Edexcel A-level Maths · Topic 6
Practice questions covering exponentials and logarithms from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Exponentials and logarithms, each with its mark scheme and worked method. Reveal the answer when you have had a go.
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(3)
(Total for Question 1 is 3 marks)
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(3)
(Total for Question 2 is 3 marks)
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(4)
(Total for Question 3 is 4 marks)
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(5)
(Total for Question 4 is 5 marks)
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(3)
(Total for Question 5 is 3 marks)
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(5)
(Total for Question 6 is 5 marks)
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Printable question pack — 6 Exponentials and logarithmsAll 70 questions across 7 specification points — 64 more than the 6 above — with the full mark scheme. Guest accessible · browser print.36 questions shown as printed · newest series first
Question and mark-scheme extracts © Pearson Education Ltd. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.
Edexcel 9MA0 · June 2024 · Paper 1 · Q7 · 8 marks
8 marks
Examiner report
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.
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.
Edexcel 9MA0 · June 2024 · Paper 2 · Q13 · 9 marks
9 marks
Examiner report
Part (a) was generally well answered with candidates commonly using the main method in the mark scheme, having made the link with y = mx + c and taking logarithms. Most candidates recognised log a = 0.81 and log b = 0.0054 with minimal preamble. When this was recognised it almost always led to correct answers. There were occasional errors with logs such as using base e instead of base 10. The misread of the 0.0054 as 0.054 was seen quite often. A common issue was candidates not reading the 3 decimal place requirement and instead writing things such as b = 1.0125 (awrt was applied so no marks were lost) or giving 3 significant figures. The method where values for t were used to obtain the values for a and b was sometimes used, but often candidates made limited progress towards obtaining an accurate answer. For part (b)(i), most candidates understood that this was the initial population but the word billion was often omitted, showing a lack of full comprehension of the model. Part (b)(ii) was less well answered than (b)(i). The proportionality or rate was often recognised but “per year” was sometimes omitted. The wording of answers also cost candidates the mark as they often talked about the “amount” or “how much” the population went up by, or they wrote correct responses followed by it increases by “1.013 billion”. Part (c) was generally well answered with the majority of candidates substituting t = 26 into their model and remembering to write billions (despite often having forgotten this in part b(i)). Occasionally candidates used t = 27 and more rarely 25. When candidates attempted to write their answer as the full form of the decimal, they often made errors in the number of digits. Part (d) was not well answered on the whole, and credit seemed to be rarely given. Many candidates focussed on an assumption that death/disease was not incorporated into the model. This showed poor understanding of modelling as the growth rate can factor things like this in when it is derived. Candidates who did make a link to extrapolation, were often unable to recognise that this was a prediction far into the future so the model may not be appropriate. Responses to this part often seemed to be rote answers that indicated a lack of understanding of the model.
The world human population P billions is modelled by P = ab^t, where a and b are constants and t is the number of years after 2004. Using estimated population figures for 2004 to 2007, a graph of log_10 P against t has points lying approximately on a straight line with gradient 0.0054 and intercept 0.81 on the log_10 P axis.
Edexcel 9MA0 · June 2023 · Paper 1 · Q6 · 6 marks
6 marks
Examiner report
This was again another accessible question with a total of 6 marks split between 3 different sub-parts. However, this topic continues to be one which candidates struggle with and in all parts of this question. It was not uncommon to see candidates attempting to use rules of indices rather than logarithms, starting off with 2a x = , 2 8 b x = + and attempting (usually incorrectly) to substitute these into the expressions given. Where this was successful candidates generally failed to remove 2 log 2 resulting in them losing one mark per part. Part (a) was a B mark for achieving the correct answer. Several candidates got as far as 2 0.5log x without substituting a into this and so gained no credit. Others could not apply the index log law to the square root of x correctly and so ended up with the answer square root of a often written as 1 2 a Part (b) surprisingly seemed to have more success. It was generally answered well with most candidates realising that factorising the argument of the logarithm was needed before applying the addition law of logarithms. There were several candidates who, having factorised correctly, applied the addition law incorrectly, resulting in an answer of a b and so lost both marks. Several candidates just wrote down the correct answer and gained both marks. If the correct answer followed the product of the relevant logs, ( ) ( ) log log 8 x x + rather than the sum, then the answer was allowed to imply the correct method, but the final mark was withheld for this incorrect working seen. The main misconception was the sight of 2 log log8 x x + which was awarded no marks, although writing log log log log8 x x + + + (possibly seen as 2 3 a + ) gained the method mark as evidence of correct addition law used at some point. Some candidates attempted to work backwards from the given log expressions for a and b with very limited success. Part (c) was the most challenging part of this question and clearly identified the stronger mathematicians of the cohort. Many responses to this part were blank, whilst the more able candidates provided very succinct solutions. For those who attempted with less success there were some rather dubious attempts at writing 64 8 x + as a single fraction. Of those who managed this correctly and progressed to applying the laws of logarithms, quite a few left in the 2 log 8 rather than simplifying to its value of 3, which meant that only the first mark was available: marks had already been awarded for the power law and addition law in previous parts of the question and here we required an understanding of reaching an answer in its simplest form. This further demonstrated the lack of logarithmic fluency for a number of candidates. Others factorised 8 from 64 8 x + but then struggled to apply the laws appropriately as they had not completed the process of writing the argument as a single fraction. Misconceptions were seen such as expressing 64 log 8 x + as 64 log8 log x + and sign errors often occurred.
a = log_2(x) and b = log_2(x + 8). Express each answer in terms of a and/or b.
