Mark Scheme (Results) Summer GCE Core Mathematics 3 (6665/01R)

Transcription

1 Mark Scheme (Results) Summer GCE Core Mathematics (6665/R)

2 Question Number Scheme Marks. (a) + ( + 4)( ) B Attempt as a single fraction (+ 5)( ) ( + ) ( + )( ) or + 5 ( + 4) M ( + 4)( ) ( + 4)( ), ( + 4) cao A, A B M A A (4 marks) Notes for Question For correctly factorising + ( + 4)( ). It could appear anywhere in their solution For an attempt to combine two fractions. The denominator must be correct for their fractions. The terms could be separate but one term must have been modified. Condone invisible brackets. Eamples of work scoring this mark are; (+ 5)( ) ( + ) Two separate terms ( + )( ) ( + )( ) Single term, invisible bracket ( + 4)( ) (+ 5) ( + ) Separate terms, only one numerator modified ( + )( ) ( + )( ) Correct un simplified answer ( + 4)( ) 6 9 If scored M the fraction must be subsequently be reduced to a correct or ( + )( ) + ( )( ) to score this mark. ( + 4)( )( ) cao ( + 4) Do Not isw in this question. The method of partial fractions is perfectly acceptable and can score full marks ( + 4)( ) B MA A

3 Question Number Scheme Marks.(a) Shape B (.5, ) B () (b) Shape B (,) B Cusp at (,) B () (5 marks)

4 Notes for Question (a) B Award for the correct shape. Look for an increasing function with decreasing gradient. Condone linear looking functions in the first quadrant. It needs to look asymptotic at the y ais and have no obvious maimum point. It must be wholly contained in quadrants and 4 See practice and qualification items for clarification. B Crosses ais at,. Accept,.5 or even, There must be a graph for this mark to be scored. marked on the correct ais. (b) B Correct shape wholly contained in quadrant. The shape to the rhs of the cusp must not have an obvious maimum. Accept linear, or positive with decreasing gradient. The gradient of the curve to the lhs of the cusp/minimum should always be negative. The curve in this section should not bend back past (, ) forming a C shape or have incorrect curvature. See practice and qualification for clarification. B The curve touches or crosses the ais at (, ). Allow for the curve passing through a point marked on the ais. Condone the point marked on the correct ais as (, ) B Award for a cusp, not a minimum at (,) Note that f ( ) scores B B B under the scheme. If the graphs are not labelled (a) and (b), then they are to be marked in the order they are presented

5 Question Number Scheme Marks.(a) 7cos + sin Rcos( α) (b) R (7 + ) 5 (5 ) B α arctan 8... awrt 8. 7 MA 5 5 cos( 8.) 5 cos( 8.) M M,A () AND MA (5) k (c ) One solution if ±, k ± 5 ft on R MAft 5 () ( marks)

6 Notes for Question (a) B R 5. Accept 5 Accept R ± 5 Do not accept R (7 + ) or the decimal equivalent 7.7 unless you see 5 or5 as well M 7 For tanα ± or tanα ±. Condone if this comes from cosα 7, sinα 7 7 If R is used then only accept sinα ± or cosα ± R R A α awrt 8.. Be aware that tan α 7 α 8. 9 can easily be mistaken for the correct answer Note that the radian answer awrt.48 is A (b) M For using their answers to part (a) and moving from 5 Rcos( ± α) 5 cos( ± α) using their R numerical values of R and α This may be implied for sight of 5. if R and α were correct M For achieving ± α awrt 45 or 5, leading to one value of in the range Note that for this to be scored R has to be correct (to sf) as awrt 45, 5 must be achieved This may be implied for achieving an answer of either 45 + their α or 5 + their α A One correct answer, either awrt 5.⁰ or.⁰ M For an attempt at finding a secondary value of in the range. Usually this is an attempt at solving their 8. 6 their 45.. A Both values correct awrt 5.⁰ and.⁰. Withhold this mark if there are etra values in the range. Ignore etra values outside the range (c ) k M For stating that their R OR k their R This may be implied by seeing k ( ± ) their R Aft Both values k ± 5 oe. Follow through on their numerical R Answers all in radians. Lose the first time that it appears but demand an accuracy of dp. Part (a) R 5 α awrt.4 Part (b) awrt.97, Accuracy must be to sf. With correct working this would score (a) BMA (b) MAAMA Mied degrees and radians refer to the main scheme

