Integration questions

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• Edexcel 9MA0 · June 2024 · Paper 1 · Q7 · 8 marks

  8 marks

  7

  
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  This question was attempted by most candidates and the majority made a strong start. Many candidates wrote full solutions and were awarded full marks in all three parts. The question allowed candidates to score marks in later question parts even if previous parts had been incorrectly answered, and a good proportion went on from an incorrect part (a) to do so. In part (a), most candidates recognised the need to integrate the expression and were able to integrate an exponential function; various methods were seen here but the majority followed the main approach in the mark scheme. A significant minority of candidates multiplied by –0.2 instead of dividing, resulting in an incorrect model and marks being lost in (b) and (c) due to having an impossible equation to solve. The vast majority of candidates remembered to include the constant of integration and use the given information to find it, likely supported by the question giving the required form of the answer as a reminder. Occasionally candidates missed out on the second method mark as they used 1.5 as the constant of integration rather than finding the constant of integration by correctly substituting H = 1.5 and t = 0, but the majority managed this with success, obtaining correct values for their constant and gaining the second method mark. A common reason for the loss of the final accuracy mark was failing to write out the final version of the model, generally omitting H = , or simply stating the values of A and B. Some candidates opted to differentiate the given answer and compare coefficients to find the values of A before going on to find B in the usual way. This verification strategy was not in line with the requirement of the question but candidates that used this method were allowed to score all but the final mark. Candidates who had a correct answer in part (a) generally performed well in part (b), with the majority able to rearrange the equation and use logarithms correctly to find the value of t. If these candidates were not awarded full marks, it was often because they were unable to convert their decimal into hours and minutes correctly, and therefore lost the final accuracy mark. Candidates who did not have a correct answer to part (a) generally gained the first method mark for making 0.2 e t A  the subject of the equation, but many were not able to achieve the second method mark due to the context of the model. Candidates with B > 1.2 reached the logarithm of a negative value which should have provided an indication that there must have been an earlier error in need of addressing but few re-traced their workings and tended to use incorrect methods to create a positive answer. Part (c) was generally less well attempted than parts (a) and (b) with some candidates failing to understand what the question was asking them to do. The most common error was for candidates to attempt to solve d 0 d H t = . A surprising proportion of candidates gave answers higher than the maximum height. For candidates with successful approaches, marks were frequently lost by forgetting units and therefore not satisfying the demand of the question and finding the height of the hole. Similarly, many decided to subtract an often correct 0.9 from the maximum height, which indicated a misunderstanding of the model.

  Text version

  A cylindrical tank has height 1.5 m and is initially full of water. Water leaks from a small hole L in the side. While leaking, the depth H metres is modelled by dH/dt = -0.12e^(-0.2t), where t is hours after the leak starts.

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

  9 marks

  10

  
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  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.

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• Edexcel 9MA0 · June 2024 · Paper 1 · Q13 · 8 marks

  8 marks

  13

  
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  This question was very successful in distinguishing the most able mathematicians from the cohort. A significant proportion of candidates struggled, with many responses either left blank or only partially completed. The nature of a show that question ensured marks were available on both parts regardless of success in part (a) but many candidates who failed to prove the result in part (a) then went on to omit part (b) and missed out on this opportunity. However, it was not uncommon to see full marks in part (a) followed by no marks in part (b), and vice versa. Most candidates who attempted part (a) recognised the requirement to find d d x  and attempted to do so, with varying levels of success. Those who differentiated successfully generally went on to substitute into the expression correctly and gained at least the first 2 marks. The flexibility in the first method mark made this accessible to most candidates who had tried at differentiation. d d 2 sin cos x a    = appeared several times and benefited from the flexibility of the first method mark. The most common incorrect derivatives were 2 cos a  and 2 cos a . In many cases the notation was particularly poor, although d d x a was very generously condoned provided the right-hand side was correct. Candidates struggled with the rearrangement of the integrand into the form required to successfully use the double angle formula. Many were not able to work with or simplify the square root terms successfully, and even those that did sometimes failed to recognise the need to use the double angle formula for sin 2. A minority of candidates progressed correctly to the final answer, while some of those that achieved the final expression still lost the final mark for failing to address the limits, or for missing the d or the integral sign. For those who did not achieve the first method mark this was commonly due to careless copying or premature simplification such as 2 sin a a  − instead of 2 sin a a  − . Those who recognised the requirement for using double angle formulae to integrate a squared trigonometric function were generally successful in part (b) of this question. Occasionally candidates struggled to deal with 2 correctly, with 2 1 cos2 sin 2 2   − = seen all too frequently. These candidates scored no marks following incorrect use of the double angle identity. The other main errors in this part were in sign errors in the double angle identity and in the integration of the trigonometric term: commonly cos4 − was integrated to 1 sin 4 4  + , and despite this still leading to the correct answer it was at the expense of the two accuracy marks. A significant proportion of candidates incorrectly attempted to increase the power and divide leading to a term in 3 sin . Some unsuccessful attempts at restarting and integrating ( ) 1 1 2 2 x a x − by parts were also seen. There was also a significant proportion of candidates who attempted integration by parts which generally gained no credit. Candidates who were successful in using integration by parts often had a succinct approach, while many reached 2 cos 2 and at this point realised the use of double angle identity would help, where it would have been more concise to use the double angle identities in the first place.

