MATHEMATICS 7357/1 Paper 1 Mark scheme June 2024

A-level MATHEMATICS 7357/1 Paper 1 Mark scheme June 2024 *246A7357/paper1/MS* 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 mathematics paper 1 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 ISW ignore subsequent working 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, penalized. 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 Circles the 1st answer 1.1b B1 –5 Question 1 Total 1 Q Marking instructions AO Marks Typical solution 2 Ticks the 1st box 1.1b B1 f 1  x  ln  x 1 Question 2 Total 1 Q Marking instructions AO Marks Typical solution 3 Circles the 2nd answer 1.1b B1 4 Question 3 Total 1 Q Marking instructions AO Marks Typical solution 4 Ticks the bottom-right box 1.2 B1 Question 4 Total 1 Q Marking instructions AO Marks Typical solution 5 Obtains sin x  1 Or Obtains cos x  0 PI by a correct value for x Condone radians and values outside of range 1.1a M1 sin x  1 x  90 , 270 Obtains 90 or 270 Or  3 Obtains both and 2 2 1.1b A1 Obtains 90 and 270 and no other values in the range. 1.1b A1 Question 5 Total 3 Q Marking instructions AO Marks Typical solution 6 Uses the chain rule to obtain 7 x3  5x6 f  x dy  7 3x2  5x3  5x6 dx Or g x3x2  5 1.1a M1 Where f  x and g x are polynomials of degree at least 1. Condone incorrect or missing brackets around their 3x2  5 Obtains 7 3x2  5x3  5x6 OE 1.1b A1 ISW Question 6 Total 2 Q Marking instructions AO Marks Typical solution 7 3  8n 1 2n  3  8n  1 2n 1 2n 1 2n  3  3 2n  8n  16n2 1 2n  3  3 2n  2 2n  4n 1 2n  3  2n  4n 1 2n  4n  3  2n 2n 1 Simplifies 8n to 2 2n 1.1b B1 1 2n Multiplies by or 1 2n 2n  1 2n  1 1.1a M1 Obtains correct single fraction with denominator of 1 2n or 2n 1 1.1b A1 Completes reasoned argument to obtain 4n  3  2n 2n  1 AG Condone eg 8n or √2n except if seen on their final line. 2.1 R1 Question 7 Total 4 Q Marking instructions AO Marks Typical solution 8(a) Obtains the correct constant term 32 1.1b B1 2  kx5  32  80kx  80k 2x2  ... Obtains 5 16kx or 10  8(kx)2 OE 5k 5  4  kx 2 PI by 2 x or 2!  2    1.1a M1 Obtains 32  80kx  80k 2x2 ... Accept list of correct terms. No ISW If more terms are given it must be 1.1b A1 obvious which are their first three terms. Subtotal 3 Q Marking instructions AO Marks Typical solution 8(b) Forms the equation 80k  4  80k 2 k  0 or 1 4 k  1 4 Since k  0 their Ak  4 their Bk2 OE 3.1a M1 May recover if x is initially included. Deduces k = 1 only 4 Or their k  their A 4B Justification of rejection k = 0 not required. 2.2a A1F Subtotal 2 Question 8 Total 5 Q Marking instructions AO Marks Typical solution 9(a) Substitutes at least one small angle identity correctly into cos 4  2sin 3  tan 2 1.1a M1 4 2 cos 4  2 sin 3  tan 2  1  2 3   2  2  1 4  8 2 Obtains a correct expression in terms of  ACF 1.1b A1 Completes argument to obtain 1 4  8 2 2.1 R1 Subtotal 3 Q Marking instructions AO Marks Typical solution 9(b) Substitutes   0.07 into their 1 4  8 2 3.1a M1 1 4  0.07  8  0.072  1.2408  1.241 Obtains AWRT 1.241 CSO 1.1b A1 Subtotal 2 Question 9 Total 5 Q Marking instructions AO Marks Typical solution 10(a) Substitutes n  300, a  7 and l  32 Into S  n a  l  n 2 Or Substitutes n  300, a  7 and d  39  3 into 299 23 S  n 2a  n 1 d  n 2 Condone n = 299 or 301 and d  AWRT 0.13 3.1a M1 S  300 7  32 300 2  3750 Obtains 3750 1.1b A1 Subtotal 2 Q Marking instructions AO Marks Typical solution 10(b) Forms an equation using S9  1260 Might see 9 2a  8d   1260  a  4d  140 2 3.4 M1 9 a  l   1260  a  l  280 2 l  6a 7a  280 a  40, l  240 Value of top prize = £240 Forms an equation using the relationship between the highest and least values. eg a  8d  6a or l  6a OE Might see l  1 a which may 6 indicate the candidate is correctly working from the highest term to the lowest term. 3.4 M1 Obtains and solves an equation in one variable having formed one equation using S9  1260 OR used the relationship between the highest and least values. 3.1a M1 Obtains £240 Must have correct units. CAO 3.2a A1 Subtotal 4 Question 10 Total 6 Q Marking instructions AO Marks Typical solution 11(a) Draws cubic graph with exactly two turning points 1.1a M1 Draws cubic graph of correct orientation passing through the origin and positive x-axis at two points. 1.1b A1 Draws fully correct sketch with x-axis intercepts correctly labelled a and 6. Ignore labelling on the y-axis. 1.1b R1 Subtotal 3 Q Marking instructions AO Marks Typical solution 11(b) Draws cubic graph of correct orientation passing through the origin and negative x-axis at two points. Or Substitutes –2x into f(x) Or Describes the reflection in the y-axis and a stretch of scale factor 1 in the x-direction 2 or a stretch scale factor  1 in 2 the x-direction. 3.1a M1 Draws fully correct sketch with x-axis intercepts correctly labelled  a and -3. 2 2.2a R1 Subtotal 2 Question 11 Total 5 Q Marking instructions AO Marks Typical solution 12(a) Substitutes u  3 into 6 1 u n PI u2  2 1.1a M1 u2  2 u3  3 u4  2 Obtains u2  2,u3  3,u4  2 Condone missing labels if order is obvious. 1.1b A1 Subtotal 2 Q Marking instructions AO Marks Typical solution 12(b) States 2 2.2a B1 2 Subtotal 1 Q Marking instructions AO Marks Typical solution 12(c) Shows that pairs of consecutive terms sum to 1 in a series Or Considers a sum of 3s and a sum of 2 s 3.1a M1 101 un  3  2  3  2  ...  3  2  3 n1 1 50 = 53 101 Deduces un = 53 n1 2.2a R1 Subtotal 2 Question 12 Total 5 Q Marking instructions AO Marks Typical solution 13(a) Substitutes x  1 into P  x 2 and obtains zero. Must see  1 bracketed 2 correctly. If bracket(s) missing must see a further step to indicate correct evaluation eg  4  8  11  4  0 or better. 8 4 2 1.1a M1  1   1 3  1 2  1  P  2   4   2   8   2   11  2   4          0 2x  1 is a factor of P  x Completes factor theorem argument by showing P  1   0 and stating  2    2x  1 is a factor of P  x OE 2.1 R1 Subtotal 2 Q Marking instructions AO Marks Typical solution 13(b) Obtains two correct coefficients of 2x2  3x  4 1.1a M1 P x  2x  12x2  3x  4 Obtains 2x  12x2  3x  4 1.1b A1 Subtotal 2