Edexcel 9MA0 · June 2023 · Paper 1 · Q11 · 7 marks
7 marks
Examiner report
This question proved to be quite challenging for a number of candidates, however those who were able to manipulate logarithms using the rules correctly, along with calculating the gradient, were often very successful. It is worth noting that there were a number of candidates who did not pay attention to the labelling on the y-axis and therefore were unable to score any marks on this question. As with many modelling questions, candidates currently lack proficiency at relating the mathematics to the context of the question. In part (a), most candidates were able to find the initial value correctly, although several lost the accuracy mark by forgetting to include units. The location of the unit was condoned e.g. 1000£ was seen on many occasions. A common reason for candidates not achieving any marks was to assume 3 V = or 10 log 3 V = , instead of 10 log 3 V = . A few candidates used natural logs instead of 10 log , resulting in 3e V = instead of 3 10 V = . Some candidates just stated the initial value was 3, not fully appreciating what information the linear graph was representing. Part (b) was discriminating, requiring some understanding of the relationship between t V ab = and the linear graph. Those candidates who were most successful with this part of the question, made the link between part (a) and the value of a without having to do additional calculations. Many candidates were able to find the gradient of the graph correctly and pick up the first method mark. However, finding a value for b was difficult for many candidates. Those candidates who did not make the link with part (a) often had to recalculate a or attempted to use simultaneous equations, often with little success. There were a lot of very convoluted incorrect solutions in this question which showed a poor grasp of logarithms and the associated rules; log t ab was commonly seen; others failed to correctly relate log log log V a t b = + to y mx c = + despite previous questions on the topic, confusing which were the variables and which were the constants. There were several cases of candidates mixing up the x, y coordinates in their substitution. It was encouraging to see that most candidates tried to give the full equation if they found values for a and b, but there was still a good number that did not and thus, lost the final mark. A few candidates were unable to score the final mark for leaving their answer as 10 log 3 0.021 V t = − . Candidates were able to gain the final mark of part (b), if the complete equation was seen in part (c), although this was rare to see. In part (c), candidates who used 24 t = generally scored the method mark even if their model was incorrect but of the correct form. The most common form used was a continuation of their answer to part (b), usually t V ab = where a was positive. The majority of candidates were able to make a valid comparison accompanied by a sufficient explanation to earn the final accuracy mark. Some calculated the percentage error to justify their thinking which was acceptable. A minority thought the small difference was too great for the model to be reliable. A few candidates used the method of substituting 24 t = into an equation of the form 10 log V p qt = + to find a value for 10 log V . The discerning candidates either compared their 10 log V with 10 log 320 , or they proceeded to find V and compared this with 320. This method was slightly more complicated, and the success rate of this method was mixed. The units in this question caught a considerable number of candidates out, when checking the suitability of the model and substituting in 2 (years) rather than 24 (months), highlighting the importance of reading the question carefully. This led to a conclusion that the model was unreliable due to the vastly different amount that was yielded. Candidates did not consider the fact that this could mean that there was an error in their own calculation.
The value V pounds of a mobile phone, t months after it was bought, is modelled by V=ab^t, where a and b are constants. A straight-line graph of log_10(V) against t passes through the points (0,3) and (10,2.79).
Edexcel 9MA0 · June 2023 · Paper 2 · Q3 · 5 marks
5 marks
Examiner report
There were very few non-attempts or zero scores here and this question proved to be a good discriminator with a spread of marks achieved by candidates. Most students managed to score at least one mark out of three in part (a) by correctly applying a single log law, usually the product, division or power rules for logs. Dealing with the constant, 2, proved to be more challenging for many. Those who wrote 2 as log2(4) and then applied the product rule for logs on the right- hand-side had most success. Others grouped the three log terms and applied the exponential rule to eliminate the logs. Sometimes there was a lot of working which wasn’t always set out as clearly as it could have been. A significant proportion of candidates did succeed, but it was not uncommon to see correct use of a log law followed by incorrect log work in an attempt to fudge the printed result. Some poor responses involved an initial ‘cancelling’ of logs before any attempt to combine log terms or to raise both sides to the base 2 correctly. There was a minority of candidates who perhaps knew how to proceed but did not show sufficient working to gain full marks in a ‘show that’ question. Candidates should be reminded that all steps should be shown in this type of question. It was a shame when fully correct log work was followed by an incomplete ‘equation’ at the end of this part, as missing the ‘= 0’ meant the loss of the final mark here. A small number of candidates who were unsure how to start part (a) used the printed answer and tried to “work backwards” but often made errors leading to incorrect solutions. In (b) part (i), almost all candidates were able to correctly state the values ‘6’ and ‘ 5 3 − ’, it was clear that in many cases candidates were making use of their calculators to solve the quadratic equation which was acceptable. Part (ii) however, was answered less reliably. Although most candidates had some understanding of the domain of a log function and knew that 5 3 x = − was not a valid solution, many were unable to provide reasoning that was precise enough for the mark here. It was common to see, for example, “a log can’t be negative”, which was not considered sufficiently clear. Others stated simply ‘because it is negative’ which was also insufficient. A range of responses which engaged specifically with the domain of a log function needing to be positive were considered acceptable such as; ‘log of a negative value is undefined/impossible/gives a math error’.