  Text version

  Let a be a positive constant and consider the definite integral from 0 to a of sqrt(x)sqrt(a-x) dx.

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• Edexcel 9MA0 · June 2024 · Paper 1 · Q14 · 9 marks

  9 marks

  14

  
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  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).

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• Edexcel 9MA0 · June 2024 · Paper 2 · Q11 · 5 marks

  5 marks

  11

  
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  Almost all candidates recognised that this question required integration and were able to correctly identify that integration by parts was needed to access any of the marks in this question. The most common method was for candidates to apply by parts twice using the formula given in the formula book, although this did frequently result in candidates getting signs wrong for their terms. A number of candidates used the “DI method” which tended to be more successful and less complicated for those candidates who tried this approach. The word LATE or ILATE written on scripts which serves to assess the priority of the first function was also seen. This seemed to be helpful for candidates to identify which function should be differentiated. A small number of candidates used a substitution, usually u = −3x, with varying degrees of success. Although this resulted in fewer minus signs to manipulate, it did not simplify the processing. For most candidates the first step of by parts was generally correct, although coefficient and sign errors were common. A few integrated the 2 8x which led to incorrect forms. Of those candidates who got the correct form for the first step most got the second stage into the correct form though many made errors with signs. It was disappointing to see how few candidates simplified double negatives. Candidates who completed the integration to a correct form generally understood the substitution of the limits although again errors in signs occasionally cost candidates the final method mark as their working did not imply use of 0e 1 = . Candidates should be encouraged to fully show their working when applying limits so that it is clear that the lower limit is being subtracted. Most responses culminated with an attempt at an answer in its exact form but candidates that obtained the correct values for A and B were in the minority. Although not penalised if the intention was clear, some notation was poor with missing dx’s and spurious or missing integral signs.

  Text version

  In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The finite region R is bounded by the curve C: y = 8x^2 e^(-3x), x >= 0, the line x = 1, and the x-axis.

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

  12 marks

  12

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

  Text version

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

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• Edexcel 9MA0 · June 2023 · Paper 1 · Q1 · 4 marks

  4 marks

  1

  
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  This was an accessible question for virtually all candidates. The fractional index had already been given to avoid any misconceptions. However, for the first question on the exam paper, this was frequently done poorly compared to previous years. Candidates may have anticipated a tougher start to the paper and wanted to demonstrate one of the more advanced integration techniques. Candidates who spotted that you needed to expand the brackets usually went on to score full marks. The most common reason for losing the final answer mark was forgetting the constant of integration. Many candidates failed to notice the simpler method in the main mark scheme and instead attempted to use integration by parts, with either part taken as u. If a candidate chose to set 2 5 u x = −, they did not always expand the brackets in their integration formula to achieve a form which was ready to integrate where the indices had been combined correctly. Candidates who set 1 2 u x = were more successful, as there was less working to do to allow access to the marks. Candidates who chose to do integration by parts did not always combine the terms in their answer, losing the final mark. Many candidates would divide both their u and 'v by 3, leading to final answers that were 1 3 as large as they should have been. Integration by substitution was occasionally seen, but rarely successfully completed. Most attempts involved setting 2 5 u x = −, and this resulted in a tougher integral that required further work; candidates rarely progressed to a point which would have been worthy of a mark. Across the various methods, a common error seen was initially to factorise out the 1 3 , work through the solution without it, but then to forget to reintroduce it. This error meant that neither of the two accuracy marks were available.

  Text version

  Find the indefinite integral of [x^(1/2)(2x - 5)]/3 with respect to x, writing each term in simplest form.

