Numerical methods questions

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• Edexcel 9MA0 · June 2024 · Paper 1 · Q3 · 6 marks

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  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.

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

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  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.

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

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

  Text version

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

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

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  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).

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• Edexcel 9MA0 · June 2022 · Paper 1 · Q8 · 8 marks

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  (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.

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• Edexcel 9MA0 · June 2022 · Paper 2 · Q5 · 6 marks

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  Part (a) of this question required candidates to use a given table of function values in an application of the Trapezium Rule. This was generally done very well with candidates knowing how to apply the rule. Most achieved h = 1.5 although a common mistake was for students to think that the number of strips was the same as the number of x values instead of being one less. Thus dividing by 5 instead of 4 resulting in h = 1.2. Most knew how to apply the trapezium rule and had the bracketing correct and so achieved the correct answer. A few left out some of the brackets followed by an incorrect calculation. 𝟏𝟏. 𝟓𝟓 𝟐𝟐(𝟏𝟏. 𝟔𝟔𝟔𝟔+ 𝟐𝟐. 𝟔𝟔𝟔𝟔) + 𝟐𝟐(𝟐𝟐+ 𝟐𝟐. 𝟐𝟐𝟐𝟐+ 𝟐𝟐. 𝟒𝟒𝟒𝟒) With or without an outer bracket either at the beginning or end – would lead them to then calculate the expression as it is written, often leading to an answer like: - 𝟏𝟏. 𝟓𝟓 𝟐𝟐(𝟒𝟒. 𝟐𝟐𝟐𝟐) + 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒= 𝟑𝟑. 𝟏𝟏𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒= 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔𝟔𝟔 There were a few who worked out the areas of individual trapezia and then summed the areas achieving a correct answer. A minority didn’t know how to apply the rule Part (b) of the question clearly stated that the result from part (a) was to be used and that the method used was to be made clear. Thus, candidates who proceeded to use the Trapezium Rule again with revised values of y received no credit. In (b)(i) many realised that the power law of logs had to be used and realised that their answer to part (a) had to be multiplied by 10, so achieving a correct answer. Many on the other hand unfortunately raised their answer to part (a) by a power of 10. Most of those who attempted (b)(ii) didn’t realise that they had to split the expression into two parts applying log laws. Part of the expression then had to be integrated and added to their answer to part (a). Often seen was the calculation 9log3(2x) or 9 ´ 13.275. Some seemed to think that log3(18x) meant that the graph/function had been subjected to a stretch of either 9 or 1/9. Some didn’t show sufficient working to be sure that they were applying the correct method whilst others showed the correct use of logs but then just added 2 to their part (a) answer, not integrating this between the limits. Even when a correct relationship of log3 18x = log3 2x + log3 9 was established, the subsequent argument often did not have sufficient detail to show where their final answer came from. The question clearly stated “making your method clear”. Very few achieved full marks on this part.

  Text version

  The table shows corresponding values for y = log_3(2x). The x-values are 3, 4.5, 6, 7.5 and 9, and the corresponding y-values, given to 2 decimal places as appropriate, are 1.63, 2, 2.26, 2.46 and 2.63.

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

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  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:

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• Edexcel 9MA0 · October 2021 · Paper 1 · Q4 · 9 marks

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  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.

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  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.

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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 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 · Q7 · 10 marks

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  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.

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• Edexcel 9MA0 · June 2019 · Paper 2 · Q2 · 4 marks

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  Q02 proved to be challenging, with a significant number of candidates struggling to realise that the length of the runway could be estimated by the area under the speed- time curve generated from the table of values given in the question. In Q02(a), only a minority of candidates estimated the length of the runway by applying the trapezium rule, which was considered the most appropriate method, with many of these giving the correct answer of 415m. While the table of values clearly shows an interval width 5, h  the application of the formula b a h n   with 6 n  instead of 5 n  led some candidates to use an incorrect strip width 25. 6 h  Other candidates made calculation errors, bracketing errors or transcription errors. Some candidates used the equations of motion with constant acceleration (i.e. the suvat equations) to estimate the length. Many of the candidates who used this method applied the calculation 2 42 (25) 2        to obtain an estimate of 550m. A few used v u at   to find the acceleration, followed by 2 1 2 s ut at   to estimate the length. Some candidates used 0 u  in their suvat calculations, in contradiction to the table of values which indicated an initial speed of 1 2 ms .  A few candidates used a method which was equivalent to the trapezium rule, of applying suvat to each of the 5 time-intervals, to give the correct answer of 415m. Other methods seen included: summing up the area of rectangles which enclosed the area above the curve for each of the 5 time intervals; summing up areas of rectangles which enclosed the area below the curve for each of the 5 time intervals; and applying (average speed)(total time). The most common methods which were deemed incorrect included 25 42 1050;   1 (25)(42) 525; 2  (speed)(time) (0)(2) (5)(5) (10)(10) ... (25)(42) 2005;        or a 6 rectangle approach (instead of 5 rectangles). As for Q02(b), while many candidates who used a trapezium rule or suvat method in Q02(a) identified their estimate of the length as an overestimate, some found difficulty in articulating a reason for this. Those who were most successful drew a diagram which showed clearly the extra area generated by the trapezium rule in relation to the curve. Some candidates sketched the speed-time curve and explained that the curve was convex or that the acceleration was continually increasing. Only a few candidates who used a rectangle method were able to give a correct reason. Some candidates answered Q02(b) by finding an alternative estimate and comparing it with their estimate in Q02(a). For example, some found 415 in Q02(a) and 550 in Q02(b), and then concluded that ‘415 must be an underestimate’. Other candidates, who also received no credit, gave physical explanations about friction, air resistance or the shape of the aircraft, while others stated that ‘the jet accelerated smoothly’.