Given that log_2(x + 3) + log_2(x + 10) = 2 + 2 log_2 x:
Edexcel 9MA0 · June 2023 · Paper 2 · Q4 · 4 marks
4 marks
Examiner report
Those scoring full marks in this question were in a significant minority of candidates. In part (a) the majority of candidates used the given information to find the value of A. Those that did not achieve this mark typically gave 85 as an answer. A few students failed to recognise the standard technique of substituting t = 0 into the expression for H to find the value of A, often using t = 1. Sometimes e0 was evaluated as e or 0 rather than 1. Part (b) had mixed outcomes for candidates. Interpretation of the given initial rate of cooling was problematic for many candidates. There was a general lack of appreciation that the rate of cooling was a continuous variable which represented d . d H t − Those candidates not recognising this and using a difference method (similar to those introduced at GCSE) between t = 0 and t = 1 were restricted to the initial B mark in part (a). The most common attempt at a response for these candidates was setting H(1) = 85 − 7.5 and working from there. Of those who recognised the need to first find d d H t and then substitute t = 0, many incorrectly used d d H t = 7.5 rather than the correct d d H t = −7.5 to express a negative increase in temperature. This led to the loss of the last mark. Another error was to forget to differentiate the + 30 to zero leaving a constant in their d d H t , only scoring at most the second method mark for substituting t = 0 and their A into their d d H t . Some differentiated incorrectly to obtain −55Bte−Bt. Several students missed out on the final mark despite correct work by not writing out the final equation. It is important that centres make candidates aware that when a question asks for the complete equation for a model that it is given as such and not just the values of any required constants.
Coffee is poured into a cup. The temperature of the coffee, H degrees Celsius, t minutes after being poured is modelled by H = Ae^(-Bt) + 30, where A and B are constants. Initially, the temperature of the coffee was 85 degrees Celsius.
Edexcel 9MA0 · June 2022 · Paper 1 · Q10 · 8 marks
8 marks
Examiner report
(Mean mark 3.8 out of 8) This was the third modelling question on the paper, this time based around an exponential model. This should have been expected given the Advance Information and candidates performed well on parts (a) and (b). Part (c) was more of a challenge with only higher achieving students scoring all 4 marks. Part (a) was either answered correctly, or the mark was not awarded because the answer stated was 265 bees not 265,000 bees. Most candidates recognised the need to differentiate in part (b), and used appropriate notation. Most did this accurately before substituting in t =10 to obtain 18.1. A number of candidates failed to secure the final mark however, for not providing a minimal conclusion relating their answer to 18,000. Common reasons for not achieving full marks were • failing to use appropriate notation often using d d y x instead of d d N t • failing to recognise the need to use differentiation often subtracting the values of b N at t =9 and 10 or else t =10 and 11 Part (c) was the most demanding part of the question with many candidates just picking up the first mark for equating the two expressions. Having equated the two expressions, the significance of the negative exponential term seemed to be poorly understood. Very common errors included • changing the 0.05 e t − term to 0.05 e t and then solving • applying 'ln' to each term in the equation to form a linear equation in t • slips in coefficients with 220 becoming 200 A number of candidates reverted to numerical methods or the use of their calculator to solve the equation rather than solve by algebraic methods. The strongest candidates recognised the quadratic equation in 0.05 e t and were able then to solve it correctly, many after making a substitution 0.05 =e t x .
A scientist models the number of bees, measured in thousands, by N_b=45+220e^(0.05t), and the number of wasps, measured in thousands, by N_w=10+800e^(-0.05t), where t is the number of years from the start of the study.
Edexcel 9MA0 · June 2022 · Paper 2 · Q2 · 4 marks
4 marks
Examiner report
This question on an exponential function saw full marks for most. Of the small number of errors that were seen in part (a), a common one was an incorrect intercept of (0, 4) from replacing 40 with 4 instead of 1. The intercept was occasionally seen as (1, 0) although this was condoned if it was positioned correctly on the sketch. A few graphs did not go into the second quadrant and others were not carefully drawn – some did not get close enough to the asymptote of y = 0, sometimes due to a poor choice of scale. Other graphs bent back on themselves in the first quadrant. Some students offered an incorrect or extra horizontal or even a vertical asymptote. Although almost all knew the correct shape, a few responses had a straight line graph. Part (b) was similarly well-answered and most used logarithms correctly to obtain a correct numerical expression for x. Those who did not use x = log4100 and arrived at log100 log 4 x = sometimes processed this incorrectly, arriving at log 96, log 25 or even 25. Some candidates who did not know how to solve the equation resorted to trial and improvement with varied success and is not a recommended strategy.
Consider the curve with equation y = 4^x.
Edexcel 9MA0 · June 2022 · Paper 2 · Q5 · 6 marks
6 marks
Examiner report
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.
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.
Edexcel 9MA0 · October 2021 · Paper 1 · Q5 · 6 marks
6 marks
Examiner report
This was yet another good source of marks with the modal mark being 6 out of 6, achieved by over 35% of the candidature. Almost every candidate gained the B1 mark in part (a) with a fairly even split between those who used 20 000×1.082 and those who first calculated the profit after Year 1 = 21600 and then Year 2 = 23328 showing each step of their working to gain the mark. Part (b) was more problematic for many candidates. There was a very good understanding of the inequality/equation to be formed, and the use of logs was almost always very good. However, a significant minority of candidates failed to gain the final A mark giving an answer of 15.3 or 16 instead of Year 17. Part (c) proved to be very well answered with the vast majority using the sum formula correctly. Errors included using n =19 instead of 20 in the formula or finding the 20th term instead of the sum. A few candidates listed all the terms before adding them which often led to accuracy errors.
