Differentiation 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 · Q3 · 6 marks

  6 marks

  3

  
  Show mark scheme

  

  Examiner report

  Many candidates were successful across all parts of this question demonstrating a good understanding of the Newton Raphson method, but this question did start to discriminate between the least confident and those who were more careful in their processing. Part (a) was attempted by almost all candidates, with the vast majority making good progress, able to find the values of f(3.6) and f(3.7) correctly and observe a sign change. A conclusion was required for the accuracy mark and had to include a reference to a sign change, continuity of the function and mention or indication of a root lying in the interval, as well as both values being correct to 1 significant figure. It appears that reference to continuity is included more frequently by candidates but there was still a good proportion of candidates who did not mention it or who made incorrect statements involving it. Common incorrect statements included “change in sign therefore continuous”, “the interval is continuous” and “ x is continuous”. Radians were required and most candidates recognised this, with very few candidates working in degrees. The differentiation in part (b) was generally well attempted with many candidates achieving both marks. The first mark was awarded for recognising that 1 tan 2 x       differentiated to 2 1 sec 2 x        which many candidates were able to do successfully. Incorrect derivatives were occasionally seen, and candidates should be advised to refer to the derivative of tan(kx) provided in the formula booklet. A common error was forgetting to differentiate the x to 1 or losing it entirely. There were some attempts at integrating and unfortunately some misread (or misinterpreted) the notation and attempted to find the inverse of the function. These responses were unable to score any marks in part (b), even if they went on to differentiate in part (c). Candidates who converted 1 tan 2 x       to 1 sin 2 1 cos 2 x x             were generally not successful at applying the quotient rule correctly. Candidates had varying degrees of success in part (c). The Newton-Raphson formula was generally well quoted or used, however, difficulty evaluating 2 3.7 sec 2       led to the accuracy mark being lost. Evidence of substitution into the formula was inconsistent across the cohort, with some clearly showing the substitution and some leaving it to the examiner to try to establish what they had done. Those that did not explicitly show the substitution of 3.7 in f(x) and ( ) f ' x often made calculation errors and thus scored no marks in this part. Candidates are encouraged to show their substitution into the formula (not just stating ( ) ( ) f 3.7 3.7 f 3.7 − ) for this reason. Those candidates who did not obtain a correct derivative in part (b) were still able to achieve the method mark for using the Newton-Raphson formula correctly with their derivative. Those that did not differentiate in part (b) were able to go on and score both marks in part (c) if they restarted the differentiation at this stage.

  Text version

  f(x) = x + tan(x/2), for pi < x < 3pi/2. Given that the equation f(x) = 0 has a single root alpha.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 1 · Q4 · 3 marks

  3 marks

  4

  
  Show mark scheme

  

  Examiner report

  This question was very similar to previous differentiation from first principles questions and many candidates scored full marks, being much better prepared than in previous series. Most candidates used ( ) ( ) f f x h x h + − to find the gradient of the chord, although a small number of candidates attempted to use ( ) ( ) f f x x h − + but failed to change the denominator to .h − Misunderstanding the definition of the derivative led to some candidates losing some or all of the marks, by writing, for example ( ) ( ) f ² f x ² x h h + − or ( ) ( ) f ² f x² x h h + − . Slips when expanding ( ) 2 x h + were condoned for the method mark but the 2x − was required. Most candidates were able to correctly expand and proceed to 2 2xh h h + or 2x h + to score the first two marks. Common errors included incorrectly simplifying the fraction, failure to cancel h correctly or an incomplete limiting argument (e.g. stopping at 0 lim2 h x → ), each of which led to the loss of the final accuracy mark. Notation was generally good, and most candidates included a reference to 0 h → somewhere in their working. A significant minority of candidates failed to include d d y x = or equivalent anywhere in their answer despite being asked to show d 2 d y x x = in the question, thereby losing the final mark.

  Text version

  Given that y = x^2.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 1 · Q5 · 6 marks

  6 marks

  5

  
  Show mark scheme

  

  Examiner report

  Most candidates attempted the differentiation in part (a) via the quotient rule and this was generally correctly applied with the majority scoring at least 2 marks out of the available 3. While many candidates were able to score full marks in this part of the question, a significant number of candidates lost the final accuracy mark due to either a sign error from expanding brackets (most commonly this was ( ) 2 2 2 3 4 6 x x x x − →− − ) or to missing the ( ) f ' x = or equivalent, a requirement for this ‘show that’ question. Candidates are advised to carefully check that their final result matches the one that they were asked to show. There were a very small number of candidates who incorrectly quoted the quotient rule and were unable to score any marks in this part of the question. A small proportion attempted to use the product rule, and while a good proportion managed this successfully, others made errors when rearranging to the required form. Some candidates broke down the original function into the sum of two fractions before differentiating and in this case a common error was a sign error between the two fractions. In part (b), candidates generally recognised the need to set their ( ) f ' 0 x  to find values of x for which ( ) f ' x is decreasing and solve their three-term quadratic to find critical values. However, not all candidates observed the demand to use algebra and consequently those who used their calculator to solve the quadratic or gave an incorrect factorisation of their quadratic forfeited the first mark. Most candidates chose to simplify their quadratic before factorising and solving to find the critical values, but lack of rigour was evident in this part. Many candidates who factorised their quadratic jumped straight from a quadratic with a 2 2x − term to factorisation of the form (𝑥 ± … )(𝑥 ± … ) (possibly having used their calculator and attempting to work backwards), and consequently lost the B mark. A few used the quadratic formula and showed their working in full thus scoring the B mark, while attempts to complete the square were rarely seen. Once critical values had been obtained, candidates were required to identify the correct region and, while many candidates were successful, there were some who did not know how to proceed. It was common to see incorrect regions selected, particularly 1 4 x −  , not realising that their original quadratic had a negative coefficient of 2x and therefore the outside region was required. Candidates who sketched the graph generally performed better in identifying the correct region but again, many drew a positive graph instead of a negative. There was confusion observed in the use of 'and' and commas instead of 'or', indicating weaknesses in formal mathematical language. It was notable how many candidates did not accurately identify the correct region for their x2 coefficient and critical values, often stemming from dividing their quadratic by –2 but forgetting or not knowing to change the direction of the inequality sign. Other observations that were common to see in this part of the question that were not required or caused complications included: • not recognising that the denominator would be positive due to the square • attempting to find a critical value for the denominator unnecessarily • trying to multiply the denominator and numerator and then work on the resulting quadratic • attempting to find turning points or second derivatives to identify the region, often unsuccessfully.

  Text version

  The function f is defined by f(x) = (2x - 3)/(x^2 + 4).

  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 · Q14 · 9 marks

  9 marks

  14

  
  Show mark scheme

  

  Examiner report

  This question involved a balloon modelled as a sphere such that its radius was increasing at a rate that was inversely proportional to the square root of the radius. Those that were unable to find a correct differential equation in part (a) were fairly limited in the number of marks that could be scored in part (b), but the given answer provided an excellent restart opportunity, and a significant proportion went on to score full marks in parts (c) and (d). In part (a), 1 mark was available for writing down a correct differential equation to model the given situation. Very often this part was answered successfully. When the mark was not awarded it was usually because candidates had simply used d 1 d r t r = rather than d d r k t r = . In part (b) candidates were given some values for the model and were required to solve their differential equation and use the given values to show the given answer solved the differential equation from part (a). For those that offered an equation without a constant of proportionality, i.e. d 1 d r t r = , it was only possible to access two out of five marks in part (b). Generally, candidates seemed well practiced in separating the variables and integrating the resulting equation correctly. When attempted, the most common error was in the index of the radius, either by using an incorrect model or by separating variables incorrectly. The most successful responses found the constant of proportionality straight away, before integrating. Some of those who did not do this at the start struggled to know to go back to find it and therefore could not progress further by substituting for t and r in their integrated model as they had both k and c still to be found. Occasionally, candidates substituted 0.9 or another value for k, instead of substituting 0.9 for d d r t . The given answer to part (b) enabled many who were unsuccessful in the early parts of this question to go on and successfully answer part (c). This question was generally very well attempted, and most candidates used correct processing to calculate the correct value. However, some failed to do so with the required accuracy or made an error in unit conversion. Many did not give the correct units, or any units, forfeiting the final mark. There was a significant minority who, having progressed to 3 2 118 r = , failed to use correct processing to solve the equation, incorrectly finding 3 2 118 and did not score the method mark as a result. A single mark was available in part (d), for suggesting a limitation to the model, and for some candidates this was the only mark gained in the question. Very many candidates scored the mark for pointing out that the balloon would eventually “pop”, or that the assumption that the balloon was perfectly spherical was unrealistic. Some candidates offered a more sophisticated response involving taking the limit for large times, although this was only acceptable if there was some reference to context such as the radius or volume. The most common incorrect response involved commenting on the “rate of inflation of the balloon being constant” as unrealistic, but this aspect had already been addressed in the model.

  Text version

  A balloon is inflated. In a simple model it is a sphere, and the rate of increase of its radius is inversely proportional to the square root of the radius. At time t seconds its radius is r cm. At t=10, r=16 and dr/dt=0.9 cm s^(-1).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q1 · 5 marks

  5 marks

  1

  
  Show mark scheme

  

  Examiner report

  This question was very accessible with the vast majority obtaining full marks on all three parts. In part (a), a few candidates integrated the expression and hence obtained no marks as they did not have a linear equation to solve in part (b), but this was a rarity. One of the more common reasons to lose a mark was the inclusion of a spurious “+ c” with one or both derivatives. A few candidates attempted the differentiation but made mistakes with the coefficients (usually resulting from dividing by rather than multiplying by the index). Another error seen was to “simplify” the second derivative from 24x − 14 to 12x − 7. In part (b) candidates could access the first mark for setting their linear second derivative equal to 0 and solving for x. There were a few candidates who lost the accuracy mark by giving a value of 24 14 instead of 14 24 and there were a surprising number of candidates who went from 24x = 14 to 1 2 x = .