7 Question Number 4.(a) f( ) Scheme Marks MA () (b) An attempt to find 4 + when M Correct answer fg() 5 A (c ) y y 4 4 y 4 M g ( ) 4 A () () (d) [ ] g ( ) ( 4 ) B gg( ) 4( 4 ) M [ ] gg( ) + g( ) A 8 ( ),.5 oe MA (5 ) ( marks)

8 Notes for Question 4 (a) M Attempt at calculating f at. Sight of is sufficient. Accept f( ) > and > for M, A f( ). Accept y, range, [, ) Do not accept f( ) >, The correct answer is sufficient for both marks. (b) M A full method of finding fg(). The order of substituting into the epressions must be correct and + must be used as opposed to + Accept an attempt to calculate + when -. Accept an attempt to put into 4 and then substituting their answer to 4 into + Do not accept the substitution of into +, followed by their result into -4 This is evidence of incorrect order. A fg()5. 4 Watch for + 5which is MA (c ) M Award for an attempt to make or a swapped y the subject of the formula. It must be a full method and cannot finish 4.. You can condone at most one arithmetic error for this method mark. + y y y is fine for the M as there is only one error 4 y 4 4 y yis fine for the M as there is only one error 4 y y + y is M as there are two arithmetic errors 4 A Obtaining a correct epression g ( ) oesuch as g ( ),g ( ) It must be in terms of, but could be epressed y or g ( ) (d) g ( ) ( 4 ). If only the epanded version appears it must be correct B Sight of [ ] M A full attempt to find gg( ) 4( 4 ) Condone invisible brackets. Note that it may appear in an equation A 6 8 Accept other alternatives such as M For factorising their quadratic or cancelling their A Bby to get value of If they have a TQ then usual methods are applicable. A Both values correct,.5 oe

9 Question Number Scheme Marks 5.(a) (b) d (cos ) sin d B vu ' uv' cos sin cos Applies to v ( ) MA sin cos d (sec ) sec sec tan d ( 6sec tan ) M 6( + tan ) tan dm 6(tan + tan ) A () (c) d sin cos dy MA dy d y cos y sin dm dy d 4 cao A () (4) Alt 5(c ) dy y dy d 4 arcsin d M Rearranging to y AarcsinBand differentiating to dm As above, but form of the rhs must be correct A Correct but un simplified answer dy d dy A d B C ( marks) MdMA A (4)

10 Notes for Question 5 (a) B Award for the sight of d (cos ) sin. This could be seen in their differential. d vu ' uv' cos M Applies to v 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 by their working, with terms written out u,u.,v.,v.followed vu ' uv' by their ) then only accept answers of the form v ± Asin cos B 4 ( ) or A Award for a correct answer. This does not need to be simplified. Alt (a) using the product rule B Award for the sight of d (cos ) sin. This could be seen in their differential. d M Applies vu ' + uv' to cos. 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 by their working, with terms written out u,u.,v.,v.followed by their vu ' + uv' ) then only accept answers of the form ± A sin B cos A Award for a correct answer. This does not need to be simplified. sin cos (b) M dm Award for a correct application of the chain rule on Sight of Csecsectan is sufficient sec Replacing sec + tan in their derivative to create an epression in just tan. It is dependent upon the first M being scored. A The correct answer 6(tan + tan ). There is no need to write µ 6 Alt (b) using the product rule M Writes sec as sec sec and uses the product rule with u' Asectan and v' Bsectanto produce a derivative of the form Asectan sec+ Bsectan sec dm Replaces sec with+ tan to produce an epression in just tan. It is dependent upon the first M being scored.

11 A The correct answer 6(tan tan ) Notes for Question 5 Continued +. There is no need to write µ 6 Alt (b) using sec cos and proceeding by the chain or quotient rule M Writes sec as (cos ) and differentiates to A(cos ) sin Alternatively writes sec as (cos ) and achieves (cos ) Acossin ( cos ) dm Uses sin tan cos and sec cos and sec + tan in their derivative to create an epression in just tan. It is dependent upon the first M being scored. A The correct answer +. There is no need to write µ 6 6(tan tan ) Alt (b) using sec + tan M dm Writes sec as + tan and uses chain rule to produce a derivative of the form Atan sec or the product rule to produce a derivative of the form Ctan sec + Dtan sec Replaces sec + tan to produce an epression in just tan. It is dependent upon the first M being scored. A The correct answer +. There is no need to write µ 6 6(tan tan ) (c) y y M Award for knowing the method that sin differentiates to cos y y correct/present. Award for sin Acos A dm The lhs does not need to be d sin y cos y. Both sides must be correct dy Award for inverting their d dy and using y y sin + cos to produce an epression for d y in terms of d only. It is dependent upon the first M being scored. An alternative to Pythagoras is a triangle. sin y cos y 4 y/

