MATHEMATICS 7357/2 Paper 2 Mark scheme June 2024

A-level MATHEMATICS 7357/2 Paper 2 Mark scheme June 2024 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardization events which all associates participate in and is the scheme which was used by them in this examination. The standardization process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardization each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardization process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. No student should be disadvantaged on the basis of their gender identity and/or how they refer to the gender identity of others in their exam responses. A consistent use of ‘they/them’ as a singular and pronouns beyond ‘she/her’ or ‘he/him’ will be credited in exam responses in line with existing mark scheme criteria. MARK SCHEME Mark scheme instructions to examiners General The mark scheme for each question shows: • the marks available for each part of the question • the total marks available for the question • marking instructions that indicate when marks should be awarded or withheld including the principle on which each mark is awarded. Information is included to help the examiner make his or her judgement and to delineate what is creditworthy from that not worthy of credit • a typical solution. This response is one we expect to see frequently. However, credit must be given on the basis of the marking instructions. If a student uses a method which is not explicitly covered by the marking instructions the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their senior examiner if in any doubt. Key to mark types M mark is for method R mark is for reasoning A mark is dependent on M marks and is for accuracy B mark is independent of M marks and is for method and accuracy E mark is for explanation F follow through from previous incorrect result Key to mark scheme abbreviations CAO correct answer only CSO correct solution only ft follow through from previous incorrect result ‘their’ indicates that credit can be given from previous incorrect result AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent NMS no method shown PI possibly implied sf significant figure(s) dp decimal place(s) AS/A-level Math’s/Further Math’s assessment objectives. AO Description AO1 AO1.1a Select routine procedures AO1.1b Correctly carry out routine procedures AO1.2 Accurately recall facts, terminology and definitions AO2 AO2.1 Construct rigorous mathematical arguments (including proofs) AO2.2a Make deductions AO2.2b Make inferences AO2.3 Assess the validity of mathematical arguments AO2.4 Explain their reasoning AO2.5 Use mathematical language and notation correctly AO3 AO3.1a Translate problems in mathematical contexts into mathematical processes AO3.1b Translate problems in non-mathematical contexts into mathematical processes AO3.2a Interpret solutions to problems in their original context AO3.2b Where appropriate, evaluate the accuracy and limitations of solutions to problems AO3.3 Translate situations in context into mathematical models AO3.4 Use mathematical models AO3.5a Evaluate the outcomes of modelling in context AO3.5b Recognize the limitations of models AO3.5c Where appropriate, explain how to refine models Examiners should consistently apply the following general marking principles: No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to students showing no working is that incorrect answers, however close, earn no marks. Where a question asks the student to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. Diagrams Diagrams that have working on them should be treated like normal responses. If a diagram has been written on but the correct response is within the answer space, the work within the answer space should be marked. Working on diagrams that contradicts work within the answer space is not to be considered as choice but as working, and is not, therefore, penalised. Work erased or crossed out Erased or crossed out work that is still legible and has not been replaced should be marked. Erased or crossed out work that has been replaced can be ignored. Choice When a choice of answers and/or methods is given and the student has not clearly indicated which answer they want to be marked, mark positively, awarding marks for all of the student’s best attempts. Withhold marks for final accuracy and conclusions if there are conflicting complete answers or when an incorrect solution (or part thereof) is referred to in the final answer. Q Marking instructions AO Marks Typical solution 1 Ticks 4th box 1.2 B1  x  12   y  22  36 Question 1 Total 1 Q Marking instructions AO Marks Typical solution 2 Circles 3rd answer 2.2a B1 125 Question 2 Total 1 Q Marking instructions AO Marks Typical solution 3 Ticks 1st box 1.1b B1 x : x  1 ∪ x : x  4 Question 3 Total 1 Q Marking instructions AO Marks Typical solution 4 Takes logs of both sides to same