Algebra and functions questions

Question and mark-scheme extracts © Pearson Education Ltd. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.

• Edexcel 9MA0 · June 2024 · Paper 1 · Q1 · 3 marks

  3 marks

  1

  
  Show mark scheme

  

  Examiner report

  This question was very accessible, familiar to candidates and of low demand, making it particularly suitable for the opening question of the paper. Candidates were required to determine the value of a constant, k, in a cubic expression, g(x), given that (x – 3) was a factor of this expression. The vast majority of candidates found this question straightforward and clearly understood that they should apply the factor theorem to the question, determining the value of k via a linear equation in k that followed from g(3) = 0. A small proportion of candidates made errors in setting up the linear equation, or in attempting to solve their linear equation, however method marks were available in these instances. Some candidates did not include an = 0 in their work, however they were able to recover and achieve the marks if the = 0 was implied by later work. Quite rarely, g(–3) = 0 was attempted but even in this case credit for attempting to solve a linear equation was available. A common alternative route was to attempt division of g(3) by (x – 3), which made this otherwise routine question significantly more challenging. A number of methods for setting out the division were seen but marks could not be awarded until their linear remainder, only in k, was set equal to zero. When algebraic division was applied it was rarely completed well enough to gain all the marks, although many made quite pleasing progress with this more complex algebra. Centres should ensure candidates consider which method is more appropriate for solving a given problem. A very small proportion of candidates attempted to factorise g(x) and then equated coefficients which only a handful of candidates completed successfully.

  Text version

  g(x) = 3x^3 - 20x^2 + (k + 17)x + k, where k is a constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 1 · Q6 · 6 marks

  6 marks

  6

  
  Show mark scheme

  

  Examiner report

  This question on the modulus function provided plenty of access and there were very few blank responses to any part, although many found part (c) challenging. The overwhelming majority of candidates answered part (a) correctly, demonstrating a strong understanding that the minimum y value must occur when 2 0 x − = . The answer was often written down without working. Those candidates that did not gain full marks often achieved the method mark for one correct coordinate. The majority of candidates who scored no marks found the y intercept of the modulus curve instead. In part (b), the most common approach was to remove the modulus signs and to attempt to solve two equations, usually ( ) 16 4 3 2 5 x x − = − + and ( ) 16 4 3 2 5 x x − = − − + . The more confident candidates had already established that the line would only intersect the right-hand branch of the modulus graph. The most common error in processing was thinking that 3 2 x − was equivalent to 3 6 x + , but, pleasingly, this was less common than in previous series. Some candidates kept the modulus signs in their equation throughout which made the equation difficult to manipulate. Disappointingly, a surprisingly large number of candidates rearranged their equations poorly by not performing inverse operations correctly. For those that lost the accuracy mark in part (b), the most frequent reason was because they failed to reject the solution 5 x = . Those using the diagram to aid visualisation of the problem were generally more successful in rejecting this extra “solution”. Some strong candidates demonstrated an excellent understanding of the relevant domains for each branch of the modulus graph and rejected x = 5 as they had established that an intersection with this branch required x < 2. Part (c) was found to be less accessible with few candidates scoring full marks for a correct range for k. The candidates that were the most successful in this part of the question were generally those who had adopted a visual approach in the earlier parts of the question and understood how to find the minimum and maximum value of k using their knowledge of gradients. There was more success in finding the correct lower critical value of 0.5 with the most efficient solutions using the coordinates of P with y = kx + 4, but many candidates failed to realise that there was an upper limit for k. There was a reluctance to use diagrammatic reasoning and most resorted to complicated algebra. Many set up simultaneous equations in x and k but did not appear to understand what they were attempting to achieve, failing to solve for k and therefore these attempts did not gain any marks. There were many erroneous attempts that fortuitously achieved k = 3 through an attempt to use a discriminant (despite starting with a non-quadratic equation), and these invalid approaches scored no marks on their own. A few attempted squaring both sides of the equation to form a quadratic equation and then using the discriminant, but this approach often failed due to not isolating an appropriate term (such as 2 x − ) before squaring. Relatively few candidates considered the gradient of the modulus graph and hence failed to obtain an upper bound for k, which meant many only achieved the method mark in this part. Overall, candidates' use of inequality notation was good and although set notation was not required, the vast majority of those that used it did so correctly.

  Text version

  Figure 1 shows a V-shaped graph with equation y = 3|x - 2| + 5. Its vertex is P and it crosses the y-axis at (0, 11). A line l has equation y = kx + 4, where k is a constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 1 · Q8 · 11 marks

  11 marks

  8

  
  Show mark scheme

  

  Examiner report

  This question was answered very well by the majority of candidates with many scoring upwards of 8 marks out of the available 11. It highlighted that candidates continue to struggle with identifying the domain and range of functions, and that in most cases candidates are not aware that they should state the domain when defining a function. There were a minority of candidates who attempted both fg(2) and gf(2) in part (a) and fg(x) and gf(x) in part (c), seemingly confused by the notation or attempting to cover both options. Candidates should be discouraged from doing this – examiners mark the final response, and, in effect, candidates are wasting valuable time by attempting both cases. In part (a), the vast majority of candidates scored full marks. Most candidates determined g(2) and then substituted the result, –1, into f(x). A few candidates attempted to substitute the expression for g(x) into f(x) to determine a general form for fg(x) before substituting x = 2 to determine the solution. Those who tried to expand the squared term before substituting often could not find the correct answer as they had errors in the expansion and/or in dealing with the negative coefficient of the squared expression. Some candidates evaluated gf(2) instead of fg(2), although this was fairly infrequent. In part (b), the majority of candidates scored 2 marks. Common errors included attempting to differentiate g(x) or attempting to find its reciprocal, both of which scored no marks, while sign errors with the 9y or 9x term meant that other candidates missed out on the first accuracy mark. The final accuracy mark was more demanding and was commonly not secured as candidates rarely attempted to write down the domain of ( ) 1 g x − , a point that has been referenced in many previous examiner reports. It was also common that candidates did not use a suitable notation for their final answer for ( ) 1 g x − . In part (c)(i), the majority of candidates scored full marks. Where errors arose, this was often a result of attempting to simplify the final solution to another expression and with the errors being algebraic rather than arithmetical, ISW was not applied. It was surprising that some candidates who were successful in part (a) treated gf(x) as ( ) ( ) g f x x  here and thus failed to score. Those who had constructed an expression in part (a) incorrectly through working out fg(x) made the same error here. Very few candidates scored the mark in part (c)(ii). There was some evidence of confusion between the domain and range and amongst those who did look for the range only one critical value was found by many candidates: often giving either gf(𝑥) ⩾ −5 or ( ) gf 0 x  rather than the full interval. A small number of candidates who managed to identify both boundary conditions used x rather than y or gf(x) and thus did not score the mark. In part (d), some candidates were able to secure marks having found the previous parts challenging. Algebraic errors were not uncommon when equating the expressions and often the = 0 was “dropped” in favour of a stand-alone expression. Candidates generally progressed to determining the discriminant, but setting the discriminant greater than or equal to zero was often seen. Some candidates attempted to find the discriminant of a quadratic set = k which scored no marks. Some of those who attempted the correct inequalities lost the final mark due to slips when simplifying such as ( ) 36 20 4 36 20 80 k k − − == − − or failure to reverse the inequality when dividing through by –20.

  Text version

  The functions are f(x) = 4 - 3x^2 for x in R, and g(x) = 5/(2x - 9) for x in R with x != 9/2. The function h is h(x) = 2x^2 - 6x + k for x in R, where k is a constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 1 · Q10 · 9 marks

  9 marks

  10

  
  Show mark scheme

  

  Examiner report

  This question involved use of straight-line graphs and integration from the AS section of the course. For the most part, it was very well attempted by the majority of the candidates, but many struggled to find a complete strategy to find the area required in part (c). Part (a) was accessible to most candidates although many did not recognise the difference between being asked to “verify” as opposed to “solve”. The most efficient method was to substitute 4 x = and verify that 0 y = and most did this succinctly. It was equally common to see candidates attempt to solve 5 2 8 0 x x − = . Correct processing to 3 2 8 x = leading to 4 x = was required for the mark. This approach was far less likely to score the mark due to errors with indices or incomplete processing. Part (b) was usually answered correctly with many candidates scoring full marks. As the instruction was to “show that”, candidates were required to present all their working and it was not acceptable to work back from the given answer. Most candidates found a correct derivative and clearly substituted 4 x = into this. It was common to see both ( ) 1 1 y y m x x − = − and y mx c = + used to find the equation of the tangent. The former provided a more efficient solution with the latter required clear working to find 48 c = which then needed to be substituted back into . y mx c = + A small number of candidates failed to get to the required form, leaving their answer in the form 12 48 y x = − + and hence lost the final mark. Others jumped directly to the given answer without evidence of a correct method for the equation of the straight-line and these candidates failed to score the method mark. The final part of the question proved more challenging and differentiated between the candidates, with many producing concise and fully correct answers. Most candidates realised it required the use of integration and often scored the B mark for integrating the curve. Many candidates failed to note that the question required an exact answer and use of algebraic integration. This meant that using a calculator to provide a decimal approximation was insufficient for the final method mark. The quality of notation in this part of the question was very varied with the dx often missing from integrals and spurious integral signs included throughout working. The majority of successful candidates used 1 2 A bh = or integration to find the area of the triangle and then subtracted the area under the curve. However, a significant proportion of candidates made the question much more difficult than necessary, using a mixture of Pythagoras and trigonometry to find the area of the triangle-this often led to errors or premature rounding. Other common errors when finding the area of the triangle included assuming the triangle was isosceles, and therefore that the intersection point was at 2 x = , or splitting the triangle into two right-angled triangles and assuming they both had the same area. Candidates attempting to use the alternative method using lines – curve were more likely to make errors with signs, brackets or limits. They were also required to correctly cancel the ( ) 7 2 2 2.4 7 terms to find an exact answer and so most often were unable to score the final method and accuracy marks. The most common invalid method was attempting to combine the lines and curve before integrating using the limits 0 and 4, and this scored no marks unless the point of intersection had been found.

  Text version

  Figure 3 shows the curve y = 8x - x^(5/2), x >= 0. It crosses the x-axis at A. The line l1 is tangent to the curve at A. The line l2 is y=8x. Region R is bounded by the curve, l1 and l2. You must show all stages; solutions relying entirely on calculator technology are not acceptable.

  Open the full paper (PDF)

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

  6 marks

  15

  
  Show mark scheme

  

  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.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q2 · 5 marks

  5 marks

  2

  
  Show mark scheme

  

  Examiner report

  Part (a) was very accessible and very well answered. The vast majority of candidates deduced the value of the common difference and then substituted this into the standard term formula for an arithmetic sequence. Those candidates that were unfamiliar with this formula were able to produce a correct calculation, or even just list the payments for all 12 months, which was acceptable. Occasionally the formula was simplified to 410 − 10n before substituting in n = 12. Part (b) was usually answered well, and candidates were able to substitute the given values into the sum formula correctly, expand brackets and obtain the required quadratic equation in N. The fact that the common difference was negative caused some candidates algebraic difficulties. Some candidates did not realise they needed to set their expression equal to 8100 but soon realised this and added it in later in their working but a few persisted with e.g. = 0 rather than = 8100. A few had bracketing and/or arithmetic errors and some missed out the ‘= 0’ in the given answer therefore losing the final accuracy mark. A small number of candidates attempted to use the term formula in this part. Another rarer approach involved using a series summation approach. These attempts were mostly successful as it required further maths knowledge and were therefore performed by most proficient mathematicians. In part (c), most candidates were able to solve the given quadratic equation and achieve the method mark. Overall, this was done with a calculator. Most chose the correct answer but quite a few left both answers or chose N = 45, appearing not to take into account the context of the question. They needed to realise that, when the loan was fully repaid after 36 months, the borrower would not continue to make loan repayments, so the second answer of 45 months needed to be rejected. This took significant effort for some - rechecking the sum for 36 and 45 terms and then finding the value of the 36th and 45th term and then rejecting 45 when the 45th term was negative - when thinking about the context would have been easier.

  Text version

  Jamie takes out an interest-free loan of £8100. Jamie makes a payment every month to pay back the loan. Jamie repays £400 in month 1, £390 in month 2, £380 in month 3, and so on, so that the amounts repaid each month form an arithmetic sequence.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q3 · 4 marks

  4 marks

  3

  
  Show mark scheme

  

  Examiner report

  This question was very effective at highlighting many candidates’ misconceptions about the relationship between function notation and graph transformations. Whilst the majority of candidates coped well with single transformations, a significant proportion demonstrated a lack of secure understanding of multi-step transformations. Part (i) presented few difficulties for most candidates. They either understood, or had memorised, that y = f (x − 2) represented a translation by two units in the positive x direction. The most common incorrect answer here was (1, −2) where candidates had confused the direction of the translation. Similarly in part (ii), a large proportion of candidates, though fewer than in part (i), knew that y = f (2x) represented a stretch of scale factor 1 2 in the x direction. Common incorrect answers were (6, −2), from multiplying the x coordinate by 2 instead of dividing, and (3, −4), which came from multiplying the y coordinate by 2. Part (iii) highlighted that many candidates are potentially guessing the net effect of a multi-step transformation rather than separately considering the stepwise effect of the individual transformations. It appears that they are mainly unsure of the order in which the transformations take place and, probably because of this, what effect each step will have, and on which variable. The variety and range of incorrect answers was perhaps surprising. Whilst some candidates provided evidence to explain their logic, most simply wrote an answer. In many cases candidates either scored no marks or both marks here. When just one out of the two marks was earned, there was often little evidence to suggest where the correct and incorrect values had come from. Where piecewise transformations were seen, it was clear, perhaps unsurprisingly that translations were more successfully handled than reflection or stretch transformations and sign errors were fairly common. Some candidates drew small sketches to help visualise the transformations which often proved helpful in achieving the correct answer. Occasionally, other candidates came unstuck by apparently misunderstanding that the transformations for each part of the question should be applied to P(3, −2) rather than sequentially applied to the result of the previous part of the question.

  Text version

  The point P(3, -2) lies on the curve with equation y = f(x), x in R. Find the coordinates of the point to which P is mapped under each transformation.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q4 · 5 marks

  5 marks

  4

  
  Show mark scheme

  

  Examiner report

  Generally, candidates found this question accessible, with many scoring full marks or nearly full marks. In part (a) most candidates used the given recurrence relation along with u1 = 6 to find u2 = 6k – 5 and then u3 = 6k2 – 5k – 5, and then set u3 = – 1 and rearranged to obtain the given answer 6k2 – 5k – 4 = 0. Algebraic errors were few, but some candidates forgot to include the ‘= 0’ in their final line and lost the second mark in this part. There were occasional attempts at using arithmetic or geometric series which gained no credit. In part (b)(i) most candidates were able to solve the quadratic equation, but not all identified that k = 4 3 was the only correct answer, or subsequently used this value, so lost the mark in this part of the question. Some candidates spent more time than they needed using the quadratic formula, where a factorisation of the quadratic would have been more direct. Use of the calculator equation solver was also permitted here. Candidates must learn to check their answers against any limitations given in the question. When errors did occur, they were most likely to appear in part (b)(ii), although most still answered it well. Most candidates identified the requirement of the question to add the first three terms of the sequence together, and correctly used k = 4 3 to calculate u2 =6 ( 4 3 ) -5 before finding 6 + 3 + (−1) = 8. Others first simplified the sum to obtain 6k or 6k2 + k − 4 before substituting the value of k. A common incorrect approach was to use u3 = 6k2 – 5k – 4 in the sum. Some also used k = 4 3 and k = 1 2 − to calculate the sum, without identifying which was the required answer. Alternatively, a good number of candidates attempted, correctly and incorrectly, to work out the values of u1 and u3 even though they were given in the question. It was also noticeable that a number of candidates used the quadratic expression from part (a) in order to work out values in (b)(ii), usually assuming u3 = k2 – 5k – 4. They, therefore, did not clearly understand what this equation represented. A few candidates incorrectly attempted to use the recurrence relation to find things like ( 4 3 )(1) – 5, ( 4 3 )(2) – 5 and ( 4 3 )(3) – 5, using these expressions to find their sum, and there were again occasional attempts using arithmetic or geometric series which also gained no credit. There were also a few attempts at integration. Overall, this was a question that allowed candidates to gain a good number of marks with confidence.

