Mark Scheme (Results) January Pearson Edexcel International Advanced Level Core Mathematics C34 (WMA02/01) January 2014 (IAL)

Transcription

1 January 1 (IAL) Mark Scheme (Results) January 1 Pearson Edecel International Advanced Level Core Mathematics C (WMA/1)

2 January 1 (IAL) Edecel and BTEC Qualifications Edecel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 15 years, and by working across 7 countries, in 1 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: January 1 Publications Code IA766 All the material in this publication is copyright Pearson Education Ltd 1

3 January 1 (IAL) 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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 January 1 (IAL) EDEXCEL GCE MATHEMATICS General Instructions for Marking 1. 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) 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 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. 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.

5 January 1 (IAL) General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic: 1. 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 the correct formula (with values for a, b and c).. Completing the square b : ± ± q ± c =, q Solving + b + c = Method marks for differentiation and integration: 1. Differentiation, leading to = Power of at least one term decreased by 1. ( n n 1 ). Integration Power of at least one term increased by 1. ( n n+1 ) 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 January 1 (IAL) Question Number Scheme Marks 1. ( ) 6 f( ) f '( ) + = = = + ( + ) ( + ) 6 f '( ) > > ( + ) Critical values 6 = =± Inside region chosen < < d (6 marks) Notes Applies the Quotient rule, a form of which appears in the formula book, to + If the formula 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, meaning that terms are written out vu ' uv' u =, v= +, u' =.., v' =... followed by their, then only accept answers of the form v ( + ) A B AB, >. Condone invisible brackets for the M. ( + ) Alternatively applies the product rule with ( ) 1 u =, v= + If the formula 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, meaning that terms are written out u =, v= ( + ) 1, u' =.., v' =... followed by their vu ' + uv', then only accept answers of the form ( ) ( ) 1 + A ± + B. Condone invisible brackets for the M. Any fully correct (unsimplified) form of f '( ) ( + ) Accept versions of f '( ) = for the quotient rule or ( + ) 1 Versions of f'( ) = ( + ) ( + ) for use of the product rule. Setting their numerator of f '( ) = or >, and proceeding to find two critical values. Both critical values ± are found. Accept for this mark epressions like >± and ±17. d For choosing the inside region of their critical values. The inequality (if seen) must have been of the correct form. Either A... B <, C... D > or < C.It is dependent upon having set the numerator > or =. Correct solution only., < and > < <. Accept ( ) Do not accept < or > or 17. < < 17.. Do not accept a correct answer coming from an incorrect inequality. This would be dma. Condone a solution < <± < < Do not accept a solution without seeing a correct f '( ) first. Note that this is a demand of the question.

7 January 1 (IAL) Question Number Scheme Marks tan + tan 5 = tan( + 5 ) = 1 tan tan = 6., (.,. ) = awrt 6.7 or 96.7 or d, + 5 =. (. ) =.. d = awrt 6.7,96.7,186.7 Notes (6 marks) tan A + tan B Uses the compound angle identity tan( A+ B) = to write the equation in the form 1 tanatanb tan( ± 5 ) =. Accept a sign error in bracket. tan( + 5 ) = d Uses the correct order of operations to find one solution in the range. Moves from tan(± 5 ) = ± 5 = arctan =... This is dependent upon having scored the first One correct answer, usually awrt 6.7, but accept any of 6.7,96.7,186.7 d Uses the correct order of operations to find a second solution in the range. This can be scored by ± 5 = 18 + their 6 or6 + their 6 =.. It may be implied by 9 + their 6.7, or 18 + their 6.7 as long as no incorrect working is seen. This is dependent upon having scored the first All three answers in the range, = awrt 6.7, 96.7,186.7 Any etra solutions in the range withhold the last A mark. Ignore any solutions outside the range 7 Radian solutions will be unlikely, but could be worth marks only if 5.87radians. tan(+ 5) = + 5 = will score dm and nothing else.