  Text version

  The curve has equation y = 4x^3 - 7x^2 + 5x - 10.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q6 · 7 marks

  7 marks

  6

  
  Show mark scheme

  

  Examiner report

  In part (a) candidates struggled with the differentiation of the exponential function and were much more confident in dealing with the natural logarithm. For the exponential function, the most successful candidates often wrote an intermediate step such as and then used the chain rule. Common incorrect responses seen in (a)(i) were 2 4 1 e x A −(A was often 4 or 8 when this was seen), ( ) 2 2 4 1 4 1 e x x − − (with or without brackets), 2 2 4 1 4 e x x −and expressions involving 2 4e . x Common incorrect responses seen in (a)(ii) were 1 8x and 1 . x In (a)(ii) a few candidates wrote 8 8ln ln x x = first and achieved the correct alternative . It was disappointing to see a sizeable number of candidates confused between the “dash” notation for differentiation and inverses. So, it was fairly common to see attempts at the inverse of each function, resulting in no marks for parts (a) or (b). Success in part (b) was dependent on having the correct forms of the derivatives in part (a). 2 4 1 u x = − 7 8 8x x There were often mistakes in using log rules (e.g. ) and some did not know what to do with 2 1 ln x . As with question 4, of those who proceeded correctly, some forgot to include the “= 0”. Generally, candidates did whatever they could to manipulate their working, and many wrote down the given answer following completely incorrect work. In part (c) many candidates scored all three marks often with no working, just by writing down the two numerical answers required. Candidates should however be advised to show their method as they risk losing both marks. In contrast, some candidates wrote down all the iterations they thought necessary, including the working for each one. Amongst those that didn’t get the final mark, the common mistakes were, stopping at too early an iterate such as x2 or x3, stating an answer of 0.6707 instead of 0.6706 and not rounding to 4dp. There were a small number of candidates who didn’t seem to realise they needed to keep applying the iteration to find the value for α and instead tried to solve the equation given in part (b). Others thought 0.6 or 0.7109 should be substituted into the equation for part (b), not realising a sequence needed to be generated.

  Text version

  Figure 1 sketches y = f(x) and y = g(x), where f(x) = e^(4x^2 - 1), x > 0, and g(x) = 8 ln x, x > 0.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2024 · Paper 2 · Q10 · 6 marks

  6 marks

  10

  
  Show mark scheme

  

  Examiner report

  Most candidates were successful in this question, showing that parametric differentiation was overall well understood by candidates. In part (a), many candidates successfully found the given equation of the tangent using parametric differentiation as specified in the question. By far, the most common approach seen involved finding the value of t and then substituting it into an expression for d d y x obtained by using d d d d d d y y t x t x =  and then using standard coordinate geometry to find the equation of the tangent. There were a number of careless slips and algebraic errors, with 2(t + 3) = 2t + 3 being disappointingly common. Once the numerical gradient had been obtained, there were usually no issues in finding the equation of the tangent in the required form, showing at least one intermediate step as this was a given answer. Most candidates achieved the equation of the tangent via the Point-Slope form although a large number went via the Slope-Intercept form. A few candidates made coefficient and/or sign errors when manipulating their equation to obtain the given answer and unfortunately lost the final accuracy mark. Those candidates who gained few, if any, marks in this question often failed to see the need to find the value of the parameter ‘t’ for the given point P (4, 2) and were unsuccessful in using the alternative method to continue to find a value for d d y x . Some candidates used an incorrect value of ‘t’, usually t = −5 rather than t = −1, as they either did not note the restrictions for t given in the question, or did not find the t value for the y coordinate of P. A few gave two equations for each value of t found. The attempted use of the Chain Rule was not always completed accurately. Some candidates converted their gradient from the use of the Chain Rule into the gradient of the normal which resulted in an inappropriate equation. When finding the value of d d y x , some candidates substituted 4 for their value of t, confusing it with the x coordinate of point P. Some candidates did not follow the instructions stating that the question had to be answered ‘using parametric differentiation’ and, consequently, scored no marks even though they achieved the required result by obtaining the Cartesian equation of the curve. Some candidates tried to complete the question by unnecessarily converting their d d d d y t t x  in terms of t to be in terms of x and/or y which often led to incorrect solutions due to algebraic and/or differentiation errors. There were many candidates who initially went wrong and restarted the question. A number of these ran out of space and incorrectly used the back pages of their answer booklet to continue this question rather than using supplementary paper. It was noted by examiners that part (b) was not attempted by a significant number of candidates. When it was attempted, it was generally not done well, with very many candidates not scoring the mark. Many achieved the correct value only to lose the mark for not stating the units. A common wrong answer was to use x = 0 in the tangent equation to give 5m. Another was to use t = −3 and then substitute this into y = 1 − t3 to give 28m, ignoring the set of values for which t was valid. Others set d 0 d y x = thinking this would give a maximum and then substituted their t value into y. Some wrongly assumed that the largest value of t for the given domain would lead to the maximum height. There was little evidence that candidates had used the sketch to help them find the maximum height.

  Text version

  A curve C has parametric equations x = (t + 3)^2 and y = 1 - t^3, for -2 <= t <= 1. The point P(4, 2) lies on C.

  Open the full paper (PDF)

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

  12 marks

  15

  
  Show mark scheme

  

  Examiner report

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

  Text version

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

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q4 · 5 marks

  5 marks

  4

  
  Show mark scheme

  

  Examiner report

  This question testing the small angle approximation for cosine and finding the equation of a straight line was accessible to many candidates with a large number scoring full marks. In part (a), some candidates formed and solved the equation but were not able to use the information that x was small to choose the correct solution. There was a very small minority of candidates who did not quote the correct small angle approximation, despite this being in the formula book. A slightly larger minority misunderstood that x being small meant that x was 0 and therefore did not include the 2x term in their solution attempt. Candidates should also pay closer attention to the question as several did not round to the required decimal places. It was also notable that some candidates at this level found it difficult to expand 2 1 1 2 2 x   −     when multiplying the fractions together. Part (b) was mostly well answered with the majority scoring both marks, with most candidates gaining these marks efficiently without unnecessary calculations. Most candidates found the gradient of the tangent by substituting 0 x = into the given derivative of f( ) x , however a small number of candidates took a longer approach by attempting to use the small angle approximation. Some candidates tried to substitute their answer from part (a) into the given derivative or found the gradient of the normal having originally found the correct gradient for tangent; these approaches lost both marks. A small number of candidates failed to realise the significance of the point (0,3) and therefore retained an algebraic gradient which they tried to substitute into y mx c = + . A number of responses were seen not giving their answer as an equation, just an expression 1 3 2 x + , or forgetting to substitute their value of c found, giving their equation as 1 2 y x c = + ; having been given the y intercept in the question, it was a requirement that for any marks to be scored that 3 c = .

  Text version

  In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The curve C has equation y=f(x), x in R. Given that f'(x)=2x+(1/2)cos x, the curve has a stationary point with x-coordinate alpha, and alpha is small.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q12 · 5 marks

  5 marks

  12

  
  Show mark scheme

  

  Examiner report

  This differentiation from first principles question was similar to the one that appeared in 2018 and candidates demonstrated a much greater confidence with the way in which they should approach this topic. The limits were provided in the question and the accessibility was evident as it was very rare to see a completely blank script, which is not always true for a differentiating from first principles question. Most candidates were able to obtain the first 3 marks of this question, with very few sign errors seen when using the addition formula. Most correctly started with the fraction for the gradient of the chord f( ) f( ) x h x h + − as given in the formula booklet. Occasionally all a candidate would attempt was the addition formula in terms of A and B, but more often than not it was in the correct format and correct. Scoring the last two marks was not as common, but still achieved more often than not. There were a variety of layouts used for the candidates to justify the replacement of terms with 0 and 1. If candidates lost marks here, it was often because they failed to separate their terms in h correctly. Some candidates introduced extra h’s, forming expressions likecos sin xh hh  . At other times, candidates simply tried to cancel terms incorrectly or stopped. Notation was generally good with limiting arguments correct. The main mistakes that resulted in lost marks were either failing to isolate the required expressions and jumping to the answer or trying to replace each of the 3 terms separately, failing to spot the common factor of sin x in two of them. There were also several candidates incorrectly splitting up the product of each term resulting in an extra h in the denominators. Incorrect notation resulted in the loss of the final mark, in particular the limit notation was often seen right through to the end, in the final answer, even after the limits had been used. A small number of candidates proceeded by using the small angle formula which was generally done well; where errors were seen it was generally not cancelling correctly. One surprising thing to note was the number of candidates who felt it acceptable to not clearly identify the numerator of the fraction by using very small fraction lines, resulting in their solution looking like only one term was being divided by h. Many candidates failed to write d cos d y x x = at the end, despite this being what they had been asked to show. If f ( )x  had not been acceptable as an answer, then many candidates would have lost the final A mark. Only a very small handful of candidates used small angle approximations.

