Proof questions

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• Edexcel 9MA0 · June 2024 · Paper 1 · Q15 · 6 marks

  6 marks

  15

  
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  Examiner report

  This question was a clear discriminator between the most secure and intermediate candidates with many making little or no progress in part (i), although many candidates made at least a respectable attempt at part (ii). The most common techniques employed in part (i) were completing the square and using the discriminant. When completing the square, candidates often stated that squared numbers are always positive and gave no indication that they could also equal 0, which was a common error the last time a similar question was asked. When finding the discriminant, candidates often obtained the answer of –4, but very little explanation followed this, and notably many seemed to think this was sufficient to conclude that the given quadratic was always positive with no reference to the coefficient of the 2 k term. It was not uncommon to see attempts analysing even and odd number scenarios, but none were seen that recovered from only considering whole numbers and extended to the entire set of real numbers. A few candidates used differentiation to find the turning point and some of these candidates calculated the second derivative to prove that their turning point was a minimum, but they usually lost the final mark as they did not write a conclusion to complete their proof. The least confident candidates simply substituted a few integer values for k into the expression and attempted to draw a conclusion from the results, scoring no marks. In part (ii), many candidates did not fully appreciate what they were being asked to do, most likely due to the unfamiliar nature of the question. As a result, some attempted algebraic manipulation of the expression, such as expanding the brackets, and were unable to progress to a solution. Other candidates believed they were being asked simply to write out the proof by adding text to the information supplied or by detailing the working for the solutions of the simultaneous equations given. However, there was a good proportion of candidates who realised that the example given was just one of many possible scenarios and proceeded to attempt to find solutions to the remaining cases. It was rare to see all of the 11 other factor pairs examined, and, in most cases, candidates only considered simultaneous equations with a positive right-hand side. The mark scheme enabled all the marks to be scored for this approach given that the simultaneous equations with negative right-hand sides could be discarded quite trivially. Often candidates only determined x or y for a couple of factors of 28 securing 1 or 2 marks. A good proportion only dealt with factor pairs ‘one way round’ and gave no justification as to why, limiting themselves to 2 marks out of the available 4. Calculators could legitimately be used to solve the simultaneous equations and candidates who did this found, almost without exception, correct values for x and y as well as saving time. Full algebraic processing of the simultaneous equations was time-consuming and prone to error. A minority of candidates realised it was possible to argue that 3𝑥+ 2𝑦  ⩾ 5 and limited themselves to examining just cases A and B from the mark scheme, but it was nice to see the strongest candidates able to distinguish themselves in this manner. Most candidates identified the reasons for rejecting factor pairs as they went along whilst a few left the rejection of the solutions to the overall conclusion, and both approaches were acceptable. Sadly, some made no comment at all on the solutions they found and were unable to score either accuracy mark as a result.

  Text version

  This question has two proof tasks. In part (ii), x and y are positive integers and the proposed equation is (3x+2y)(2x-5y) = 28. A student's contradiction proof begins by assuming such x,y exist and checking the factor pair 3x+2y=14, 2x-5y=2, which gives x=74/19 and y=22/19, not integers.

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• Edexcel 9MA0 · June 2024 · Paper 2 · Q15 · 12 marks