8 January 1 (IAL) Question Number Scheme Marks tan + tan 5 = + = 1 tan tan5 (alt 1 ) tan tan 5 ( 1 tan tan 5 ) tan5 tan = = (.9...) 1+ tan5 = 1.5 = awrt 6.7 d = 19.5 (7.5 ) =.. d = awrt 6.7,96.7,186.7 Notes Cross mulitiplies, collects terms in tan and makes tan the subject. Allow for tan =... tan5 Accept tan = or the decimal equivalent tan = awrt tan5 d Correct order of operations to find one solution tan =... = arctan.. =.. This is dependent upon having scored the first One correct solution usually awrt 6.7 d Uses the correct order of operations to find another solution in the range. This can be scored by Either = 18 + their 1. or6 + their 1. =.. Or 9 + their 6.7, or 18 + their 6.7 This is dependent upon having scored the first All three answers in the range, = awrt 6.7, 96.7,186.7 Any etra solutions in the range withhold the last A mark. Ignore any solutions outside the range 7 (6 marks) Question Number Scheme Marks (alt ) tan Similar to alt 1 but additionally uses tan = 1 tan ( tan 5) tan + ( + tan 5) tan + (tan 5 ) = tan = awrt 118., 89. = 6. 7, 96. 7, d d

9 January 1 (IAL) Question Number Scheme Marks (a) 17 8 ( )( ) A+ B + + C+ D Compare terms: A= B1 Compare terms: B= B1 Compare either term or constant term: A+C =17 or B+D =8 C =..or D =.. C = 1, D= () (b) d = + + d = + + +, ln( ) 1, 1 1 = ln() ln(5) 1 = 6 + ln d 5 = 6 + ln ( ) (6) (1 marks)

10 January 1 (IAL) (a) Notes for Question B1 States that A=. It may be implied by writing out the rhs as ( )( ) B1 States that B=. It may be implied by writing out the rhs as ( )( ) + B + + C+ D A+ + + C+ D Compares either the or constant terms and proceeding to find a numerical value of either C or D. This mark may be implied by a correct value of either C or D. Both values correct C = 1, D= Alternatively can be scored via division B1, B1 for sight of the and the in quotient + for proceeding to get a linear remainder The correct linear remainder. Accept. If their division is unclear accept the answers in the correct place in part b (b) For using their answers to part (a) to rewrite the integral in the form ' C' ' D' d ' A ' ' B' d = with numerical values for A,B,C, and D. + + We will condone a restart to this question in (b) but it would not score marks in (a) For the correct method of integrating the ' A ' + ' B' part. Follow through on their A and B A ( A + B) Accept + B or. It cannot be scored for an attempt to integrate (' A ' + ' B' )( + ) ' C ' For the correct method of integrating +. Accept ln( + ) constant ln( + ) or constant d. ' C' This can be scored for values of D as long as + ' D' ' C' ' D' d = d+ d Correct integral + + ln( + ) + ( c). There is no requirement for +c. For putting in limits, subtracting and correctly collecting terms in ln s using subtraction law. 1 1 It is dependent upon having scored the previous M. Allow, for eample, ln ln 5 = ln CSO and CAO 6 + ln ( )

11 January 1 (IAL) Question Number Scheme Marks (a) fg(1) = f () = 7 (b) Either g()= or g ( ).5.5 < g( ) (c) Attempt change of subject of + 9 y = y(+ ) = y = 9 y 9 y (y 1) = 9 y = y 1 d 1 9 g ( ) =, 1.5<, B1 ft () () () (d) Attempts f() = +5=11 k 11 Or f() = +5=5 k > 5 5< k 11 () (11 marks)