base PI by x  2  log 71570 5 1.1b B1 log5 5x2   log5 71570 x  2  1570 log5 7 x  2  1570 log5 7  1900.23 Uses log An  n log A PI by x  2  log 71570 5 1.1a M1 Completes a reasoned argument using logarithms to 2.1 R1 obtain AWRT 1900 Question 4 Total 3 Q Marking instructions AO Marks Typical solution 5 Differentiates x3 and sin x to obtain 3x2 and cos x OE 1.1b B1 dy  3 x2 sin x  x3 cos x dx sin2 x Uses the quotient rule and obtains numerator Ax2 sin x  Bx3 cos x Condone any denominator Or Writes as a product and applies the product rule to obtain Ax2cosec(x)  x3cosec(x) cot(x) 3.1a M1 3 x2 sin x  x3 cos x Obtains sin2 x ACF No ISW 1.1b A1 Question 5 Total 3 Q Marking instructions AO Marks Typical solution 6 Expands brackets, at least one completely correct 1.1a M1 4 sin2   12 sin cos  9 cos2  36 sin2  12 sin cos  cos2   30 40 sin2   10 cos2   30 40 sin2   10 1 sin2    30 30 sin2   10  30 sin2   2 3 sin   6 3 since  is obtuse sin  6 3 Expands both brackets correctly 1.1b A1 Uses sin2   cos2   1 correctly to eliminate cos2  or sin2  from their equation PI by 30 sin2   10  30 3.1a M1 Obtains an equation of the form sin2   k or cos2   k where 0 k 1 PI by sin  k or sin   k 1.1a M1 Obtains sin   2 3 1.1b A1 Completes reasoned argument to obtain sin  2 and 3 explains why sin  2 3 Must come from sin    2 3 FT their sin2   k where 0 k 1 2.4 R1F Question 6 Total 6 Q Marking instructions AO Marks Typical solution 7(a) Obtains 50 1.0023  50 1.0022  50 1.002 with 50  1.0022  50  1.002  50 1.002 or better seen 2.1 R1 T  50 1.0022  50 1.002  501.002 3  50 1.0023  50 1.0022  50 1.002 Subtotal 1 Q Marking instructions AO Marks Typical solution 7(b)(i) Models the total as the sum to n terms of a geometric sequence Evidence for this could include at least two of a, r or n substituted into sum formula a = 50.1, r = 1.002, n = 120 Condone a = 50 or n = 10 or 119 PI by AWRT 6724, 6737, 6801 or 505.53 3.3 M1 50.111.002120  T120  11.002  6787.1595... Total in account = £ 6 787 Forms the correct expression for the correct total   50.111.002120  T120  or 11.002 120  50 1.002x x1 3.3 A1 Obtains £6 787, £6 787.15 or £6 787.16 3.2a A1 Subtotal 3 Q Marking instructions AO Marks Typical solution 7(b)(ii) Makes a reasonable comment in context. For example: The interest rate is unlikely to remain fixed. Or May have needed to withdraw some amount. Or May change the monthly payments. 3.5b E1 The interest rate is unlikely to remain fixed for the whole 10 years Subtotal 1 Question 7 Total 5 Q Marking instructions AO Marks Typical solution 8(a)(i) Obtains 12  a  b log10 24 ISW 3.4 B1 12  a  b log10 24 Subtotal 1 Q Marking instructions AO Marks Typical solution 8(a)(ii) Eliminates a to obtain an equation in b 3.1a M1 12  a  b log10 24  (6.4  a  b log10 3) 5.6  b log10 24  b log10 3  b log 24 10 3  b log10 8 b  5.6 log10 8 Obtains b log10 h from b log10 their 24  b log10 3 or b log their 24 10 3 where 3h = their 24 1.1a M1 Completes a reasoned argument to show b  5.6 log10 8 Must include log 24 OE or 10 3 log10 24  log10 8  3 AG 2.1 R1 Subtotal 3 Q Marking instructions AO Marks Typical solution 8(a)(iii) Obtains AWRT 3.44 1.1b B1 a = 3.44 Subtotal 1 Q Marking instructions AO Marks Typical solution 8(b) Substitutes a value for 0  x  0.25 into the model with their a and b = AWRT 6.2 PI by correct negative y-value Or Substitutes x = 0 into the model with their a and b = AWRT 6.2 and states that the value for y is undefined Or Substitutes y = 0 into the correct model with and gets x = AWRT 0.28 3.4 M1 When x  0.25 y  3.44  6.2 log10 0.25  0.29 The model predicts a negative median mass for a monkey that is one week old, therefore it is unsuitable for use with monkeys 1 week old or less. Completes reasoned argument to find a correct median mass for their value of x and concludes that the model cannot be used to predict the median mass of monkeys less than one week old. Condone that the model cannot be used to predict the median mass of monkeys for their value of x where 0  x  0.25 Condone omittance of median or use of weight throughout 3.5a R1 Subtotal 2 Question 8 Total 7 Q Marking instructions AO Marks Typical solution 9(a)(i) Obtains 123x2 1 13x  2! OE with at least two terms correct 1.1a M1  1    123x2 1 3x  1 1 3x  2!  1 3x  9x2 Obtains 1 3x  9x2 1.1b A1 Subtotal 2 Q Marking instructions AO Marks Typical solution 9(a)(ii) Writes fraction as 2  3 x1 PI by 1  3 x  9 x2 2 4 8 1.1b B1 1  2  3x1 2  3x   3 x 1  21 1      2  1   3x  12  3x 2   1  1        2   2  2!  2   1  1  3 x  9 x2 2  3x 2 4 8 1 , 3 x and 9 x2 form a geometric 2 4 8 sequence with common ratio 3 x 2 Factorises to obtain the form 21 1 Ax1 PI by 1  3 x  9 x2 2 4 8 1.1a M1  3x 1 Expands 1  to obtain  2   3x  12  3x 2 1 (1)         2  2!  2  OE Condone one sign error 1.1a M1 Completes a correct argument to show 1  1  3 x  9 x2 2  3x 2 4 8 2.1 R1 States that the common ratio is 3 x 2 2.2a B1 Subtotal 5