  Text version

  A sequence u_1, u_2, u_3, ... is defined by u_(n+1) = ku_n - 5, with u_1 = 6, where k is a positive constant. Given that u_3 = -1:

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q9 · 7 marks

  7 marks

  9

  
  Show mark scheme

  

  Examiner report

  There was a very mixed response to this question. There was a fairly even split between the two alternative methods of using a model of the form 2 H ax bx c = + + and the completed square form of ( ) 2 9 H A x B = − + . Most candidates recognised the need for a coefficient of 2x (other than 1) and were therefore able to access some, if not all, the marks for this part. Using 2 H ax bx c = + + , most candidates accessed the first mark by identifying that c = 2 and correctly using the point (20, 0.8) to form an equation in a and b. The majority of candidates then used x = 9 and d 0 d H x = to form a second equation in a and b. A very small number of candidates formed a second equation by using another point which the curve passed through, either (−2, 0.8) or (18, 2), identified by using the symmetry of the curve. Unfortunately, not all candidates realised that the value of c was the H intercept of 2 and so lost all marks in this part. Using ( ) 2 9 H A x B = − + , the most common error was for candidates to incorrectly assume that B = 2. Otherwise, candidates had frequent success using this format for the model. The vast majority of responses correctly used (0, 2) and (20, 0.8) to form 2 equations in A and B. A small number of candidates used an equation of the form ( ) ( ) 2 9 81 H A x B = − − + and some attempted a linear transformation of the curve. These were less successful. Some candidates switched their model from 2 H ax bx c = + + to ( ) 2 9 H b a x = + − and then confused their a’s and b’s assuming they were the same value. Very few errors were made once correct equations in a and b (or A and B) had been found so via both approaches, candidates were successful in correctly solving their simultaneous equations and achieving a fully correct equation for the model. Whilst the vast majority of candidates left their final answer as an equation, some lost the final mark for failing to have H = f (x) and instead giving their answer as y = f (x). When no marks were scored in this part, the most common error was to assume B = 2 for the form ( ) 2 9 H A x B = − + . A surprising number of candidates also attempted a linear model and also scored no marks in this part. In part (b), most candidates were able to identify a limitation of the model, so this part was well answered, often independently to the rest of the question. Most successful responses referred to the fact that air resistance hadn't been considered. Other common responses achieving the mark included “the spin of the ball has not been considered” or “the ground is unlikely to be completely horizontal/flat”. A small number of responses did correctly refer to the fact that H is unlikely to be a perfect quadratic function in x or that the ball is unlikely to travel in a vertical plane therefore adding a third dimension to the motion. The most common incorrect responses involved candidates referring to the ball outside its flight path e.g. “H will become negative” or “the ball will go below the ground”. Some candidates did give vague single word answers like “air resistance” or “spin” which did not gain them any credit as there was no link to the model. In part (c), the vast majority of candidates recognised the need to substitute x = 16 into their model and candidates who achieved an incorrect quadratic model in part (a) were still able to achieve the method mark. When candidates achieved full marks in part (a) they usually went on to achieve full marks in part (c) too. It was extremely rare for H = 2.96 to be achieved and an incorrect conclusion to follow as most candidates knew to conclude correctly, fully linking back to the question. Very few candidates attempted the alternative approach of substituting in H = 2.5 into the model to create and solve a quadratic equation in x. This approach did then require candidates to understand that the (larger) solution of x = 17.0 meant that Chandra could not catch the ball. This seemed a more challenging approach than just substituting x = 16 in the model.

  Text version

  A graph models the path of a small ball travelling in a vertical plane above horizontal ground, with a fixed origin O on the ground directly below the throwing point. H metres is the height above the ground and x metres is the horizontal distance from O. The ball is thrown from A(0, 2), caught at B(20, 0.8), and reaches its maximum height when x = 9. A quadratic function linking H with x is used to model the path.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q12 · 12 marks

  12 marks

  12

  
  Show mark scheme

  

  Examiner report

  This question tested the candidates’ skill in various areas, i.e. finding partial fractions, integration, rules of logarithms, manipulating equations to eliminate logarithms using exponentials, indices, and commentating on the expected behaviour of a variable as another tends to infinity. This was a challenging question which in a number of instances, candidates struggled with after part (a). Those however, who were well prepared, were able to score full marks often with relatively concise solutions. Part (a) was well answered. Most candidates were familiar with the process of rewriting expressions using partial fractions and obtained a correct answer such as ( ) 1 1 25 25 25 V V + − . There were occasional sign slips in the coefficients, and some candidates failed to write down their final answer, or use it in part (b), losing the second mark here. In part (b) a number of correct forms of integration were possible. Many candidates appreciated that an attempt at separation of variables and integration was required, and many reached correct equations such as ( ) 1 1 1 ln ln 25 25 25 10 V V t c − − = + or ( ) 1 1 1 ln 25 ln 25 25 25 25 10 V V t c − − = + . There were however, many slips with coefficients, and the omission of the minus sign in front of the ln(25 – V) or ln25(25 – V) term was common. Additionally, some omitted a constant of integration entirely, which meant that no further progress could be made. A surprising number of candidates did not see any link with part (a) and instead, integrated a quadratic in V, or did not integrate at all. Most candidates who attempted integration on both sides and included a constant of integration were able to score 2 method marks by using t = 0 and V = 20 to find the value of their constant and then substituting V = 24 to find the time taken. Most who got this far converted their final answer into minutes as required by the question although some candidates left the answer as 0.7167 which lost the final accuracy mark in this part. A few candidates used the alternative method of applying limits of 20 and 24 to their integral with respect to V and 0 and T for their integral with respect to t. Many candidates did not reach part (c), and some who did were not able to gain marks there because they either had not established a numerical constant of integration, or because their integrated expression was not of a correct form to earn credit. Successful attempts started with expressions such as ( ) 1 1 1 1 ln ln 25 ln 4 25 25 10 25 V V t − − = + to obtain expressions such as ( ) 1 1 ln 25 4 25 10 V t V = − and then eliminated the log to obtain ( ) 2.5 e 4 25 t V V = − and made V the subject. Some candidates obtained a correct expression for V but failed to reach the form required by the question. Incorrect log work, however, was common and expressions such as ( ) 2.5 e 4 4 25 t V V = + − were seen frequently. Candidates often made sign or coefficient errors but were awarded the method marks in (c) for a correct algebraic approach for the required rearrangement. A significant number didn’t reach the required form, leaving their answer with two 2.5 e t terms rather than the required single 2.5 e t − term. In part (d), candidates sometimes identified that L = 25, even when little else of credit was presented throughout, but usually those who had not attempted part (c) also made no attempt at (d). The reasons given were often not rigorous enough to gain credit, even when 25 had been established as the required value of L. When correct explanations were seen they usually stated that L = 25 because 2.5 e 0 t − → as t → or that L = 25 because d 0. d V t =

  Text version

  The volume V microlitres of a plant cell t hours after the plant is watered is modelled by dV/dt = (1/10)V(25 - V). The plant cell has an initial volume of 20 microlitres.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q14 · 8 marks

  8 marks

  14

  
  Show mark scheme

  

  Examiner report

  A common marking profile in this question was to score all three marks in (a) followed by zero marks in (b). In part (a) the most common approach was to complete the square and very often it was successful although there were a variety of incorrect answers for the centre – usually variations of ( ) 3, 7 .   Quite a few knew the strategy for finding the radius although not all of them knew how to use it correctly and 91 r = was a frequently seen incorrect answer from attempting ( ) ( ) 2 2 3 9 7 49 33 0. x y − + + + + + = A surprising number of candidates evaluated 72 as something other than 49. In part (b), it was common for no marks to be scored and there were a large number of non- attempts. The most successful candidates were those who drew diagrams and looked at the problem geometrically. However, the most common approach was an algebraic one. This typically involved solving the 2 circle equations simultaneously, but the majority of such attempts were aborted before any marks were gained. There was also some poor use of algebra with things such as ( ) ( ) 2 2 3 7 25 3 7 5 x y x y − + + =  −+ + = seen a surprising number of times and such attempts gained no credit. Many candidates did recognise that there would be a discriminant involved so they often arrived at a point when they thought that it would be appropriate to calculate one. However, this was usually prematurely. Despite this, it was noted that some examiners did report the occasional successful algebraic attempt. Some candidates calculated the equation of the line passing through the centres but then did not realise what they had to do with it. Most who were able to find the correct distance between the centres recognised they had to add and subtract the radius found earlier to give the limits of the solution set. Candidates who made it this far sometimes lost the final accuracy mark due to the use of incorrect notation. For candidates who got as far as the correct interval for k, the most successful were those who gave the final answer as a single set. Candidates who broke the answer into two sets, often used the incorrect symbol between them.

  Text version

  The circle C_1 has equation x^2 + y^2 - 6x + 14y + 33 = 0. A different circle C_2 has centre (-6, -8) and radius k, where k is a constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q2 · 6 marks

  6 marks

  2

  
  Show mark scheme

  

  Examiner report

  This question was accessible and generally well answered, with a large number of candidates gaining full marks with concise solutions. Some, however, did not pay heed to the instruction to show all stages of their working. The vast majority of candidates correctly identified that part (a) related to an application of the factor theorem, substituted in a and were able to reach the required result. However, many lost marks for not including " = 0" until the final line. In a small number of cases, " = 0" was omitted completely and, in these cases, candidates were awarded no marks. Amongst the unsuccessful candidates, the majority had attempted algebraic long division using ( ) x a − or carried out an attempt at factorisation via inspection, but had often abandoned their attempts, presumably unsure of how this would lead to the required result. Very few successfully attempted long division and set the remainder equal to 0. Factorisation using inspection then solving two simultaneous equations by equating like terms was rarely seen as a method. In part (b)(i), most candidates had no problems solving the cubic, getting the values 0, 3 4 and 2 − for a. However, many did not take note of the fact that a was positive, and that the question asked for the value of a, so lost the mark for not rejecting 𝑎= 0 and 𝑎= −2. There was confusion as to the reasons for rejecting values. Some thought that a had to be an integer so rejected 3 4 . Some chose 2 − as their solution giving spurious reasons such as “a is a real number.” Of those who did not identify 3 4 a = as the only valid answer, many recovered by opting to use only 3 4 a = in part (b)(ii). Part (b)(ii) was well done by those who showed every step of their working, but some failed to reject the values of a that were not needed. Once the cubic was obtained by substituting 3 4 a = and setting f( ) 3 x = , those who solved the cubic without factorising to 2 (4 5 10) x x x + − lost the final A mark. Candidates were asked to show all stages of their working and not proceeding as far as the quadratic factor was penalised. Many gave the exact solutions followed by rounded decimal solutions, or even just the decimal solutions, so were not paying heed to the question asking for exact solutions. There were some who were unable to solve the cubic or the quadratic, having taken out a factor x outside the brackets. These were mainly cases where they tried to complete the square or their value for a meant that they could make no further progress. Some did not collect terms on one side with 0 on the other side, and having taken out a factor of x, solved their quadratic ignoring the constant on the other side. The solution 𝑥= 0 was omitted by many candidates due to division by x instead of using factorisation. Others rejected this particular solution, if it was found, possibly confusing the restriction that a was positive rather than x. Those who did reach the correct quadratic factor were usually successful in finding the two exact roots for the A1 mark. A common incorrect approach was to find f(3) rather than setting f( ) 3 x = . This approach gained no credit.

  Text version

  In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. f(x) = 4x^3 + 5x^2 - 10x + 4a for x in R, where a is a positive constant. Given that (x - a) is a factor of f(x).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q5 · 6 marks

  6 marks

  5

  
  Show mark scheme

  

  Examiner report

  This was a lovely twist on a usually very standard question using the trapezium rule that was overall very well-answered. It was very rare to see a blank response to this question. Candidates usually scored full marks in part (a). Where marks were lost the most common error was working out “h”, the width of each strip, as (4 3) 1 6 6 − = where 6 was the number of ordinates given, rather than (4 3) 1 5 5 − = , where 5 was the number of strips required. Very few candidates failed to give “h” a value. Most applied the standard trapezium rule correctly, although a minority failed to use the area of 17.59 to form an equation. Only a handful of solutions used the method of adding the areas of individual trapezia, which is much more time consuming and does not demonstrate as effectively an understanding the trapezium rule, which is provided in the formula booklet. The brackets were dealt with correctly in most cases although quite a number of candidates expanded the brackets before collecting terms, creating a more difficult equation to cope with. Others multiplied both sides by 10 which made the manipulation much easier. Several candidates had invisible brackets and recovered appropriately but lost the A mark as the answer was given. Although many values needed to be copied from the given table of x and y values, there were very few copying slips seen. Part (b) was dealt with very well overall, with a variety of different methods used to solve the equations simultaneously, usually via the use of a calculator. Some arithmetic errors were seen in forming the second equation, 28 a b + = , which resulted in the wrong values for a and b. Again, despite requiring the copying of many values from the given table of x and y values to form the required equation, there were few transcription errors. It is also worth noting that nearly all candidates used the given equation in (a) to correctly solve for values of a and b. Although it was made possible for them to gain the method mark for using “their” equation, this was very infrequently utilised, or required. Furthermore, it was pleasing to see candidates who made no attempt at (a) still proceeded to use the given answer to attempt and generally gain full marks in (b).