A company made a profit of GBP 20,000 in Year 1. A model predicts that yearly profit increases by 8% each year, so the yearly profits form a geometric sequence. Show all stages of working; solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · October 2021 · Paper 1 · Q8 · 9 marks
9 marks
Examiner report
This was the first significant modelling question on this paper. As with other series, candidates find aspects of modelling difficult. The barriers to success here were parts (b) and (c). In part (a) many candidates made progress and achieved a complete equation for the model with k given exactly as ln(2)/5 or rounded to 0.139. A few students rounded more accurately giving k as 0.1386. Some rounded k to 0.14 and lost the accuracy mark. A fair number of students who failed to find the correct value of ‘A’ evaluated e0 as e. In part (b) many students used an incorrect method to find the rate of increase in the number of bacteria. Some substituted t =8 into their expression for N, which just found the number of bacteria after 8 hours; others then went on to divide this number by 8 or even 1000 to try and find the rate of increase. Those who realised that the expression for N needed to be differentiated to find the rate of increase often scored at least the method mark. For part (c) the majority of candidates who obtained the correct equation of the model in (a) were able to use both models given to set up an appropriate equation in t or T to gain the first method mark. Processing the resulting equation using logs to obtain a linear equation in t or T posed more of a challenge and many made no further progress. A large proportion who did attempt to make progress often used incorrect log work: some multiplied the logs rather than add them; others did not find the log of all the components, usually the ‘2’. Some students combined the exponential functions incorrectly by dividing the powers rather than subtracting them. Those scoring all three marks here were usually the most able candidates.
A first bacteria population is modelled by N=Ae^(kt) for t>=0, where A and k are positive constants and t is hours from the start. Initially there are 1000 bacteria and the population takes exactly 5 hours to double. A second population is modelled by M=500e^(1.4kt), using the same value of k.
Edexcel 9MA0 · October 2021 · Paper 2 · Q3 · 3 marks
3 marks
Examiner report
This 3 mark question was attempted by almost all students and differentiated between those who could apply laws of logarithms and those who could not. It was generally a well answered question and most students were able to get some marks and a large majority scored full marks. A few students showed an invalid line of before recovering the error and continuing to the correct value of 4/39 for y, so only scored 1 of the 3 available marks. A few students were unable to apply the subtraction law but they were often able to score the B mark for 9 seen. This was a straight forward question on logarithms so it was nice to see the candidates were generally prepared and those that correctly removed the logs were, on the whole, able to go on get full marks. A few students make slips when solving their linear equation, obtaining 39/4 instead of 4/39. A few also tried to use base e incorrectly, and a minority tried expanding the logs, hence getting no marks. ( ) ( ) ( ) ( ) 3 3 3 3 log 12 5 log 12 5 log 1 3 2 2 log 1 3 y y y y + + − − = ⇒ = −
Using the laws of logarithms, solve the equation log_3(12y + 5) - log_3(1 - 3y) = 2.
Edexcel 9MA0 · October 2021 · Paper 2 · Q10 · 6 marks
6 marks
Examiner report
The majority of candidates scored both marks for part (a). A common error was omitting base 10 when writing their final equation. In “show that” questions, we need to see the equation as stated in the question. Part (b) was demanding for many candidates, with confusion as to how to correctly deal with the log terms. Most candidates who were successful found the value of by finding the gradient of the line. A less popular but usually successful approach was using simultaneous equations to eliminate . Solving the equation to find the value of caused some problems. The question required a complete equation for the model, so just stating the correct values of and without the equation written explicitly meant candidates scored 3 out of 4 marks, which was a very common error. Part (c) was very rarely completed successfully. Common errors were thinking was the intercept on the T axis or the time for the initial swing of the pendulum.
The time T seconds that a pendulum takes to complete one swing is modelled by T = al^b, where l metres is the pendulum length and a and b are constants. A straight-line graph of log_10(T) against log_10(l) passes through (-0.7, 0) and (0.21, 0.45).
Edexcel 9MA0 · October 2021 · Paper 31 · Q6 · 7 marks
7 marks
Examiner report
There is a requirement in the specification for questions occasionally to draw together knowledge from different areas of 9MA0 and this was a question aiming to do that. Many students realised that to start the problem they could use the fact that the sum of the probabilities must equal 1 but fewer than half of the students could move beyond that mark. Those who realised they had a sum of logarithms were often able to arrive at the equation abc = 36 but few could devise a strategy to deduce the values of a, b and c. Some stated that all of the values had to be greater than 1 though it was not always clear that they deduced this from the given bullet points in the question. Around a quarter of the students found the correct set of values with varying degrees of explanation on the way. Part (b) was often left blank though the first mark was available without values for a, b and c provided brackets were used correctly for the expression. Nearly 10% managed to score full marks for the question.
The discrete random variable X has the probability distribution P(X = a) = log_36(a), P(X = b) = log_36(b), P(X = c) = log_36(c), where a, b and c are distinct integers with a < b < c and all three probabilities are greater than zero.
Edexcel 9MA0 · October 2020 · Paper 1 · Q2 · 3 marks
3 marks
Examiner report
As with question 1 there were many fully correct answers here. Of these, most took logs or lns of both sides, then used the power law on each side and the correct order of operations before proceeding to the correct answer of 81.6. There were a few candidates who failed to put brackets around (3p -1) but they were usually able to recover. There were several cases of candidates failing to apply the power rule to both sides, usually not dealing with the ‘210’. Some candidates misapplied the power rule, multiplying first for example by 3, rather than 3p -1. Common errors were • after applying the power rule correctly changing log5/log 4 to log(5/4). • taking either log4 of one side or log5 of the other without any application of the power law
By taking logarithms of both sides, solve 4^(3p-1)=5^210.