12 January 1 (IAL) Notes for Question (a) Full method for their answer to g(1) being subbed into f. The order must be correct. 9 Can be scored for seeing 1 being substituted into cso 7. Do not accept multiple answers. Just 7 would score both marks as long as no incorrect working is seen (b) Calculates the value of g at either end. Sight of or.5 is sufficient..5 < g( ). Accept.5 < y Also accept variations such as (.5,], All values (of y) bigger than.5 but less than or equal to. Do not accept this in terms of. (c) For an attempt to make (or a switched y) the subject of the formula. For this to be scored they must cross multiply and get both (or switched y) terms on the same side of the equation. Allow slips. d This is dependent upon the first M being scored. In addition to collecting like terms they must factorise and divide. Condone just one numerical/sign slip. Accept being given as a function of y. 1 9 g ( ) = 1 or g ( ) =. Accept the form y =, g 1 = instead of g ( ) but the function 1 must be in terms of. B1ft Accept either.5 <, ( 5., ] or follow through on the candidates range to part (b). The domain cannot be epressed in terms of y. Do not follow through on y R R (d) Attempts to find either f() or f(). Evidence could be seeing + 5or + 5 or the sight of 5 or 11. For a range with both ends, for eample5< k < 11, 5 k 11, 5 k < 11or alternatively getting one end completely correct k 11, or k > 5. Accept y 11etc cao5< k 11. Accept (5,11], k 11and k > 5 Do not accept k 11or k > 5 or 5< y 11for the final mark

13 January 1 (IAL) Question Number Scheme Marks 5 (a) Sets y = and takes ln of both sides to get ln y= ln Differentiates wrt to get 1d y ln d y = =.. y d d d Rearranges to achieve d y ln d cao * () (b) Differentiates wrt d d y y 6y 6y ln d d oe, B1, Substitutes (, ) AND rearranges to get d y d dy dy 16ln + 1 = 16ln = ( =.758) d d 1 Alt 1 5 (a) Writes Find equation of tangent using (, ) and their numerical d y d d (16 ln )( ) y = 1 oe Accept y = (6) e ln = d e ln e ln ln ln Differentiates wrt to get ( ) d (9 marks) = = cao d * Alt 5 (a) Sets y = and takes ln of both sides to get ln y ln y= = ln y= ln ln () ()

14 January 1 (IAL) Notes for Question 5 (a) Sets y =, takes ln of both sides, then uses inde law to get ln y= ln d Differentiates wrt to get 1d y ln y d = and then proceeds to d y.. d = Alternatively differentiates wrt y to get 1 ln d = y d y dy * ln d =. This is a given answer. All aspects of the proof must be present (Alt a ) ln Writes = e d ln ln d Differentiates wrt to get ( e ) e = A d d ln ln * ( e ) e ln = = ln. This is a given answer. All aspects of the proof must be present d (b) dy Uses the product rule to differentiate y. Evidence could be sight of Ay + d If the rule is quoted it must be correct. It could be implied by u=.., u =.., v=.., v =.. followed by their vu +uv. For this M to be scored y must differentiate to d y, it cannot differentiate to 1. d dy B1 Differentiates + y + 6y d 1 1 dy Watch for people who divide by first. Then + y + y d A completely correct differential. It need not be simplified. Accept the form d + 6yy d + 6y d + dy = ln d for the first three marks dy dy dy Note that = + 6y + 6y+ = lnis A but they can recover if their intention is clear. d d d Substitutes =, y = into their epression, and rearranges to find a numerical value for d y d d Uses their numerical value to d y and (, ) to find an equation of a tangent. d y dy Accept = Numerical d If y = m+ cis used then a full method must be seen to find c using both (, ) and a numerical m. (16 ln )( ) y =. 1 y (16ln ) y (ln16 1) Accept alternatives such as =, = 1 6 y As the form of the answer is not required accept awrt dp y = , =.76

15 January 1 (IAL) Question Number 6 (a) 6 (9 A ) + Scheme 1 1 A = 6(9 + A ) = or B1 Marks 1 1 A A = A A = A A = Compare to B + C B = B1 A = 6 B1 1 1 C = ' ' ' A ' = 18 d (b) Coefficient of A = 5 =! 9 7 (7 ) () (9 marks)