  12 marks

  15

  
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  This question provided a challenge to most candidates and was in general a good discriminator. In part (a), it was perhaps disappointing, although not unexpected, to see that so many candidates’ instinctive first step was to multiply out the brackets rather than realising that a simpler, quicker and more accurate method was available. In part (a), the use of the chain rule and factorisation could greatly reduce the required effort and the potential for making mistakes. Those candidates with a good understanding of the chain rule and implicit differentiation were able to use this efficient method quickly to accumulate the 5 marks available. However, a common error of omitting the 1 in (1 + d d y x ) was seen many times. Furthermore, a significant proportion of candidates who differentiated the left-hand side using the chain rule, often chose to expand the derivative, nonetheless, thus complicating the remainder of the question. Unfortunately, the majority of candidates chose to first expand ( ) 3 x y + before attempting to differentiate. Whilst this was, in many cases, completed correctly there was of course an inherent time penalty. A significant proportion of candidates made errors when expanding and displayed a lack of algebraic fluency and poor application of the binomial expansion. Common errors that were seen included missing out terms in the expansion and poor understanding of indices. Once the brackets had been expanded, there were clearly some gaps in understanding of how to successfully carry out the implicit differentiation, particularly for terms that also required application of the product rule. It was quite common to see for example 2 3x y going to d 6 d y x x and 2 3xy going to d 6 d y x . Overall, the right-hand-side was dealt with correctly although the constant term −2 was sometimes retained following differentiation and sometimes d 6 d y y x was seen. When candidates attempted to move all terms over to one side of the equation, there were sometimes sign errors or slips in copying terms. Upon completion of the process of differentiation, almost all candidates seemed to understand that the task was to gather the d d y x terms together, factorise and divide in order to isolate the d d y x , and the majority who reached this stage were able to earn credit here provided their differentiation had been sufficiently structurally correct to produce sufficient terms in d d y x . It was very rare to see a spurious d d y x term retained which has been an issue in previous sessions. In general, those candidates who were able to successfully complete part (a) found part (b) straightforward. However, the accuracy mark in this part was not available to those candidates who had not managed to achieve the correct derivative in part (a). Due to the nature of the function, setting x = 1 and y = 0, gave a ‘correct’ value of d 1 d 2 y x = in many cases even when d d y x was incorrect. As a result, candidates were often not alerted to the fact they had an incorrect d d y x . Disappointingly, candidates who did not obtain 1 2 due to errors in their differentiation, often attempted to manipulate their work to obtain the value that they could see was required from the printed answer and this often meant that the method mark was out of reach, which was a shame. Part (c) provided a challenge for many candidates and was a good discriminator. It is possible that some candidates were running out of time by this part of the paper, and this seemed to lead to simple algebraic and sign errors. However, many candidates were able to make at least some progress and most recognised the need to substitute the equation from part (b), which had been provided, into the equation from part (a). Most were able to expand to obtain a cubic in x albeit with frequent errors and slips, in particular a common error of cancelling the constant terms leading to a cubic with only three terms was often seen. The alternative elimination of x to lead to a cubic in y was rarely seen. Having obtained a cubic equation, many candidates made little creditworthy further progress and the majority seemed not to have made the link to earlier parts of the question and so did not attempt to take (x − 1) out as a factor of their cubic. This was perhaps understandable when the cubic was incorrect although it seemed that even candidates with the correct cubic did not consider using the fact that x = 1 was a root. Many candidates used their calculator to obtain the three roots of their cubic and others employed dubious techniques either attempting to find a ‘discriminant’ by ignoring the 3x term or attempting to take out a factor of x whilst ignoring the constant term. Those candidates who had a three-term cubic rather than a four-term cubic were unable to progress here. The final two marks of the paper were only accessible to those who were very secure in their algebraic accuracy and knowledge as these marks were contingent on both a correct cubic and a correct quadratic factor. Most candidates who met these criteria successfully used the discriminant to demonstrate that the quadratic had no roots although some used more elaborate reasoning involving the gradient of the cubic or the minimum point of the quadratic. Unfortunately, a significant proportion of candidates who got this far, did not manage to earn both marks because their justification or conclusion was insufficient, and it was sometimes disappointing to see otherwise well-answered solutions lose the final mark because they concluded with ‘no roots’ rather than also relating this back to the question to conclude that there could be ‘no further points of intersection’.

  Text version

  The curve C has equation (x + y)^3 = 3x^2 - 3y - 2. The point P(1, 0) lies on C.

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• Edexcel 9MA0 · June 2023 · Paper 1 · Q14 · 4 marks