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

  5 marks

  5

  
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  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:

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

  7 marks

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

  Text version

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

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

  10 marks

  11

  
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  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).

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• Edexcel 9MA0 · June 2023 · Paper 2 · Q13 · 13 marks

  13 marks

  13

  
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  Most candidates attempted at least part (a) of this question, but some found the work more demanding in (b) and (c) and made little progress in either part. Despite this, there were many fully correct or nearly fully correct attempts at the whole question. In part (a), most candidates successfully factorised out 2 3−and attempted the correct resulting expansion of 2 1 3 x −   +     . There were some arithmetic slips, but generally candidates often reached the correct answer. A common error, however, was failing to factorise properly, reaching ( ) 2 2 3 1 . x − − + There were a number of attempts at direct expansion. Some of these used non-standard binomial coefficients involving negative integers. To achieve credit for using these, candidates needed to demonstrate the meaning of their non-standard notation. Very occasionally candidates multiplied their correct simplified expansion by 9 or 27 at the end. This still received full marks as examiners could ignore subsequent working following seeing a correct simplified answer. In part (b), many candidates correctly multiplied their expansion by 6x and then integrated using the limits of 0.2 and 0.4 reaching 0.03304, gaining full marks. A common error, however, was not giving the final answer to the required accuracy of four significant figures, thus losing the final accuracy mark. Those candidates who had expanded part (a) to more than three terms also lost this final accuracy mark as it led to a different approximation. A more complicated and unnecessary method involving integration by parts was used by a small minority of candidates. Candidates also often tried inappropriate approaches like dividing 6x by their expansion followed by a logarithmic integral. A minority of candidates, despite the warning at the start of the question, clearly used calculator technology to obtain 0.032865, rather than the answer of 0.03304 required. Others made inappropriate attempts at the trapezium rule, thus gaining no credit as algebraic integration was required. In part (c), the commonest methods of integration adopted were substitution, using either u = x + 3 or u = (x + 3)2 , integration by parts, integration after applying partial fractions and integration using the reverse chain rule after writing the integrand in appropriate form. Candidates had varying success with all of these approaches, gaining a variety of marks. Integration by parts seemed to lead to most arithmetical mistakes. Errors resulted from using parts ‘the wrong way round’ and others integrated the ( ) 2 3 x − + term incorrectly. When integrating by substitution some students incorrectly thought that 2 18 u − would integrate to a natural logarithm. There was some confusion with the limits; some candidates did not change the limits to “u” values and other who did evaluate new limits for their substitution switched back to an expression in x before substituting their changed limits. When using partial fractions some students struggled with the initial format for the partial fractions and it was not uncommon to see partial fractions of the form 3 3 A B x x + + + . There was a number of candidates who either did not attempt this part of the question at all or who lacked a correct strategy. There were responses where students attempted to integrate numerator and denominator separately, e.g. 6x to 3x2 and ( ) 2 1 3 x + to a variation of ( ) 2 ln 3 x + , and proceeded to multiply both results. A number of candidates were awarded just the first mark as their overall problem solving strategy was correct despite the fact they were unable to fully implement it. There were also many candidates who could carry out the work in this part completely successfully.

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

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• Edexcel 9MA0 · June 2022 · Paper 1 · Q4 · 3 marks

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  (Mean mark 1.9 out of 3) This was a generously worded question with plenty of guidance to help candidates throughout. As a result, it was generally well attempted, with many candidates scoring full marks. There were a number of candidates who made no attempt at part (a) despite the guidance to express the given sum as an integral. Where candidates had used the information given in the question the correct answer was relatively easy to deduce. Common errors seen in part (a) were omitting the limits or the dx, and writing dx as dx (although this was condoned). Candidates who had attempted part (a) generally went on to attempt part (b). Most candidates provided a fully correct solution and gained both marks. Where errors were made, this was usually at the last mark. Common errors seen were; • incorrectly integrating the expression to ln x or ln 2x • differentiating 2 x rather than integrating it • leaving the answer as 2ln3 rather than ln9

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  Let L=lim_(delta x -> 0) sum_(x=2.1)^(6.3) (2/x) delta x.