  Text version

  A continuous curve has equation y=f(x). The table gives corresponding values: x = 3, 3.2, 3.4, 3.6, 3.8, 4 and y = a, 16.8, b, 20.2, 18.7, 13.5 respectively, where a and b are constants. The trapezium rule is used with all the y-values to find an approximate area under the curve from x=3 to x=4. This area is 17.59.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q7 · 8 marks

  8 marks

  7

  
  Show mark scheme

  

  Examiner report

  Overall, this was an accessible question on functions which allowed candidates to attempt all sections of it even when they found parts (a) or (b) particularly challenging. Part (c) had the lowest number of errors out of the entire question and was one of the most successfully answered parts over the entire paper. In part (a), a large majority of candidates were able to score the B mark available, with nearly all attempting to give the range. Although the mark scheme was generously accepting notations such as f 3  or range 3  , some candidates lost the mark for unacceptable labels such as 3 x  or f( ) 3. x  In general, candidates correctly rearranged the formula and interchanged x and y successfully in part (b). A small number of candidates made some errors in applying the correct order of operations and this resulted in an incorrect expression for the inverse function. A common error was seen with candidates who incorrectly manipulated 3 2 y x = + − to obtain 2 9 2 y x = + − . The candidates who lost the A mark were often due to labelling the inverse as y = instead of e.g. 1 f ( ) ... x − = . A small minority of candidates misread 1 f ( ) x − as f ( )x  and so they differentiated the function instead of finding the inverse of it. Candidates should be advised to read the question and the labels carefully to check which skill is required. A common error costing candidates the B mark was failing to record the domain of the inverse function. This mark was rarely scored across the entire cohort, despite 3 marks possibly indicating that more was required by candidates than just rearranging to find 1 f ( ) x − . Candidates who paid careful attention to how f and g were defined in the question would be able to attempt similarly for the inverse function. Part (c) was well attempted by many candidates even when they could not attempt parts (a) or (b). Most candidates scored both marks available in this part. Finding f(6) first and then substituting this result into g was more common than substituting 6 x = into the composite function gf(x). This second approach caused some errors due to incorrect algebraic manipulation leading to various incorrect answers for the composite function. A large number of candidates scored the first mark in part (d) as they correctly formed the equation ( 3)(3 ) 15 a a − + = and proceeded to a quadratic in a. Several candidates failed to spot that 2 a a = , or that 2 2 2 2 a a + − = . Some candidates that found 2 6 a =  did not reject the negative solution losing the last A mark. Some candidates possibly rejected the negative solution by seeing that the domain of f had to be greater than 2 and fortuitously scored the final mark. However, others may have considered that the ( ) 2 2 6 2 − + also satisfied this domain resulting in ( ) ( ) 2 f 2 6 2 0 − +  , but that ( ) g 2 6 0 −  so this negative solution had to be rejected. Other errors included writing decimal answers only. On several occasions 2 9 15 a − = incorrectly became 2 6 a = . A number of candidates were not able to obtain the quadratic equation needed (hence, did not score any marks) because they got confused by 2 3 a + and not realising that this was actually3 a + . Others tried to manipulate the equation leading to a quartic equation and usually solved via a calculator (and successfully in a few cases), however those who had proceeded along this route typically were expected to apply algebraic division or the factor theorem to find the quadratic factor, having introduced other solutions from squaring and the complexity of their expressions meant it was rare for the method mark to be scored.

  Text version

  The function f is defined by f(x)=3+sqrt(x-2), for real x with x>2. The function g is defined by g(x)=15/(x-3), for real x with x!=3.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q10 · 9 marks

  9 marks

  10

  
  Show mark scheme

  

  Examiner report

  Candidates were typically able to make good progress in part (a) with the method of completing the square being familiar to many when dealing with the equation of the circle. Part (b) was focused on using the discriminant and solving a quadratic inequality, however, the manipulation proved to be a lot for some, resulting in a wide range of marks. Again, the access was good on this question such that candidates who could not do part (a) were not restricted from attempting part (b), although those with errors from part (a) often found that this caused issues in part (b) and resulted in losing many of the marks. In general candidates obtained ( 3 , ) k k − as the centre of the circle in part (a). Common errors included (3 , ) k k − and ( 3 , ) kx ky − . If candidates were successful at completing the square, they could find the correct radius of the circle. When attempting to complete the square a common error was to subtract 2 3k instead of( ) 2 3k . Occasionally, candidates would find 2 2 10 7 r k = − but did not proceed to take the square root. Those who collected terms on the right-hand side of the equation in order to form the square of the radius made fewer sign errors than those who did not. Others tried to manipulate the equation mentally, so writing down a wrong radius with no evidence of its development (and thus no method mark). One error was to write 2 10 7 10 7 k k − = − . Sometimes there were sign slips such as writing the 7 − as +7 when rearranging the equation. In part (b), whilst a number of candidates had little idea how to proceed, the vast majority did realise what was required and attempted to produce an equation in x by substitution. Many lost their way in the algebraic complexity, or did not know what to do with the equation (sometimes trying to solve it to find a value for k). A significant number of candidates did produce a three-term quadratic for x, with coefficients in terms of k. Centres should note that the scheme required the quadratic in x to be of the form specified, and that it be set out in a way which identified the coefficients (possibly implied by later use in the discriminant). So, for example, “ (2 4) k − x” was required rather than “ 2 4 kx x − ” as this provided evidence of recognising that the coefficients were the key to solving this problem. Quite a few candidates had a constant term which was not of the required form which meant that no marks could be scored. In general candidates who found critical values for their discriminant by solving 2 4 0 b ac − = were more successful than those who expressed it as an inequality. Often the inequality was simply carried forward to a statement about k; those who identified the two critical values by solving an equation perhaps thought more carefully about where the region was. The “need” to write a single inequality was seemingly pressing for many, so 7 85 7 85 k +   − was seen. A few candidates did not use exact values of k and lost the final accuracy mark. Some candidates used 'and' with their two regions, however, most recognised that this needed to be 'or', either stating this explicitly or just using a comma between their two answers. A number of capable candidates brought some interesting approaches to bear on the problem, adopting a variety of techniques from Further Maths specifications.

  Text version

  A circle C has equation x^2 + y^2 + 6kx - 2ky + 7 = 0, where k is a constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q13 · 7 marks

  7 marks

  13

  
  Show mark scheme

  

  Examiner report

  A significant number of candidates found this question very challenging, with a fair proportion of blank answers, and another significant number working with various substitutions in (a) that were quickly abandoned. Those who were able to work with both the quadratic and trigonometric models confidently were able to score highly and demonstrated a good understanding of how mathematics can be used to model real life contextual problems. In part (a), there were two common approaches that led to successfully scoring full marks. One was applying the knowledge that 2 ( 20) 0 b t − = when 60 H = which therefore meant 60 a = . This then enabled the candidate to substitute 2, 0 H t = = and find a complete equation. It was clear when candidates were able to apply this knowledge, and knowing this gave them a huge advantage in this question. Noticing that b was positive was pivotal when taking this approach. The other approach that led to successful answers involved multiplying out the brackets, and differentiating with respect tot. This enabled the candidate to find that 20 t = when d 0 d H t = , thus overcoming a shortfall in knowledge over when geometrically the maximum occurs in a negative quadratic. This second approach led to success less often, and candidates often ended up with a page of various quadratic expressions and equations that looked costly in terms of time and did not yield marks. An overwhelming majority of candidates who scored full marks in (a) successfully found H to be 2 m in part (b). Some candidates still managed to find this value with either an incorrect answer in (a) or no answer at all, by perhaps noticing that due to the symmetry of the quadratic, the height after 40 seconds was the same as at the start, which they are told was 2 m in the question. In part (c), a significant number of candidates had no real idea how to start this. Many attempts involved fruitlessly expanding cos(9 ) t  + but most candidates who successfully found  and  did not do this. The candidates who realised that when 60 H = , cos(9 ) 1 t  + = which meant 31 = had the most success. Candidates who did this typically then substituted 20 t = into cos(9 ) 1 t  + = to find a value of . Candidates who found one value typically found the other. There were a number of candidates who found the two values that did not ‘find a complete equation’ as the question required. However, there were a pleasing number of fully correct solutions, and almost all candidates who found and  continued to give the full model. In part (d), the most popular correct answers involved an identification of the cyclic nature of the trigonometric function, or realising the first model would become very negative. A significant number of incorrect answers involved candidates saying the original model would keep increasing over time, which showed a lack of understanding of the negative quadratic model. Others made reference to the alternative model being continuous which was too vague. Some candidates managed to score this mark without scoring any other mark on this question which was, again, pleasing to see candidates attempting later parts of questions even if they have made little progress on earlier parts.

  Text version

  On a roller coaster ride, passengers travel in carriages around a track and complete multiple circuits. The maximum vertical height of a carriage above the horizontal ground is 60 m, and a carriage starts a circuit at a vertical height of 2 m. The height H metres, t seconds after the carriage starts the first circuit, is modelled by H=a-b(t-20)^2, where a and b are positive constants.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q3 · 5 marks

  5 marks

  3

  
  Show mark scheme

  

  Examiner report

  There were very few non-attempts or zero scores here and this question proved to be a good discriminator with a spread of marks achieved by candidates. Most students managed to score at least one mark out of three in part (a) by correctly applying a single log law, usually the product, division or power rules for logs. Dealing with the constant, 2, proved to be more challenging for many. Those who wrote 2 as log2(4) and then applied the product rule for logs on the right- hand-side had most success. Others grouped the three log terms and applied the exponential rule to eliminate the logs. Sometimes there was a lot of working which wasn’t always set out as clearly as it could have been. A significant proportion of candidates did succeed, but it was not uncommon to see correct use of a log law followed by incorrect log work in an attempt to fudge the printed result. Some poor responses involved an initial ‘cancelling’ of logs before any attempt to combine log terms or to raise both sides to the base 2 correctly. There was a minority of candidates who perhaps knew how to proceed but did not show sufficient working to gain full marks in a ‘show that’ question. Candidates should be reminded that all steps should be shown in this type of question. It was a shame when fully correct log work was followed by an incomplete ‘equation’ at the end of this part, as missing the ‘= 0’ meant the loss of the final mark here. A small number of candidates who were unsure how to start part (a) used the printed answer and tried to “work backwards” but often made errors leading to incorrect solutions. In (b) part (i), almost all candidates were able to correctly state the values ‘6’ and ‘ 5 3 − ’, it was clear that in many cases candidates were making use of their calculators to solve the quadratic equation which was acceptable. Part (ii) however, was answered less reliably. Although most candidates had some understanding of the domain of a log function and knew that 5 3 x = − was not a valid solution, many were unable to provide reasoning that was precise enough for the mark here. It was common to see, for example, “a log can’t be negative”, which was not considered sufficiently clear. Others stated simply ‘because it is negative’ which was also insufficient. A range of responses which engaged specifically with the domain of a log function needing to be positive were considered acceptable such as; ‘log of a negative value is undefined/impossible/gives a math error’.

  Text version

  Given that log_2(x + 3) + log_2(x + 10) = 2 + 2 log_2 x:

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q10 · 7 marks

  7 marks

  10

  
  Show mark scheme

  

  Examiner report

  Part (a) was generally done well with most candidates not deterred by the occurrence of the “k” in the numerator. The majority of students correctly formed the partial fractions and set up the identity in terms of A and B. Once there, most of them correctly went forward by substituting x as 2 and  4 and finding A and B in terms of k. Some set up their partial fractions in the form 4 2 B C A x x + + + − but very rarely managed to achieve A = 0. The method of comparing coefficients and solving two equations in A and B simultaneously was seen very occasionally. Errors were made where candidates had assigned the wrong values for A and B in their final partial fractions. A common sign error was made when substituting in x = − 4, leading to solving −12k − 18 = −6A to obtain A = 2k − 3, instead of A = 2k + 3. Some did not fully simplify their expressions for 𝐴 or 𝐵, leaving them as fractions. Another error at this stage generally involved not substituting for x in the 3kx term, leaving A and B in terms of x. Some students transposed the x  2 and x + 4 at the beginning so lost the B mark but generally followed through correctly and scored the method mark. In a small number of cases, the last mark was lost after candidates correctly found the numerators in terms of 𝑘, but did not write down the correct partial fractions. Very few candidates managed full marks in part (b), although the majority did gain the first method mark for integrating to obtain …ln(x + 4) or …ln(x  2). Those who did not recognise the logarithmic form for the integration were unable to score any marks in this part. Some achieved the correct form after first using a formal substitution. Of those who integrated their partial fractions successfully, the vast majority were far from strict in their use of the modulus symbol and many lost marks due to lack of appreciation of its importance. Use of modulus notation for integrating reciprocal functions should be picked up by centres as a teaching point with future cohorts. Students who had written the incorrect term ( ) ( ) 3 ln 5 k − − − or even the correct term ( ) 3 ln 5 k − − − after applying the limit 3 x = −, went on to indicate that they believed this term was zero or could just be ignored. Another example of poor practice was missing brackets around the coefficients e.g. ( ) ( ) 2 3ln 4 3ln 2 k x k x + + + − − being surprisingly common. The “invisible” brackets were sometimes recovered but in many cases were not. There were also examples where the brackets around the (x + 4) and the (x  2) were also missing. Those with a systematic well-laid out approach for the substitution of limits in both terms and identifying clear subtraction of the lower limit, went on to score 3 or 4 marks in part (b) and commonly full marks for the question. Many candidates however were hampered by their own layout making the correct interpretation of ln (5) as ln |5| even more challenging. The perhaps unfamiliar, exact form needed for k also proved a challenge for even the best candidates. There were some elaborate attempts to achieve an expression for k, some using exponentials, but many gave up along the way. There were, however some very elegant and correct solutions to this question and some different forms of the exact equivalent answer were seen, for example 5 21log e 6 − but most correct answers were given as 21 6. ln5 −

  Text version

  The function f is defined by f(x) = (3kx - 18)/[(x + 4)(x - 2)], where k is a positive constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q12 · 10 marks

  10 marks

  12

  
  Show mark scheme

  

  Examiner report

  In part (a), most candidates were successful in understanding the requirement to substitute zero into NA and NB. There were some processing errors resulting in incorrect values for NA and NB. A significant number failed to score the method mark by not subtracting their values. There were a few candidates that did not recognise the question was scaled in thousands and left a final answer of 5. In part (b), a large majority of candidates were able to deduce that T = 3 and most explanations were around the idea of the values increasing after this point. The next most common explanation centred on the graph being a minimum or having “turning point” at this location. There were also candidates who stated the gradient has increased or was positive. In part (c), a significant number of candidates failed to identify either of the correct equations. Many were able to identify and solve correctly at least one equation to find one critical value for t. A common error often followed their correct equation of 5 = 3t when they wrote the value of t as 3/5 and similarly, but less common, 13 = 3t leading to 3 13 t = . Some students kept the modulus signs within their equations and ignored them at various stages in their working which often led to incorrect equations. There was an array of methods seen trying to negotiate the modulus signs, and A B A B + = + was seen a few times and 2 2 2 A B k A B k + =  + = was also seen. Those who found two values for t were generally able to score the 2nd follow through accuracy mark for choosing the outside region for their critical values. The biggest challenge was the final accuracy mark as quite a few candidates did not write the inequalities using set notation. Those who attempted set notation usually were unable to write their answer correctly in that form, some confusing  with . A common answer was 5 13 3 3 t t         which was condoned for this final mark. Those who attempted the squaring method in this part had minimal success depending on when they chose to square. Many candidates were successful in part (d) even if they had not progressed with the other parts of the question. The majority of candidates correctly referenced the subscribers will become negative to obtain this B mark. Where a candidate attempted this part but did not obtain this mark, it was usually due to not making a reference to the number of subscribers becoming negative, saying it goes to zero or that it is linear or decreasing. Several candidates thought it couldn’t go above 8000 subscribers.

