a [s] B Substitute values in v ω ( ): 7 (8g/a){( a) ( a) } M A Solve to find numerical value of a: 9 (8g/) 7a, a 9 A Cambridge International Eaminations

42 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar EITHER: Find tension at top from F ma vertically: T mu /a mg B OR: Use energy at e.g. θ to upward vertical: ½ mv ½ mu mga ( cos θ ) Find tension T by using F ma radially: T mv /a mg cos θ Eliminate v : mu /a mg ( cos θ ) Find T at top by taing θ : T mu /a mg (B) Find u min by requiring T at top [or T > ]: u /a g so u min ag A.G. B Find v at bottom from conservation of energy: ½mv ½mu mg a M v ag ag, v (ag) A Find new speed V from conservation of momentum: m V mv with m m ¼m M V v/ (ag/) or (/) (ag) AEF A Find w at angle θ from conservation of energy: ½ m w ½ m V (condone m instead of m here since cancels out) m ga ( cos θ ) M A [ w ag (6/ cos θ )] S.R. Invalid energy method (ma /): ½ m w ½ mu [gives T (mg/)( cos θ ) ] mga ( cos θ ) ¼mga ( cos θ ) (B) Find tension T there by using F ma radially: T m w /a m g cos θ B Eliminate w : m V /a m g ( cos θ ) A Substitute for m and V: (mg/)(6/ cos θ ) AEF or mg/ (/) mg cos θ A Find cos θ when string becomes slac from T : cos θ ⅓ 6/ / or M A S.R. Allow if found from T mg (6/ cos θ ) Find or use sample mean 8 / 8 and estimate population variance: s / 9 (allow biased here: or 6 ) 8 or / or 677 M Find confidence interval (allow z in place of t) e.g.: 8 ± t ( 8 / ) M A Use of correct tabular value: t 9,.97 6[] A Evaluate C.I. correct to s.f.: ± 8[] or [ 8, 8] A Cambridge International Eaminations

43 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar 6 Find prob. p of head from mean variance: /p ( p)/p, p ⅔ A.G. M A (i) Find P(X ) (denoting p by q [ ⅓]): P(X ) q p /8 or 7 B (ii) (iii) Find or state P(X > ): P(X > ) [ ( q q q ) p ( q ) ] q /8 or M A Formulate condition for N: q N > 999, [ (⅓) N < ] M Tae logs (any base) to give bound for N: N > log / log ⅓ M Find N min : N > 6 9, N min 7 A (N < 6 9 or N 6 9 earns M A) 8 7 Find F() for : F() ( )/6 B Find G(y) from Y X for : G(y) P(Y < y) P(X < y) P(X < y / ) F(y / ) (result may be stated) (y / )/6 M A Find g(y) for corresponding range of y: g(y) y / / A.G. A (i) (ii) Find or state corresponding range of y : y 6 A.G. B Find median value m of Y: (m / )/6 ½ m / M A Find E(Y) [or equivalently E(X )]: E(Y) y g(y) dy y / dy / [y / ] 6 / / / or 9 7 M A 9 Cambridge International Eaminations

44 Page 6 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar 8 Find mean of sample data [for use in Poisson distn.]: λ / B State (at least) null hypothesis (AEF): H : Poisson distn. fits data or λ B Find epected values λ r e -λ /r! (to d.p.): (ignore incorrect final value here for M) M A Combine last two cells so that ep. value : O i : E i : 7 M* Calculate value of χ (to d.p.; A dep M*): χ (allow 7 9 if d.p. ep.values used) 7 99 M A State or use consistent tabular value (to s.f.): cells: χ, cells: χ, (correct) 7 cells: χ,.9 7 B State or imply valid method for conclusion e.g.: Accept H if χ < tabular value M Conclusion (AEF, requires both values correct): Distn fits or λ A Not combining cells [so χ 8 6] can earn B B M A M M B M (ma 7) 9 Calculate gradient b in y y b ( ) : S y /8 79 S /8 b S y / S 68 ( s.f.) M A Find regression line of y on : y /8 b ( 7/8) M A 68 ( 9) 68 Find y when 7: 7 9 or 8 Allow use of regression line of on y (since neither variable clearly independent): S yy 6 /8 6 b S y / S yy (M A) 7/8 b (y /8) (M A) y 6 8 Y 6 or 6 (A) Cambridge International Eaminations