  Text version

  The speed of a small jet aircraft was measured every 5 seconds from the time it turned onto a runway until it left the ground. Time (s): 0, 5, 10, 15, 20, 25 Speed (m s^-1): 2, 5, 10, 18, 28, 42

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

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  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.

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• Edexcel 9MA0 · June 2018 · Paper 1 · Q4 · 4 marks

  4 marks

  4

  
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  This was perhaps the most poorly attempted question on the whole of the paper. Part (a) should have been familiar to centres but one which required the students to set up their own function, usuallyf( ) 2ln(8 ) , x x x    before substituting in the x values of 3 and 4. This was well done, but also quite commonly, the only mark gained in the question. To show that 3 4,    a student needed two correct calculations, give a valid reason, which required both a change in sign and a mention of continuity, as well as giving a brief deduction. Explanations rarely mentioned the fact that the function needed to be continuous in this interval. Fully correct solutions to part (b) were very rare. Many students assumed that this was a question on continued iteration and gave all values of x from 1x to 8x without any consideration of the demand of the question. Students were required to ''use the graph'' to show whether or not the iteration formula could be used to find an approximation for .  Hence, to satisfy this demand, a cobweb diagram starting at 1 4 x  was required followed by an explanation that ''it can be used'' because ''the cobweb spirals inwards towards . 

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  The curve y=2ln(8-x) meets the line y=x at a single point whose x-coordinate is alpha. The following coordinate-consistent graph is supplied. A student uses x_(n+1)=2ln(8-x_n), for positive integer n, starting with x_1=4. Supplied graph (not to scale): Legend: / is y=x; * is y=2ln(8-x); X is their intersection. y | // | //// | //// | /// | //// | //// | /// | //// | //// |********** /// | ********* //// | *******X/// | ///****** | //// ****** | //// ***** | /// ***** | //// **** | //// **** | /// *** | //// *** |//// ** +-------------------------------+------------------------ 0 4 x

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• Edexcel 9MA0 · June 2018 · Paper 2 · Q5 · 6 marks

  6 marks

  5

  
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  This was a well-answered question with part (a) and part (c) providing most of the discrimination. In part (a), most students substituted a defined 3 2 f( ) 2 1 n n n x x x    and a correct 2 f ( ) 6 2 n n n x x x    into the Newton-Raphson formula 1 f( ) . f ( ) n n n n x x x x   Many students used correct algebra to arrive at the given result. Some students moved directly from 3 2 1 2 2 1 6 2 n n n n n n x x x x x x       to 3 2 1 2 4 1 6 2 n n n n n x x x x x      without showing the correct intermediate step of rationalising the denominator. There were bracketing errors, sign errors and the error of omitting ‘ 1 nx ’ to write only the right-hand side of the Newton-Raphson formula. In part (b), nearly all students wrote down a correct 2 3 3 2 , . 4 3 x x   In part (c), only a few students gave an acceptable explanation that detailed why the Newton- Raphson method did not work for this question. Some of these students stated ‘because there is a stationary point at 0, x  then the method could not work’ while others explained that a tangent to the curve at 0 x  would never meet the x-axis. The explanation, ‘not being able to divide by zero in the formula’, by itself, was not sufficient to gain credit in part (c).

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  The equation 2x^3 + x^2 - 1 = 0 has exactly one real root.

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

  6 marks

  8

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

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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 · specimen · Paper 1 · Q1 · 6 marks

  6 marks

  1

  
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  The curve y = x/(1 + sqrt(x)), x >= 0, bounds a finite region R with the x-axis and the lines x = 1 and x = 3. Corresponding values are: x = 1, 1.5, 2, 2.5, 3; y = 0.5, 0.6742, 0.8284, 0.9686, 1.0981.

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• Edexcel 9MA0 · specimen · Paper 2 · Q6 · 9 marks

  9 marks

  6

  
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  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.

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