16 January 1 (IAL) Notes for Question 6 For this question (a) and (b) can be treated as one whole question. Marks for (a) can be gained in (b) (a) B1 For taking out a factor of 1 9 Evidence would seeing either or 1 6 or before the bracket. 1 A For the form of the binomial epansion with n = and a term of 9 1 A To score it is sufficient to see just either the first two terms. ie or the first term and a later term if an error was made on term two. Condone poor bracketing If the 9 has been removed incorrectly accept for this M mark 1 1 ( ka ) 1 1+ = 1+ ( ka ) + ( ka ) Any (unsimplified) form of the binomial epansion. Ignore the factor preceeding the bracket. 1 1 A A is acceptable. 9 9 The bracketing must be correct but it is OK for them to recover. B1 For writing down B = Note that this could be found by substituting = into both sides of epression B1 For writing down A = 6 d For substituting their numerical value of A into their coefficient of (involving A or A ) to find C. The coefficient does not need to be correct but the previous must have been scored For C =. Accept equivalents like C = =. 18 (b) 6 For a correct unsimplified term in of the binomial ep with 1 their ' A' n = and term = A Sight of with or without the factor of with their numerical A! 9 Accept with or without correct bracketing but must involve A. 5A 5A 5A You may have to check or = if there is little working Allow this mark if a candidate has a correct unsimplified fraction in part a, but then uses an incorrect simplification in calculating the term in 6. This incorrect simplification must include...a 5 or other eact correct equivalents such as ,. Accept with the

17 January 1 (IAL) Question Number Scheme Marks 7 (a) Applies vu ' + uv' with u=+ and v=ln or vice versa (b) Sets ( ) ( ) ( ) ( ) f '( ) = ln( ) ln( ) = and makes ln the subject 1+ ln( ) = = e d* () () 1+ (c) 1 + Subs =.6 i nto = e 1 = awrt.675, = awrt.68 = awrt.685, () (d) A= (.7, -1.) () (11 marks) Alt 7 (a) Writes f( ) = ln+ ln and applies vu ' + uv' 1 1 f'( ) = ln( ) + ln + + ()

18 January 1 (IAL) Notes for Question 7 (a) Fully applies the product rule to (1 + )ln. This can be achieved by Either setting u=+ and v= ln. 1 If the rule is quoted it must be correct and ln when differentiated. If the rule is not quoted, it can be 1 implied by u = +, v= ln, u' = A+ B, v' = followed by their vu +uv only accept answers of the form f '( ) ln( ) ( A B) ( ) = Or writing f( ) = ln+ ln and attempting to apply the product rule to both parts. It must be seen to be correctly applied to at least one of the products. See above for the rules. 1 Again insist on ln when differentiating. Or applying the product rule to a triple product. 1 Look for epressions like d ( uvw) = u ' vw+ uv ' w+ uvw' d Two out of the four separate terms correct (unsimplified). All four terms correct (unsimplified) f '( ) ln( ) ( ) ( ) = or if two applications f'( ) = ln( ) + ln( ) + + (b) Sets or implies that their f'( ) = and proceeds to make the ln term the subject of the formula, To score this f'( ) does not need to be correct but it must be of equal difficulty. Look for more than one term in ln and two other 'unlike terms'... d Dependent upon the last. For moving from ln =... = e * CSO = e. All aspects of the proof must be correct including the position of the minus sign and the bracketing. (b) Alt working backwards By taking ln s proceeds from = e to ± ln (1 + ) =± (1 + ) d Dependent upon the last. Moves to ± ln ( + ) ± ( + ) = For a statement that completes the proof. Accept hence f'( ) = solution is the coordinate of A. All aspects must be correct including their f'( ). (c) For an attempt to find 1 from the value of by substituting.6 into e Possible ways this can be scored could be sight of e or awrt.7 1 =awrt.675. = awrt.68 = awrt.685 (d) For either = 7., or y = 1. as a result of using =.6 truncated or =.7 rounded from part c. Alternatively for substituting their answer for in part (c) either to dp or rounded to dp into f( ) to find the y coordinate of A. Accept sight of (1 + )ln A= (.7, -1.). +