12 Notes for Question 5 Continued Candidates who write d y do not score the mark. d cos arcsin BUT d y d sin arcsin does score M as they clearly use a correct Pythagorean identity as required by the notes. A dy d 4. Epression must be in its simplest form. Do not accept dy d 4 or dy d 4 for the final A

13 Question Number 6.(i) (ii) Scheme cosec sin M sin cos M cosec sec λ A sec θ secθ tan θ sec θ secθ (sec θ ) + + M Marks () sec θ + secθ + (secθ + )(secθ + ) M secθ, A cosθ.5, M π 4π θ,, π AA (6) (9 marks) ALT (ii) sin θ cos θ cosθ cos θ sec θ + secθ tan θ + + cosθ sin θ cosθ ( cos θ ) + M cos θ + cosθ + ( cosθ + )(cosθ + ) cosθ.5, MA π 4π θ,, π M,A,A (6) (9 marks)

14 Notes for Question 6 (i) M Uses the identity cosec sin M Uses the correct identity for sin sin cos in their epression. Accept sin sin cos + cos sin A λ following correct working (ii) M Replaces tan θ by sec θ ± ± to produce an equation in just secθ M Award for a forming a TQ in secθ and applying a correct method for factorising, or using the formula, or completing the square to find two answers to secθ If they replace secθ it is for forming a TQ in cosθ and applying a correct method for finding two cosθ answers to cosθ A Correct answers to secθ, or cos θ, M Award for using the identity secθ and proceeding to find at least one value for θ. cosθ If the TQ was in cosine then it is for finding at least one value of θ. A Two correct values of θ. All method marks must have been scored. Accept two of, 8, 4 or two of π, 4π, A All three answers correct. They must be given in terms ofπ as stated in the question... Accept 6. π,. π, π Withhold this mark if further values in the range are given. All method marks must have been scored. Ignore any answers outside the range. Alt (ii) M Award for replacing sec θ with cos θ, secθ with cosθ, sin θ tan θ with multiplying through by cos θ cos θ (seen in at least terms) and replacing sin θ with ± ± cos θ to produce an equation in just cosθ M Award for a forming a TQ in cosθ and applying a correct method for factorising, or using the formula, or completing the square to find two answers to cosθ A cos θ, M Proceeding to finding at least one value of θ from an equation in cosθ. A Two correct values of θ. All method marks must have been scored Accept two of, 8, 4 or two of π, 4π, A All three answers correct. They must be given in terms ofπ as stated in the question.

15 .. Notes for Question 6 Continued Accept 6. π,. π, π All method marks must have been scored. Withhold this mark if further values in the range are given. Ignore any answers outside the range

16 Question Number 7.(a) f Scheme ( ) + + Marks ± 5 awrt -.8, -.68 MA () (b) Uses vu ' + uv' f'( ) ( ) ( ) e e MAA () (c ) e e ( + ) + ( + + ) { } e M ( + 4) ( + ) M ( ) ( ) + + A* () (d) Sub (n + ).4 into n+ ( n + ) awrt.4, awrt.47 awrt.4 MA,A () (e) Sub -.45 and -.45 into f'( ) and start to compare signs f '(.45) +.4, f '(.45) 5. M Change in sign, hence f'( ) in between. Therefore α -.4 (dp) A () ( marks) Alt 7.(c) ( + ) ( + ) { } ( + ) ( + ) when f'( ) e 6 4 ( + ) Hence the minimum point occurs when ( + 4) M M A