  Text version

  The numbers of subscribers, in thousands, to music-streaming companies A and B are modelled for t >= 0 by N_A = |t - 3| + 4 and N_B = 8 - |2t - 6|, where t is the time in years since monitoring began. A sketch in the original paper shows both model graphs over the first 5 years. Use the equations to answer all parts.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q1 · 4 marks

  4 marks

  1

  
  Show mark scheme

  

  Examiner report

  (Mean mark 3 out of 4) This was an accessible question for virtually all candidates. It asked for the point to which P was mapped, and most candidates gave this as a coordinate pair. Part (a) was frequently correct; candidates identified the translation and applied it successfully. The attempts at part (b) were less successful. Most solutions used the modulus concept but applied it to both coordinates. Part (c) was a 3-stage transformation and required the candidates to identify both the transformations and the order in which they were to be applied. This was more of a challenge. In particular, the stretch parallel to the y-axis and final translation were mixed up; those who did, generally received some credit as they obtained x = 0. Question % of candidates scoring full marks 1 42% 2 69% 3 39% 4 49% 5 34%

  Text version

  The point P(-2, -5) lies on the curve with equation y=f(x).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q2 · 3 marks

  3 marks

  2

  
  Show mark scheme

  

  Examiner report

  (Mean mark 2.5 out of 3) This was a familiar and accessible question with most candidates scoring 2 or 3 marks. The most common and successful approach was via the use of the factor theorem. Following on from setting f (-2) = 0, most candidates were able to set up and solve the linear equation in k. Errors seen resulting in the loss of marks were: • having (10k) not (10+k) in the linear equation • expanding -6(10+k) to ±60±k, or making a sign error on the final stage to get k=17 • setting f (2) = 0 • expanding the expression before substituting, often leading to sign errors or the loss of the 42 There was a small proportion of candidates who tried to use algebraic division to solve the problem, but most did not score many marks since the working needed here was much more complicated. Often the division was left incomplete, or there was not a linear remainder in k to set equal to zero. Even fewer candidates attempted factorisation by inspection; only a handful of candidates gained marks using this method.

  Text version

  f(x)=(x-4)(x^2-3x+k)-42, where k is a constant. Given that (x+2) is a factor of f(x).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q5 · 6 marks

  6 marks

  5

  
  Show mark scheme

  

  Examiner report

  (Mean mark 4.8 out of 6) This question involved solving simultaneous equations set within the context of the height of a tree modelled over a number of years. The question was very accessible for many candidates. In part (a) the majority of candidates gained the first three marks for forming and solving two correct simultaneous equations. Many went to a great deal of effort to solve the simultaneous equations using algebraic methods, whereas some just put the equations into a calculator and wrote down the values of the constants a and b. Some candidates used the 3 significant figure value of a to subsequently find the value of b, resulting in finding b as 1.94 rather than 1.95 and thus losing the final A mark. A significant number of candidates correctly found the values of the constants a and b, but then failed to state the equation of the model explicitly anywhere in their answer, losing the final A mark. In part (b) most candidates realised they had to substitute either t =20 or h =7 into their model and compare the value obtained with the given values. However, some candidates did not know how to evaluate the model with several coming to the wrong conclusion that the model was poor, because it didn’t give the exact values for h or t. There were quite a few candidates who correctly calculated the percentage error as an impressive comparison. It is very important for candidates to give a reason for their answer when the question explicitly demands it.

  Text version

  The height h metres of a tree, t years after being planted, is modelled by h^2=at+b for 0<=t<25, where a and b are constants. The height was 2.60 m exactly 2 years after planting and 5.10 m exactly 10 years after planting.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q6 · 6 marks

  6 marks

  6

  
  Show mark scheme

  

  Examiner report

  (Mean mark 2.1 out of 6) There was a very mixed response to this question on a cubic function with only one in eight candidates scoring all 6 marks. Part (a) was for finding the set of values of x for which f ' (x) < 0. Candidates generally knew what to do and most scored this B mark. Marks were lost when candidates gave the inequality as 2 6 x   or else 2 f ( ) 6 x ′ < < . In part (b) the request to give the answer in set notation proved a challenge for many candidates, with many only gaining the M mark for correctly identifying at least one of the correct inequalities. A minority of candidates gained both marks by giving their answer in acceptable set notation but there was widespread misapplication of ∩ rather than . ∪ Part (c) required the candidates to find the equation of the curve. A very common response, which gained only the first mark, was to give their answer as 2 ( 6) y x x = − without any appreciation of the stretch factor. Candidates who did appreciate that it could be 2 ( 6) y ax x = − could easily use the coordinate (2, 8) to find the value of a and hence the equation of the curve. A sizeable minority of otherwise good candidates lost the final mark by incorrectly writing down the equation of the curve as 2 1 ( 6) 4 C x x = − . There were many more time consuming and long winded methods seen here. A significant number of candidates formed the equation of a cubic, i.e. 3 2 y ax bx cx d = + + + and then attempted to use the points on the curve to find the values of a, b, c and d. This approach was rarely fully correct with many failing to identify that d = 0 or being able to form only 2 equations. A small minority of candidates did attempt to use the turning points at x = 2 and x = 6 to form the equation d ( 2)( 6) d y x x x = − − without again, an appreciation of the need for a stretch factor. It was disappointing to see how many candidates got bogged down in pages of algebraic working without noticing that this part of the question was only worth 3 marks.

  Text version

  A sketch of the cubic curve C, with equation y=f(x), shows that C passes through the origin, has a maximum turning point at (2,8), and has a minimum turning point at (6,0).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q11 · 7 marks

  7 marks

  11

  
  Show mark scheme

  

  Examiner report

  (Mean mark 4.3 out of 7) This question was based around the intersection of a cubic and a quadratic equation. It proved to be very accessible with many candidates able to make progress in both parts. In fact 30% of all candidates scored full marks on this question. Part (a) asked candidates to verify that the curves intersect at 1 . 2 x = The easiest and most common approach was to substitute 1 2 x = into both equations, obtain y values of 41 4 in each and then conclude that the curves did intersect. Candidates using this approach generally scored the method mark, but some lost the accuracy mark either from failing to show full details of the calculations, or from failing to state the conclusion that the curves intersect. Other methods seen included • setting the equations of the curves equal, rearranging to form a cubic, and then showing that 1 2 x = satisfied the resulting equation. Again, most gained the method mark, but again some lost the accuracy mark due to a lack of a conclusion. • setting the equations of the curves equal, rearranging to form a cubic, using the calculator to solve the cubic equation and then writing down the three solutions, indicating that one of the solutions was 1 . 2 x = This approach could only achieve the method mark here. In part (b) most candidates made some progress with weaker students scoring only the first mark for setting up the cubic equation. The majority were then able to divide by either 1 2 x   −       or ( ) 2 1 x − to form a quadratic factor and then equation which could then be solved to find the positive root. Marks lost were usually as a result of • slips in establishing the cubic equation or else errors in the division • not giving the exact answer of 4 33 −+ but instead its decimal equivalent • using a calculator to solve the cubic equation 2 3 2 15 42 17 0 x x x + − + =

  Text version

  For x>0, curve C_1 has equation y=2x^3+10 and curve C_2 has equation y=42x-15x^2-7. The curves intersect at x=1/2 and again at a point P.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q13 · 7 marks

  7 marks

  13

  
  Show mark scheme

  

  Examiner report

  (Mean mark 3.8 out of 7) This question on arithmetic sequences and series proved to be very discriminating. The proof of the sum of an arithmetic sequence is standard, but was not answered well by many candidates. Most candidates however were able to then made progress with the rest of the question, having been given the required formula. Good candidates often did well in this question however, with 17% of the cohort achieving full marks. In part(i) those who had learned the standard proof for the sum of the first n terms of an arithmetic series often gained full marks. It was however, unusual to see a well set out, fully correct proof. Candidates often failed to write a correct expression for Sn at the start of their proof, or sometimes having an incorrect first or last term or extra terms (most often a+nd) within their sum. Other errors included not displaying a sufficient number of terms, not writing S= or Sn =, and using commas instead of + sign between the terms. In part(ii) most realised that they could make headway even if they hadn’t attempted part (i). In (ii) (a) many of the candidates used a=20, d = -0.8 and Sn = 64 in the sum formula to score the M mark. Poor notation such as (n-1)-0.8 was frequently seen, but candidates usually recovered from this. Following a correct equation most were able to proceed correctly to the given quadratic for the accuracy mark. A common error was where candidates used +0.8 as the common difference rather than -0.8 Part (ii) (b) was generally answered without issues. Almost all candidates managed to get B1 for getting the two values of n, either by factorising or by using a calculator. Part (ii)(c) was not answered very well with most candidates identifying 10 as the valid solution but not adequately justifying this choice. The reasons provided were often too vague or not related to why 10 was better than 16. More tended to score the mark by explaining that at 16 weeks he would be saving a negative amount rather than using the fact that 10 was the smaller number or that he would have had enough by 10 weeks so would not continue saving until 16 weeks.

  Text version

  In an arithmetic series, the first term is a and the common difference is d.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q1 · 4 marks

  4 marks

  1

  
  Show mark scheme

  

  Examiner report

  The first question on the paper provided a good start for many candidates who were able to score full marks. A fairly small minority solved by squaring both sides of the equation. This method was generally very accurate when used. Most candidates however, attempted the alternative strategy of setting up two separate equations. Of those who did not get full marks, most were able to identify that they should be solving 3 − 2𝑥𝑥 = 7 + 𝑥𝑥 but a common error was in the incorrect manipulation of the equation to get 𝑥𝑥 = −3/4. Others lost minus signs along the way or deliberately removed them, perhaps because this was a question involving modulus signs and they held the misconception that values should therefore be positive. Many candidates were also able to identify that a second solution came from 2𝑥𝑥 – 3 = 7 + 𝑥𝑥 but some incorrectly assumed no negatives were permitted and instead used 2𝑥𝑥+ 3 = 7 + 𝑥𝑥. As with the first solution, the most common error, once identifying the right equation, was due to incorrect rearrangement to give 𝑥𝑥 = 4. Some students however, only set up and solved one equation failing to recognise that two would be required. Although not required by the question, some candidates annotated their graphs; clearly constructing the line 𝑦𝑦 = 7 + 𝑥𝑥 to “show” the 2 solutions. Others unnecessarily calculated y coordinates, perhaps misunderstanding what was required by a ‘solution’. A small number of candidates obtained the correct answers, but then rejected 𝑥𝑥= −4/3. Other candidates incorrectly gave two positive solutions. Candidates should be advised to refer back to the graph provided in the question as a means of checking the validity of their solutions. In this case that one was clearly positive and one negative. A substantial minority of candidates demonstrated a lack of understanding of the modulus function and simply changed negative signs to positive signs or assumed that the modulus function is distributive i.e. |a + b| = |a| + |b|. A small number of candidates included inequalities in every step of their attempt and others maintained the presence of modulus signs throughout their entire attempt.

  Text version

  In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q10 · 8 marks

  8 marks

  10

  
  Show mark scheme

  

  Examiner report

  Parts (a) and (b) were generally answered well by most candidates, however parts (c) and (d) proved to be good discriminators with many failing to score any marks. Part (a) required candidates to find f −1( 3 2). The most common approach seen was finding the inverse function then substituting in x = 3 2. This was usually carried out successfully with a few candidates losing a mark for arithmetic slips when rearranging. Very few attempted the more straightforward method of solving f (x) = 3 2. Of those gaining no marks, the common misconception was substituting x = 3 2 into the reciprocal of the function or evaluating f ( 3 2). Several substituted x = 3 2 into f′(x). In part (b) most students seemed confident about how to start and the majority of those who attempted it achieved both marks. Candidates either equated the numerator to achieve 8x + 5 = A(2x + 3) + B or used the long division method. There were some errors in calculating A after finding the correct value for B. Those who used the long division method made fewer mistakes than those who compared coefficients. A few candidates attempted to manipulate the numerator and expressed it in terms of the denominator with a good rate of success. A few candidates lost the final accuracy mark due to substituting the correct A and B into the wrong place to give −7 + 4/(2x + 3) In part (c) it was common to see the correct boundary values given but many lost the mark by using incorrect notation. Common errors seen were 0 ≤ x ≤ 4, 0 ≤ g(x) ≤ 4, or even 0 ≤ f(x) ≤ 4; using < instead of ≤; or missing out the lower boundary value altogether. Many did not realise that the values of the range of g-1(x) are the same as the domain of g(x) and found g-1(x) first but then failed to give the correct range, gaining no marks. Giving the range as 0 to 16, instead of 0 to 4 was a common incorrect answer. A significant number obtained only one of the correct values. Part (d) proved to be demanding for most candidates. Some candidates realised they could simply evaluate f(0) and f(4) to find the boundary values but most started by trying to find an expression for fg-1(x). Many were able to find g-1(x) = √(16 − x) and used this to obtain fg-1(x) = (8√(16 − x) + 5)/(2√(16 − x) + 3) or fg-1(x) = 4 − 7/(2√(16 − x) + 3) (using their answer to (b)). Some made no further progress from here and thus scored no marks. Of those that did carry on, common errors from this point included using 4 as the upper value rather than 16 or changing √(16 − x) to an incorrect expression such as 4 − √x. Since one of the boundary values was 0 some candidates found 37/11 from incorrect work and did not score the method mark. Some candidates, who did find the correct values, lost the final mark by using < instead of ≤ in their inequality. It was not uncommon to see part (d) left blank.

  Text version

  The function f is defined by f(x) = (8x + 5)/(2x + 3), x > -3/2. The function g is defined by g(x) = sqrt(16 - x^2), 0 <= x <= 4.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q12 · 6 marks

  6 marks

  12

  
  Show mark scheme

  

  Examiner report

  Part (a) was generally well answered. The quotient rule was usually applied correctly for the first two marks and, given the quadratic factor was given, most candidates were able to arrive at the correct expression for g(𝑥𝑥). Some used the product rule instead and usually gained the first two marks although it was a little more difficult to get the final answer and some algebraic errors were made. Although most candidates gained the M mark in (a) their brackets were very often incorrect or missing. This was usually recovered in later steps allowing full marks to be obtained. When differentiating 4x2 + k, some candidates left k in the answer, or differentiated to 8x + 1, gaining no marks. Occasionally the terms on the numerator were reversed and these attempts gained no marks in (a) as in incorrect method was implied. Some missed the 3 when differentiating e3x using the chain rule. Most candidates understood there was no need to expand the denominator. Where they did so, they generally did it correctly and no marks were lost. This was a “Show that” question so “f′(x) =” or “ d d y x = ” needed to be seen and was sometimes missing, losing the final A mark. Candidates found part (b) more challenging. Although most knew to set f′(x) to zero for stationary points, there were many marks lost due to insufficient rigour in the presentation of method when looking for one or more roots. Generally candidates understood that it was the quadratic expression that needed to be used and not any part of g(x). Overall, most realised that the number of solutions for a quadratic was determined by the discriminant although a significant number did not. The majority attempted b2 – 4ac with the appropriate values but candidates often set 𝑏𝑏2 −4𝑎𝑎𝑎𝑎= 0 rather than 𝑏𝑏2 −4𝑎𝑎𝑎𝑎≥0. The B mark was usually gained by the use of the discriminant > or ≥ 0, it was rarely gained by making an appropriate statement. Some stated g(𝑥𝑥) ≠0 and a few g(𝑥𝑥) > 0 which may have been required if full justification had been required. Most candidates gained the M mark for calculating the value of the discriminant using the correct values for a, b and c. Other methods included completing the square to find the least value of the quadratic and then considering the y coordinate. Some differentiated the quadratic to find the minimum turning point and again considered the y value. These alternative methods were often incomplete and unsuccessful in most cases seen. Few realized the range had a lower limit, but this was condoned for the A mark as the k was defined as positive in the question. A common error was to give the final inequality as 4 9 k < .