  Text version

  y=sin x, where x is measured in radians. You may use without proof the formula for sin(A +/- B), and assume that as h tends to 0, sin h / h tends to 1 and (cos h - 1)/h tends to 0.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 1 · Q15 · 13 marks

  13 marks

  15

  
  Show mark scheme

  

  Examiner report

  This question brought together a variety of topics and was appropriately placed at the end of the paper. This was very good at discriminating between candidates at the higher end, but it was also pleasing to see a good number of weaker candidates who were still able to attempt later parts such as part (c) and part (d) to a pleasing degree of success. In part (a) a significant proportion of candidates understood that they needed to use the product rule on 7 ex x and most candidates who did, applied it correctly. A common error was just to write 7ex or 7 ex x . Most candidates were unable to achieve the B1 mark, with most not applying the chain rule and so missing terms or making errors with powers. Candidates usually used their expressions correctly with the quotient rule, but few candidates scored the final method mark due to not having the correct expressions. Most candidates chose to use the quotient rule method, but those who used product rule instead, and initially rewrote the denominator as ( ) 1 3 2 e 2 x − − , were generally successful in achieving the first four marks. Some, however, omitted brackets in their expression which lost the final two marks. The final A1 mark was achieved by very few candidates as most did not complete the rearrangement process. Some candidates adjusted their answer by using the given answer in part (b) to work out what the values of A and B should be. Whilst candidates should not typically use later information in earlier parts, this was condoned, as in some cases it would have been too difficult to determine whether this had been done so the benefit of the doubt was given. To achieve the correct form in part (a) from correct differentiation deserved full credit, regardless of the route to the final form required. It should be noted, however, that there were a few who integrated instead of differentiating. In part (b) very few candidates attempted a solution having not completed a solution to part (a). Some determined candidates attempted a solution using A and B and so were able to achieve the method mark. Some candidates attempted a solution by using the given solution to work out what the values of A and B should be and then attempted a solution with those values. A minority of candidates who attempted a solution did not show all the steps required for a show that question. In part (c), the majority of candidates scored the mark for a correct diagram, although some started their initial line higher up than the x-axis, which was condoned on this occasion. Several candidates used a cobweb diagram, and for others there were often lines drawn to the left of 1 x = . Only a handful of candidates put arrows on their lines, but their omission was, again, condoned. Part (d), for many candidates who had not scored in or attempted earlier parts, was an opportunity to score full marks relatively easily. This was very well done. Occasionally, rounding 2 x to 1.50 rather than 1.502 was seen. However, those who wrote down an unrounded answer which rounded to 1.502 first were able to get the A mark. Candidates who showed the value embedded in the formula and then did not achieve 1.502 were able to get the M mark, too. A minority of candidates skipped 1.502 to go to the next term, 1.873 but this still was able to score the method mark as this was still evidence that the iterative formula had been used. The correct answer 1.968 = was widely achieved for part (b)(ii), although 1.967 was a common incorrect answer. The final mark was also lost for a small number of candidates who failed to round to 3 decimal places. Part (e) was the final part on the paper and for many this resulted in no marks. Most of the candidates who attempted it found a suitable interval but substituted the values in the iterative formula instead of a valid equation set equal to zero. Many candidates wrote the required conclusion with no supporting evidence in an attempt to gain a mark. Some candidates, who used a valid function correctly, failed to score the final mark for not commenting that their function was continuous. A significant proportion of candidates attempted an iterative process with a different iterative formula, including attempting the Newton-Raphson method. The question required used of a suitable interval and so these methods scored no marks. Overall, whilst this question was demanding, there were some very impressive solutions demonstrating an excellent grasp of calculus, strong algebraic skills, and the ability to provide a rigorous argument to prove given answers and results.

  Text version

  A curve has equation y=f(x), where f(x)=7xe^x/sqrt(e^(3x)-2), for x>ln(cube root of 2).

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q1 · 4 marks

  4 marks

  1

  
  Show mark scheme

  

  Examiner report

  This question provided an accessible start to the paper for the vast majority of candidates, most of whom picked up at least some of the marks available. In part (a) most candidates were able to differentiate f (x) twice and were able to earn both marks. If candidates did lose marks it was mainly due to arithmetic or processing errors such as differentiating 3x2 to 5x. The most common error was a slip in the final coefficient of x. In (b) part (i), most candidates were able to make a successful attempt to solve f ′′(x) = 0 for x. The mark in this part was sometimes lost due to sign errors when rearranging. It was surprisingly common to see 6x + 4 = 0 leading to 2 3 x = . A significant number of candidates did not simplify their answer, but this was condoned as were values which followed correctly from an incorrect second derivative of the correct form. Part (ii) proved to more of a challenge and many responses highlighted a lack of understanding or awareness of the condition for a function to be concave. A number of blank responses were seen here. Many who attempted this part of the question deduced, or guessed, that it was necessary to consider the sign of the second derivative but were often confused about which sign was required. Others attempted incorrect calculations setting f ′(x) = 0 and solving the resulting quadratic in x. Occasionally, candidates attempted sketches of f (x) and/or f ′(x) in order to determine the range of f (x) which was not required. Others attempted to demonstrate a sign change in attempt to prove that the solution from part (i) was an inflection point. For those employing the correct method, both strict and non-strict inequalities were accepted.

  Text version

  The function f is defined by f(x) = x^3 + 2x^2 - 8x + 5.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q4 · 4 marks

  4 marks

  4

  
  Show mark scheme

  

  Examiner report

  Those scoring full marks in this question were in a significant minority of candidates. In part (a) the majority of candidates used the given information to find the value of A. Those that did not achieve this mark typically gave 85 as an answer. A few students failed to recognise the standard technique of substituting t = 0 into the expression for H to find the value of A, often using t = 1. Sometimes e0 was evaluated as e or 0 rather than 1. Part (b) had mixed outcomes for candidates. Interpretation of the given initial rate of cooling was problematic for many candidates. There was a general lack of appreciation that the rate of cooling was a continuous variable which represented d . d H t − Those candidates not recognising this and using a difference method (similar to those introduced at GCSE) between t = 0 and t = 1 were restricted to the initial B mark in part (a). The most common attempt at a response for these candidates was setting H(1) = 85 − 7.5 and working from there. Of those who recognised the need to first find d d H t and then substitute t = 0, many incorrectly used d d H t = 7.5 rather than the correct d d H t = −7.5 to express a negative increase in temperature. This led to the loss of the last mark. Another error was to forget to differentiate the + 30 to zero leaving a constant in their d d H t , only scoring at most the second method mark for substituting t = 0 and their A into their d d H t . Some differentiated incorrectly to obtain −55Bte−Bt. Several students missed out on the final mark despite correct work by not writing out the final equation. It is important that centres make candidates aware that when a question asks for the complete equation for a model that it is given as such and not just the values of any required constants.

  Text version

  Coffee is poured into a cup. The temperature of the coffee, H degrees Celsius, t minutes after being poured is modelled by H = Ae^(-Bt) + 30, where A and B are constants. Initially, the temperature of the coffee was 85 degrees Celsius.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q5 · 5 marks

  5 marks

  5

  
  Show mark scheme

  

  Examiner report

  Part (a) of this question was accessible for the majority of candidates. Most understood that the derivative at a stationary point is zero and were able to achieve both the method mark and the accuracy mark, substituting x = 3 into the given derivative and setting it equal to zero. Occasionally the accuracy mark was lost due to insufficient working following substitution e.g. 2(3)3 – 9(3)2 +5(3) + k =0 followed by k = 12 thus omitting a required intermediate step. In a ‘show that’ question candidates need to ensure that they have completely justified the given answer. Occasionally students showed the substitution of x = 3 into the derivative, processed the powers and simplified to obtain k – 12 and then jumped to k = 12. These responses lacked the appreciation that the correct answer had been achieved because the gradient equals 0 at x = 3. There were some rare instances of candidates incorrectly substituting x = 3 and d 10 d y x = − to find the value of k. Some candidates automatically integrated the given derivative but then realised that the differentiated expression was required in order to answer part (a) which they then successfully did. Other candidates integrated the expression and then substituted in the value for k = 12 to achieve an answer for part (b). Occasionally they then used the answer to part (b) to find k = 12, so were unable to access the marks available for this part of the question. A rarely seen alternative was to use algebraic division to divide the given polynomial by the linear factor (x – 3) to achieve a remainder of 12 + k, and then equate that remainder to zero. This approach was met with varying degrees of success and students were often unable to complete the algebraic division successfully. In part (b), candidates generally understood that they were required to integrate the expression and the straightforward integration meant the first method mark was commonly achieved. Most realised that they then needed to find the value of their constant of integration using x = 3 and y = 10 to determine the point where the curve crossed the y-axis and achieved the second method mark. The majority of students were successful in achieving full marks in this part, with most stating their answer in coordinate form (0, −28) though some candidates left their answer as c = 28, not making the link with the y intercept explicit. Very few candidates made numerical slips in their processing to find the value of their constant of integration. Where mistakes were made, a common reason for not scoring the second method mark was due to substituting in a value of 0 instead of 10 for y in the integrated expression. Where the constant of integration had been omitted candidates went directly to substituting x = 0. Less commonly, some candidates differentiated rather than integrated and tried to find the roots of the second derivative rather than the y intercept.

  Text version

  The curve C has equation y = f(x). The curve passes through P(3, -10) and has a turning point at P. Given that dy/dx = 2x^3 - 9x^2 + 5x + k, where k is a constant:

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q7 · 7 marks

  7 marks

  7

  
  Show mark scheme

  

  Examiner report

  Almost all candidates attempted to use implicit differentiation with good levels of success. It was somewhat surprising that a significant number of candidates were let down by failing to differentiate the constant term correctly despite, in many cases, managing to correctly carry out some of the more complex steps in the implicit differentiation. It was not uncommon though to see errors in either the product rule for 2xy (often obtaining ( ) d d 2 2 d d y xy x x = or ( ) d d 2 2 d d y xy x x x = or errors in the application of the chain rule when differentiating 3y2. A surprising number of candidates set their first line of working equal to d d y x and fortunately this was often ignored in later working and so was condoned. Some candidates however, attempted to incorporate the extra d d y x term into their manipulations to find an expression for d d y x and this was costly. A few candidates struggled to deal with the negative terms in their numerator (or denominator) which led to sign errors in part (b). Others made slips in the rearrangement which lost the final mark. In some cases, there were issues with clarity of notation for the final statement of d d y x in terms of the positioning of the minus sign and candidates should be advised to take care when drawing the vinculum, particularly when the starting term of the numerator is negative to make sure the sign is in the correct place. In part (b), the majority of candidates were able to use the correct approach to find the equation of the normal and candidates were often able to earn at least some of the marks here even if they had been unsuccessful in part (a). The least successful approach for calculation of the gradient of the normal tended to involve an attempt to rearrange the expression for d d y x into d d x y − prior to substitution and this seem to be more error-prone than the alternative of substituting in values as a first. Some candidates did not spot that point P on the curve had been given in the question and so didn’t use this point in order to find the equation of the line. In some cases, candidates found the equation of the tangent rather than the normal whilst others made arithmetical slips and lost the accuracy mark. Occasionally candidates did not state their equation in the required form which was a shame when it followed correct work.