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• Edexcel 9MA0 · June 2022 · Paper 1 · Q12 · 5 marks

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  (Mean mark 2.8 out of 5) This was a very straight forward question on integration by parts with 40% of the candidates scoring full marks. On the reverse side there were also a sizeable number, some 30%, who did not know how to begin and failed to score any marks. Those candidates who knew the method nearly always scored 3, 4 or 5 marks. Reasons for these candidates dropping 1 or 2 marks nearly always centred around • a failure to integrate the term 4 1 4 x x × correctly, many times to 4 12 x • failing to correctly apply the limits, usually achieving 8 7 1 e 16 16 − • achieving a correct answer of 8 7 1 e 16 16 + then inexplicably ''simplifying'' this to 8 7e 1 + Low scoring candidates who attempted the question, usually resorted to integrating each term of the integrand or else having made a good start, disappointingly integrating the 4 1 4 x x × term to 5 ln 20 x x ×

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

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

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  (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.

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  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.

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• Edexcel 9MA0 · June 2022 · Paper 2 · Q8 · 6 marks

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  Most students knew that this question was to be solved by integration although a very small number thought that the question was about partial fractions, differentiation or a mixture of differentiation with integration. There were multiple approaches to solving this question including integrating a function with simplified indices, integrating by parts and integrating by substitution. The most common and the simplest method seen was multiplying out and simplifying indices. For the candidates who recognised that 4√x in the denominator should be rewritten as ¼ x -½ in the numerator there was a lot of success on this question. However, there were many candidates who brought the 4 to the top and left it as 4 rather than ¼ and earned a maximum of 3/6. Some wrote 4x½ in the numerator multiplying out to give incorrect powers and losing the first two method marks. Finding the correct limits and substituting them in a changed function was done by most leading to quite a few correct answers. Quite a few gave part or all of the final answer as a decimal, this was not surprising as most were using their calculators to do the final evaluation and calculator use did not leave the result of e.g. (√2)5 as a surd. Some candidates recognised that the area needed to be positive, however the final answer was permissible even with a negative area. There were often errors in solutions constructed using integration by parts. There were mistakes made using the formula correctly, some bracketing errors, and some attempted parts the other way round getting harder expressions in the integral which mostly resulted for them a maximum of 1 mark (3rd M mark) obtained. There were a few incorrect substitutions selected for integration by substitution approaches, however candidates who chose u = √x or u = 4√x were generally successful. A few candidates who set e.g. u = x – 2 in a substitution method scored few marks, if any, as such attempts did not progress the candidate towards a function they could integrate. In substitution method a few candidates made arithmetic errors in finding the new limits or did not attempt to change them at all.

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  In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The curve has equation y = ((x - 2)(x - 4))/(4sqrt(x)) for x > 0. The finite region R is bounded by the curve and the x-axis between x = 2 and x = 4.

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

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

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

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• Edexcel 9MA0 · October 2021 · Paper 1 · Q11 · 8 marks

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  The modal mark here was 3, most likely due to the accessibility of part (a). Marks in part (b) were only really awarded to the highest scoring candidates on this exam. Most candidates successfully answered part (a) applying the Trapezium Rule using all five values of y from the given table. Some students, who found the correct estimate of the area of R, failed to round their answer as requested, or occasionally rounded incorrectly, and lost the accuracy mark. Errors, although rare, were usually as a result of an incorrect strip width. In part (b) many candidates failed to recognise that integration by parts was needed to make progress. Several students, who made progress and correctly applied ‘integration by parts’ twice, unfortunately lost the final accuracy mark due to sign or coefficient errors that had crept in. Most who integrated appropriately applied the correct limits and applied ln(4) = 2ln(2). Errors were many and varied here and included • Candidates who believed that ( ) ( ) 2 3 1 d 3 ln x x ln x = ∫ • Candidates who believed that ( ) 2 2 ln x ln x = Candidates who used the substitution ln u x = seemed to more easily recognise 2e d u u u ∫ as integration by parts twice.

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  The finite region R is bounded by y=(ln x)^2, x=2, x=4 and the x-axis. The supplied table is: x=2,2.5,3,3.5,4 and corresponding y=0.4805,0.8396,1.2069,1.5694,1.9218.

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

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  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).

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  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.