45 Page 7 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find product moment correlation coefficient r: r 79 / ( 6) or ( 68 ) 797 M A* State both hypotheses (B for r ): H : ρ, H : ρ B State or use correct tabular two-tail r-value: r 8, % 77 B* State or imply valid method for conclusion e.g.: Reject H if r > tab. value (AEF) M Correct conclusion (AEF, dep A*, B*): There is non-zero correlation A 6 A Find MI of lamina about Q: I lamina ⅓m{(a) (a/) } m(9a/) M A [ (/ 8/) ma ma ] State or find MI of rod about Q: I rod (⅓ ) M (a/) [ Ma ] B Sum to find MI of object about Q: I ma Ma (8m M) a A.G. A Find MI of object about mid-point of PQ: I (/ ) ma ⅓ M (a/) Use eqn of circular motion to find d θ/dt for ais l : [ ]I d θ/dt (/) ma ¾ Ma ¾ (7m M) a A.G. M A mg (9a/) sin θ Mg (a/) sin θ M A [ (9m/ M/) ga sin θ ] [Approimate sin θ by θ and] find ω in SHM eqn: ω (m M)g / (8m M) a M Find period T for ais l from π/ω : (AEF) T π {(8m M) a / (m M)g} A Use eqn of circular motion to find d θ/dt for ais l : [ ]I d θ/dt mg a sin θ M [Approimate sin θ by θ and] find ω in SHM eqn: ω mg / (7m M) a M Find period T for ais l from π/ω : (AEF) T π {(7m M) a / mg} A Verify that T T when m M: (AEF) T π (8 a / g) T B [Taing m M throughout nd part can earn: M A M A M M A B (ma 6/8)] 8 Cambridge International Eaminations

46 Page 8 Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar B State hypotheses (B for ), e.g.: H : µ X µ Y, H : µ X µ Y B State assumption (AEF): Distributions have equal variances B Find sample means and estimate popln. variances:, y 8 s X (8 /) / 9 (allow biased here: 6 or 6 ) or 6 s Y (8 7 6 /) / (allow biased here: 8 or 668 ) 6 or / or 67 M Estimate (pooled) common variance: s (9 s X s Y ) / (AEF) (note s X and s Y not needed eplicitly) or (8 / /) / or 66 M A Calculate value of t (to s.f.): [-] t ( y ) / s (/ /) M A State or use correct tabular t value: t, [allow 9] B* (or can compare y 6 with 86) Correct conclusion (AEF, on t, dep *B): t > 9 so mean masses not same B S.R. Implicitly taing s X, s Y as popln. variances: z ( y ) / (s X / s Y /) (may also earn first B B M) 6 / ( 78 Comparison with z.97 and conclusion: > 96 (can earn at most /9) so mean masses not same (B) 9 State hypotheses (B for ), e.g.: H : µ X 8, H : µ X > 8 or H : µ X µ Z, H : µ X > µ Z B Calculate value of t using s X from above: t ( 8) / (s X / ) M A State or use correct tabular t value: t 9,.9 8 [allow 8] B* (or can compare with 67) Correct conclusion (A.E.F., on t, dep *B): t > 8, so claim is justified or mean mass of Royals > mean mass of Crowns B Cambridge International Eaminations

47 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November series 9 FURTHER MATHEMATICS 9/ Paper, maimum raw mar This mar scheme is published as an aid to teachers and candidates, to indicate the requirements of the eamination. It shows the basis on which Eaminers were instructed to award mars. It does not indicate the details of the discussions that too place at an Eaminers meeting before maring began, which would have considered the acceptability of alternative answers. Mar schemes should be read in conjunction with the question paper and the Principal Eaminer Report for Teachers. Cambridge will not enter into discussions about these mar schemes. Cambridge is publishing the mar schemes for the October/November series for most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademar of Cambridge International Eaminations.

48 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find independent equations for T, R A, R B : Resolve horizontally: R B T cos α M A Resolve vertically: R A W T sin α M A Tae moments about A: R B a sin θ W (a/) cos θ (a may be omitted from moment eqns) T a(sin α cos θ cos α sin θ ) or T a sin (α θ ) or T a cos θ sin α M A Tae moments about B: Tae moments about C: Tae moments about D: R A a cos θ W (a/) cos θ T a(sin α cos θ cos α sin θ ) or T a sin (α θ ) or T a sin θ cos α (M A) R A a cos θ W (a/) cos θ R B a sin θ (M A) R A a cos θ W (a/) cos θ R B a sin θ (M A) Solve for T, R A, R B (AEF in W and α): T W / sin α or ½W cosec α B R A W / B R B W / tan α or ½W cot α B 9 9 For A & B use conservation of momentum, e.g.:mv A mv B mu (allow v A v B u) Use Newton s law of restitution (consistent signs): v B v A eu M M Combine to find v A and v B : v A ( e) u/, v B ( e) u/ A, A Find e from v A v B with v B [ ] v B : ( e) 8 ( e), e / M A EITHER: Equate times in terms of reqd. distance : (d )/ v A d/v B /v B (AEF) M A [speeds need not be found: v A v B u/9, v B u/9] Use v A v B v B to solve for : d d, d M A OR: Find dist. moved by A when B reaches wall: d A (d/v B ) v A d (M A) Find reqd. distance : ½ (d d A ) d (M A) Cambridge International Eaminations