Edexcel 9MA0 · October 2020 · Paper 1 · Q8 · 2 marks
2 marks
Examiner report
Candidates were expected to realise that ''when the rate of change is proportional to the y value, an exponential model should be used.'' The question seemed to take a great many candidates by surprise and many assumed that 'n was proportional to n' or that 'n was proportional to t', not that the rate of growth of 'n was proportional to n'. Those that did interpret the situation correctly could write down an equation very quickly whereas others attempted to solve a differential equation. Equations scoring one of two marks were common and included e , e t kt n A n = = and e kt n A b = +
A company monitors the total number n of a new smartphone sold t days after release. During the monitored period, the rate of increase of n is proportional to n.
Edexcel 9MA0 · October 2020 · Paper 2 · Q3 · 5 marks
5 marks
Examiner report
This was a fairly accessible question for prepared candidates and provided an early source of marks for many. Generally, most candidates recognised the need to use the power law for logs and correctly obtained 2 log(4 −𝑥𝑥) = log (4 −𝑥𝑥)2, then proceeded to remove the logs. In fact, many candidates jumped immediately to (4 −𝑥𝑥)2 = (𝑥𝑥+ 8) with no sign of logs whatsoever in their solution. This was condoned although it must be noted that combining steps of working in a ‘show that’ question can be risky. Some candidates diligently wrote out 10log (4−𝑥𝑥)2 = 10 log(𝑥𝑥+8) before removing logs, others subtracted log (4 −𝑥𝑥)2 from both sides and used the log subtraction law to combine the log terms before removing the log. Both approaches were often successful although the latter tended to be the riskier of the two as some candidates forgot that 100 = 1 and instead obtained (4−𝑥𝑥) (𝑥𝑥+8) = 0 which was then fudged to give the required solution. Those candidates that did not recognise the need to employ the power law for logs early on in their solution made very little progress. Disappointingly, some candidates failed to earn the final mark in part (a) as their final line was missing the “= 0”. Candidates should take care to check their final line matches the printed answer in a ‘show that’ question. The first part of part (b) was straightforward and gave an opportunity to re-enter the question for those who had come unstuck in part (a). Indeed, for some candidates this was the only mark attained in this question. It was surprising that a number of candidates left their ‘solution’ as (𝑥𝑥−1)(𝑥𝑥−8) or gave ‘solutions’: (𝑥𝑥−1) and (𝑥𝑥−8). Only a very small number of candidates incorrectly obtained 𝑥𝑥= −1 or 𝑥𝑥= −8. The second part of part (b) was more discriminating and a careful explanation was required here. Most candidates were able to identify 𝑥𝑥= 8 as the value which was not a solution. However, in order to gain credit, it was also necessary for candidates to make specific reference to the fact that log(4 −8) is undefined. Too many candidates simply stated ‘logs can’t be negative’ which was incorrect and quite different to ‘you can’t take logs of a negative number’. Candidates should be reminded that reasoning should be precise and specific.
The equation 2log(4 - x) = log(x + 8) has real solutions, where the logarithms have the same base.
Edexcel 9MA0 · October 2020 · Paper 2 · Q5 · 4 marks
4 marks
Examiner report
Nearly all candidates combined the equations correctly and gained the first mark. Many then struggled to progress further. A few made mistakes with the indices such as changing 3(2x) into 6x but the most common mistake was attempting to take logs incorrectly, usually log15 - log(2(2x)) or log(15 - 2x+1 ) being wrongly expanded to arrive at log15 - (x + 1)log2. Those who realised 2x+1 = 2(2x) usually went on to solve correctly, giving a correct exact answer and therefore gain full marks. Candidates need to understand that seeing an equation with an unknown index does not automatically mean that it is a question involving logarithms.
The curves with equations y = 3(2^x) and y = 15 - 2^(x + 1) intersect at one point.
Edexcel 9MA0 · October 2020 · Paper 2 · Q9 · 6 marks
6 marks
Examiner report
This question proved to be quite challenging for many candidates and awarding full marks was rare. Nonetheless, most candidates recognised the need to set up simultaneous equations in A and B and were often able to use the information provided in the question to establish at least one equation linking A and B. Usually problems arose initially with the 18 = 𝐴𝐴 − 𝐵𝐵𝑒𝑒0 which was frequently incorrectly simplified to 18 = 𝐴𝐴 or sometimes 18 = 𝐴𝐴 − 1. Indeed, some candidates assumed immediately that A = initial temperature = 18 as is sometimes the case in modelling questions. Such errors proved costly as candidates were then limited to one mark from four in part (a). It was pleasing to see that most candidates who managed to set up the two equations correctly were often unphased by the presence of e−0.7as a coefficient of B. Often this was rewritten as a decimal approximation before solving for B. This was acceptable although some candidates did not provide sufficiently accurate approximations and so lost a mark for values of A and B that were not accurate to 3sf (51.7 and 69.7 were sometimes seen). Candidates also lost a mark here for being too accurate and giving their final values of A and B to 3dp for example. Candidates should be advised to check the question carefully for the required degree of accuracy. Unfortunately, it was quite common for candidates to find correct values for A and B but then fail to state the equation of the model in full which led to the loss of one mark. Part (b) was quite discriminating. Occasionally, candidates noted that the maximum temperature according to the model is 69.6°C but the most common approach taken by candidates was to set 𝜃𝜃(𝑡𝑡) = 78 and attempt to solve for 𝑡𝑡. Some candidates were unable to solve the equation but allowed their work to peter out without comment. However, a good number of candidates correctly noted that a solution was not possible - recognising that the model would not reach the boiling point of ethanol which was sufficient for one mark. It was also common however to see minus signs conveniently disappear in order to obtain a solution for 𝑡𝑡 at all costs. Such candidates usually obtained 𝑡𝑡 = 25.9 and concluded that the model would be appropriate up to this time; or obtained 𝑡𝑡= −25.9 noting that ‘time cannot be negative’. A significant number of responses gave general comments relating to the model being inappropriate once ethanol’s boiling point was reached - such responses were not creditworthy. Unfortunately, errors in the calculation of A and B from part (a) sometimes resulted in a version of 𝜃𝜃(𝑡𝑡) = 78 which could legitimately be solved which precluded the marks in (b). It was extremely rare to award the second mark in part (b) as candidates almost always failed to acknowledge that the model was inappropriate because the maximum value of 69.6°C is significantly lower than the boiling point of 78°C.