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• Edexcel 9MA0 · October 2021 · Paper 2 · Q12 · 7 marks

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  This question discriminated well with weaker students generally able to pick up 2 marks and many able students were able to gain full marks. Part (a) was a relatively straightforward substitution question and as such many students made a very positive attempt at this part. There were a mixture of methods used to answer part (a) with most students correctly identifying du/dx or dx/du. Those that found the second of these found the substitution very straightforward and generally went on to achieve full marks. Students who found du/dx instead sometimes struggled with the indices and how to deal with the negative power when dividing and hence made mistakes here. There were a small number of students who simply failed to make any attempt at this question, and generally these students gave up on the question entirely. A few students lost the accuracy mark for not stating the limits or omitting du. In part (b) it was good to see many students recognised that “hence” meant that the answer in part (a) must be used and proceeded with the correct method of expanding the cubic and dividing by u, proceeding correctly to the correct answer. If this was done correctly these students would usually go on to successfully apply the limits correctly and gain full marks. Unfortunately, a number of students attempted a method of integration by parts or some kind of product rule and therefore gained no marks here. Unfortunately a few did not take heed of the “hence” and proceeded to integrate the original expression scoring no marks in this part. Some weaker students split the integral into an incorrect sum of 2 fractions eg 2 𝑢𝑢+ (𝑢𝑢−1)3 𝑢𝑢 and then used integration by parts which they tended to give up on half way through. The method most successful was to take the 2 outside the integral, then integrate. Those that multiplied by 2 were more likely to make slips on the value of the coefficients for the 3rd and 4th term. Several students lost the final 2 marks as they used incorrect limits either 16 and 0 or 5 and 0.

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  Consider the definite integral from x = 0 to x = 16 of x/(1 + sqrt(x)) dx.

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

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  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.

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  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.

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• Edexcel 9MA0 · October 2020 · Paper 1 · Q10 · 10 marks

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  This question proved to be more discriminating than expected with many failing to notice the partial fractions in part (b), thereby losing 6 marks. Part (a) was well answered by many candidates, particularly those who opted to calculate dx/du rather than du/dx, as the latter proved harder to substitute correctly into the expression in terms of u. Some students lost the final accuracy mark for not showing a correct intermediate line with integral signs and some did not finish their working with an expression showing the given integral. A few candidates made arithmetic slips in finding the limits or did not attempt to change them at all. Of those who did use partial fractions in part (b), most found the values of "A" and "B" correctly. Unfortunately many integrated B/(3 + 2u) to B ln(3 + 2u), failing to divide the coefficient by 2. The final Method mark was still available for these candidates, but a number failed to simplify their numerical logs to an expression of the required form and so did not gain this mark. A pleasing number of well prepared candidates did gain full marks in this question.

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  Consider I=integral from 5 to 10 of 3/[(x-1)(3+2sqrt(x-1))] dx.

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• Edexcel 9MA0 · October 2020 · Paper 1 · Q14 · 10 marks

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  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.

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  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.

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• Edexcel 9MA0 · October 2020 · Paper 2 · Q1 · 5 marks

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  Part (a) was generally answered well, with the majority of candidates using h = 0.5 and finding an area rounding to 1.50, but a number completed it with a final answer of 1.505… and consequently lost the final A mark for part (a). The most common error seen was the use of h= 2 5 which candidates are arriving at because of a misunderstanding in the ‘n’ seen in the formula in 𝑏𝑏−𝑎𝑎 𝑛𝑛 , they presumably think n is the number of entries in a row of the table ignoring that the first column represents x0 and y0. Those candidates who calculated the width of the trapezia independent of the formula 𝑏𝑏−𝑎𝑎 𝑛𝑛 were usually more successful. The formula was set up correctly on most occasions, with the most common error being the lack of exterior brackets, often resulting in 4.95 as an answer. There were some candidates using a variation of the formula, with the 1 2 absorbed into the brackets, and a few worked out the areas of four separate trapezia, usually successfully. Success in part (b) was mixed, with around half the candidates multiplying their answer by 3 and the others multiplying by 9. A minority of candidates restarted, losing access to marks in this question. In part (c), the mark was only available to those who arrived at an answer rounding to 4.5 in part (b). Answers were often concise and clear, stating that their answer was correct to 2s.f. or 1d.p. and therefore quite accurate. Others went for a percentage error approach, which was also acceptable and usually completed successfully. The main error in this part was that candidates misunderstood what was being asked of them, and instead arguing that the answer was an underestimate, confusing this with another common trapezium rule question.