  4 marks

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  Candidates were required to prove the statement: 3 3 ( 1) n n + − is odd for all natural numbers, n. This was similar to some of the proof questions that have been on previous papers, so it was pleasing to see the majority of candidates were able to make more progress on this type of question this time. A large number of candidates tried to start correctly by substituting 2 n k = and 2 1 n k = + (or 2 1 n k = −) for even and odd numbers; this was the most likely method to achieve full marks. Candidates would often produced the correct expressions which were written in the form 2(......)+1 and then conclude appropriately. The final mark was often lost in many of the complete attempts as at least one element of the proof was either missing or had errors. Some candidates only proved one case (odd or even only) and some candidates did not include an overall conclusion, which was necessary for the final mark. Most common errors seen in expansions were from 3 3 (2 ) 2 k k = and this often led to candidates achieving a cubic expression rather than a quadratic. Some used 2n and 2 1 n + which was penalised by losing the final mark. Errors in expanding and simplifying the algebra were common: expanding triple brackets was the most common approach, although the binomial expansion was also seen, on occasion. The other common approach was from an attempt of algebra with logic. It was common to see candidates expand and simplify the expression to achieve a three-term quadratic but only a few went on to factorise their quadratic as required, and even fewer explained the logic for why ( 1) n n + was even correctly. A significant number of candidates tried to expand and simplify the given expression and usually achieved 2 3 3 1 n n + + scoring the first 2 marks, but very few got beyond this without making any logical arguments at all or did not factorise the expression. e.g. 3 ( 1) 1 n n + + which gained 3 marks. Success was also seen when the given expression was expanded and simplified to 2 3 3 1 n n + + and then candidates substituted 2 n k = and 2 1 n k = + to much the same result as the main method seen in the mark scheme. A number of candidates tried proof by contradiction, but these almost all fell into being marked either from the main scheme or by an expanding and factorising attempt. Whilst the question did require candidates to use algebra, credit was given for those candidates who did just try to apply logic with a maximum score of 2 marks. It was pleasing, however, to see that most candidates had appreciated the demand of the question and attempted some manipulation. A few candidates tried the Further Maths Method “proof by induction”, but most could not complete the proof correctly.

  Text version

  Let n be a natural number.

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• Edexcel 9MA0 · June 2023 · Paper 2 · Q15 · 3 marks

  3 marks

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  A number of candidates lost all marks on this question by not attempting any solution. Most candidates correctly expanded the given expression and used sin2 x + cos2 x = 1 to simplify and score the first two marks. A small number incorrectly used (sin x − cos x)2 = sin2 x − cos2 x, scoring no marks. Others attempted to substitute values of 90 and 180 into the given expression instead of using algebraic manipulation. Whilst most candidates scored the first two marks, only the very best were able to score the final mark, with most candidates failing to offer a convincing reason for a contradiction. Candidates who drew a graph of sin x and cos x or sin 2x to support their understanding tended to score well. Successful candidates typically explained that 2x must be between 180 and 360, and hence sin 2x must be negative, which is a contradiction with their inequality. Those that worked with sin x cos x were generally more successful in achieving this mark than those who chose to use sin2x. However, it was clear that not all students understood what an obtuse angle was. Another typical correct conclusion involved candidates stating if x is obtuse, sin x must be positive and cos x must be negative, leading to a negative product and a contradiction. Of those who were able to offer a reason for the contradiction, some still lost the final A mark for failing to then reach a conclusion that the original statement was true or because of errors in the proof such as missing x’s or mixed variables.

  Text version

  A student attempts this question: Given that x is an obtuse angle, use algebra to prove by contradiction that sin x - cos x >= 1. The proof begins: Assume that sin x - cos x < 1 when x is obtuse. Therefore (sin x - cos x)^2 < 1.

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• Edexcel 9MA0 · June 2022 · Paper 1 · Q7 · 5 marks

  5 marks

  7

  
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  (Mean mark 1.6 out of 5) This question, on two different kinds of proof, proved more demanding than expected. As a result, there were very few completely correct solutions to both parts of this question. In part (i) candidates were often unable to set up a proof by contradiction using appropriate language, failing to use words such as 'assume', 'there exists' or 'let' . Others simply set up a contradiction such as 'assume p and q are odd' without any reference to pq being even. Many candidates also failed to use different variables for their odd numbers p and q, implying that p and q were either the same odd number, or two consecutive odd numbers such as 2n+1 and 2n+3. Candidates who managed to write p and q using different variables were generally able to multiply out pq correctly and write it in the form 2(.....)+1. If this stage was reached, the conclusion was generally correct and an acceptable sentence was written down. A mark of 011 was possible from the mark scheme. In part (ii) most candidates expanded the left hand side and cancelled terms getting to 2xy < 8x2 or equivalent for the method mark. Many then lost the second mark by failing to identify the point at which the inequality reverses, simply writing down y < 4x followed by 'so y > 4x'. Other candidates did reverse the inequality at the right point but didn't give an explanation as to why, and so lost the final mark.

  Text version

  This question concerns integer proofs.