19 January 1 (IAL) Question Number Scheme Marks 8 (a) 1 coseca cot A= coseca = sin A tan A sin A B1 cosa = sin Acos A sin A cos A = sin Acos A (1 cos A) sin A sin A = = = tan A sin Acos A sin Acos A cos A * () (b)(i) cosecθ cot θ = tan θ = arctan θ = = Accept awrt.5 6 (ii) 5 tanθ + cotθ = 5 cosecθ = 1 θ = arcsin = awrt.6,1.7 5 d (6) (1 marks) Alt 8 (a) Alt 8b(ii) SC 8b(ii) 1 cosec cot tan tan sin A tan A A B1 cosa sin A = sin Acos A sin A cos A sin Acos A cos A= sin A (1 cos A) = sin A sin A= sin A QED(minimal statement must be seen) * () sinθ cosθ 1 tanθ + cotθ = 5 + = 5 = 5 sin θ = cosθ sinθ 1 sin θ 5 This can now score all of the marks as it is effectively using part (a) 1 tanθ + cotθ = 5 tanθ + = 5 tan θ 5tanθ + 1= tanθ θ = awrt.6,1.7 1,1,, This is not using part (a) and is a special case with one mark per correct answer. One answer =1 Two answers=11

20 January 1 (IAL) (a) B1 Notes for Question 8 Writes coseca as sin A cos A Uses the double angle formula for sin A (see below) and cot A = to write the given epression in sin A terms of just sin Aand cos A For the double angle formula accept sight of sin Acos A or sin Acos A+ cos Asin A Condone sin A = sin cos for this method within their solution. For writing the given epression as a single fraction in terms of just sin Aand cos A The denominator must be correct for their fraction and at least one numerator must have been modified. cos A sin A sin Acos A Accept - not the lowest common denominator. sin Acos A sin A sin Acos A 1 cosa sin A cos A Accept Incorrect fraction but denominator correct sin Acos A sin A sin Acos A * A completely correct proof. This is a given solution and there must not be any errors. For this mark do not condone epressions like sin A = sin cos The s must be cancelled at some point. It is OK for them to just disappear sin A The 1 cos A term must be replaced with sin A. sin Acos A The epression sin A cos A must be clearly seen before being replaced by tan A but sin A is OK sin Acos A (b (i)) Uses part (a) to write given equation in form tan θ = and proceeding to θ =.. This may be implied by A = θ as long as they proceed to θ =.. Accept a restart as long as they get to the line tan θ = θ =. Accept awrt.5 (radians) but not. Ignore etra solutions outside the range. 6 Withhold this mark if etra solutions are given inside the range. The answer without working does not score any marks. The demand of the question is clearly stated Hence solve meaning that you should see an equivalent statement to tan θ = (b (ii)) Uses part (a) to write given equation in form cosecθ = C, where C is a constant. It may be implied by sin θ =.. d Dependent upon the first method and is scored for the correct order of operations. 1 1 The mark is scored for cosec θ C sin θ arcsin.. C θ = = = C θ = It can be implied by a correct answer only if the line cosecθ = C is present. One correct solution awrt.6 or 1.7. Accept awrt.656,.6. Remember to isw here. Accept awrt 11.8 or 78. if the mark in (bi) had been lost for Both solutions correct and no etras inside the range. See above for alternatives. The correct answers without working does not score any marks. Special case where they produce answers from a quadratic in tanθ can score 11 if they get both answers and no others inside the range.

21 January 1 (IAL) Question Number Scheme u = = u = ( u) du 9(a) ( ) d Marks d ( u)du 8 du = u = u + = 8lnu+ u ( + c) d ( ) = 8ln + ( + c) oe (b) Height increases when d h h = > ( < ) h < 16 dt (c) B1 dh h dh dt = = dt h t 8ln( h) + ( h) = + c (6) () Substitute t=, h=1 8ln+ 6 = c d t 8ln( h) + ( h) = 8ln+ 6 oe t Substitute h=1 into 8ln( 1 ) + ( 1 ) = 8ln + 6 dd t = awrt 118 (years) (7) (15 marks) Alt (c) B1 dh h dh dt = = dt h ( h) ( h) t 8ln + = ( h) ( h) h= 1 t 8ln + = h= 1 t= T ( 8ln ( 1 ) + ( 1 )) 8ln ( 1) + ( 1) ( ) = T d,dd, T = awrt 118 (years)