  Text version

  The function f is defined by f(x) = e^(3x)/(4x^2 + k), where k is a positive constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q14 · 10 marks

  10 marks

  14

  
  Show mark scheme

  

  Examiner report

  In part (a) a large proportion of candidates demonstrated a good understanding of how to find the partial fractions and the methods seen were often very clear. Substitution of 𝑥𝑥 = −1 and 𝑥𝑥 = 1/2 was much more common than comparing coefficients. Generally the only mistakes made on this part of the question were arithmetic. A small number lost the final mark for failing to write out the complete partial fractions. A minority of candidates used 𝐴𝐴 + B( 𝑥𝑥 + 1) + 𝐶𝐶 (2𝑥𝑥−1) thinking it was an improper fraction but many proceeded correctly to found 𝐴𝐴 = 0 . The most challenging aspect of part (b) for the candidates was to separate the variables correctly. A large proportion of candidates did not make the link to part (a) and often divided both sides by 3V as a starting point, i.e. ∫ 1 3𝑉𝑉. 𝑑𝑑𝑑𝑑= ∫ 1 (2𝑡𝑡−1)(𝑡𝑡+1) . 𝑑𝑑𝑑𝑑 Where candidates did this they often would incorrectly replace the right hand side with the partial fraction they had found in part (a). For the candidates that got as far as this, the integration of the partial fractions they put in place was generally completed well with the key …ln(2t − 1) - …ln(t + 1) being found more often than not although a common error was leaving a multiplier of 2 when integrating the first fraction to give 2ln(2𝑡𝑡−1). Regardless of the integration a candidate had achieved, there were a large number of candidates that did not consider the constant of integration and tried to jump straight to the answer at this stage. Generally, if a candidate remembered their integration constant, they were more often than not successful in using the given limits to obtain a value which they could use to find an equation for V. A few attempts at solving the first order linear differential equation using an Integrating Factor were seen. Those employing this method were usually reasonably successful. In part (c) the solution of 30mins was only obtained by the strongest candidates although many did not attempt this part of the question or possibly failed to see it. Of those that did attempt this part, a common answer was to obtain ½ but not recognise the units of the question were in hours, some giving it incorrectly as 0.5 minutes As the equation for the volume was a given in this question most candidates were able to attempt this in spite of their attempt at part (b). Candidates gave the correct 6m3 although a substantial number thought that 3m3 was the answer. Although candidates were not required to provide units, many did so correctly and marks were rarely lost for stating incorrect units although a few instances of mm3 were seen.

  Text version

  When chemicals A and B are mixed, oxygen is produced. The total volume V cubic metres of oxygen produced t hours after mixing is modelled by dV/dt = 3V/((2t - 1)(t + 1)), with V >= 0 and t >= k, where k is a constant. Exactly 2 hours after mixing, V = 3.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q16 · 12 marks

  12 marks

  16

  
  Show mark scheme

  

  Examiner report

  This question proved to be a good differentiator between candidates. It provided a challenge for many candidates but equally there were a number of accessible marks here. Most candidates attempted part (a), many attempted part (b) and at least partially attempted part (c), usually attempting to find the lower limit for k. There were, however, very few completely correct answers to this question and a substantial minority made no attempt at any part, perhaps having run out of time by the time they reached the last question on the paper. Part (a) was mostly approached correctly with a good number scoring full marks. Candidates generally noticed the instruction to use parametric differentiation. Whilst d𝑥𝑥 d𝑡𝑡 was most often correct, d𝑦𝑦 d𝑡𝑡 proved a challenge for many who struggled to differentiate sec2𝑡𝑡. Many did not seem to know that these functions were standard derivatives presented in the formula booklet and instead there were some long-winded and overly complex attempts to convert to sine or cosine functions and such approaches often led to errors. However, even some of the most inefficient approaches managed to persevere through lengthy algebra to obtain a correct d𝑦𝑦 d𝑥𝑥 eventually. Almost all who attempted d𝑥𝑥 d𝑡𝑡 and d𝑦𝑦 d𝑡𝑡 knew and attempted d𝑦𝑦 d𝑥𝑥= d𝑦𝑦 d𝑡𝑡 ÷ d𝑥𝑥 d𝑡𝑡. Those who made errors when determining d𝑦𝑦 d𝑥𝑥 often exhibited an over-reliance on trying to match the printed result and thus ‘adjusted’ their d𝑦𝑦 d𝑥𝑥 in an attempt to obtain the required value for the normal gradient. Often this meant the loss of a method mark. Candidates would be better advised to utilise a correct method rather than incorrectly manipulating their answers to match the result they are attempting to achieve and caution should always be advised when working backwards from a given answer. Nonetheless, application of negative reciprocals and equations of lines was generally successfully attempted, and candidates were clearly confident in these areas. There were a few attempts at a Cartesian approach in part (a) which gained zero marks as the question had specified the use of parametric differentiation. In part (b), there were a multitude of approaches seen and candidates were often at least partially successful. Most candidates recognised the need to begin with a trigonometric identity, usually sec2𝑡𝑡= 1 + tan2𝑡𝑡 and rearranged the expressions for x and y to obtain expression for sec2𝑡𝑡 and tan2𝑡𝑡 before substituting into the identity to obtain an equation in 𝑥𝑥 and 𝑦𝑦 which could in turn be rearranged to the required form. However, sometimes candidates took a far lengthier approach, beginning with e.g. sin2𝑡𝑡+ cos2𝑡𝑡= 1 and using 𝑥𝑥 and 𝑦𝑦 to find expressions for sin𝑡𝑡 and cos𝑡𝑡. An alternative and popular, though inelegant, method was to start from the target Cartesian equation and substitute in the trigonometric forms for 𝑥𝑥 and 𝑦𝑦 before rearranging to obtain an identity. Provided a conclusion was given, this approach was credited with both marks. Occasionally candidates used the equation for 𝑥𝑥 to obtain an expression for 𝑡𝑡 in terms of arctangent before substituting into the expression for 𝑦𝑦 which, if correctly done, was an acceptable approach. Several students just substituted the coordinates of points such as P into the given Cartesian equation, stating that they satisfied the equation. Such attempts did not constitute a proof and so gained no marks. Part (c) provided a particular challenge and usually only the lower limit was attempted. Most candidates appeared to completely miss the need to find an upper limit. Of those who did manage to find both limits, only a small minority of candidates got to the final solution with correct inequality interpretation. It is worth noting that candidates achieving full marks for this part very often had clear construction lines on their graph that enabled them to see the correct approach. Most often though, candidates gained the method mark for the lower limit, 43 8 , for identifying that two intersections were required and set up a correct inequality for the discriminant for the quadratic equation obtained by equating the equations of the line and the curve. Common errors in manipulation, particularly of the coefficient of ½ from the quadratic meant that a number of candidates did not earn the accuracy mark for 43/8. Many candidates stopped after obtaining a lower limit for k, but others explored the parameter limits at 𝑡𝑡= − 𝜋𝜋 4 and 𝑡𝑡 = 𝜋𝜋 3 to identify 𝑘𝑘 values of 13 2 and 23 2 −√3 for the line passing through the associated points. Very few, however, recognised 13 2 as an upper limit and a number of candidates assumed incorrectly that 13 2 < 𝑘𝑘 < 23 2 −√3. A minority of candidates incorrectly thought that setting d𝑦𝑦 d𝑥𝑥 = 0 or d𝑦𝑦 d𝑥𝑥 =− 1 2 would provide limiting criterion for k values. In summary this was an interesting question which gave rise to a wide variety of responses in all three parts. There were instances of both succinct and ingenious approaches seen, particularly in part (c) and it was pleasing to see tenacity in some response where candidates had inadvertently opted for less efficient approaches but had persevered nonetheless to obtain the required results.

  Text version

  The curve C has parametric equations x = 2tan(t) + 1 and y = 2sec^2(t) + 3, where -pi/4 <= t <= pi/3. The line l is the normal to C at the point P where t = pi/4.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 1 · Q1 · 3 marks

  3 marks

  1

  
  Show mark scheme

  

  Examiner report

  This was a very familiar and accessible question, and hence proved to be a suitable first question. A significant majority of candidates gave a fully correct response to this question, with the most common and successful approach being via the use of the factor theorem. Very few candidates tried to evaluate f (-1). Following on from correct use of f (1)=0, there were occasional errors in solving the resulting simple linear equation. One (perhaps unexpected) error seen several times was to simplify a(1)3 as a3. Algebraic division was attempted by some candidates, but the working needed here was much more complicated and so often led to errors (mainly with signs). This does highlight the importance of choosing the best method to a problem, as well as carrying it out accurately.

  Text version

  Let f(x)=ax^3+10x^2-3ax-4. Given that (x-1) is a factor of f(x), find the value of the constant a. You must make your method clear.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 1 · Q2 · 4 marks

  4 marks

  2

  
  Show mark scheme

  

  Examiner report

  This focussed on the ability of a student to complete the square and proved to be as accessible as the first question. The vast majority of responses seen were fully correct. In part (a) most candidates were able to score full marks. The form (x-2)2 + .... was almost always achieved, although occasionally an incorrect (x+2)2 +…. was offered. The most common error was having +9 instead of +1. In response to part (b) both P and Q were often stated correctly with some candidates able to score the follow through mark for their Q from their value of ‘b’ from part (a). Occasionally a mark was lost if they had P =5 instead of y =5 when x=0. Very few sketches were seen, which if correctly labelled would have also scored full marks in (b). It was relatively common in (b)(ii) to see differentiation used, rather than the completed square expression.

  Text version

  Given f(x)=x^2-4x+5 for real x. The curve y=f(x) meets the y-axis at P and has a minimum turning point at Q.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 1 · Q7 · 9 marks

  9 marks

  7

  
  Show mark scheme

  

  Examiner report

  The first few marks in this question were very accessible but the final part of the question proved to be a useful discriminator especially around grade C and below. Part(a) The majority of candidates achieved full marks in this part. Sign errors were responsible for most errors, although failure to halve was also seen with either or both of (x ±10)2 and (y ± 4)2 being seen. Having achieved (x-5)2+(y+2)2 = 18, it rare to see the centre or radius not found correctly. Part (b) This part was found to be discriminating with complete solutions seen from only the most competent of candidates. The most common approach was to substitute y = 3x + k into the circle equation to find a quadratic in x (with coefficients in terms of k) for the points of intersection. Errors mostly arose when (3x + k +2)2 was expanded incorrectly or the 2x + 6kx terms not being gathered together. Many solutions then stopped after the substitution and expansion step, as candidates did not seem to appreciate what to do with an equation of this complexity in two variables. Application of b 2 = 4ac to find equal roots was used on many occasions but errors frequently occurred. Only a very few solutions involved attempts via b2 - 4ac > 0. It was not always obvious to see how a student had solved their quadratic (formula not quoted and values for a, b and c not seen) but a limited number of correct solutions were seen. Using a calculator quadratic solver was frequent and acceptable. Several attempts started by using implicit differentiation on the circle equation. Although this is a possible method, it did not often go beyond using dy/dx = 3 to get x+3y+1= 0. Another method seen several times was to find the equation of the normal ( i.e. the radius ) to try to make progress towards finding values of k, but this method also seldom went further than the equation of the line.

  Text version

  The circle C has equation x^2+y^2-10x+4y+11=0. The line l has equation y=3x+k, where k is a constant.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 1 · Q9 · 11 marks

  11 marks

  9

  
  Show mark scheme

  

  Examiner report

  This question was answered with varying degrees of success. Around 28% of the candidates failed to score any marks here at all, which was very suprising. There were however, many well formed and accurate solutions by careful candidates. In part (a) most candidates managed to write down a correct identity and use an appropriate method to find values for B and C. To prove A =0 was generally done via equating coefficients of x2 or by setting x = 0 or x=1. A few candidates wrote down three equations for the coefficients and constants, but rarely was a correct set of solutions gained from this method. It was also not uncommon to see the RHS expanded either with or without the terms involving A in an attempt to prove that A = 0. Errors included having an incorrect identity from the start which resulted in no marks. In part (b)(i) it was rare to see fully correct expansions. Good candidates often lost the final mark due to an error in adding the required terms. A common error was forgetting to use their value for C and only adding the expansion for (1-2x)-1 rather than C(1-2x)-1. A few were unable to extract 2-1 correctly from (5x+2)-2 (usually having this as 2). Other errors seen involved an expansion of (1 + 2/5 x)-2 or else (1 + 5/2 x) 2 rather than (1 + 5/2 x) -2. Very few correct answers were seen for part (b)(ii) and this was very often not attempted. Those who did write a correct range sometimes also gave a second range which therefore did not gain the B mark here.

  Text version

  Let f(x)=(50x^2+38x+9)/((5x+2)^2(1-2x)), where x!=-2/5 and x!=1/2. Given that f(x) can be expressed as A/(5x+2)+B/(5x+2)^2+C/(1-2x), where A, B and C are constants:

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 1 · Q12 · 9 marks

  9 marks

  12

  
  Show mark scheme

  

  Examiner report

  This was the second modelling question on this paper. It proved to be a significant discriminator between candidates, with a significant minority making little or no progress. It certainly seems to be a topic that students find hard to access. Part (a) The most common approach which led to either partial or complete success was to start with H=ax2+bx+c and use (0,3) to find that c =3. Many candidates then used (120,27) to find an equation connecting a and b, but sometimes no more progress was made. Of those who did make further progress, the most common approach was to use differentiation and the knowledge that the gradient was zero when x=90 to get a second equation in a and b. Candidates using this approach often went on to get (a) fully correct and then also to gain most if not all of the marks in (b) as well. It was only rarely that there was a misunderstanding about H=0 rather than H=3 when x=0. There were other methods used in (a) and any of these could have led to success if applied correctly. For example, using the fact that the quadratic has a turning point when x =90 and so H = a(x-90)2+c was used correctly by some candidates. Alternatively, using additional implied points such as (60,27) or (180,3) due to the symmetry of quadratic curves, was used appropriately by candidates. Part (b) Both parts were done well by candidates who had achieved a correct answer to (a). Part (b) (i) There were only a very few who omitted the units here. A wrong answer to (a) would have had made it difficult to gain the accuracy mark here. Part (b)(ii) A wrong answer to (a) could have scored a method mark here if the candidate’s quadratic had been solved correctly. Part (c) A wide range of comments were quoted here– some of those that did not gain the B1. Many of these focussed on what happened to the ball after hitting the ground, or picked out a relatively trivial aspect such as the height of the tee being inaccurate. Others mentioned the weight of the ball, or that it depended on the force of the contact. There were many acceptable comments. These should have focussed on why one of the four modelling assumptions referred to in the question may not have been valid. Acceptable responses included the fact that the ball is not a particle, the ground may not be horizontal, the path may not be a quadratic (parabola) or it might not travel in a vertical plane (Spin was mentioned in this context).