  Text version

  The table gives values of y = sqrt(x/(1 + x)), correct to 4 significant figures. x: 0.5, 1, 1.5, 2, 2.5 y: 0.5774, 0.7071, 0.7746, 0.8165, 0.8452

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

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

  Text version

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

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

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

  Text version

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

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• Edexcel 9MA0 · October 2020 · Paper 2 · Q12 · 11 marks

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  Many candidates left the entire question or parts of this question completely blank and clearly found it difficult to access the content being assessed here, perhaps due to a lack of familiarity with integration using parametric equations. Many tried to establish the Cartesian equation to answer various parts. That said, there were many candidates who provided clear and concise solutions to each part. In part (a)(i), most candidates recalled the formula for integrating parametric equations and knew to find 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑. Awareness of the double angle formula for sin2t was good. The main problem caused for these candidates was a lack of “dt” throughout their work, and only included this in the final line, losing the final accuracy mark. There was a significant proportion of candidates who attempt ed some differentiation using the chain rule: 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑= 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 x 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 , scoring no marks. In part (a)(ii), many knew that the integral was a multiple of cos3 t but had trouble getting to the printed answer of 20 with correct steps. As the answer was given, it was a requirement to see the limits calculated separately as 0 – ( -20 ) = 20. Integration by parts was a common approach for unsuccessful candidates, while others attempted to use a substitution, often not using the necessary u = cosx. Those candidates with an understanding of the integration of f’(x)[f(x)]n were generally very successful here. Some lost the penultimate accuracy mark usually due to a sign error. In part (b) many candidates who had found part (a) difficult gave up and lost the opportunity of scoring some easier marks. Many candidates who scored no marks elsewhere were still able to access at least the first two marks in part (b) by setting sin2t = 4.2 5 and solving, frequently getting at least one correct angle; but too many of these found their second value of t by subtracting their first value from π instead of calculating π – 𝑠𝑠𝑠𝑠𝑠𝑠−1( 0.84) before dividing by 2. The following mark was for finding two values of x, but many simply substituted their first value in to find x, thinking that this gave them the width of the walkway. Too many marks were lost by approximating too soon and hence not having values with the required accuracy. A few lost the last mark because they omitted the units.

  Text version

  A curve C is defined parametrically by x = 6sin(t), y = 5sin(2t), for 0 <= t <= pi/2. The region R lies between C and the x-axis. The same curve models the cross-section of a dam. A horizontal walkway at height y = 4.2 meets C at M and N; its width is the horizontal distance MN.

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• Edexcel 9MA0 · June 2019 · Paper 1 · Q8 · 10 marks

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  (Mean Mark 4.9 out of 10) The first six marks in this question relating the definite integral with the area under the curve were straightforward. Only the best candidates were able to access the last four with many struggling to explain the significance of the root 5.442. Most students gained all 4 marks in part (a). Some unnecessarily complicated their proof by attempting integration by parts but in the main, all calculations were accurately performed. As with all proofs, part (b) proved to be very discriminating. Many candidates picked up the first mark for setting 4 3 2 1 2 20 4 4 3 3 b b b    but there was much confusion as to the sign of 20 3 on the right hand side. Most students attempted to proceed by the method shown in the scheme. It should be noted that division by 2 4 4 b b   was much easier and faster than dividing by ( 2) b  twice. Fully correct solutions to (b) were rare with many picking up 2 out of the 4 marks. In part (c) many candidates realised that 5.442 was to the right of 4 on the x- axis and shaded an appropriate area, scoring one mark. Full explanations as to its significance were rare and confined to the best candidates. Any suitable statement that alluded to the fact that the area above the curve is equal to the area below the curve (between 2  and 5.442) was acceptable.

  Text version

  The curve y=x(x+2)(x-4) crosses the x-axis at x=-2, 0 and 4. Region R1 is between the curve and the negative x-axis for -2<=x<=0. Region R2 is between the curve, the positive x-axis and x=b, where 0<b<4. The areas of R1 and R2 are equal.

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

  11 marks

  13

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

  Text version

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

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• Edexcel 9MA0 · June 2019 · Paper 2 · Q5 · 3 marks

  3 marks

  5

  
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  Q05 proved challenging for many candidates and some candidates were unfamiliar with the notation that was used. Most candidates scored either all three marks or zero marks in this question. There were a substantial number of blank responses. It is noted that section 8.4 (guidance) of the Pure Specification states ‘Recognise 0 f( )d lim f( ) b b x a x a x x x x        ’. It is also noted that a similar question had not appeared on the 9MA0 June 2018 papers, the SAMs, the specimen papers or the mock papers. Those candidates who recognised that 9 0 4 lim x x x x     was another way of writing 9 4 d x x  generally produced a correct solution to score all 3 marks. Only a few made errors in their integration of x or in the application of the limits 4 and 1. Most candidates, however, made no creditable progress. Some applied the trapezium rule with 1, h  but did not link their trapezium rule attempt with a stated 9 4 d . x x  Other candidates attempted to sum x using integer values from 4 x  to 9, x  resulting in 4 5 6 ... 9 ( 15.1597...).      Some candidates mistakenly interpreted ‘ x  ’ as a trigger to differentiate , x with a few of these candidates attempting to differentiate x from first principles. A few candidates applied the incorrect method   9 4 4 5 6 ... 9 d x x      to give an answer of 2.49 to 3 significant figures.