22 January 1 (IAL) Notes for Question 9 (a) d Scored for an attempt to write in terms of u and differentiating to get either d in terms of du or du in d terms of u. Accept for the M incorrect epressions like = 16 u = u du The minimum epectation is that the epression in u is quadratic and the derivative in u is linear. du.5 Alternatively uses u = to find = C and attempts to get d = f( u) or similar. d du Either d = ( u) or d ( )d du = u u or ' = ( u ) or equivalents. Condone d = 8 + u Accept these within the integral for both marks as long as no incorrect working is seen. A An attempt to divide their d by u to get an integral of the form B (d u ) u +. A fully correct integral in terms of u. Condone the omission of du if the intention is clear. 8 Accept forms such as du u + 1 8u + du 1 u d For integrating 1 ln u u and increasing the power of any other term(s). It is dependent upon the previous method mark. There is no need to set the answer in terms of. There is no need to have +c For a correct answer in terms of with or without +c. There is no requirement for modulus signs. Accept either = 8ln + ( ) ( + c) or = 8ln( ) ( + c) oe (b) Setting d h... or d h = h = > h<... Accept h=16 for this mark. dt dt Stating either h < 16 or < h < 16 or h < 16or all values up to 16. A correct answer can score both marks as long as no incorrect working is seen. (c) B1 dh dt Writing h = or equivalent. It must include dh and dt but could be implied For an attempt to integrate both sides, no need for c Follow through on their answer to part (a) for or u with h on the lhs with At on the rhs. t A fully correct answer with +c 8ln( h) + ( h) = + c d Substitute t =, h = 1 in an attempt to find c. Minimal evidence is required. Accept t =, h = 1 c =.. The previous M must have been awarded. t A correct equation 8ln( h) + ( h) = 8ln+ 6oe. t Accept 8ln( h) + ( h) = + awrt.79 dd Substitute h=1 into their equation involving h, t and their value of c in an attempt to find t. It is dependent upon both M s being scored in this part of the question. Again accept minimal evidence Awrt 118 (years). The answer without any correct working scores marks. Condone and h being interchanged in this part of the question. (c Alt) Setting the limits is equivalent to understanding that there is a constant. d Using limits of 1 and 1 and subtracting dd Using limits of 't' and and subtracting A fully correct epression involving just t Awrt 118 (years). No working = marks

23 January 1 (IAL) Question Number 1 (a) Scheme Marks 1 1+ λ = µ 5 + λ 1 = + µ 1 5+ λ = µ any two of λ = 1 + 5µ Full method to find both λ and µ d () +() 1 = 1 + µ µ = 1 Sub µ = 1into () 5+ 1λ = ( 1) λ = Check values in rd equation 1+ ( ) = ( 1). B1 1 Position vector of intersection is OR + ( 1) 1 = dd, (6) (b) (c) 1. 1 = = 1 5 Scalar product =, lines are perpendicular Let X be the point of intersection OX = 7 8 AX = OX OA = () 8 11 OB = OX + AX = + = () (11 marks)

24 January 1 (IAL) Notes for Question 1 (a) For writing down any two equations that give the coordinates of the point of intersection. Accept two of 1+ λ = µ, 5+ λ = µ, 5 λ = 1+ 5µ There must be an attempt to set the coordinates equal but condone slips. d A full method to find both λ and µ. Don t be overly concerned with the mechanics of the method but it must end with values for both. It is dependent upon the previous method Both values correct µ = 1 λ = B1 The correct values of λ and µ must be substituted into both sides of the third equation with some calculation (or statement) showing both sides are equal. This can also be scored via the substitution of µ = 1λ = into both equations of the lines resulting in the same coordinate. dd Substitutes their value of λ into l 1 to find the coordinates or position vector of the point of intersection. It is dependent upon having scored both methods so far. Alternatively substitutes their value of µ into l to find the coordinates or position vector of the point of intersection. It may be implied by out of correct coordinates. Correct answer only. Accept as a vector or a coordinate. Accept (-,, 7) (b) A clear attempt to find the scalar product of the gradient vectors You must see an attempt to multiply and add. Eg or Allow for slips. The above method must be followed by a reason and a conclusion. The scalar product must be zero. Accept =, hence perpendicular. Accept =, therefore proven An attempt to find the vector AX where X is their point of intersection using AX = OX OA This is scored if the difference between the vectors or coordinates are attempted Attempts to find the coordinate or vector of B using OB = OX + AX or OB = OA + AX Allow a misunderstanding on the direction of AX 11 Correct answer only. OB = 1 or 11i 1j + 11k 11 Do NOT accept the coordinate for this mark. Correct answer with no working scores all marks. The correct coordinate would score out of. (c Alt) Using values of λ A= (5,7,) Attempts to find λ on l 1 at X ( λ = ) and A ( λ = ) Uses the difference between λ values to find λ (-6) at B X= (-,,7) cao B