  Text version

  In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A curve has equation x^3 + 2xy + 3y^2 = 47.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q9 · 7 marks

  7 marks

  9

  
  Show mark scheme

  

  Examiner report

  Most candidates found part (a) challenging but it was usually attempted. There were not many blank responses. Of the four methods listed on the mark scheme, one common successful method was to substitute t in terms of x into y = 6ln(t + 3) via “completing the square” (way 1). When substituting into y = 6ln(t + 3), if a correct t equation was created, then this usually led to full marks. Another common approach was to rearrange t in terms of y, usually successfully as 6 e 3 y t = − and substitute into x = t2 + 6t  16 (way 3). When using this method some candidates had difficulty squaring 6 e y meaning they were unable to earn full marks. Some candidates incorrectly set 25 5 x x + = + thus also losing the accuracy mark. A significant number of candidates factorised x = t2 + 6t  16 to give x = (t  2)(t + 8) and then wrongly thought x = t  2. This then often led to y = 6ln(x + 5) which scored no marks in this part. Candidates were generally more successful in part (b) than part (a). Many candidates found the y intercept by substituting t = 2 into the parametric equation for y rather than substituting x = 0 into the Cartesian equation. The first method mark was more challenging. Some candidates attempted the derivative from the Cartesian equation and they usually reached either ( ) 3 x A + or 3 x whilst others attempted to use the chain rule approach. Some candidates obtained the correct derivative but failed to substitute x = 0. If a candidate achieved the first method mark, they usually went on to achieve the dependent method mark for attempting to find the equation of the tangent. There were very few cases where the candidate used the negative reciprocal of their gradient. The most successful way candidates achieved full marks was differentiating their Cartesian equation. Those that differentiated parametrically were often caught out by arithmetic and algebraic errors or substituted t = 0 instead of t = 2 and failed to achieve the gradient correctly. For the final answer, some candidates had failed to realise the form of equation the question was asking for so left their answer as “y =” or they didn’t have 𝑎, 𝑏 and 𝑐 as integers. Many candidates who had incorrectly answered (a), were able to go on to achieve full marks in (b) by using parametric differentiation to work out the gradient. There were very few blank responses for this part of the question.

  Text version

  The curve C has parametric equations x = t^2 + 6t - 16 and y = 6 ln(t + 3), where t > -3.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2023 · Paper 2 · Q11 · 10 marks

  10 marks

  11

  
  Show mark scheme

  

  Examiner report

  Many candidates found this question challenging and there were a significant number of blank responses. Candidates found part (a) particularly challenging, with many attaining either no marks or the first mark only. The concept of connected rates of change is one that candidates frequently struggle with. A good number found an expression for V and successfully differentiated to get d 200 d V h = , although some confused the volume of water with the volume of the tank. Many candidates failed to recognise that d 1 d V t h  and hence d d V k t h = . Of those who had some concept of the rate being d d V t , many misunderstood the idea of inverse proportion and others missed the constant of proportionality. Even where candidates had correct expressions for both d d V h and d d V t , they often failed to use the chain rule correctly to find d d h t . Many candidates who did link their expressions correctly, achieving d 1 d 200 h k t h =  or equivalent, stated that this could be expressed as h  , with 200 k = and so gained all three marks. Others made no reference to 𝜆 and so lost the accuracy mark. A less common, but correct, approach used ( ) d 200 d d d h V t t = Approaches to part (b) of the question were very mixed. Many candidates failed to realise that, as the question said, ‘use the model to find an equation’, they were expected to, for example, separate the variables and use calculus to solve the differential equation. As they were asked to give their answer in a specific form, a large number started from this and found the constants A and B using the given values of t and h. Fortunately, they were able to gain two marks from the special case. Candidates who used calculus rarely had problems integrating h , although errors were seen later when substituting into 3 2 h or in dealing with the 2 3 when rearranging the equation. Candidates who started this part with the 𝜆 were generally more successful. Some candidates correctly found c and 𝜆, and then rearranged. Others rearranged before making substitutions, stating 3 2 A = and 3 2 c B = . A minority lost the second accuracy mark as they did not give the final equation in the correct form. Some candidates who had correctly found d 1 d 200 h k t h =  solved the differential equation using 200 k in place of 𝜆, in most cases successfully, although the algebraic manipulation was more tricky. Some candidates who had made an incorrect attempt at (a) continued with their answer, usually d 1 1 d 200 h t h =  , in this part. As they only had one constant, the ‘c’, in their expression, they could only attain the first two method marks. It was rare to see the constant of integration missed and basic arithmetical errors when finding either 𝜆 and c or A and B were also relatively infrequent. The alternative approach, rearranging to give d d t h h  = or 200 h k was seen on a few occasions. The majority of candidates who had a correct form of the equation in part (b) made a good attempt at part (c) and were generally able to gain the method mark. Relatively few lost the accuracy mark due to missing the units in their final answer. A number of candidates used an incorrect value for h, most commonly 1000, and so lost both marks.

  Text version

  A cuboid tank has a 20 m by 10 m base and height 5 m. At time t minutes after water started flowing into the tank, the water height is h m and its volume is V m^3. In a model, the tank sides have negligible thickness and dV/dt is inversely proportional to sqrt(h).

  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 · Q8 · 8 marks

  8 marks

  8

  
  Show mark scheme

  

  Examiner report

  (Mean mark 4.1 out of 8) This was the second modelling question on the paper. The given equation modelled the speed of a car as it travelled between a pair of traffic lights. Many weaker candidates missed out part (b) and scored marks in the more accessible parts (a) and (c). Most candidates managed to reach the correct answer of 25 in part (a). A common incorrect answer was 10, possibly from sight of the 10 in the given equation (10 0.4 )ln( 1) v t t = − + The first two marks in part (b) were often achieved by those who attempted it. Using the product rule to find d d v t and then setting this = 0 gave them one form of the given equation. The final two marks were only achieved by the best of students. In reality, all that was required was to make t the subject of the formula, but it was beyond most. Part (c) was the most successful part of the question and familiar to most candidates. Many were able to carry out the iteration process correctly, and write down 7.298 without any problems. In part (c) (ii), the omission of the units, “seconds”, was the most common slip.

  Text version

  A car stops at two sets of traffic lights. It takes T seconds to travel between them. Its speed v m s^(-1), t seconds after leaving the first lights, is modelled by v=(10-0.4t)ln(t+1) for 0<=t<=T; the car has speed 0 at t=0 and again at t=T.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q10 · 8 marks

  8 marks

  10

  
  Show mark scheme

  

  Examiner report

  (Mean mark 3.8 out of 8) This was the third modelling question on the paper, this time based around an exponential model. This should have been expected given the Advance Information and candidates performed well on parts (a) and (b). Part (c) was more of a challenge with only higher achieving students scoring all 4 marks. Part (a) was either answered correctly, or the mark was not awarded because the answer stated was 265 bees not 265,000 bees. Most candidates recognised the need to differentiate in part (b), and used appropriate notation. Most did this accurately before substituting in t =10 to obtain 18.1. A number of candidates failed to secure the final mark however, for not providing a minimal conclusion relating their answer to 18,000. Common reasons for not achieving full marks were • failing to use appropriate notation often using d d y x instead of d d N t • failing to recognise the need to use differentiation often subtracting the values of b N at t =9 and 10 or else t =10 and 11 Part (c) was the most demanding part of the question with many candidates just picking up the first mark for equating the two expressions. Having equated the two expressions, the significance of the negative exponential term seemed to be poorly understood. Very common errors included • changing the 0.05 e t − term to 0.05 e t and then solving • applying 'ln' to each term in the equation to form a linear equation in t • slips in coefficients with 220 becoming 200 A number of candidates reverted to numerical methods or the use of their calculator to solve the equation rather than solve by algebraic methods. The strongest candidates recognised the quadratic equation in 0.05 e t and were able then to solve it correctly, many after making a substitution 0.05 =e t x .

  Text version

  A scientist models the number of bees, measured in thousands, by N_b=45+220e^(0.05t), and the number of wasps, measured in thousands, by N_w=10+800e^(-0.05t), where t is the number of years from the start of the study.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q15 · 10 marks

  10 marks

  15

  
  Show mark scheme

  

  Examiner report

  (Mean mark 5.6 out of 10) This question, based on the minimum surface area of a children's toy proved to be a useful source of marks so late in the paper. In fact 25% of the cohort achieved full marks. Many candidates who could not access part (a) were able to use their skills in differentiation to find an appropriate value for the radius and prove that their value gave a minimum surface area. Unsurprisingly, part (a) was the most demanding part of this question. The key was to find an expression (using the volume of 240 cm3) for h in terms of r. Unfortunately some solutions featured the volume of a sphere, which proved a costly error, from which there was no return. Candidates produced their formula for surface area in a variety of ways – sometimes leaving it to the last line to substitute in the value for the angle, occasionally replacing 0.8 by 4/5. As with other “show” questions, sufficient detail has to be written to convince the reader that progress is genuine. Part (b) well generally well done. Many candidates scored full marks for reaching either 3 1050 or 10.2. Occasionally some candidates mistakenly wrote down 1050 and subsequently simplified their answer to 5 42 . Very few candidates who attempted this part failed to score marks. The main reason for this was usually an inability to differentiate 1680 r which usually became1680 1 or 1680ln r Most candidates were able to find the second derivative correctly in part (c). How it should be used was confused in much of the work seen. Common reasons for a loss of marks here were; • candidates solving the equation 2 2 d d S r = 0 • use of incorrect notation including 2 2 d d S r and 2 2 d d y x • incomplete solutions with calculations or conclusions missing

  Text version

  A solid toy is a right prism shaped like a sector of a circle. Its congruent end faces ABC and DEF are sectors of radius r cm and angle BAC=0.8 radians, with centre A on the first face. Corresponding edges AD, CF and BE are perpendicular to the sector faces and each has length h cm. Thus the three lateral faces have widths r, r and the arc length 0.8r. The volume of the toy is 240 cm^3.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 1 · Q16 · 9 marks

  9 marks

  16

  
  Show mark scheme

  

  Examiner report

  (Mean mark 1.9 out of 9) The final question on the paper proved to be a challenge for many candidates and there were several blank responses seen. It is possible that this was due to time management. Some candidates attempted unsuccessfully to change to a Cartesian equation, suggesting that possibly parametric integration had not been covered. In part (a), candidates who recognised the need to use parametric integration were usually able to differentiate correctly to find d d x t , multiply this by y and then use the double angle formulae to find an acceptable expression for d d x y t . It was common for candidates to score the first two marks. The next two marks were more elusive, and it was common to see candidates perform a lot of onerous manipulation without making any significant progress. Those who were most likely to be successful were the ones who had an expression in terms of 2 sin 2t and then spotting the need to use the double angle formula to form a term in cos 4t. For many candidates the mark for the upper limit was all that was achieved in part (a). However, some failed to achieve it as they left it as 4 (value for x not t) or 45 (degrees). It was good to see students who were unsuccessful at part (a) still attempting part (b), and in some cases scoring full marks. Most candidates were able to integrate the first two terms correctly, but a significant proportion were unable to integrate the third term to get an expression in 3 sin .t The majority of correct solutions came from the use of the reverse chain rule for the final term. Those who attempted to use integration by substitution or by parts were usually unsuccessful. Candidates who had successfully completed the integration often went on to score full marks.