  Text version

  A golf ball is modelled as a particle moving in a vertical plane above horizontal ground. Its height H metres is plotted against horizontal distance x metres from where it was hit. The ball starts from a platform 3 m high, reaches maximum height after travelling 90 m horizontally, and is 27 m above ground after 120 m. H is modelled as a quadratic function of x.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 1 · Q14 · 4 marks

  4 marks

  14

  
  Show mark scheme

  

  Examiner report

  This question proved equally challenging and very few fully correct solutions were seen. The modal mark achieved here was 2, scored by over 40% of the candidates, most often for a fully correct attempt at dy/dx. Most solutions seen used the quotient rule although incorrect application of the rule often meant that no marks could be awarded. The most common errors were subtracting the wrong way round on the numerator, or else adding the two expressions on the numerator. Differentiation of √x suprisingly caused a few problems as well. Product rule expansions seldom got past the first expression. The ability to manipulate the ensuing algebraic expression was often lacking, with fractional terms within a fraction, the main cause for failure. Alternative methods which might have simplified the algebra were seen, but only rarely. Only a handful spotted that a difference of two squares method applied to the numerator would have made this a very simple problem.

  Text version

  Given y=(x-4)/(sqrt(x)+2) for x>0:

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 2 · Q2 · 5 marks

  5 marks

  2

  
  Show mark scheme

  

  Examiner report

  This question was well answered with the modal mark being full marks. Part (a) was the least well answered part where and it is clear that students have difficulty with the concept of range. Incorrect answers included y < 7and y  7. Some candidates attempted to use the graph’s x-intercepts in their work, so incorrect answers such as 14 14 4 4 x − < < appeared occasionally. In contrast, part (b) was well answered, with most candidates gaining all of the marks. The most common approach was to find f (1.8) = 0.52 and then g (x) = 0.975. Some candidates correctly established the composite function gf (x) and then calculated gf (1.8). There were occasional instances of finding fg (1.8) by mistake. Similarly (c) was well answered. Most candidates knew the process of finding the inverse function by rearranging to make x the subject and switching the x’s and y’s (or by switching first and then rearranging) and many successfully reached the correct answer of ( ) 1 g 5 3 x x x − = − .

  Text version

  The functions f and g are defined by f(x) = 7 - 2x^2 for x in R, and g(x) = 3x/(5x - 1) for x in R with x != 1/5.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 2 · Q11 · 10 marks

  10 marks

  11

  
  Show mark scheme

  

  Examiner report

  In general part (a) was poorly answered, with many candidates seemingly unfamiliar with sketching modulus functions. The majority of candidates scored at least one B mark for their upturned V but a V shape was occasionally seen, typically scoring no marks. Of those with the correct, many could not find the correct intersections and maximum point. Finding the correct y intercept seemed to be less challenging than finding the x intercept or the maximum point. Part (b) was better answered than part (a). Many candidates made correct attempts to find the second critical value of , gaining the M mark, but was often not found. The ‘squaring’ method was occasionally seen, but as this led to a 3TQ in x and k and which was hard to factorise this was rarely successful. The final A mark was often lost due to lack of correct set notation. There were few fully successful answers to part (c) but some candidates could write down one correct co-ordinate. There was occasional success seen by candidates who used algebra by writing the transformed function as .

  Text version

  Let k be a positive constant. Figure 4 is an upright V-shaped sketch of y = |2x - 3k|, with its vertex on the positive x-axis. Define f(x) = k - |2x - 3k|.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 1 · Q4 · 5 marks

  5 marks

  4

  
  Show mark scheme

  

  Examiner report

  This was a relatively straightforward test on function notation. In part (a) most candidates opted to find the inverse function 1 f ( ) x − before substituting in 7. x = Very few attempted the more straightforward method of solving f ( ) 7. x = Most candidates achieved some success, with some slips being made by those attempting to find 1 f ( ) x − . Of those gaining no marks, the majority came from a misunderstanding of the notation with some substituting 7 x = into f ( ) x ′ and others into [ ] 1 f ( ) x − In part (b) most candidates understood the notation and gained the first mark by forming an expression of the correct form. Many were then able to multiply both numerator and denominator by ( ) 2 x − to form a single fraction of the required form. This part is where most errors crept in; some having problems with the bracketing but most with the process of multiplying all terms on both numerator and denominator by the (x – 2) term.

  Text version

  The function f is defined by f(x)=(3x-7)/(x-2), for real x with x!=2.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 1 · Q7 · 5 marks

  5 marks

  7

  
  Show mark scheme

  

  Examiner report

  A very significant majority of the candidates did not seem to know how to define a region using inequalities. It was common to see a response where the area of region R was attempted using integration. It was only a small minority of candidates who made a correct inequality statement at the end, even with the allowance of follow through on their quadratic curve and their line equations. It was often the case that this final mark was lost due to a candidate using the letter R rather than y in their inequality. Finding the equations of the boundaries of the region is essential when defining the inequalities, so candidates who went on to find an area were mostly still trying to do this and if successful would have only lost the final mark. A very small minority of candidates found the area below the line by using a trapezium area formula and so did not manage to gain the marks for the line equation. Finding the line equation correctly was achieved by a majority of candidates, although a slip with the line gradient did occur occasionally even if the diagram showed a convincingly positive gradient. Finding the equation of the quadratic curve was found to be more difficult, with a common error being f(x)=(x+2)2+13. Occasionally a candidate spotted that this was not consistent with the curve going through (0,25) and tried to adjust their answer, but not often in a correct way. Attempts at using f (x)=a(x+2)2+13 were often successful at finding a, whilst attempts at using f (x)=ax2+bx+c were much less often correct.

  Text version

  A quadratic curve C, y=f(x), meets a straight line l at (-2,13) and (0,25). The point (-2,13) is the minimum turning point of C. The finite region R lies above C and below l between the intersection points.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 1 · Q9 · 9 marks

  9 marks

  9

  
  Show mark scheme

  

  Examiner report

  Parts (a) and (b) of question 9 were standard bookwork involving the product rule of differentiation using an exponential function. Well rehearsed candidates were very successful here scoring the majority of marks. The diagram should have helped in determining the maximum and minimum points of the function. Unsurprisingly candidates found part (c) more challenging, especially (c) (ii) where the significance of x=0 was rarely understood. It was however a good discriminator for stronger candidates.

  Text version

  Let f(x)=4(x^2-2)e^(-2x) for real x. Define g(x)=2f(x) for real x, and h(x)=2f(x)-3 for x>=0.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q3 · 5 marks

  5 marks

  3

  
  Show mark scheme

  

  Examiner report

  This was a fairly accessible question for prepared candidates and provided an early source of marks for many. Generally, most candidates recognised the need to use the power law for logs and correctly obtained 2 log(4 −𝑥𝑥) = log (4 −𝑥𝑥)2, then proceeded to remove the logs. In fact, many candidates jumped immediately to (4 −𝑥𝑥)2 = (𝑥𝑥+ 8) with no sign of logs whatsoever in their solution. This was condoned although it must be noted that combining steps of working in a ‘show that’ question can be risky. Some candidates diligently wrote out 10log (4−𝑥𝑥)2 = 10 log(𝑥𝑥+8) before removing logs, others subtracted log (4 −𝑥𝑥)2 from both sides and used the log subtraction law to combine the log terms before removing the log. Both approaches were often successful although the latter tended to be the riskier of the two as some candidates forgot that 100 = 1 and instead obtained (4−𝑥𝑥) (𝑥𝑥+8) = 0 which was then fudged to give the required solution. Those candidates that did not recognise the need to employ the power law for logs early on in their solution made very little progress. Disappointingly, some candidates failed to earn the final mark in part (a) as their final line was missing the “= 0”. Candidates should take care to check their final line matches the printed answer in a ‘show that’ question. The first part of part (b) was straightforward and gave an opportunity to re-enter the question for those who had come unstuck in part (a). Indeed, for some candidates this was the only mark attained in this question. It was surprising that a number of candidates left their ‘solution’ as (𝑥𝑥−1)(𝑥𝑥−8) or gave ‘solutions’: (𝑥𝑥−1) and (𝑥𝑥−8). Only a very small number of candidates incorrectly obtained 𝑥𝑥= −1 or 𝑥𝑥= −8. The second part of part (b) was more discriminating and a careful explanation was required here. Most candidates were able to identify 𝑥𝑥= 8 as the value which was not a solution. However, in order to gain credit, it was also necessary for candidates to make specific reference to the fact that log(4 −8) is undefined. Too many candidates simply stated ‘logs can’t be negative’ which was incorrect and quite different to ‘you can’t take logs of a negative number’. Candidates should be reminded that reasoning should be precise and specific.

  Text version

  The equation 2log(4 - x) = log(x + 8) has real solutions, where the logarithms have the same base.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q6 · 7 marks

  7 marks

  6

  
  Show mark scheme

  

  Examiner report

  As would be hoped, most candidates scored some marks on this question. Many found part (a) quite straightforward and there were a good proportion of candidates who gained full marks. Candidates used various approaches to find A, B and C, but the most common tended to be long division. There were a significant number of candidates who performed a completely correct division, but were unable to relate their answer to A, B and C and so lost the accuracy marks. There were many more errors seen when students attempted to multiply (Ax + B) by (x + 2) and compare the numerators, and when candidates selected values of 𝑥𝑥 to substitute in, any errors usually cost both accuracy marks. In comparison, those who used long division were more likely to make an error only in the remainder (−3 − +12 = 9 being a common slip) and so secure the first accuracy mark. In part (b), a pleasingly high number of candidates who had found values for A, B and C were able to successfully integrate to gain the first two marks, often aided by the follow through mark that was available. Candidates generally substituted the limits correctly and there was also a relatively good use of the laws of logarithms to combine their terms, albeit with occasional sign errors. The most common being “−15ln 8 −(−15 ln 2) = −15ln 8 − 15 ln 2”. It was not uncommon for a candidate to then incorrectly combine this to get " − 15ln 8 2”. Most success was seen where candidates changed the ln (8) into ln (23) and then 3 ln(2) before combining the two ln 2 terms.

  Text version

  For x != -2, (x^2 + 8x - 3)/(x + 2) is identically equal to Ax + B + C/(x + 2), where A, B and C are constants.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q8 · 6 marks

  6 marks

  8

  
  Show mark scheme

  

  Examiner report

  This question was very well answered, with many candidates gaining full marks. Candidates who used the factor theorem (solving f(-4) = 0 to find a = 6, as in the main scheme) were more successful than those who attempted long division, which invariably went wrong and in some cases resulted in a remainder containing ‘x’. Typical mistakes included: solving f(4) = 0 instead of f(-4) = 0 (usually forfeiting 2 marks), mistakes in integrating the ‘ax’ term and forgetting the constant of integration or adding it in at the end. It was noted that some candidates were able to score full marks in this question without explicitly finding a i.e. by considering (x + 4) (2x2 ….....-3) and deducing the x coefficient of the quadratic directly.

  Text version

  The function f satisfies f'(x) = 6x^2 + ax - 23, where a is a constant. The graph of y = f(x) has y-intercept -12, and x + 4 is a factor of f(x).

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q11 · 7 marks

  7 marks

  11

  
  Show mark scheme

  

  Examiner report

  Part (a) of this question was generally well answered, with most candidates being able locate P correctly. A few candidates launched into some lengthy algebra, in some cases resulting in values for x and y bearing no relation to the position of P on the diagram. Answers to part (b) from this cohort of candidates were very disappointing and displayed an overreliance on algebra and poor understand of modulus graphs. Very few candidates identified x = -10.6 as the solution to the equation. Instead most answers wrote down the solution to 3x+40 = 2(x+4) – 5 {x =-37} as well as x = -10.6, scoring M1A0. A few candidates used a calculator to directly write down the correct value of x to gain both marks – perhaps not a bad strategy for students who are weak at modulus graphs. A simple check on a calculator would also have established that only x = -10.6 was a solution to the printed equation. Part [c] of this question was one of the most challenging parts of this paper, with hardly any candidates gaining all 3 marks. Candidates seemed unable to ‘visualise’ the question and see directly that there would be an intersection if a was greater than 2, so the B mark was hardly ever gained (and consequently the A mark was immediately lost). The M mark for considering P and attempting a = 5/4 was occasionally gained, although as above this often entailed long winded algebraic methods e.g. solving simultaneous equations to find a.

  Text version

  The graph of y = 2|x + 4| - 5 has vertex P.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q13 · 6 marks

  6 marks

  13

  
  Show mark scheme

  

  Examiner report

  Part (a) was often answered correctly and candidates knew to set the denominator equal to zero and solve to obtain a value for 𝑘𝑘. Unfortunately, candidates sometimes left their answer as 𝑥𝑥= e2 which was, of course, incorrect. Some candidates gave the inexact answer of 𝑘𝑘 = 7.39 which was condoned. Commonly seen incorrect answers included 𝑘𝑘 = 0, or 𝑘𝑘 = 1. A number of candidates gave the value 𝑘𝑘= 3 suggesting that the expression in the denominator had been incorrectly interpreted as ln(𝑥𝑥−2) rather than ln(𝑥𝑥) −2 as was stated. A small number of candidates wrote “there is no 𝑘𝑘 in the question” perhaps failing to spot its presence in the domain. Part (b) was attempted by most candidates although it was not uncommon to see attempts to find the inverse of g(𝑥𝑥) instead of differentiating. The differentiation in part (b) was however correctly carried out by many candidates, with the majority using the quotient rule. Those who stated the correct formula and the individual components of u, v, u’ and v’ were often more successful. A few used ቀ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑ቁ 2 instead of 𝑣𝑣2 in the denominator, while others made slips on expanding the bracket 3(𝑙𝑙𝑙𝑙(𝑥𝑥) − 2). Some attempted the product rule, usually with less success. Many candidates did not attempt to simplify the numerator of g′(x). Often those that did made slips and it was frustrating to see candidates losing marks due to missing brackets and sign errors when expanding. It was common to see numerators incorrectly simplified to –13 𝑥𝑥 or 5 𝑥𝑥 although many did manage the correct simplification to 1 𝑥𝑥. Some candidates unnecessarily expanded the denominator, sometimes incorrectly writing ln (𝑥𝑥2) instead of (ln 𝑥𝑥)2. Candidates who managed to correctly simplify their expression for g′(x) argued correctly why g′(x) > 0 for all 𝑥𝑥 in the domain, but some simply stated this with no reason or gave an incomplete argument of “𝑥𝑥> 0 so g′(𝑥𝑥) > 0” and lost the final mark in this part. In part (c), most candidates only gained one out of two available marks as they usually considered only the set of values of 𝑎𝑎 for which the numerator is positive by solving 3𝑙𝑙𝑙𝑙𝑙𝑙−7 = 0 to obtain 𝑎𝑎> 𝑒𝑒 7 3. Usually no consideration was given to the case where the numerator and denominator are both negative and the few that did solve 𝑙𝑙𝑙𝑙𝑙𝑙= 2 usually failed to state 0 < 𝑎𝑎< 𝑒𝑒2.