49 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar Find by equating equilibrium tensions: mg (a/)/a mg (a/ a) / a M A (vertical motion can earn M only) ½ /, 6/ or A Apply Newton s law at general point, e.g.: m d /dt mg (a/ )/a (lose A for each incorrect term) mg (a/ a ) / a or m d y/dt mg (a/ y)/a mg (a/ a y) / a M A Simplify to give standard SHM eqn, e.g.: d /dt ( /)g / a S.R.: B if no derivation (ma /) 8g/a A State or find period using π/ω with ω (8g/a): T π (a/8g) or π (a/g) ( on ω) or 8 (a/g) or a [s] B Substitute values in v ω ( ): 7 (8g/a){( a) ( a) } M A Solve to find numerical value of a: 9 (8g/) 7a, a 9 A Cambridge International Eaminations

50 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar EITHER: Find tension at top from F ma vertically: T mu /a mg B OR: Use energy at e.g. θ to upward vertical: ½ mv ½ mu mga ( cos θ ) Find tension T by using F ma radially: T mv /a mg cos θ Eliminate v : mu /a mg ( cos θ ) Find T at top by taing θ : T mu /a mg (B) Find u min by requiring T at top [or T > ]: u /a g so u min ag A.G. B Find v at bottom from conservation of energy: ½mv ½mu mg a M v ag ag, v (ag) A Find new speed V from conservation of momentum: m V mv with m m ¼m M V v/ (ag/) or (/) (ag) AEF A Find w at angle θ from conservation of energy: ½ m w ½ m V (condone m instead of m here since cancels out) m ga ( cos θ ) M A [ w ag (6/ cos θ )] S.R. Invalid energy method (ma /): ½ m w ½ mu [gives T (mg/)( cos θ ) ] mga ( cos θ ) ¼mga ( cos θ ) (B) Find tension T there by using F ma radially: T m w /a m g cos θ B Eliminate w : m V /a m g ( cos θ ) A Substitute for m and V: (mg/)(6/ cos θ ) AEF or mg/ (/) mg cos θ A Find cos θ when string becomes slac from T : cos θ ⅓ 6/ / or M A S.R. Allow if found from T mg (6/ cos θ ) Find or use sample mean 8 / 8 and estimate population variance: s / 9 (allow biased here: or 6 ) 8 or / or 677 M Find confidence interval (allow z in place of t) e.g.: 8 ± t ( 8 / ) M A Use of correct tabular value: t 9,.97 6[] A Evaluate C.I. correct to s.f.: ± 8[] or [ 8, 8] A Cambridge International Eaminations

51 Page Mar Scheme Syllabus Paper Cambridge International A Level October/November 9 Question Number Mar Scheme Details Mar 6 Find prob. p of head from mean variance: /p ( p)/p, p ⅔ A.G. M A (i) Find P(X ) (denoting p by q [ ⅓]): P(X ) q p /8 or 7 B (ii) (iii) Find or state P(X > ): P(X > ) [ ( q q q ) p ( q ) ] q /8 or M A Formulate condition for N: q N > 999, [ (⅓) N < ] M Tae logs (any base) to give bound for N: N > log / log ⅓ M Find N min : N > 6 9, N min 7 A (N < 6 9 or N 6 9 earns M A) 8 7 Find F() for : F() ( )/6 B Find G(y) from Y X for : G(y) P(Y < y) P(X < y) P(X < y / ) F(y / ) (result may be stated) (y / )/6 M A Find g(y) for corresponding range of y: g(y) y / / A.G. A (i) (ii) Find or state corresponding range of y : y 6 A.G. B Find median value m of Y: (m / )/6 ½ m / M A Find E(Y) [or equivalently E(X )]: E(Y) y g(y) dy y / dy / [y / ] 6 / / / or 9 7 M A 9 Cambridge International Eaminations