  Text version

  The curve C has parametric equations x=8sin^2(t) and y=2sin(2t)+3sin t for 0<=t<=pi/2. The region R is bounded by C, the x-axis and the vertical line x=4; it is the area under the first part of C from the origin to x=4.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q4 · 3 marks

  3 marks

  4

  
  Show mark scheme

  

  Examiner report

  This question on differentiation from first principles saw good scoring and the method was widely recalled. A small number of slips with the algebra were seen and some replaced f (x + h) with 2x2 + h or (2x + h)2 or 2(x2 + h) instead of 2(x + h)2. There was the occasional poor squaring and mishandling of the “2”s e.g. expanding 2(x + h)2 to get 2x2 + 2xh + h2 was surprisingly common. Some leeway was afforded to the students with notation here although use of h = 0 instead of h 0 was penalised if no indication of the limiting process was seen. A small number of responses did not have an “f′(x) =” or equivalent and so forfeited the final mark.

  Text version

  Given that y = 2x^2:

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2022 · Paper 2 · Q6 · 7 marks

  7 marks

  6

  
  Show mark scheme

  

  Examiner report

  Part (a) was generally well done with many candidates scoring 3 out of 4 but very few scoring 4 out of 4. Candidates generally understood that the question required an initial differentiation and this was done correctly on the whole. However, incorrect attempts at differentiation included: • 16cosቀ 1 2 𝑥𝑥ቁ−3 • 8cosቀ 1 2 𝑥𝑥ቁ−3 • −4cos ቀ 1 2 𝑥𝑥ቁ−3 Candidates also understood that they had to set their derivative to zero which required a combination of manipulating the algebra, applying arccos, and doubling their answer. The difficulty encountered in this part was the recognition that it was the 3rd solution which was required and most gave either the 1st or 2nd turning point. As a diagram was given in the question, candidates should use this to help them understand what is required. Some students who gave the 1st turning point also checked by further differentiation that their point was a maximum. A few candidates unnecessarily found the corresponding y value for their x value. In establishing their value of x, it was disappointing to see all of the following: • candidates halving instead of doubling • using degrees • arriving at a correct answer but leaving it as 14.01 instead of 14.0 In part (b), most candidates only referred to the change of sign and didn’t mention continuity. Occasionally a candidate wanted to allude to continuity but was unable to articulate this correctly. For example incorrect wording including referring to the graph as “constant”. Some candidates who referred to continuity used incorrect statements such as “because x is continuous” or “because the interval (or data) is continuous”. Part (c) was answered well with many candidates scoring full marks. Amongst those who didn’t score full marks, many did not heed the demand in the question to “Show your method” and just quoted their calculator’s solution without any explanation whilst others just wrote down a formula but did not show an application with values embedded. Others made rounding mistakes such as giving a final answer of 4.805 or 4.81. Premature rounding of f (5) and f′(5) also led to a final answer of 4.81. A significant number of candidates erroneously used degrees but could score the method mark if appropriate working was shown. Some candidates were able to “recover” minimal working due to the accuracy of their answer 4.804...., and gained the two marks. Some candidates also evaluated f (5) incorrectly even though it was given in the question. It was disappointing to see students misquoting The Newton-Raphson formula, typically with f(x) and f′(x) the wrong way round, even though the formula was given in the formula book.

  Text version

  The curve has equation y = f(x), where f(x) = 8sin(x/2) - 3x + 9 for x > 0, with x measured in radians. A sketch shows the first positive x-axis crossing at x = alpha and the later local maximum P. Given, to 3 decimal places, f(4) = 4.274 and f(5) = -1.212:

  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 · 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 · Q4 · 9 marks

  9 marks

  4

  
  Show mark scheme

  

  Examiner report

  Aspects of this question proved to be more demanding than expected, but overall good marks were scored by candidates with most achieving over half marks in this question. Part(a) Almost all candidates identified the need to differentiate and set their expression equal to zero. Although the derivative was often accurately found – the chain rule aspect of the log expression was omitted by some so that the expression 4x-4 was not seen as the numerator of the required fraction. Some numerators seen included 2x-2, 2x- 4 or most often 1. Manipulating their equation into the desired form was also an issue for some, with sign errors causing problems, whilst others could not properly convert their equation involving fractions into one not involving fractions. Part (b) A significant majority of candidates obtained both required answers to this part. Occasionally only x2 and x3 were found. There were just a few occasions where a candidate had misread the 1/7 in the iterative formula as ½. If the working was shown for this situation, then a method mark could still have been awarded. Part (c) Whilst many candidates realised that a sign change of a function between 0.3405 and 0.3415 was required, it was also common to see an attempt at further iteration. This method was not acceptable as a proof of the location of the root. A few chose to substitute into an inappropriate function such as f (x), or the iterative formula, and some failed to specify a function at all. For those who did use (typically) the expression from (a), most selected correct values to substitute in and identify the change of sign, finishing with a clear conclusion. It was very common, however, for the final mark to be lost in this part because there was no mention that the chosen function was continuous.

  Text version

  The curve y=f(x), where f(x)=x^2+ln(2x^2-4x+5), has a single turning point at x=alpha. The iteration x_(n+1)=(2+4x_n^2-2x_n^3)/7 is used to approximate alpha, starting with x_1=0.3.

  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 · Q5 · 7 marks

  7 marks

  5

  
  Show mark scheme

  

  Examiner report

  This question was accessible to the majority of students however part (b) proved to be a good discriminator with only the most able gaining all 7 marks. The majority of students gained the first three marks of this questions by correctly differentiating the expression twice. A few arithmetic errors were seen occasionally, although if this was the case then students would only lose 1 out of the three marks. Part (b) differentiated out the students to a greater extend. Most students could identify the correct method to find the stationary point, although there were a disappointing number of students who failed to follow this up with a concluding statement or preamble. A number of students got confused between parts (i) and (ii) and tried to use the second derivative to identify stationary points. The most common method to identify the stationary point was using the first derivative and substituting the value x = 1, although there were a significant number who chose to deduce the value x = 1 by factorising their cubic expression. Both approaches tended to lead to a positive outcome, and full marks. Part (ii) was much more mixed in response. A significant number of students gained no marks on this part as they assumed that when both first and second derivative equal zero then the point is a point of inflection. Of those students who successfully answered this question there were an equal number of students who considered the first derivative as those that considered the second derivative. Most of these students considered an appropriate intervals to gain full marks here, however a significant number of students considered inappropriate intervals such as 0 and 2 or 1.5 and so were unable to gain the accuracy mark.

  Text version

  The curve C has equation y = 5x^4 - 24x^3 + 42x^2 - 32x + 11 for x in R.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 2 · Q7 · 9 marks

  9 marks

  7

  
  Show mark scheme

  

  Examiner report

  Part (a) was generally well answered, with most students able to differentiate correctly and understand the need to use the first derivate to find the gradient of the tangent line. A few tried to use the diagram to obtain the gradient, even after differentiating correctly so were only able to gain the first B mark, but they were in a minority. The vast majority of candidates who correctly differentiated (a few integrated, hence lost the marks), were able to go on to find the correct equation of the line, often using y = mx + c rather than 𝑦𝑦−𝑦𝑦1 = 𝑚𝑚(𝑥𝑥−𝑥𝑥1). A small minority displayed poor algebraic skills when rearranging their linear equation to find c, and a few used a gradient of 1 2 instead of 2. In part (b), most students understood the need to show that the y-intercepts were the same for both the line and the curve, but often candidates struggled to write an adequate conclusion to verify they meet again, and hence lost this mark. Most students were able to gain some marks in part (c) but a score of two out of four was common. Those who got this, generally integrated the curve, obtaining a correct area of 455/12, but then failed to go any further. Very few students used the area of a trapezium formula, instead preferring to integrate the (2x – 23) or to split the area into a rectangle and triangle. Of those who attempted the area of the trapezium a number struggled to get the correct dimensions. Fully correct solutions using a trapezium were rare. A few students translated the curve and line upwards correctly by 23 units and often scored full marks using this method. Those students who attempted the curve minus the line, often struggled with the constant 23, or incorrectly collected like terms (e.g. an x term with an 𝑥𝑥2 term).

  Text version

  In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The curve C has equation y = x^3 - 10x^2 + 27x - 23. The point P(5, -13) lies on C, and the line l is tangent to C at P. The sketch shows that l also meets C at x = 0. The finite region R is bounded by C above and l below between x = 0 and x = 5.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 2 · Q8 · 9 marks

  9 marks

  8

  
  Show mark scheme

  

  Examiner report

  Many candidates gained some credit in this question but few gained full marks. Part (a) was relatively well answered. Most candidates identified that implicit differentiation was required and made good progress by correctly differentiating the given equation. When mistakes happened at this stage, they usually appeared in the differentiation of rather than in the differentiation of or . There were a few candidates, however, who clearly were not able to employ implicit differentiation and who gained no marks. Once a differentiated expression had been obtained, many were able to rearrange correctly to make the subject, obtaining the required answer of . A common mistake, however, was to make sign errors in the rearranging process and incorrect answers like and were surprisingly common following correct differentiation. In (b) most candidates gained some marks but few used a fully correct method to achieve the correct answer of and . Most were able to rearrange the equation of the normal, and identified that its gradient was . They then identified the gradient of the tangent as and used their result from (a) at and to attempt to form a correct equation in and such as . A number of candidates incorrectly equated the numerators and denominators to form the equations and and solved these to fortuitously arrive at the correct solution, losing most marks in (b). Another incorrect approach was to form the equation of the normal of the form and to rearrange to the form followed by 'comparing coefficients' with (an incorrect method). Many candidates successfully substituted and into the equation of C to get . This step was missed, however, by quite a few candidates, who could not then gain full marks for the question.