A quantity of ethanol is heated. Its temperature theta degrees C at time t seconds after heating begins is modelled by theta = A - Be^(-0.07t), where A and B are positive constants. Initially the temperature is 18 degrees C. After 10 seconds it is 44 degrees C.
Edexcel 9MA0 · October 2020 · Paper 2 · Q13 · 6 marks
6 marks
Examiner report
Part (a) was often answered correctly and candidates knew to set the denominator equal to zero and solve to obtain a value for 𝑘𝑘. Unfortunately, candidates sometimes left their answer as 𝑥𝑥= e2 which was, of course, incorrect. Some candidates gave the inexact answer of 𝑘𝑘 = 7.39 which was condoned. Commonly seen incorrect answers included 𝑘𝑘 = 0, or 𝑘𝑘 = 1. A number of candidates gave the value 𝑘𝑘= 3 suggesting that the expression in the denominator had been incorrectly interpreted as ln(𝑥𝑥−2) rather than ln(𝑥𝑥) −2 as was stated. A small number of candidates wrote “there is no 𝑘𝑘 in the question” perhaps failing to spot its presence in the domain. Part (b) was attempted by most candidates although it was not uncommon to see attempts to find the inverse of g(𝑥𝑥) instead of differentiating. The differentiation in part (b) was however correctly carried out by many candidates, with the majority using the quotient rule. Those who stated the correct formula and the individual components of u, v, u’ and v’ were often more successful. A few used ቀ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑ቁ 2 instead of 𝑣𝑣2 in the denominator, while others made slips on expanding the bracket 3(𝑙𝑙𝑙𝑙(𝑥𝑥) − 2). Some attempted the product rule, usually with less success. Many candidates did not attempt to simplify the numerator of g′(x). Often those that did made slips and it was frustrating to see candidates losing marks due to missing brackets and sign errors when expanding. It was common to see numerators incorrectly simplified to –13 𝑥𝑥 or 5 𝑥𝑥 although many did manage the correct simplification to 1 𝑥𝑥. Some candidates unnecessarily expanded the denominator, sometimes incorrectly writing ln (𝑥𝑥2) instead of (ln 𝑥𝑥)2. Candidates who managed to correctly simplify their expression for g′(x) argued correctly why g′(x) > 0 for all 𝑥𝑥 in the domain, but some simply stated this with no reason or gave an incomplete argument of “𝑥𝑥> 0 so g′(𝑥𝑥) > 0” and lost the final mark in this part. In part (c), most candidates only gained one out of two available marks as they usually considered only the set of values of 𝑎𝑎 for which the numerator is positive by solving 3𝑙𝑙𝑙𝑙𝑙𝑙−7 = 0 to obtain 𝑎𝑎> 𝑒𝑒 7 3. Usually no consideration was given to the case where the numerator and denominator are both negative and the few that did solve 𝑙𝑙𝑙𝑙𝑙𝑙= 2 usually failed to state 0 < 𝑎𝑎< 𝑒𝑒2.
The function g is defined by g(x) = (3ln(x) - 7)/(ln(x) - 2), for x > 0 and x != k.
Edexcel 9MA0 · June 2019 · Paper 1 · Q7 · 7 marks
7 marks
Examiner report
(Mean Mark 4.2 out of 7) It was pleasing to see that most candidates attempted to form an exponential model using a suitable equation. This is one of the new topics in the specification and candidates who didn't gain any marks on this question, generally did so because they did not attempt it, or that they tried to use a non-exponential model. In part (a) t V Ar or kt V Ae were the most popular types of correct models that candidates opted for. However, some methods were often muddled or poorly shown, but many candidates who used a correct model were able to proceed to obtain correct values for their constants. Those candidates who used the model kt V Ae frequently gave the constant k as a log and errors sometimes occurred with the sign of this constant. Errors in calculating constants with the model t V Ar often occurred when candidates used 1 t V Ar instead. In part (b) candidates who correctly answered part (a) usually understood that for this section they needed to substitute 10 t into their model, and most were able to do this. Many candidates then proceeded to make a sensible comparison of the value gained by their model with the actual value in order to comment on the model’s reliability (sometimes calculating and commenting on % error). The final mark, however, was frequently lost because answers were too vague, without a clear comparison or assessment of reliability being made. Some candidates appeared to think that a model was only acceptable if it gave the exact real value. Most candidates who attempted part (c) understood which of their two constants would need to be altered, although marks were lost either because the answer given just said the constant needed to be changed or because they wrongly identified whether the constant needed to be decreased/increased. This was more of a difficulty for those candidates who had used the model kt V Ae , because candidates were often dealing with negative constants and/or constants given as logs so the phrases used were not always mathematically correct. An example was often useful to convey what the candidate meant where there was ambiguity. Candidates who had gone with model t V Ar were less likely to make an error in judging whether their constant should increase or decrease.
In a simple model, the value GBP V of a car depends on its age t years. Car A is worth GBP 20,000 when new and GBP 16,000 after one year. Its actual value after 10 years is GBP 2,000. Car B has the same new value but depreciates more slowly.