17 Question Number Alt 7(e) Scheme Sub -.45 and -.45 into cubic part of f'( ) and start to compare signs Adapted f '(.45) +.6, f '(.45).4 M Marks Change in sign, hence f'( ) in between. Therefore α -.4 (dp) A () Alt 7 (e) Sub -.45, -.4 and -.45 into f( ) ( + + ) e and start to compare sizes f (.45) 4., f (.45) 4., f (.4) 4. M f (.4) < f (.45), f (.4) < f (.45). Therefore α -.4 (dp) A () Notes for Question 7 (a) M Solves + + by completing the square or the formula, producing two non integer answers. Do not accept factorisation here. Accept awrt -.4 and -.6 for this mark A Answers correct. Accept awrt -.8, Accept just the answers for both marks. Don t withhold the marks for incorrect labelling. (b) M Applies the product rule vu ' + uv ' to ( + + ) e. 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 dy f'( ) e ( A + B) + ( + + ) Ce d A One term of f'( ) e (+ ) + ( + + ) e correct. There is no need to simplify A A fully correct (un simplified) answer f'( ) e (+ ) + ( + + ) e (c ) M Sets their f'( ) and either factorises out, or cancels by e to produce a polynomial equation in M Rearranges the cubic polynomial to A + B C + Dand factorises to reach ( A + B) C + D or equivalent ( + ) A* Correctly proceeds to. This is a given answer ( + )

18 Notes on Question 7 Continued (c) Alternative to (c) working backwards ( + ) M Moves correctly from to ( + ) M States or implies that f'( ) A Makes a conclusion to tie up the argument For eample, hence the minimum point occurs when ( + ) ( + 4) (d) M Sub n.4 into n+ n ( + ) ( + ) (.4 + ) This may be implied by awrt -.4, or n + (.4 + ) A Awrt..4. The subscript is not important. Mark as the first value given A awrt.47 awrt.4 The subscripts are not important. Mark as the second and third values given (e) Note that continued iteration is not allowed M Sub -.45 and -.45 into f'( ), starts to compare signs and gets at least one correct to sf rounded or truncated. A Both values correct (sf rounded or truncated), a reason and a minimal conclusion Acceptable reasons are change in sign, positive and negative and f'( a) f'( b) < Minimal conclusions are hence α.4, hence shown, hence root Alt using adapted f'( ) (e) M Sub -.45 and -.45 into cubic part of f'( ), starts to compare signs and gets at least one correct to sf rounded or truncated. A Both values correct of adapted f'( ) correct (sf rounded or truncated), a reason and a minimal conclusion Acceptable reasons are change in sign, positive and negative and f'( a) f'( b) < Minimal conclusions are hence α.4, hence shown, hence root Alt using f( ) (e) M Sub -.45, -.4 and -.45 into f( ), starts to compare sizes and gets at least one correct to 4sf rounded A All three values correct of f( ) correct (4sf rounded ), a reason and a minimal conclusion Acceptable reasons are f (.4) < f (.45), f (.4) < f (.45), a sketch Minimal conclusions are hence α.4, hence shown, hence root

19 Question Number 8.(a) (b) Scheme 8 t P + 7 cao Marks MA 8 t P 8 B 8 (c ) t, P k B 8 (d) Sub t into P P t + 7e e k. e (...) oe M,A 7. k ln awrt.86 MA 7 cao MA dp 8 kt (e) 7ke kt M,A d t (+ 7 e ) dp Sub t 46 A dt t () () (5) () () ( marks)

20 (a) M Sets t, giving k e Notes for Question 8. Award if candidate attempts 8 8, A Correct answer only. Accept for both marks as long as no incorrect working is seen. (b) B 8. Accept P < 8. Condone P 8 but not P > 8 (c ) 8 B Sets both t, and P 5 5 k + 7e This may be implied by a subsequent correct line. k M Rearranges the equation to make the subject. They need to multiply by the ± k proceed to e A, A> e ± k + 7e term, and k. A The correct intermediate answer of e, or equivalent. Accept awrt k 5 Alternatively accept e,.8.. or equivalent. ± k M Proceeds from e A, A> by correctly taking ln s and then making k the subject of the formula. k ln( A) Award for e A k ln( A) k k k ln C If e was found accept e C k lnc k As with method, C > A Awrt k.86 dp (d) M 8 Substitutes t into P 7e with their numerical value of k to find P A ( P )697 or other eact equivalents like 697. (e) dp C M Differentiates using the chain rule to a form e kt d t (+ 7 e ) kt (+ 7 e ) C e Accept an application of the quotient rule to achieve kt (+ 7 e ) dp 8 kt A A correct un simplified 7ke. kt d t (+ 7 e ) The derivative can be given in terms of k. If a numerical value is used you may follow through on incorrect values. A Awrt 46. Note that M must have been achieved. Just the answer scores kt kt