  Text version

  The curve C has equation px^3 + qxy + 3y^2 = 26, where p and q are constants.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 2 · Q13 · 6 marks

  6 marks

  13

  
  Show mark scheme

  

  Examiner report

  This question involving differentiation of parametric equations was found to be challenging by the majority of students, particularly the differentiation of 𝑦𝑦= cosec3 𝜃𝜃 and part (b). The majority of candidates knew what was expected in part (a) and proceeded to attempt to use the chain rule to find d𝑦𝑦 d𝑥𝑥, however they found differentiating 𝑦𝑦= cosec3 𝜃𝜃 challenging. Some did not check in the formula book that the derivative of cosec θ was negative, or omitted the constant 3, and this cost 3 of the 6 marks available in the whole question. For those that attempted to write 𝑦𝑦= sin−3 𝜃𝜃 and use either the chain rule or the quotient rule success was mixed. Other common incorrect derivatives included d𝑦𝑦 d𝜃𝜃= (cosec3 𝜃𝜃cot3 𝜃𝜃) and variants of this. Differentiating 𝑥𝑥= sin 2𝜃𝜃 proved less troublesome, with most doing so correctly. There was a proportion of candidates that rewrote sin 2𝜃𝜃 as 2 sin 𝜃𝜃cos 𝜃𝜃 and used the product rule, creating more work but ultimately doing so successfully in most cases. The method mark was scored by the majority of candidates as a result, although there were a few candidates who tried to write 𝑦𝑦 in terms of 𝑥𝑥 and differentiate, with no marks being scored for this approach. A number of candidates, having found a correct derivative, proceeded to attempt to tidy this up, unfortunately using time that could have been spent elsewhere in the paper. Part (b) proved more challenging for many candidates, with some unable to make a start as they were unable to process 8 = cosec3 𝜃𝜃 to arrive at either a value for sin 𝜃𝜃 or 𝜃𝜃. Of those that did arrive at a value for sin 𝜃𝜃 or 𝜃𝜃, many were able to work in exact form using their value to find an exact value for their 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑, others were unable to proceed or attempted to give a decimal answer. 24√3 and 8√3 were common incorrect final answers. There were, however, some excellent and concise responses to this question.

  Text version

  The curve C has parametric equations x = sin(2theta) and y = cosec^3(theta), where 0 < theta < pi/2.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2021 · Paper 2 · Q14 · 12 marks

  12 marks

  14

  
  Show mark scheme

  

  Examiner report

  This question proved challenging for the majority of candidates, with many not answering it at all. For those that did attempt a response, part (a) was reasonably well answered, with many candidates gaining at least one of the B marks but quite a few then failing to correctly apply the chain rule and complete the proof. Some weaker candidates recognised they had to consider "flow in - flow out” but didn’t identify this as d d V t , with d 0.48 0.1 d h h t = − sometimes seen. It was also not uncommon to see d d V t = 0.48 or d d V t = 'uncombined' in some responses. Some candidates even started with and “differentiated” to get the printed result. Responses to part (b) were mixed. Some candidates failed to separate the variables correctly or did not recognise the logarithmic form for the integration. It was not unusual for the “+c” to be missed completely, but if it was included candidates were able to gain the 2nd method mark despite poor integration. Some candidates changed to , making their following working much harder. There was some occasional use of Further Maths methods (e.g. integrating factors), potentially gaining full marks. Candidates who did not answer (a) rarely attempted (b), perhaps not realising (b) could be answered independently using the printed differential equation. Part (c) was attempted by many candidates, even if they hadn’t attempted (a) and (b). Students who progressed well in (b) often interpreted their formula correctly to obtain marks in (c). A minority of students who didn’t complete (b) still attempted (c) with some success, e.g. recognising that d d h t was zero at the maximum height or using a ‘ flow in = flow out’ argument.

  Text version

  Water flows at a constant rate into a cuboid tank whose base is 8 m by 3 m and whose height is 5 m; all sides have negligible thickness. A tap T is at the bottom. At time t minutes after the tap is opened, the water depth is h metres. Water enters at 0.48 m^3 per minute and is modelled as leaving through the tap at 0.1h m^3 per minute.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 1 · Q8 · 2 marks

  2 marks

  8

  
  Show mark scheme

  

  Examiner report

  Candidates were expected to realise that ''when the rate of change is proportional to the y value, an exponential model should be used.'' The question seemed to take a great many candidates by surprise and many assumed that 'n was proportional to n' or that 'n was proportional to t', not that the rate of growth of 'n was proportional to n'. Those that did interpret the situation correctly could write down an equation very quickly whereas others attempted to solve a differential equation. Equations scoring one of two marks were common and included e , e t kt n A n = = and e kt n A b = +

  Text version

  A company monitors the total number n of a new smartphone sold t days after release. During the monitored period, the rate of increase of n is proportional to n.

  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 1 · Q14 · 10 marks

  10 marks

  14

  
  Show mark scheme

  

  Examiner report

  Students found aspects of this question difficult despite the fact that the differential equation to be solved in (b) was not too demanding. Only high scoring candidates tended to make any significant progress in this question. In part (a) many candidates failed to use the chain rule d d d d d d V V r t r t = × with an acceptable d d V t and d . d V r It was possible to proceed to the given answer without knowing the formula for a sphere, using 3 V r = λ but errors and blank solutions were common. Common errors were; • stating d d V k t = −, and then failing to provide a minimal explanation of how their k/4𝜋𝜋 was replaced with k. • quoting an incorrect formula for the volume of a sphere. Attempts included V = π r3 , 2πr3 , 4πr3 and even 4πr2. In (b) students who attempted to solve the given differential equation often separated the variables correctly and integrated both sides with both indices correct. Marks were lost when students failed to give a constant of integration and as a result could not apply the given conditions correctly to their equation in r and t. In part (c) many students were confused by the wording and did not realise that a limiting value for t was required. Many attempts merely stated that t could not be negative or that t >0. Those candidates who did proceed to a limiting value for t generally had the right idea and most of these obtained a value of t >5 and gained the A mark when their equation form part (b) was followed through.

  Text version

  A large spherical balloon is deflating. At time t seconds it has radius r cm and volume V cm^3. The model states that V decreases at a constant rate. Initially r=40; after 5 seconds r=20; the constant volume decrease continues until the balloon is empty.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 1 · Q15 · 7 marks

  7 marks

  15

  
  Show mark scheme

  

  Examiner report

  Although fully correct solutions to this question were very rare, many candidates were able to score marks from both parts. In part (a) the first two marks for the implicit differentiation were very accessible for the prepared candidate. Proceeding to the required result for the second two marks proved a little more challenging. Only high achieving students noticed the relationship could be proven via the trigonometric identity 2 2 1 tan sec . y y + = In (b) candidates often scored the M1 for an unsimplified second derivative or a correct structure with a sign incorrect. Where the first A1 was scored in (b), it was usually awarded for a correct simplified 2 2 d d y x , or solving 2 2 d 0 d y x = to get 4 27 x = Very few candidates seemed aware that they needed to show a change in sign of 2 2 d d y x either side of 4 27 x = to show a point of inflection.

  Text version

  The curve C has equation x^2 tan(y)=9, where 0<y<pi/2.

  Open the full paper (PDF)

• Edexcel 9MA0 · October 2020 · Paper 2 · Q7 · 10 marks

  10 marks

  7

  
  Show mark scheme

  

  Examiner report

  In part (a), most candidates used a correct method to differentiate, the majority using the quotient rule. A few applied the rule the wrong way round (uv’ – vu’)/v2. Candidates who initially wrote the quotient as 2𝑥𝑥 3 2 + 1 2 𝑥𝑥 1 2 were often more successful with the differentiation. A few candidates forgot about the 4lnx term but most dealt with this correctly. Errors from missing brackets or when simplifying their differentiated expression, in particular dealing with the powers, meant that many did not gain full marks in this part. In part (b), although almost all candidates made dy/dx = 0 and reached 12x2 + x - 16√x = 0 the majority then failed to make progress and did not gain any marks. Several made no attempt to rearrange this equation and others made x2, x or √x the subject but did not make the key step of dividing by √x first. Part (c) was generally well done, with most candidates getting part (i) correct. Some candidates left out part (ii) and a few made an error with their accuracy here, 1.15651 and 1.1565 being seen a few times.

  Text version

  The curve C has equation y = (4x^2 + x)/(2sqrt(x)) - 4ln(x), x > 0. The curve has a minimum point 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 · June 2019 · Paper 1 · Q3 · 5 marks

  5 marks

  3

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 2.8 out of 5) This was an accessible question for most students, but the second half of part (a) did discriminate at all levels. In part (a) students generally recognised the need to use the quotient or product rule and were successful at differentiating for the first 2 marks. The most common mistake was forgetting to square the denominator, with incorrect order of terms in the numerator being less common. Students should also be reminded to quote a formula before attempting to use it, as they may still gain method marks for this, even if there are slips in its application. Omission of brackets in terms in the numerator was also common, but often recovered in working. However, it would benefit students to be careful in the use of brackets. Having achieved a derivative, attempting to simplify to the required form proved difficult for many students. Many did not realise that they could factorise the (x+1) out early and instead expanded fully, simplified then factorised again to cancel in the last step. This wasted valuable time and it was during this process that errors usually occurred. The alternative method of simplifying the fraction before differentiating was less common, with usually poor attempts at partial fractions being applied leading to error. Those who successfully simplified the fraction almost always went to achieve the correct answer. Nearly all students who were correct in part (a) achieved the mark for part (b), including many who had failed to cancel the (x+1). Indeed, there were many cases where a correct answer to part (b) was seen following incorrect work in (a), presumably due to the use of a calculator to plot and inspect the graph.