25 January 1 (IAL) Question Number Scheme Marks 11(a) 6sin t = sin t =.5 t = 6 B1 (b) dy dy dt 6cost d d sint dt 6cos dy 6cost 6 Sub t = into = = = 6 d sint 1 sin y 1 Uses normal gradient with (5, ) = 5 y= 1 1 * (1) (6) Sub = 1cos t, y= 6sin t, into y= sint = 1cos t 1 18sin t = 1(1 sin t) 1 sin t+ 18sint 59 =, (sint 1)(1sin t+ 59) = 59 sin t = ( t = ) 1 Using either their t or sint to find either coord of B Hence B has co-ordinates (.8, -.5). These are eact values, The equivalent fractional answers are , 5 5 (8) (15 marks)

26 January 1 (IAL) Notes for Question 11 (a) B1 t = Accept awrt.56 (dp). Answers in degrees is B 6 (b) dy dy Uses dt d = d and differentiates to obtain a gradient function of the form Acost Bsin t dt dy 6cost cost = =. There is no requirement to simplify this. d sin t 1sin t Substitutes their value of t (from part a) into their d y to get a numerical value for the gradient. d Accept also their t being substituted into d dy or d dy Achieves a correct numerical answer for d y d = d 1 d 1 1 = or = dy dy. It needs to be attributed to the correct derivative. Do not accept d dy = 1 This may be implied by the correct value in the gradient of the normal. Award for a correct method to find the equation of the normal. d y 5 d They must use (5, ) and their numerical value of '' ''. Eg = dy dy d If y = m+ cis used then it must be a full method using (5, ) and with m = '' '' as far as c =.. dy * cso. This is a proof and you must be convinced of all aspects including the sight of an intermediate y 1 line between = and y= An attempt to substitute both = 1cost and y= 6sin t into y= 1 1forming a trigonometrical equation in just the variable t. Uses the identity cos t = 1 sin t and rearranges to produce a quadratic equation in sin t. If the identity cos t = cos t sin tor cos t = cos t 1 is used instead, one further step, using the identity cos t = 1 sin t, must be seen before the mark can be awarded. A correct TQ= insin t. Look for sin t+ 18sin t 59 = or equivalent. For a correct attempt at solving the TQ= (usual rules) in sin t. Accept a correct answer (from a graphical calculator) as justification. 59 Award for either sin t = oe or t = Using either their t or sint to find either the or y coordinate of B. Accept as evidence sight of 1cos ' t ' or 6sin' t ' or one correct answer (awrt dp). One coordinate both correct and eact. These are eact answers (.8, -.5). Both coordinates correct and eact. Cso and cao (.8, -.5).

27 January 1 (IAL) Alternative solution to parts b and c using the Cartesian equation of C Question Number Scheme Marks Alt11(b) 5 1(1 sin t) 1 y 9 = = d 1y 1 1 = = = dy = 9 9 y y 1 = y = (6) 5 5 Alt(c ) Sub = 1 y into y= 1 1 y = 1 1 y y + 7y 51= ( y )(5y+ 177) = y =... 5 Substitutes y =... into = y 1 9 Alt(c ) Sub =.8, y = y = 6 into y= = 1 1 (8) 1 6 = 1 1 ( ) = (5 1519)( 5) = Substitutes =... into 1 y = 6 =.8, y =.5 (8)