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• Edexcel 9MA0 · June 2022 · Paper 2 · Q11 · 4 marks

  4 marks

  11

  
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  Examiner report

  Most students made some attempt at this proof question, and there was a mixture of responses that achieved the full range of possible marks. It was striking that a number of students attempted to answer the question without doing any algebra, despite the clear instruction within the question. Most candidates who used algebra recognised that a strategy of considering odd and even numbers could be used to prove the statement with most using 𝑛𝑛= 2𝑘𝑘 and 𝑛𝑛= 2𝑘𝑘+ 1, although a minority only attempted one of these. Within the algebra, a common misconception was that (2𝑘𝑘)2 = 2𝑘𝑘2. Occasionally there were a few candidates that tried to consider the cases of 3k – 1, 3k, and 3k + 1 which caused those candidates difficulty in progressing the proof. Although students knew that they needed to show the expression was a multiple of 2 to conclude that it was even, few seemed to be looking for the earliest opportunity to do this, and therefore did more algebraic manipulation than was necessary. Some also presented expressions like 8𝑘𝑘3 + 12𝑘𝑘2 + 16𝑘𝑘+ 6 and declared them to be even without explaining why this is the case. A handful used 𝑛𝑛= 2𝑛𝑛+ 2 for the even case, again having to do more algebraic manipulation than was needed. Most candidates who attempted this approach gained the method marks but the accuracy marks were sometimes lost for algebraic slips or more commonly for failing to give a final statement that the result was even for both odd and even cases, or simply stating that the statement was a multiple of 2 without concluding that it was therefore even. It was surprising for the odd case how many candidates had a preference for fully expanding their expressions into a cubic before factorising, i.e. 8k3 + 12k2 + 16k + 6 = 2(4k3 + 6k2 + 8k + 3) rather than simply factoring (2k + 1)(4k2 + 4k + 6) into 2(2k + 1)(2k2 + 2k + 3) for example which is algebraically easier to achieve. Where algebra was not used, the most common approach was to use a logic argument involving the rules of combining odd and even numbers. Where the argument was convincing students were sometimes able to gain the method marks. A few candidates attempted a proof by contradiction or by induction, but usually produced an argument that was incomplete and made little progress.

  Text version

  Let n be a natural number.

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• Edexcel 9MA0 · October 2021 · Paper 1 · Q15 · 6 marks

  6 marks

  15

  
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  Candidates have not been too successful in answering questions on the topic of ‘Proof’ within this new specification, and this question proved to be no different. Part (i) The meaning of “proof by exhaustion” was not fully understood by all. Many candidates failing to consider all four of the possible cases mentioned. A common response was to test just n =4 and n =5 and conclude that it was true for n = 4 but not n = 5 so it was proven for n ≤ 4. Some candidates who did consider all values of n = 1,2,3 and 4, left some expressions unevaluated. Occasionally the eight required values were found but there was no evidence that they were being compared. It was essential that a proper conclusion is made at the end of a proof, and despite having convincing working and reasons, some candidates lost the final accuracy mark here because they did not have a conclusion saying that the required result was indeed true. Part (ii) Very many candidates did score the first mark here for saying “Suppose that m is odd”. However, a common error was to state “let m3+5 be odd” instead leading to 0 marks. It was common to see 2p+1 or 2k+1 being used to represent an odd number and there were many occasions where (2p+1)3+5 was seen in a candidate’s working. In some cases there were algebraic errors in attempting to expand the cube of (2p+1), but most candidates did achieve the required four term cubic 8p3+12p2+6p+6. When this cubic was correct, those candidates very often went on to gain the next two accuracy marks by proving, by factorisation, that m3+5 was even, and then stating that there is a contradiction (to the assumption that m is odd) and so m must be even. .

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  Let N denote the natural numbers.

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• Edexcel 9MA0 · October 2020 · Paper 1 · Q16 · 4 marks

  4 marks

  16

  
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  Marks on this question were only scored by above average candidates and many made little or no progress towards a proof. There were many blank responses. Candidates who attempted to factorise 2 2 4p q − were more successful than by other methods, although the large majority of these scored only the first mark, by stating the assumption and factorising to ( )( ) 2 2 p q p q + − , but making no further progress. Of those that did consider the factors of 25, many did set up and solve two relevant equations simultaneously, and so gained 3 marks, but very few considered both of the required pairs of equations and so did not all marks. There were many discussions of odd/even integers but this was commonly followed by working in the same variable for p and q, which gained no marks

  Text version

  Let p and q be positive integers.