  Text version

  Let y=(5x^2+10x)/(x+1)^2, where x!=-1.

  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 · Q14 · 7 marks

  7 marks

  14

  
  Show mark scheme

  

  Examiner report

  (Mean Mark 2.4 out of 7) A number of candidates did not attempt this question. Whether this was due to lack of understanding or time constraints is unknown. Part (a) was generally well done with a surprising proportion using implicit differentiation. A good number who found d d y x correctly failed to substitute in y =0 to find the value of d d y x at the origin. Candidates who tried to expand sin2y using the double angle formulae and then differentiating using the product rule often went wrong. The small angle approximation in (b) part (i) was also well handled. Most settled for the form 8 x y  although some rearranged to y = . Marks were generally lost when sin 2y was replaced by 2 instead of 2y. Many candidates failed to see the link between (a) and (b)(i) even though they had both answers correct. They were unable to see that the gradients were the same or else failed to explain themselves coherently. Only a small percentage of candidates were able to secure full marks for part (c) and it was not uncommon to see the whole question left unanswered. This type of question was common on Core 3 and usually well answered.

  Text version

  The curve C is defined by x=4sin(2y), with -pi/4<y<pi/4, and passes through the origin O.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 2 · Q11 · 11 marks

  11 marks

  11

  
  Show mark scheme

  

  Examiner report

  In general, Q11 discriminated well between candidates of all abilities. Q11(a) was demanding with most candidates scoring no more than one mark. Q11(b), Q11(c) and Q11(d) proved accessible to candidates of all abilities. In Q11(a), most candidates took logs, as advised by the question, and many obtained either ln ln y x x  or log log . y x x  At this point a significant number of candidates made no further creditable progress. Some candidates used a complete method of applying the product rule on ln x x and implicit differentiation to obtain a correct 1 d 1 ln . d y x y x  Most of these candidates then applied d 0 d y x  to obtain 1 e . x   Some candidates, who received partial credit, used a ‘ 10 log ’ method to obtain 0.1, x  after finding 10 1 d 1 log . d y x y x  A few candidates used logarithms without reference to a base. Those who proceeded to obtain 0.1 x  received partial credit and those who obtained 1 e x   were allowed full marks. There were some alternative methods employed by a few candidates in Q11(a). A few candidates rewrote x y x  as ln e x x y  and applied ln d (1 ln )e 0 d x x y x x    to give a correct 1 e . x   Others differentiated both sides of the equation ln ln y x x  with respect to y and usually proceeded to find 1 e . x   A very few candidates applied the quotient rule to the equation ln , ln y x x  but were rarely successful with this method. In Q11(b), most candidates evaluated either xx or 2 xx  at both 1.5 x  and 1.6. x  Many candidates compared their values with either 2 or 0 (depending on whether they evaluated xx or 2 xx  ). Although many of these concluded that 1.5 1.6,    only a few made any reference to the curve being continuous in this interval. A few candidates, who made no creditable progress, used values of x outside the given range, e.g. 1.4 x  and 1.6. x  Very few candidates used values within the range e.g. 1.51 x  and 1.59, x  which meant that they were able to earn both marks. In Q11(c), most candidates found a correct 4 1.673, x  although a few did not state 4x correct to 3 decimal places. Some wrote down the iterates 1 2 , x x and 3x as part of their solution, while others stated 4x with no intermediate work. In Q11(d), some candidates gave correct descriptions such as ‘sequence oscillates between 1 and 2’ or ‘the sequence is periodic with period 2’. Other descriptions such as ‘alternates between 1 and 2’, ‘keep getting 1 and 2’ or ‘fluctuates between 1 and 2’ were condoned. Some credit was given for partial explanations such as ‘sequence is non-convergent’, ‘sequence oscillates’ or ‘sequence is divergent’.

  Text version

  The curve C has equation y = x^x for x > 0. The point P(alpha,2) lies on C.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2019 · Paper 2 · Q13 · 10 marks

  10 marks

  13

  
  Show mark scheme

  

  Examiner report

  Q13 discriminated well between the medium and higher ability candidates. Lower ability candidates, however, struggled to gain any creditable progress. It was obvious that some candidates had been well prepared for this ‘optimisation’ question and the quality of their responses reflected this. candidates who attempted this question generally found Q13(b) and Q13(c) accessible but some struggled to complete the proof in Q13(a). In a few cases, candidates who scored zero marks in Q13(a) went on to achieve full marks in the remainder of Q13 by using the given result for the surface area. There were, however, many instances where candidates who struggled with Q13(a) then gave up and did not attempt the remainder of the question. In Q13(a), successful candidates followed a correct strategy of forming an equation for the volume of the tank, rearranging their equation to give an expression for h in terms of r and substituting their expression for h into their formula for the surface area of the tank. Some candidates did not halve the given sphere formulae so that they could use these for the hemispherical shell. This led to errors in both their volume equation and surface area expression. Other candidates did not include the area of the circular base in their surface area expression. These errors led to an incorrect volume equation 2 3 4 6 3 r h r     and incorrect surface area expressions 2 5 2 A r rh     and 2 2 2 . A r rh     In other cases, bracketing errors, manipulation errors, or using incorrect formulae for the curved surface area of a cylinder, prevented candidates from achieving the given result. In Q13(b), many candidates who attempted this part applied a complete method to find their value of r for which d 0. d A r  Most successful candidates found 1.05 r  metres, with a few giving an exact answer 3 18 5 r   metres, which was condoned. Some candidates differentiated A incorrectly to give d 10 12ln , d 3 A r r r    while others used incorrect algebra when solving the correct equation 2 12 10 0. 3 r r     A few candidates used the incorrect method of finding a value of r for which 2 2 d 0. d A r  In Q13(c), most candidates who attempted this part correctly substituted their value of r, found from solving d 0, d A r  into the model with equation 2 12 5 , 3 A r r    with many obtaining a correct minimum surface area of 2 17m . A few candidates, who made no creditable progress, substituted their value for r, found from solving 2 2 d 0, d A r  into the model for A.

  Text version

  A storage tank is modelled as a hollow circular cylinder closed at one end, with a hemispherical shell at the other. The walls have negligible thickness. The cylinder has radius r metres and height h metres; the hemisphere also has radius r. The tank's volume is 6 m^3. A sphere of radius r has volume (4/3)pi r^3 and surface area 4pi r^2.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 1 · Q2 · 7 marks

  7 marks

  2

  
  Show mark scheme

  

  Examiner report

  Part (a) of this question allowed the demonstration of good skills in differentiation and was well answered by the majority of students. A small minority made careless errors on the indices, for example writing x as 2 x, leading to the loss of both marks as their derivative was not of the required form. Others dealt with the index correctly, but made arithmetic errors on the coefficient when differentiating the term. The second derivative was almost always either correct, or a correct follow through, from the students first derivative. In part (b) a significant number of students were seen attempting to solve the equation dy/dx = 0 although most quickly realised that they could substitute x = 4 into dy/dx to obtain 0. To be awarded the final mark of part (b), students were required to interpret their value of dy/dx at x=4 and conclude that the point was a stationary point of the curve. Many students omitted this final step thus failing to fully answer the question. For part (c) there was a little more variation. Most students were aware of the need to substitute 4 x  into their second derivative and evaluate its sign, although some failed to do this and instead considered the sign of either y, dy/dx and/or d2y/dx2 either side of 4 x  (usually at 4.1 and 3.9). A follow through mark enabled students with an incorrect negative or fractional index from part (a) to achieve both marks in the last part of the question. A significant proportion of students seemed unable to interpret what they had proven from their calculations, and many seemed to think they were looking for an inflection point, or else determining that the curve was either convex or concave. As with part (b), the most common source of lost marks was therefore a lack of correct reasoning before making a conclusion.

  Text version

  A curve C has equation y=x^2-2x-24sqrt(x), where x>0.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 1 · Q5 · 5 marks

  5 marks

  5

  
  Show mark scheme

  

  Examiner report

  The majority of students knew to use the quotient rule in this question, although the slips on signs and bracketing in the terms of the numerator were commonplace. We would encourage all students to state the formula being used, with their expressions for ‘u’ ‘v’ ‘du’ and ‘dv’, to ensure that their method is made clear to examiners. Getting a correct derivative into the required form using trigonometric identities proved to be more challenging than the differentiation in this question. Students recognised the need to use the Pythagorean identity for sine and cosine to simplify their numerator and denominator but were less confident when applying the double angle formula for sine, and it was not uncommon for a final answer to be left as 3 2 2sin 2  , or for there to be errors in an attempt to write this fraction in the required form.