28 January 1 (IAL) (b) Notes for Question 11 Uses the double angle formula cos t = 1 sin t to get the equation of C in the form y 5 The correct equation is obtained. That is = 1 y or equivalent 1 y = Differentiates wrt y (usual rules) and subs y= to get a numerical value to d dy d 1 = This may be implied by the correct value in the gradient of the normal. dy Award for a correct method to find the equation of the normal. d y 5 d They must use (5, ) and their numerical value of. Eg = dy dy d If y = m+ cis used then it must be a full method with (5, ) and with m = '' '' as far as c =.. dy * cso. This is a proof and you must be convinced of all aspects including the last line of = f( ). y 1 = 5 y= 1 1 (c) 5 Sub their = 1 y into y= 1 1 to produce an equation in y 9 Alternatively subs their y = 6 1 into y= 1 1 to produce an equation in Forms a quadratic equation in y (or ) For achieving 5y + 7y 51= / = or equivalent For a correct attempt at solving the TQ= in y (or ). If you see the answers you can award this. We are accepting answers from a calculator. Correct factors. If the correct y (or ) is given then this mark is automatically awarded. Substitutes their y = into their = f( y ) =.. or vice versa One correct, either =.8 or y =-.5 The values must be eact Both correct. =.8 and y = Accept =, y =, 5 5

29 January 1 (IAL) Question Number 1(a) ( ) (b) Scheme y = (sin + cos ) = (sin + cos ) = = = (sin + cos + sin cos ) ( ) = 1+ sin * V = y d= (1+ sin ) d V = (1 + sin ) d= ( + sin ) d d= = = + d sind OR d = = sin d= ± B cos ± C cos d cos = + cos d = ± B cos ± Csin ± Dsin d d cos sin cos = + + sin d= B cos Csin D cos 1 ± ± ± = 8 V = (1 + sin ) d= d+ sin d = oe dd, Marks () (9) (1 marks)

30 January 1 (IAL) (a) Notes for Question 1 For squaring y AND attempting to multiply out the bracket. The minimum requirement is that y = ( sin + cos +...). There is no need to include ' ' for this mark. Using sin + cos = 1and sin cos= sin to achieve y = ( 1 + sin ) There is no need to include ' ' for this mark. You may accept sin + sin cos + cos = 1+ sin * It must be stated or implied that V = y d. (b) d y by ( ) It may be implied by replacing (sin + cos ) A correct proof must follow involving all that is required for the previous The limits could just appear in the final line without any eplanation. Note that this is a given answer For splitting the given integral into a sum and integrating There is no need for limits at this stage d= = For integrating or. There is no need to simplify this. Accept sin dor to A. d = = sin dby parts. The integration must be the correct way around. There is no need for limits. If the rule is quoted it must be correct, a version of which appears in the formula booklet. Accept for this mark epressions of the form cos sin d = + cos d OR sin d=± B cos ± Ccos d cos sin d= + cosd A second application by parts, the correct way around. No need for limits. See the previous for how to award. It is dependent upon this having been awarded. Look for sin d=± B cos ± Csin ± Dsin d A fully correct answer to the integral of sin d cos sin cos = + + dd For substituting in both limits and subtracting. The two M s for int by parts must have been scored. Either of linked to first M or linked to ddm. Accept in the form Correct answer and correct solution only. Accept eact equivalents 1 V =

31 January 1 (IAL) Alt way ()- Where candidate does not split up first. (b) V = (1 + sin ) d = ( ± Acos ) B ( ± Acos ) d nd cos cos = d = ( ± Acos ) B( C ± Dsin ) ± E ± Fsin d rd d cos sin sin = + d V cos sin cos = st 6 cos sin cos = dd =, , (9) 1 ST, Seen after two (not necessarily) correct applications of integration by parts, it is for integrating the term nd It is for the first attempt at an application of integration by parts on ( 1+ sin d ) Look for ( ± cos ) ( ± cos ) A B A d for the method rd d It is for a further attempt at an application of integration by parts the correct way around. It is dependent upon the first method having been awarded. = ± cos ± sin ± ± sin A B C D E F d Look for ( ) ( )

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