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• Edexcel 9MA0 · October 2020 · Paper 2 · Q16 · 4 marks

  4 marks

  16

  
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  This was a very poorly answered question, with the majority of candidates not writing anything at all. A fairly common attempt was to only split the natural numbers into odd/even cases and square 2n, 2n+1, which gained no marks. Many students who obtained any marks on this question seemed to stumble on these more by chance than by considering the problem. The problem mentioned that the square was to be a multiple of 3 or one more than a multiple of 3 and many students were able to write this as 3n and 3n + 1. Some went on to square these and show that the result was also a multiple of 3 or one more than a multiple of 3. Few students realised that they needed a third case (e.g. 3n - 1 or 3n + 2) to cover all integers, and as a result two marks was the most that could be achieved. It was very rare to see full marks scored in this question. Candidates who did gain full marks typically presented model solutions with thorough explanations and conclusions.

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  Let n be any natural number.

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• Edexcel 9MA0 · June 2019 · Paper 1 · Q10 · 6 marks

  6 marks

  10

  
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  (Mean Mark 1.7 out of 6) The idea of proof is another new one in this updated specification. It was not well known and a disproportionate number of blank or barely started solutions indicated a lack of expertise in this topic. Candidates who scored full marks for part (i) most often did so via algebraic proof, letting n=2m (for even numbers) and then n=2m+1 (for odd numbers), although several candidates missed out on the final accuracy mark as they failed to insert a conclusion for all n. Weaker candidates tried to use m + 1 instead of 2m + 1 to represent odd numbers. Also many candidates only considered either odd or even numbers, but not both. Candidates who set out to use proof via contradiction tended to score few marks as they did not show an understanding of how to use that method of proof. Proofs via logic were rare and tended to appear as part of a mix of different methods where the candidates seemed to be unclear of the best way to proceed. Some candidates seemed to think that using random numbers to show that the expression is not divisible by 4 amounted to proof (receiving 0 marks). Students need more practice on this style of proof. In part (ii) more candidates approached the problem algebraically than graphically. They were generally able to set up the two equations / inequalities required, but errors often occurred in their use of algebra when attempting to solve them. Some candidates only set up one of the two required inequalities. Many students attempted the question by just substituting numbers in to the expression, with most of these deducing that the statement was always true. Where candidates drew a graph, they often did not describe why their graph indicated that the statement is sometimes true. Many graphs showed a line and a V shape but some were incorrect in the relative positioning of these shapes.

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  Let N denote the natural numbers.

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• Edexcel 9MA0 · June 2018 · Paper 2 · Q3 · 5 marks

  5 marks

  3

  
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  This question discriminated well between students of all abilities with some students struggling to access the final mark in part (b). In part (a), most students provided a complete solution to disprove the statement given in the question. Common errors included: choosing inappropriate values for , m n (e.g. 2 and 3 which were sometimes seen); choosing two rational numbers; attempting to disprove the general statement by using algebra; and not giving a conclusion, despite choosing appropriate values for , . m n In part (b)(i), most students drew a V-shaped graph in the correct position with many labelling 3 or (0, 3) as the y-intercept. In part (b)(ii), those students who superimposed the graph of 3 y x   over the graph of 3 y x   were usually more successful in explaining why 3 3 x x    for all real values of x. Otherwise, a variety of explanations were seen, of which some were detailed, usually containing at least one correct statement. Common errors included explanations which did not consider the case 0 x  or explanations including the statement that the graphs of 3 y x   and 3 y x   were parallel for all 0. x  A significant number of students who only tested the statement 3 3 x x    for specific values of x (usually one positive and one negative) did not receive any credit in this part.

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  Consider the statement: If m and n are irrational numbers, where m != n, then mn is also irrational.

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• Edexcel 9MA0 · sample · Paper 2 · Q6 · 6 marks

  6 marks

  6

  
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  For each statement, state whether it is always true, sometimes true or never true, giving a reason.

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• Edexcel 9MA0 · specimen · Paper 2 · Q14 · 8 marks

  8 marks

  14

  
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  Answer both statements with complete justification.

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