  Text version

  Given y=(3sin(theta))/(2sin(theta)+2cos(theta)), where -pi/4<theta<3pi/4:

  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 2 · Q9 · 5 marks

  5 marks

  9

  
  Show mark scheme

  

  Examiner report

  This was a well-answered question considering that this was the first time that differentiation from first principles has been tested on the new specification. Most students wrote down a correct expression for the gradient of the chord joining the points ( , cos )   and ( , cos( )) h h     for a small value h. Many students simplified their gradient expression to give a correct sin cos 1 sin cos h h h h           and applied the limiting arguments given in the question to deduce that   d cos sin . d     Some students lost the final mark for not applying limiting arguments correctly. Common errors included: stating that sin cos 1 1, 0 h h h h    ; not using a limiting argument; or applying the limiting argument inconsistently. Other students lost the final mark for the use of inconsistent notation in their proof. These students, who usually referred to both x and  in their proof, wrote incorrect statements such as   d cos sin dx    or f( ) cos . x  

  Text version

  The angle theta is measured in radians. You may assume the formula for cos(A +/- B) and the limits sin(h)/h -> 1 and [cos(h) - 1]/h -> 0 as h -> 0.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 2 · Q10 · 8 marks

  8 marks

  10

  
  Show mark scheme

  

  Examiner report

  This question discriminated well between the average and more able students. There waere a significant minority of students, however, that made no creditable progress. In part (a), many students struggled to translate the description of the model into a mathematical statement. Some incorrect statements, such as 2 d 1 , d r t r  2 d d r r t  or d 1 , d r t r  did not contain a constant of proportionality; while other incorrect statements 2 d , d r kr t  2 d d r r t k  or d d r k t r  contained a constant of proportionality. Many students used a correct method to separate the variables and most proceeded to integrate their 2 d d r r k t    to give an equation linking r and t which contained a constant of integration. At this stage, some students incorrectly integrated 2r to give 2 .r The majority of students applied the boundary conditions 0, 5 t r   and 4, 3 t r   (or 240, 3 t r   ) to find both constants and proceeded to give an equation linking r with t. Some students who worked in minutes obtained correct answers of either 3 250 49 , 2 t r   3 49 125 2 r t   or 3 250 2 . 49 r t   A few students who worked in seconds obtained correct answers of either 3 15000 49 , 120 t r   3 49 125 120 r t   or 3 15000 120 . 49 r t   Many students lost the final mark in this part by not including sufficient detail in their definition of their variables. Some students defined r as the radius of the mint, in mm, but it was rare to see the time, t, defined from a correct starting point (i.e. from when the mint was placed in the mouth) and with the correct unit of time used. In part (b), many students substituted 0 r  in their equation from part (a) and rearranged to find a value for t, with a significant number obtaining the correct answer, 5 minutes 6 seconds. Some students lost the final mark by giving their final answer as either 5.10 minutes, 250 49 minutes or 306 seconds. Part (c) was generally well-answered with a variety of valid limitations given. Students commented on the shape of the mint while it was being sucked; on how the mint was sucked; or that the model predicted a negative radius for the mint after a time of 5 minutes 6 seconds after the mint was placed in the mouth. Some students who stated that ‘the radius of the mint may not decrease at a constant rate’ did not receive any credit because the radius of the mint in the original model was not decreasing at a constant rate.

  Text version

  A spherical mint of radius 5 mm is placed in the mouth and sucked. Four minutes later its radius is 3 mm. In a simple model, the rate of decrease of the radius is inversely proportional to the square of the radius.

  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 · June 2018 · Paper 2 · Q13 · 10 marks

  10 marks

  13

  
  Show mark scheme

  

  Examiner report

  This question, despite being unstructured, proved accessible, with many students scoring at least 7 out of the 10 marks available. Those students who annotated the diagram were generally more successful in formulating a correct strategy to find the area of the region R. Most students differentiated ln y x x  by using the product rule to give the correct d 1 ln . d y x x  A few students found the correct d ln d y x x x x   and simplified this incorrectly to give d ln . d y x x  Many students used a complete method to find the x-value where the normal to C at (e, e) cuts through the x-axis, with the majority finding the correct 3e. x  Most students integrated ln y x x  correctly by using integration by parts to give 2 2 1 1 ln . 2 4 x x x  Many students applied the correct limits of e and 1 to obtain a correct 2 1 1 e 4 4  with some making a bracketing error when applying the limit of 1 to obtain the incorrect 2 1 1 e . 4 4  The majority of students attempted to find the area of the triangle by using the formula 1 (base)(height). 2 While some students applied the formula correctly to give 1 (3e e)(e), 2  there was a considerable number that applied a base length of 3e by writing 1 (3e)(e). 2 A few students used the equation of the normal to find the area of the triangle by applying the correct 3e e 1 3 e d , 2 2 x x          but not many evaluated this to arrive at the correct answer of 2e . Several incorrect strategies were seen such as trying to evaluate e 0 ln d , x x x  3e 1 ln d x x x  or even 3e 1 1 3 ln e d . 2 2 x x x x                 Only a minority of students applied the calculation 2 2 1 1 e e 4 4   to achieve the correct 2 5 1 e . 4 4 

  Text version

  The curve C has equation y = x ln(x), x > 0. Point P = (e, e) lies on C. The line l is the normal to C at P. Region R is bounded by C from its positive x-axis intercept to P, by l from P to its x-axis intercept, and by the x-axis.

  Open the full paper (PDF)

• Edexcel 9MA0 · June 2018 · Paper 2 · Q14 · 10 marks

  10 marks

  14

  
  Show mark scheme

  

  Examiner report

  This was the most demanding question on the paper. Parts (a) and (d) were accessible to many students. Most students made some creditable progress in part (b), but many students found it difficult to access the demanding material in part (c). In part (a), most students evaluated or deduced that there were 90 mice at the start of the study. In part (b), most students used either the chain rule or the quotient rule in their attempt to find an expression for d . d N t There were many incorrect attempts at the chain rule such as 0.25 0.25 2 d 6300e (3 7e ) d t t N t      or 0.25 2 d 1575(3 7e ) d t N t     while the most common incorrect attempt at the quotient rule was 0.25 0.25 0.25 2 d (3 7e )(1) 1575e . d (3 7e ) t t t N t        Although some impressive algebra was seen in manipulating a correct 0.25 0.25 2 d 1575e d (3 7e ) t t N t     to achieve d (300 ) d 1200 N N N t   , many students were unsuccessful in their attempts at eliminating t or the 0.25 e t  term. Some students substituted 0.25 900 3 7e t N    into d (300 ) d 1200 N N N t   and used algebra to give 0.25 0.25 2 d 1575e d (3 7e ) t t N t     , and provided some indication that their proof was complete. Some students differentiated N by using implicit differentiation while other students attempted to rearrange the given equation for N to make t the subject before finding an expression for d . d t N A few students were successful in applying an alternative method of solving the differential equation d (300 ) d 1200 N N N t   to give 0.25 900 . 3 7e t N    In part (c), most students misinterpreted the question and instead applied a method to find the time T for the maximum value of N, rather than the time T at which the rate of growth d d N t was a maximum. Many students, therefore, used 300 N  and manipulated 0.25 900 300 3 7e T    to give 0.25 e 0. T   A few students differentiated d , d N t with most writing 2 2 d , d N t which was condoned, and solved the equation 1 (300 2 ) 1200 N  to give 150. N  It was rare to see students using the symmetry of the quadratic function (300 ) 1200 N N  to deduce that 150. N  Although some students gave their answer to part (c) as 150, T  most students who progressed this far solved the equation 0.25 900 150 3 7e T    to give the correct value for T. In part (d), many students deduced 300 P  with only a few students deducing 299, P  which was also accepted.

  Text version

  A population of mice is modelled by N = 900/[3 + 7e^(-0.25t)], where N is the number of mice t months after the start of the study and t >= 0.

  Open the full paper (PDF)

• Edexcel 9MA0 · sample · Paper 1 · Q1 · 7 marks

  7 marks

  1

  
  Show mark scheme

  

  Text version

  The curve C has equation y=3x^4-8x^3-3.

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

• Edexcel 9MA0 · sample · Paper 1 · Q10 · 5 marks

  5 marks

  10

  
  Show mark scheme

  

  Text version

  The angle theta is measured in radians. You may assume the formula for sin(A+B) and that, as h tends to 0, sin(h)/h tends to 1 and [cos(h)-1]/h tends to 0.

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

• Edexcel 9MA0 · sample · Paper 1 · Q13 · 13 marks

  13 marks

  13

  
  Show mark scheme

  

  Text version

  The curve C has parametric equations x=2cos(t) and y=sqrt(3)cos(2t), where 0<=t<=pi. Point P lies on C where t=2pi/3, and the line l is the normal to C at P.

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

• Edexcel 9MA0 · sample · Paper 1 · Q15 · 8 marks

  8 marks

  15

  
  Show mark scheme

  

  Text version

  For 0<=x<=pi, f(x)=4sin(2x)/e^[sqrt(2)x-1]. The curve y=f(x) has a maximum turning point P followed by a minimum turning point Q.

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

• Edexcel 9MA0 · sample · Paper 2 · Q3 · 4 marks

  4 marks

  3

  
  Show mark scheme

  

  Text version

  Given y = x(2x + 1)^4, show that dy/dx = (2x + 1)^n(Ax + B), where n, A and B are constants to be found.

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

• Edexcel 9MA0 · sample · Paper 2 · Q5 · 4 marks

  4 marks

  5

  
  Show mark scheme

  

  Text version

  The mass m grams of a radioactive substance, t years after observation begins, is modelled by m = 25e^(-0.05t).

  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 · Q15 · 10 marks

  10 marks

  15

  
  Show mark scheme

  

  Text version

  The curve C has equation y = 5x^(3/2) - 9x + 11 for x >= 0. The point P = (4, 15) lies on C, and the line l is tangent to C at P. Region R is bounded by C, l and the y-axis. Solutions based entirely on graphical or numerical methods are not acceptable.

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

• Edexcel 9MA0 · specimen · Paper 1 · Q9 · 5 marks

  5 marks

  9

  
  Show mark scheme

  

  Text version

  The curve C has equation y = 2x^3 + 5 and passes through P(1, 7).

  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 · Q6 · 9 marks

  9 marks

  6

  
  Show mark scheme

  

  Text version

  The curve y = f(x), where f(x) = (8 - x)ln(x) for x > 0, cuts the x-axis at A and B and has a maximum turning point Q.

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

• Edexcel 9MA0 · specimen · Paper 2 · Q8 · 7 marks

  7 marks

  8

  
  Show mark scheme

  

  Text version

  A bowl is modelled as a hemispherical shell. Initially it is empty and water flows into it. When the vertical water depth is h cm, the volume is V = (1/3)pi*h^2(75 - h) cm^3, for 0 <= h <= 24. The inflow rate is 160*pi cm^3 s^(-1) for 0 <= h <= 12.

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

• Edexcel 9MA0 · specimen · Paper 2 · Q12 · 7 marks

  7 marks

  12

  
  Show mark scheme

  

  Text version

  A curve C is given by sin(x) + cos(y) = 0.5, where -pi/2 <= x < 3pi/2 and -pi < y < pi. A point P lies on C and the tangent to C at P is parallel to the x-axis.

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