Edexcel 9MA0 · June 2019 · Paper 1 · Q9 · 5 marks
5 marks
Examiner report
(Mean Mark 2.6 out of 5 ) There were many good attempts to this question, though complete answers to part (b) were achieved only by the very best. In part (a) most students gained the B mark by writing log log log a a b b , though a few did attempt to rearrange first and used other equivalents. Students who gained this mark generally were successful at 'undoing' the logs and making a the subject. Students who failed to recognise the logarithmic rule for subtraction or addition often made no progress with the question. Part (b) was more challenging with many only scoring the first mark. Many students focused solely on the exclusion of b = 1 from the denominator and paid no heed to the condition that a >0. Few students were able to provide a full explanation including that as a >0, 2 0 1 b b and so b > 1. Most candidates knew that the denominator of a fraction cannot be zero, though few of them used the words “undefined” or “infinity” in their answers, but would more often refer to 'math error' or 'can’t didide by 0'.
Let a>b>0 and suppose log(a)-log(b)=log(a-b), with all logarithms in the same base.
Edexcel 9MA0 · June 2019 · Paper 2 · Q8 · 6 marks
6 marks
Examiner report
Q08 was found challenging by many candidates, with Q08(i) more successfully answered than Q08(ii). The most successful candidates were the ones who listed the first few terms of the series. In other words, those candidates that wrote 4 5 6 1 1 1 20 20 20 ... 2 2 2 and 5 5 5 3 4 5 log log log ... 2 3 4 gained a better understanding of the series that they were required to sum. There were, however, many attempts that made no creditable progress and a significant number of blank responses to either one or both parts of Q08. In Q08(i), many candidates identified that a sum to infinity was required in their solution and attempted to apply the formula . 1 a r Most used 1 2 r in this formula, although some used incorrect values such as 1 4 r or 1 . 2 r Disappointingly, a significant number of candidates used 4 r from the expression given in the question and applied it to the formula , 1 a r even though the formula only works when 1 1. r There were a number of correct strategies that candidates could use. The most successful strategy was to apply 4 1 1 20 , 2 2 a r to , 1 a r and many candidates who used this strategy achieved the correct answer of 2.5. Other complete strategies included applying 3 1 1 1 1 20 20 2 2 r r r r or 3 0 0 1 1 20 20 2 2 r r r r which led to (20 17.5) or (40 37.5) respectively. Some candidates applied an incorrect overall strategy by subtracting an incorrect number of terms from their sum to infinity. These candidates gave incorrect answers such as (20 18.75), (40 38.75), (40 35) or (40 17.5). A few candidates obtained the correct answer of 2.5 by using their calculator. Only a few candidates produced a correct solution to Q08(ii). There were two main correct methods that were used by candidates in equal measure. One was to list the first few terms and the last few terms of the series and use the addition law of logarithms to achieve 5 3 4 50 log ... . 2 3 49 Many candidates used cancelling to complete their proof correctly by writing 5 5 50 log log (25) 2. 2 The other method was to use the subtraction law of logarithms to give 48 5 5 1 (log ( 2) log ( 1)) n n n and list terms to give 5 5 5 5 5 5 (log 3 log 4 ...... log 50) (log 2 log 3 ...... log 49). Again many candidates used cancelling and completed their proof by writing 5 5 5 log (50) log (2) log (25) 2. A few candidates progressed by writing 5 5 log (3 4 ....... 50) log (2 3 ....... 49) and used factorials to give 5 5 5 5 50! 50! log log 49! log log (25) 2. 2 2(49!) Some candidates incorrectly assumed an arithmetic series and attempted to find its sum by using ( ) 2 n a l or [2 ( 1) ] 2 n a n d with 5 5 3 50 48, log , log 2 49 n a l and 5 5 4 3 log log . 3 2 d A few candidates incorrectly assumed a geometric series and attempted to find its sum. Some candidates, who made no creditable progress, resorted to listing their terms as decimals, or attempted to add all 48 terms as decimals by writing 0.2519... 0.1787... 0.1386... ...... 0.0125... 2.
This question concerns an infinite geometric sum and a telescoping logarithmic sum.
Edexcel 9MA0 · June 2019 · Paper 2 · Q9 · 9 marks
9 marks
Examiner report
In general, Q09 discriminated well between candidates of all abilities. Q09(b) and Q09(c) were more accessible to candidates than Q09(a). Some candidates left Q09(a) blank while others struggled to make any creditable progress. In Q09(a), many candidates wrote down a valid log equation as part of their explanation. Those who used mathematical reasoning to progress from 10 10 10 log log log d k n V to n d kV (or vice versa) usually showed an intermediate step in their working and so were successful in their explanation. A few candidates made errors in progressing between these two equations, such as an incorrect method of combining logarithms (e.g. 10 10 log log d n kV ), or an incorrect use of powers (e.g. 10 ). c n d V Those who made a direct comparison of their log equation with y mx c usually did so clearly enough but sometimes did not allude to n d kV (if using 1.77 rather than 10 log k in their log equation). A significant number of candidates did not apply 1.77 10 or use logarithms to show that 0.017. k In Q09(b), the majority of candidates substituted 30 V and 20 d into n d kV and applied logarithms correctly, leading to the correct value of 2.08. n A few candidates used the incorrect method of simplifying 20 0.017(30)n to give 20 (0.51) . n Some candidates substituted in V and d the wrong way around, usually leading to 2.50, n and a few used both V and d as 20. A small minority substituted 30 V and 20 d into a correct log equation, while a few others correctly found n by finding the gradient between the known points (0, 1.77) and 10 10 (log 30, log 20). The most common mistake was not writing a complete equation 2.08 (0.017) d V or 10 10 log 1.77 2.08log d V after finding a correct 2.08. n In Q09(c), most candidates applied 60 V to their exponential model and found the braking distance. Some then used the incorrect method of comparing this breaking distance with the overall stopping distance. The majority attempted to calculate the thinking distance and either added it to their breaking distance d or subtracted it from 100. The main error was due to the units. Some candidates did not attempt to convert 1 60 kmh to 1 ms and simply multiplied 60 by the thinking time of 0.8 seconds to obtain a thinking distance of 48 metres. Others only managed a partial conversion, with some using 1 75 metres 1 instead of 13.3m km 75 as their thinking distance. In some of the answers, candidates used an incorrect method of combining their values to give a dimensionally incorrect quantity. For example, they might add their breaking distance to either a speed or a time. Candidates who obtained a thinking distance of 13.3m usually reached a correct answer and conclusion. A few compared the maximum speed needed to stop at the puddle with the given 60, while even fewer compared the maximum value of n needed to stop at the puddle with 2.08; n and with correct conclusions these were valid solutions.