  Text version

  The curve is y = sqrt(x). A point P(x,y) lies on it, and a typical rectangle under the curve has height y and width delta x.

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• Edexcel 9MA0 · June 2019 · Paper 2 · Q14 · 15 marks

  15 marks

  14

  
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  Q14 discriminated well between the higher ability candidates. Lower and medium ability candidates, however, found this question demanding with most of these candidates scoring no more than two marks. There were a few candidates who made no attempt at this question, either because the question was too demanding, or because their time had run out. Most higher ability candidates made good progress in Q14(a) and Q14(c) but they often struggled to make progress in Q14(b). In Q14(a), most candidates who attempted this part differentiated 4 u h   to give a correct 1 2 d 1 , d 2 u h h   with a few finding a correct d 2(4 ) d h u u   and some differentiating incorrectly to give 1 2 d 1 . d 2 u h h   At this stage, some candidates failed to make further creditable progress. Many candidates struggled to apply the substitution completely to give an integral of the form (4 ) d , k u u u   with some making sign errors in the numerator of this integral. Those who progressed this far generally went on to divide each term in the numerator by u to give an integral of the form d , A B u u         with many of them obtaining a correct 8 2 d .u u          Most integrated correctly and applied 4 u h   to obtain a correct 8ln 4 2(4 ) . h h c      Some candidates could not deal with the transition from 8 c  to k to achieve the given 8ln 4 2 , h h k     and a large number of these candidates incorrectly stated 8. k  Those few candidates who applied integration by parts on 2 8 d u u u   rarely progressed to give a correct solution to this part. In Q14(b), some candidates correctly set d 0 d h t  but only a few deduced the equation 4 0. h   Some candidates used 16 h  to state a correct allowable answer such as 0 16, h   0 16 h   or 16. h  A common mistake was to solve 4 h  to give 2. h  Material for Q14(c) was sometimes seen in Q14(b). Most candidates who attempted Q14(c) separated the variables correctly. Some candidates integrated 0.25 t incorrectly to give 0.25 0.75 , t t   or 1.25 5 4 t while others integrated 1 (4 ) h  without reference to the result from Q14(a). Many candidates applied 0, 1 t h   to find their constant of integration. Common errors at this stage included not applying a constant of integration and using either 0, 0 t h   or 1, 1 t h   to find their constant of integration. Most candidates who progressed this far applied a complete process of substituting their constant of integration and 12 h  into their integrated equation and solving this equation to find their value for t. In many cases, sign errors, bracketing errors, manipulation errors or an incorrect method of solving a correct 1.25 221.2795202..., t  prevented candidates from obtaining the correct answer 75.2 t  years.

  Text version

  A tree's height h metres at time t years after planting is modelled by dh/dt = t^0.25(4 - sqrt(h))/20. Time t is non-negative.

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• Edexcel 9MA0 · June 2018 · Paper 1 · Q7 · 7 marks

  7 marks

  7

  
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  Both parts of this question were answered well, allowing students to demonstrate their confidence in integrating fractional expressions in x and working with unknown constants. The majority of students failed to make any statement regarding the integral in (a) being independent of k, or in (b) inversely proportional to k. Whilst not a requirement for the final A mark in either part, the logical completion of the question would have been to make these conclusions and show an understanding of these mathematical terms. Common method errors in this question included integrating to obtain a natural logarithm in part (b) as well as part (a). Additionally, despite the successful implementation of limits, some careless and disappointing algebra was seen in simplifying their final expressions in an attempt to answer the question. A more unusual but not uncommon error was related to the fact that ‘k’ occurred both in the integral and in the limits. Some students integrated with respect to k whilst others substituted the limits for the term in k rather than the term in x. Other unusual attempts were seen by students attempting to use partial fractions within part (b).

  Text version

  Let k be a positive integer.