  Text version

  The function g is defined by g(x) = (3ln(x) - 7)/(ln(x) - 2), for x > 0 and x != k.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q14 · 7 marks

  7 marks

  14

  
  Show mark scheme

  

  Examiner report

  Part (a) presented some thinking challenges for many candidates, in particular spotting that a = b = r in their equation for the circle. About half of the candidates were not able to deduce the correct equation of the circle. The most commonly seen incorrect equations were 𝑥𝑥2 + 𝑦𝑦2 = 𝑟𝑟2 , (𝑥𝑥+ 𝑟𝑟)2 + (𝑦𝑦+ 𝑟𝑟)2 = 𝑟𝑟2 and (𝑥𝑥 ± 𝑎𝑎)2 + (𝑦𝑦 ± 𝑏𝑏)2 = 𝑟𝑟2. Some of these candidates managed to gain the method mark for using 𝑦𝑦= 12 – 2𝑥𝑥, but a disappointingly high number didn’t progress. Relatively few candidates who initially used 𝑎𝑎 and 𝑏𝑏, correctly replaced them with 𝑟𝑟 to attain at least one mark. Some candidates tried to use a sketch to help them. This did allow some to identify the centre as (𝑟𝑟, 𝑟𝑟), but many put the circle in the wrong quadrant and proceeded to use (𝑟𝑟, −𝑟𝑟) or (−𝑟𝑟, 𝑟𝑟). The majority of candidates who gained the B1 and M1 expanded their expressions correctly to give the correct equation and attain the A1, although the algebra involved in squaring the 3-term expression (𝑟𝑟2 −24𝑟𝑟+ 144)was not always carried out correctly. Candidates were generally more successful when they expanded (𝑦𝑦−𝑟𝑟)2 first before substituting 𝑦𝑦= 12 − 2𝑥𝑥. Part (b) proved to be more accessible to most candidates. Many recognised the need to use the discriminant and equate to zero, though poor algebraic manipulation often cost them the accuracy marks. Almost all candidates used a calculator to solve the quadratic, though there were examples of the quadratic formula being used, mainly correctly. A small number resorted to calculus to find a gradient, but very few successfully attempted to differentiate their equation and set the gradient to -2. A few tried to use the idea of gradient or even equate the equation with 12 – 2𝑥𝑥.

  Text version

  A circle of radius r lies in the first quadrant and touches both coordinate axes. The line 2x + y = 12 intersects the circle.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q15 · 8 marks

  8 marks

  15

  
  Show mark scheme

  

  Examiner report

  A number of the responses were completely blank suggesting perhaps that candidates had run out of time to complete this question. In part (a), very high proportion of the candidates had some idea what the layout of the proof looked like yet only around 50% were able to complete it accurately and rigorously. There were many solutions containing the final the term of Sn as arn rather than arn - 1. Candidates who had the correct initial sequence regularly went on to complete the proof correctly. Only a small number prematurely factorised. Some were trying to apply the method of proving the sum of an arithmetic series instead, hence losing the majority of the marks. In part (b), the first method mark was gained by the majority of students. Some didn’t divide by “a” or wrongly cancelled r10 and r5 in their equation, so lost the first accuracy mark. However, many of the students who gained the first two marks continued to gain the last two marks by dealing with the quadratic and achieving the correct solution for “r”. A few candidates used (n - 1) in the formula rather than n and a number of candidates got confused with n and r, replacing r = 10 and 5 in the denominator of each fraction. Only around 50% realised that they had ended up with a quadratic equation in r5 and candidates who did not cancel the "a" out early often ended up with an overcomplicated equation which they could not solve. Very few did not eliminate r = 1 if they found it, although a few solved the quadratic and then forgot to cube root the answer and so had r = 3.

  Text version

  In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A geometric series has first term a and common ratio r, where a != 0 and r != 1. Its sum to n terms is S_n.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 1 · Q1 · 3 marks

  3 marks

  1

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 2.6 out of 3) This proved to be a very suitable start to the paper. Most candidates used the factor theorem and produced an equation in a by setting f (-3) = 0. Most could then solve their linear equation in a to find its value. Candidates who used this method generally went on to score full marks. Common mistakes tended to be arithmetic or sign errors, for example not cubing (-3) accurately in the first term. Other methods were seen but were generally less efficient and not as effective. Those candidates that started with (x+3)(ax2+bx +c) and equated coefficients were generally successful, however those candidates that tried to solve this problem using long division generally made errors and did not score highly. It was rare to see students scoring zero on this question.

  Text version

  Let f(x)=3x^3+2ax^2-4x+5a, where a is a constant. Given that (x+3) is a factor of f(x):

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 1 · Q2 · 5 marks

  5 marks

  2

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 2.5 out of 5) This question proved to be more challenging for a number of candidates although many were still able to score full marks. In part (a) it was important for candidates to pay attention to the scale of the graph. Most candidates who did draw a straight-line graph were able to make the link between one point of intersection and one real root to the equation, however, a significant minority of these were unable to construct the correct straight-line graph with enough accuracy to gain both marks for this part of the question. In part (b) almost all candidates were able to write down the small angle approximation for cosx even if they were unable to use it in order to answer the question. Most candidates did substitute correctly into the equation in order to gain a quadratic equation, although some did contain a mixture of variables. Sign errors sometimes occurred at this stage and the weakest candidates were unable to deal correctly with resulting fractions in order to simplify the equation. Many candidates appear to be using calculators in order to solve quadratics and in this question, this usually resulted in full marks. The final mark was lost by some candidates who correctly solved the quadratic equation but gave both roots as their final answer.

  Text version

  The equation cos(x)-2x-1/2=0 can be interpreted by comparing the curve y=cos(x), with x measured in radians, with a suitable straight line.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 1 · Q5 · 10 marks

  10 marks

  5

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 6.3 out of 10) This question on completing the square and the quadratic function was accessible to all, especially the first 6 marks. Part (c) was more demanding but there were some excellent responses here. Part (a) was very straightforward and many students were able to write down the answer without any difficulty. Although errors were rare, one common incorrect answer was 2 2( 1) 8 x   . Part (b) was equally straightforward, even for candidates who did not achieve the correct result in part (a). Many used the form 2 ( ) a x b c   or their graphical calculators to produce accurate and well drawn curves. Marks were lost here when candidates drew V shaped curves or had incorrectly placed turning points. As stated earlier, part (c) was a lot more demanding. Rather strangely part (i) was found more accessible than part (ii). Many candidates sketched the new curve and compared the positions of the turning points noting that   1,7  had ''moved'' to   1,3 . Disappointingly a rather sizeable majority could only describe this ''movement'' as a ''transformation'' rather than ''translation'' . Fewer students could use the form 2 21 h( ) 2( 1) 7 x x    to find the range of the function. Many resorted to substituting 0 x  producing a range with 21 9 y  as one of the limits. Common incorrect answers for the range of h scoring one mark were h( ) 3 x or 0 f ( ) 3 x  .

  Text version

  Define f(x)=2x^2+4x+9 for real x, g(x)=2(x-2)^2+4x-3 for real x, and h(x)=1/(2x^2+4x+9) for real x.

  Open the full paper (PDF)

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

  6 marks

  10

  
  Show mark scheme

  

  Examiner report

  (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.

  Text version

  Let N denote the natural numbers.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 1 · Q12 · 10 marks

  10 marks

  12

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 4.1 out of 10) This question discriminated well between students of all abilities with part (a) and (b) more accessible to students compared to part (c) and (d). In part (a) most students were successful at using the product rule and setting dy/dx equal to 0 (though the latter was implied rather than being explicit in many cases). Missing the −0.25 factor when differentiating the e(-0.25x) was fairly common, but the differentiation was mostly correctly carried out. A few students failed to use the product rule correctly, instead multiplying the derivatives of each term, but these were in the minority. Having attempt the derivative the majority did set the result equal to zero (possibly implied) and proceeded to cancel the exponentials. Reasoning for the cancellation was usually not given, but in this instance there was no requirement for such reasoning. However it might be well noted that such explanations may be required under the new specification. Many students lost the final accuracy mark as they did not state or show division by cos x. Students need to be clear that in a ‘show that’ question all steps need to be shown. Part (b) was the most successfully answered part of the question. Students realised that the graph of |f(x)| should be above the x-axis. Where students did not gain full credit, this was usually due to poor drawing of loops (not decreasing heights, or rounding at the cusps). A few students chose to draw over the original figure which made it difficult in many cases to determine which pieces formed their graph. They would be well advised to sketch separate graphs in similar questions. There was less success in part (c). Attempts at this part were varied, with a sizeable proportion not making the connection with part (a) at all and attempting to find the value of H(t) at π /2 or some other value, or failing to substitute into H(t) at all. Many solved tan x = 4 to find the acute angle but did not then look for the further solution which was needed for the height between the first and second bounces and so achieved only the first mark. Also, use of degrees instead of radians was a common error, but this was allowed the first method mark. The magnitude of their resulting answer should have alerted them to the fact that something had gone very wrong. Part (d) was another part with a varied array of answers. Common incorrect responses referred to negative time, and the times between each bounce would be longer or that the ball would not bounce forever. Many students gave a rote answer of “the effect of air resistance has not been taken into account” or “energy loss” or similar vague reasons rather than relate to the context of the question. Since the model may have taken air resistance into account (the contributing factors to the model were not listed in the question), such an answer was unacceptable.

  Text version

  For x>0, let f(x)=10e^(-0.25x)sin(x). Also define H(t)=|10e^(-0.25t)sin(t)| for t>0. The graph of H consists of non-negative arches with zeros at t=pi,2pi,3pi,... and decreasing heights. H(t) models a ball's height in metres t seconds after it is kicked; consecutive zeros represent bounces.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 1 · Q13 · 11 marks

  11 marks

  13

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 5.5 out of 11) Only the strongest candidates scored all 11 marks here, with the majority losing at least 1 mark, usually the first mark in (a) or the last mark in (b). In part (a) (i) many candidates realised that x = 2 was linked to establishing that q = 4 but they found difficulty explaining this by referencing the asymptote, or the denominator being zero. Almost all candidates provided good explanations in (a)(ii) even when they had provided inadequate explanations on part (i) In part (b) the vast majority of students knew they had to use partial fractions to attempt the question with many achieving the correct expression and correct integration. The most common error was in simplifying the final logarithmic expression with many candidates leaving some of their answers in terms of ln (3/4) or ln (4/3) for example. It was uncommon therefore to see a fully correct solution in the required form.

  Text version

  The curve C has equation y=(p-3x)/[(2x-q)(x+3)], for real x with x!=-3 and x!=2. Constants p and q are such that C passes through (3,1/2) and has vertical asymptotes x=2 and x=-3. Region R is bounded by C, the x-axis and x=3; its other x-boundary is where C meets the x-axis.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 2 · Q1 · 3 marks

  3 marks

  1

  
  Show mark scheme

  

  Examiner report

  Q01 proved challenging for a significant number of candidates. This was probably because of the more open-ended nature of the question, allowing candidates to select their own problem-solving strategy to express y as a function of x. It was not apparent to most candidates that y could be expressed as a linear function of x. More able candidates, however, were able to provide a concise solution. Other candidates were still able to achieve full marks but expressed y as a more complicated function of x. A few candidates attempted to express x as a function of y, suggesting a lack of understanding of the terminology. It appeared that most candidates failed to realise that Q01 could be solved purely using indices. Instead, most candidates chose to immediately take logarithms as their first step towards a solution. A correct first step, e.g. 1 log(2 4 ) log , 2 2 x y         was rewarded by the mark scheme. Unfortunately, some candidates combined taking logarithms with an incorrect application of the addition law for logarithms as their first step, therefore gaining no marks. E.g. 2 4 log2 log4 . x y x y    Most candidates who had a correct first step of work went on to achieve full marks, with some only failing to do so by making an arithmetic error. Q01 highlighted some candidates’ weaker algebraic, indices and logarithms skills. Common errors included 2 4 8 , x y xy   2 4 8 , x y x y    1 1 , 2 2 2 4 2 x x  or 1 2 2 4 4 . 2 2 2 2 x x y y     The most common correct answers were   1 2 2 log log(2 ) , log4 x y   4 1 log 2 2 2 x y        or   1 2 2 2 ln . ln 4 x y  Only a minority of candidates obtained a correct 1 3. 2 4 y x  

  Text version

  Given 2^x * 4^y = 1/(2sqrt(2)).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 2 · Q6 · 10 marks

  10 marks

  6

  
  Show mark scheme

  

  Examiner report

  Q06 discriminated well between candidates of all abilities, with only a few candidates gaining full marks. There were some marks in Q06(b) and Q06(d) which were accessible to the majority of candidates. In Q06(a), many candidates struggled with the split domain for g( )x (i.e. for 2, x  2 g( ) ( 2) 1 x x    and for 2, g( ) 4 7) x x x    and only a minority gained both marks. Many candidates found a correct g(0) 5,  but a significant number used the incorrect method of substituting ‘5’ to give 2 gg(0) (5 2) 1 10.    Only a minority used a correct method and found gg(0) 4(5) 3 13.    Occasionally candidates found an algebraic expression for gg( ), x although this was often incorrect. Very few obtained a correct answer of 13 by substituting 0 x  into 2 4(( 2) + 1) 7 x   or 2 4( 2) 3. x   In Q06(b), many candidates attempted to solve the inequality g( ) 28 x  by finding the critical values for x. Most solved the linear equation to find a correct critical value 35 , 4 x  and many solved the quadratic equation to find 2 3 3. x   A minority rejected 2 3 3 x   and deduced that 2 3 3 x   was the second critical value. Only a few candidates used the critical values and the diagram in Q6 to write down the correct solution 35 2 3 3 . 4 x x     Some candidates, however, incorrectly gave 35 , 4 x  2 3 3 x   or gave 35 2 3 3 4 x x     as their final answer. A few candidates made no creditable progress in Q06(b) by using a method of equating both parts of g( )x and attempting to solve the equation 2 ( 2) 1 4 7. x x    Although Q6(c) required a comment for both functions, f and g, some candidates only wrote a comment about one of the two functions. A significant number of candidates believed that g did not have an inverse due to its being defined in ‘two parts’. Candidates who were successful in Q6(c) usually gave a reason such as, ‘g is a many- one function and h is a one-one function’. A few correctly stated that the inverse of g is one-many, which is not a function, whereas the inverse of h is one-one, which is a function. A few candidates incorrectly described g as a ‘one-many function’. In Q06(d), only a few candidates applied the method of finding x by applying 1 h . 2        As this method was less prone to error, most candidates who used it scored full marks. Most candidates applied a complete method of finding their inverse 1 h ( ), x  followed by using their inverse to form and solve the equation 1 1 h ( ) . 2 x   Many candidates who found a correct answer 7.25 x  lost the final mark in Q06(d). This was because they used the incorrect inverse 1 h ( ) 2 1 x x     (instead of the correct 1 h ( ) 2 1 x x    ) which led to their ‘solving’ the incorrect 1 1 2 x  (which has no solutions) to give 7.25. x  A few candidates in Q06(d) incorrectly believed that 1 h ( )x  referred to the reciprocal of h, or even to the first derivative of h with respect to x.

  Text version

  The function g is defined by g(x) = (x - 2)^2 + 1 for x <= 2, and g(x) = 4x - 7 for x > 2. The function h is defined by h(x) = (x - 2)^2 + 1 for x <= 2.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 2 · Q7 · 7 marks

  7 marks

  7

  
  Show mark scheme

  

  Examiner report

  Q07 discriminated well between the medium and higher ability candidates. Lower ability candidates struggled to make much progress with this modelling question and most of them scored no more than one mark. In Q07(a), most candidates wrote down a correct equation for the model in the form , y kx c   with constants k and c. A minority struggled to give a suitable form of the linear equation, with some omitting the fixed cost constant, c. Some candidates gave incorrect answers, such as ; y x  equations involving exponential models; or differential equations. Q07(b) proved challenging for those candidates who struggled to grasp the concept ‘profit sales cost   ’, with many of them incorrectly assuming that the model from Q07(a) was for the profit made. Many of these candidates used (800, 500) and either (300, 80) or (300, 80)  to form two linear equations, with some erroneously finding a gradient of 0.84. Those candidates who understood the link between profit and cost usually found the total cost to the factory of £1100 and £680 for 800 and 300 bars of soap respectively. Most used (800, 1100) and (300, 680) to form two linear equations and proceeded to find the gradient, 0.84, and y-intercept, 428. A large proportion concluded by stating the given equation 0.84 428. y x   A few candidates, however, did not complete their analysis with a suitable conclusion, and so lost the final mark. Some candidates used (800, 1100) and (300, 680)to verify the model 0.84 428. y x   Many of these were successful but often failed to give a suitable conclusion stating that the given model was true. Q07(c) gave a significant number of candidates their only creditable access to Q07. Some candidates gave a correct interpretation such as 0.84 represented ‘the cost of making an extra bar of soap’. Incorrect explanations included that 0.84 represented ‘the selling price of a bar of soap’ or ‘the profit per bar of soap made’. There was a mixed response to Q07(d). The most popular approach was to form and solve a correct 2 0.84 428 n n   or 2 0.84 428, n n   which mostly led to the correct answer 369. n  A final answer left as an inequality such as 368 n  or 369, n  without reference to the statement that 369 was the number of bars required, was not sufficient for the final mark. Some candidates used trials of both 368 n  and 369 n  in a complete trial and error method leading to a correct conclusion of 369. n 

  Text version

  A small factory makes bars of soap. On any day, its total cost y pounds for making x bars is modelled as a fixed cost plus a cost proportional to x. Bars sell for 2 pounds each. When 800 bars are made and sold the profit is 500 pounds; when 300 bars are made and sold the loss is 80 pounds.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 1 · Q9 · 10 marks

  10 marks

  9

  
  Show mark scheme

  

  Examiner report

  Question 9(a) was well answered. Students generally demonstrated a good understanding of implicit differentiation as well as the chain and product rules. Most reached the given result scoring all 4 marks. In question 9(b) however, many made the numerator of the fraction equal zero (i.e. y = x) instead of the denominator (x = 3y). This kind of error usually resulted in only two of the five marks being scored. There were however many full and accurate solutions, with very few students unable to pick out the coordinates of the point P from the two solutions to their simultaneous equations. Only a small number of students gained the B1 mark in question 9(c). They simply needed to state that the point furthest north can be found by substituting y= x into the equation of the curve and picking out the positive solution. Explanations tended to be incorrect or incomplete, with many stating that it could be found by putting either 0 x  or 0. y 

  Text version

  The curve x^2-2xy+3y^2=50 models a cycle track, with x and y measured in km. Take positive x as east and positive y as north. Points P and Q are respectively the points furthest west and furthest east of the origin O.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 1 · Q14 · 10 marks

  10 marks

  14

  
  Show mark scheme

  

  Examiner report

  The final question on the paper was a challenge to all. Part (a) was straightforward with most recognising the need to use the identity 2 cos2 1 2sin t t  in some form to prove the required result. This was one of the more successful ''show that'' parts of the paper. Only stronger students, however, were able to understand and explain why the curve C did not include all points of the curve   2 6 3 . y x    Many drew the whole of the parabola in part (b) scoring one out of the three marks. Those who managed to sketch the correct part of the parabola rarely were able to explain why the domain (and range) were restricted. Stating that ''because 0 2 t   '' was not enough to explain why for instance 1 5 x   . The best students were able to explain that as 1 sin 1 t   then 3 2sint  would have a maximum value of 3 2 5   and a minimum value of 3 2 1.   Part (c) was also demanding and better students annotated their graphs to help their understanding of the question (See Figure 1). Most however used the discriminant condition 2 4 0 b ac   for two roots and were able to proceed to two values. Arithmetic errors were quite common using this method. Most students who proceeded to the answer were well versed in writing their answer using set notation. Figure 1

  Text version

  A curve C has parametric equations x=3+2sin(t), y=4+2cos(2t), for 0<=t<2pi.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 2 · Q1 · 6 marks

  6 marks

  1

  
  Show mark scheme

  

  Examiner report

  This question proved accessible to all, but only the most able students scored full marks. In part (a), most students found the value of g(5) and substituted this value back into g( ), x with nearly all of them being successful in finding the correct 40 gg(5) . 9  A minority found gg( )x algebraically and substituted 5 x  into the result. In part (b), only a few students stated the correct range 15 2 g( ) . 2 x   The most common incorrect answers included 15 15 15 g( ) , g( ) , g( ) 2 2 2 x x x    and 15 2 . 2 x   In part (c), most students used a correct process to find 3 5 2 x y x    with some finding 11 3. 2 y x    A few students, who misunderstood the notation 1 g ,  tried to either differentiate g( )x or write down the reciprocal of g( ). x Some students lost the final mark by stating neither 1 3 5 g ( ) 2 x x x     nor its domain in correct notation. Some students made no attempt to write down the domain of 1 g . 

  Text version

  The function g is defined by g(x) = (2x + 5)/(x - 3), with domain x >= 5.

  Open the full paper (PDF)

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

  5 marks

  3

  
  Show mark scheme

  

  Examiner report

  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.

  Text version

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

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 2 · Q6 · 6 marks

  6 marks

  6

  
  Show mark scheme

  

  Examiner report

  This was a well-answered question with part (b) and part (c) providing most of the discrimination. In part (a), nearly all students found f (2) 0  and many used this result to deduce that ( 2) x  was a factor of f( ). x The method of long division was the most popular and successful method to find the quadratic factor of f( ). x Those who used the more time-efficient method of comparing coefficients were slightly more prone to making sign or manipulation errors. Most students arrived at the correct answer 2 f( ) ( 2)( 3 2 5), x x x x      with some writing the equivalent 2 f( ) (2 )(3 2 5). x x x x     There were a few students who found both correct factors but did not express their final answer as a product of two algebraic factors. In part (b), many students used part (a) to write down 2 4 2 ( 2)( 3 2 5) 0 y y y      and most deduced the two real solutions 2. y  Surprisingly a few students solved 2 2 y  to give 2. y  A significant number of students stated, without proof, that 2 ( 3 2 5) x x    or 4 2 ( 3 2 5) y y    where 2, x y  had no real solutions. Some students applied ‘ 2 2 4 (2) 4( 3)( 5) 56 0 b ac        ’ with an appropriate conclusion to show that 4 2 ( 3 2 5) 0 y y     had no real solutions. Common errors included just stating ‘ 2 4 0 b ac   ’ with no reference to a discriminant calculation or the omission of ‘ 0  ’ after a correct discriminant calculation. Only a minority of students were able to bring the two strands of the proof together to show that the given equation had only two real solutions. In part (c), some students correctly deduced that the given equation had 3 real solutions for 7 10 ,      with a few students incorrectly deducing that there were 6 real solutions. Most students misunderstood the requirement of the question by finding 23.1, 26.2, 29.4,  with those students who did not go on to state that there were 3 real solutions losing the mark for this part.

  Text version

  Let f(x) = -3x^3 + 8x^2 - 9x + 10 for real x.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 2 · Q8 · 7 marks

  7 marks

  8

  
  Show mark scheme

  

  Examiner report

  This question discriminated well between the average and more able students. There was a significant minority of students, however, that made no creditable progress. In part (a), a minority of students struggled to write down a suitable form of the quadratic model. Some students, who tried to model the height of the rugby ball as a negative quadratic cutting the x-axis at 0 x  and 40, x  wrote down either ( 40) H x x   or (40 ), H x x   but struggled, due to the lack of a constant, to fit the condition (20, 12) to their model. Those who wrote (40 ) H Ax x   usually applied (20, 12) to their model to give the correct 3 (40 ). 100 H x x   Other students who adopted a model of the form 2 H ax bx c    were usually successful in using the conditions (0, 0), (40, 0) and (20, 12) to give the correct 2 0.03 1.2 . H x x   Only a few students used the model 2 12 ( 20) H x     with condition (40, 0) to find a correct 2 3 12 ( 20) . 100 H x    A few other elaborate methods were seen, including one where (40, 0) and , 12 2 b a        were applied to 2 , H ax bx   which usually resulted in a correct 2 0.03 1.2 . H x x   In part (b), many students substituted 3 H  into their quadratic model and most applied a correct method for solving their resulting 3-term quadratic equation or their quadratic equation of the form 2 ( 20) , 0. x k k    Many students found the largest solution in the range (0, 40) and gave their answer, usually to 3 significant figures, in metres. Students gave a variety of acceptable limitations for the model in part (c) with many stating that ‘the model assumes that there is no wind or air resistance acting on the ball’. No credit was given for answers that discussed what might have happened to the rugby ball after it had hit the ground.

  Text version

  A rugby ball is kicked from ground level at O and travels in a vertical plane. Its height H metres is plotted against horizontal distance x metres. The trajectory is a downward-opening parabola through (0, 0) and (40, 0) and reaches a maximum height of 12 metres. A horizontal rugby-post bar, perpendicular to the path, is 3 metres above the ground.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 2 · Q11 · 7 marks

  7 marks

  11

  
  Show mark scheme

  

  Examiner report

  Part (a) was accessible to most students, whereas some students struggled to gain access to part (b). In part (a), most students used the given 2 1 11 6 ( 3)(1 2 ) ( 3) (1 2 ) x x B C A x x x x          to write down a correct identity 2 1 11 6 (1 2 )( 3) (1 2 ) ( 3). x x A x x B x C x          Some students used a method of long division to deduce that 3 A and formulated the correct identity 10 10 (1 2 ) ( 3). x B x C x       Only a few students wrote down the incorrect identity 2 1 11 6 (1 2 ) ( 3). x x B x C x       Most students used either a method of substituting values into their identity or a method of comparing coefficients with most finding the correct values for A, B and C. In part (b), a minority of students struggled to make any creditable progress. These students usually evaluated f ( )x for increasing x-values and concluded that the corresponding values for f ( )x were continually decreasing. The majority used the chain rule to differentiate f( ). x A common error at this stage was for students to differentiate 1 2(1 2 )x    to give 2 2(1 2 ) , x    while other students tried to apply an incorrect method of showing f ( ) 0. x   Only a few of the students who obtained a correct 2 2 f ( ) 4( 3) 4(1 2 ) x x x        provided an acceptable explanation detailing why f ( ) 0. x  

  Text version

  For x > 3, f(x) = (1 + 11x - 6x^2)/[(x - 3)(1 - 2x)]. It is given that (1 + 11x - 6x^2)/[(x - 3)(1 - 2x)] is identically A + B/(x - 3) + C/(1 - 2x).

  Open the full paper (PDF)

• Edexcel 9MA0 · sample · Paper 1 · Q5 · 3 marks

  3 marks

  5

  
  Show mark scheme

  

  Text version

  A curve C has parametric equations x=2t-1 and y=4t-7+3/t, where t!=0.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 1 · Q8 · 6 marks

  6 marks

  8

  
  Show mark scheme

  

  Text version

  For x>2.5, f(x)=ln(2x-5)+2x^2-30.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 1 · Q11 · 9 marks

  9 marks

  11

  
  Show mark scheme

  

  Text version

  An archer shoots an arrow. Its height H metres above the ground is modelled by H=1.8+0.4d-0.002d^2 for d>=0, where d metres is the horizontal distance from the archer. The arrow travels in a vertical plane until it hits the ground.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 2 · Q1 · 3 marks

  3 marks

  1

  
  Show mark scheme

  

  Text version

  The function f is defined by f(x) = 2x^3 - 5x^2 + ax + a, where a is a constant.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 2 · Q4 · 5 marks

  5 marks

  4

  
  Show mark scheme

  

  Text version

  The functions f and g are defined by f(x) = e^x for x real, and g(x) = 3ln(x) for x > 0.

  Pearson does not publish a direct link to this question paper.

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

  6 marks

  6

  
  Show mark scheme

  

  Text version

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

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 2 · Q9 · 5 marks

  5 marks

  9

  
  Show mark scheme

  

  Text version

  The constant A satisfies integral from 1 to 4 of (3sqrt(x) + A) dx = 2A^2.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 2 · Q11 · 6 marks

  6 marks

  11

  
  Show mark scheme

  

  Text version

  The function f is defined by f(x) = 2|3 - x| + 5 for x >= 0. Its unlabelled graph is V-shaped on this domain.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 2 · Q14 · 9 marks

  9 marks

  14

  
  Show mark scheme

  

  Text version

  A closed cylindrical drinks can has radius r cm, height h cm and volume 500 cm^3. The material has negligible thickness. Its total surface area is S cm^2.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · sample · Paper 2 · Q16 · 12 marks

  12 marks

  16

  
  Show mark scheme

  

  Text version

  A meerkat population is modelled by dP/dt = (1/22)P(11 - 2P), for t >= 0 and 0 < P < 5.5, where P is the population in thousands and t is the time in years since the study began. Initially there were 1000 meerkats, so P = 1 when t = 0.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · specimen · Paper 1 · Q5 · 10 marks

  10 marks

  5

  
  Show mark scheme

  

  Text version

  Let f(x) = x^3 + ax^2 - ax + 48, where a is a constant. It is given that f(-6) = 0.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · specimen · Paper 1 · Q6 · 6 marks

  6 marks

  6

  
  Show mark scheme

  

  Text version

  A road-tunnel entrance is measured as 6 m wide at its horizontal base and 5 m high at its centre. In a coordinate model the base midpoint O is the origin, the base endpoints are A(-3, 0) and B(3, 0), and the highest point is C(0, 5). A quadratic curve BCA models the entrance. A coach is 4.1 m high and 2.4 m wide.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · specimen · Paper 1 · Q10 · 13 marks

  13 marks

  10

  
  Show mark scheme

  

  Text version

  The function f is defined by f(x) = (3x - 5)/(x + 1), for real x with x != -1. The function g is defined by g(x) = x^2 - 3x on the domain 0 <= x <= 5.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · specimen · Paper 1 · Q11 · 10 marks

  10 marks

  11

  
  Show mark scheme

  

  Text version

  A cubic curve C, with equation y = f(x), crosses the x-axis at A, O and B in that left-to-right order, where O is the origin. It is given that f'(x) = k - 4x - 3x^2, where k is constant, and that the horizontal distance AB = 4sqrt(2).

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · specimen · Paper 2 · Q4 · 6 marks

  6 marks

  4

  
  Show mark scheme

  

  Text version

  Consider the graph with equation y = |2x - 5|.

  Pearson does not publish a direct link to this question paper.

• Edexcel 9MA0 · specimen · Paper 2 · Q5 · 5 marks

  5 marks

  5

  
  Show mark scheme

  

  Text version

  The line l has equation 3x - 2y = k, where k is a real constant. The line intersects the curve y = 2x^2 - 5 at two distinct points.

  Pearson does not publish a direct link to this question paper.