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• Edexcel 9MA0 · June 2018 · Paper 1 · Q10 · 8 marks

  8 marks

  10

  
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  This was another modelling question, this time modelling the height of a roller coaster above the ground by a differential equation. In part (a), the majority of students were able to separate the variables of H and t and thus integrate both sides. Integrating cos0.25t was done well with most dividing by 0.25 or multiplying by 4. Almost all students were then able to proceed with ease to   ln 0.1sin 0.25 H t  . The last two marks proved harder to score. Quite a sizeable proportion of students failed to include the constant of integration in   ln 0.1sin 0.25 H t c   and so were unable to apply the boundary condition 0, 5 t H   in an attempt to find c. The final mark was the hardest to gain. The answer was given and many students merely copied down the result on the examination paper. There had to be clear evidence that the constant they found was processed correctly and that their log work was sound. The requirement of at least one correct line of working between their working and the given equation, with no incorrect working, was fair to those who showed a clear argument. Part (b) was found by almost all students, even those who could not attempt part (a). In this case the exact answer 5e 0.1 was accepted as well as 5.53 metres. Fewer students scored both marks in (c), but many were aware that it would occur when sin(0.25 ) 1. t  The most common errors were to find the first value using , 2  or else use 450 rather than 5 , 2 thus arriving at the incorrect answer of 1800 seconds.

  Text version

  The height H metres of a roller-coaster passenger is modelled by dH/dt=[H cos(0.25t)]/40, where t is seconds from the start. Initially H=5.

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• Edexcel 9MA0 · June 2018 · Paper 1 · Q13 · 7 marks

  7 marks

  13

  
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  This was an open ended question on integration. The three most common methods seen are shown the mark scheme, two methods using substitution and one using integration by parts. Integration by parts was popular and well understood by the majority of students. Most knew that   3 2 2 2d 2 3 x x x     and were able to integrate to an expression of the form     3 5 2 2 2 2 Ax x C x    Slips in the values of the constants A and C were rare. However, as with earlier questions, there was a lack of clarity in moving from     2 3 5 2 2 0 4 8 2 2 3 15 x x x          to the given answer of   32 2 2 15  Substitution was also well understood with 2 u x   more common than 2. u x   Errors here included the omission of finding dxin terms of d ,u poor expansion of expressions such as 2( 2) u u  within the integral, and, as with parts, a lack of acceptable working shown when proceeding to the given answer.

  Text version

  Consider the definite integral from 0 to 2 of 2x sqrt(x+2) dx.

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

  8 marks

  10

  
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  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.

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• Edexcel 9MA0 · June 2018 · Paper 2 · Q13 · 10 marks

  10 marks

  13

  
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  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.

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• Edexcel 9MA0 · sample · Paper 1 · Q4 · 4 marks

  4 marks

  4

  
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  Given that a is a positive constant and the integral from a to 2a of (t+1)/t with respect to t is ln(7):

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• Edexcel 9MA0 · sample · Paper 1 · Q14 · 10 marks

  10 marks

  14

  
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  The curve C has equation y=[x^2 ln(x)]/3-2x+5 for x>0. The finite region S is bounded by C, x=1, the x-axis and x=3. Corresponding values are: x=1,1.5,2,2.5,3 and y=3,2.3041,1.9242,1.9089,2.2958, with y values given to 4 decimal places as appropriate. In part (c), solutions based entirely on graphical or numerical methods are not acceptable.

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

  5 marks

  9

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

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• Edexcel 9MA0 · sample · Paper 2 · Q15 · 10 marks

  10 marks

  15

  
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  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.

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

  12 marks

  16

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

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• Edexcel 9MA0 · specimen · Paper 1 · Q7 · 5 marks

  5 marks

  7

  
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  The curve y = 2e^(2x) - xe^(2x), for real x, bounds a finite region R with the x-axis and y-axis.

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

  10 marks

  11

  
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  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).

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• Edexcel 9MA0 · specimen · Paper 1 · Q12 · 7 marks

  7 marks

  12

  
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  Consider the definite integral from theta = 0 to theta = pi/2 of sin(2theta)/(1 + cos(theta)) dtheta.

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• Edexcel 9MA0 · specimen · Paper 2 · Q7 · 12 marks

  12 marks

  7

  
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  A bacterial culture has area p mm^2 at time t hours after it is placed onto a circular dish. A scientist models the rate of increase of the area as proportional to the area of the culture. For measurements during the first 24 hours, a graph of ln(p) against t lies close to a straight line with gradient 0.14 and vertical intercept 3.95.

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• Edexcel 9MA0 · specimen · Paper 2 · Q10 · 9 marks

  9 marks

  10

  
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  A curve C has parametric equations x = ln(t + 2), y = 1/(t + 1), for t > -2/3. A finite region R is bounded by C, the line x = ln(2), the x-axis and the line x = ln(4).

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