A research engineer models a car's braking distance d metres in wet conditions when the brakes are applied at speed V km h^-1. A graph of d against V is curved and contains the point (30,20). A graph of log_10(d) against log_10(V) is a straight line with vertical intercept (0,-1.77).
Edexcel 9MA0 · June 2019 · Paper 2 · Q11 · 11 marks
11 marks
Examiner report
In general, Q11 discriminated well between candidates of all abilities. Q11(a) was demanding with most candidates scoring no more than one mark. Q11(b), Q11(c) and Q11(d) proved accessible to candidates of all abilities. In Q11(a), most candidates took logs, as advised by the question, and many obtained either ln ln y x x or log log . y x x At this point a significant number of candidates made no further creditable progress. Some candidates used a complete method of applying the product rule on ln x x and implicit differentiation to obtain a correct 1 d 1 ln . d y x y x Most of these candidates then applied d 0 d y x to obtain 1 e . x Some candidates, who received partial credit, used a ‘ 10 log ’ method to obtain 0.1, x after finding 10 1 d 1 log . d y x y x A few candidates used logarithms without reference to a base. Those who proceeded to obtain 0.1 x received partial credit and those who obtained 1 e x were allowed full marks. There were some alternative methods employed by a few candidates in Q11(a). A few candidates rewrote x y x as ln e x x y and applied ln d (1 ln )e 0 d x x y x x to give a correct 1 e . x Others differentiated both sides of the equation ln ln y x x with respect to y and usually proceeded to find 1 e . x A very few candidates applied the quotient rule to the equation ln , ln y x x but were rarely successful with this method. In Q11(b), most candidates evaluated either xx or 2 xx at both 1.5 x and 1.6. x Many candidates compared their values with either 2 or 0 (depending on whether they evaluated xx or 2 xx ). Although many of these concluded that 1.5 1.6, only a few made any reference to the curve being continuous in this interval. A few candidates, who made no creditable progress, used values of x outside the given range, e.g. 1.4 x and 1.6. x Very few candidates used values within the range e.g. 1.51 x and 1.59, x which meant that they were able to earn both marks. In Q11(c), most candidates found a correct 4 1.673, x although a few did not state 4x correct to 3 decimal places. Some wrote down the iterates 1 2 , x x and 3x as part of their solution, while others stated 4x with no intermediate work. In Q11(d), some candidates gave correct descriptions such as ‘sequence oscillates between 1 and 2’ or ‘the sequence is periodic with period 2’. Other descriptions such as ‘alternates between 1 and 2’, ‘keep getting 1 and 2’ or ‘fluctuates between 1 and 2’ were condoned. Some credit was given for partial explanations such as ‘sequence is non-convergent’, ‘sequence oscillates’ or ‘sequence is divergent’.
The curve C has equation y = x^x for x > 0. The point P(alpha,2) lies on C.
Edexcel 9MA0 · June 2018 · Paper 1 · Q10 · 8 marks
8 marks
![Edexcel 9MA0 · June 2018 · Paper 1 · Q10 · 8 marks — 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.](/api/paper-crops/9ma0/2018-06/paper-1/q10/question-r1-d3f41347.webp)
Examiner report
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.
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.
Edexcel 9MA0 · June 2018 · Paper 1 · Q12 · 10 marks
10 marks
Examiner report
This question, based upon the formula , t V ap modelled the value of a car in £'s against the time in years after 1st January 2001. Part (a) was very well answered by the majority of students. Most chose the correct values of t and V and successfully solved a pair of simultaneous equations. Common errors involved using incorrect values for t or failing to show that A was approximately 24 800. It is important in answering questions such as (b) that students are precise about the language and words that they choose. For example, for (b)(i) the statement was required to reference ''the car'', ''its value'' and the ''initial time''. In part (b)(ii), ''the amount the value of the car was increasing by'' was common and scored 0 marks. For this part, a clear answer would be ''it is the rate at which the cars value is increasing each year'' or ''the cars value is increasing by 6.6% a year''. In part (c), students generally scored the first 3 marks by correctly using logarithms and proceeding to t =21.8 or 21.9. Most students then thought that the year must be 2023 instead of 2022 as they had rounded 21.8 to 22 and added it to 2001.
The value V pounds of a vintage car t years after it was first valued on 1 January 2001 is modelled by V=Ap^t, where A and p are constants. The car was worth 32000 pounds on 1 January 2005 and 50000 pounds on 1 January 2012.
Edexcel 9MA0 · June 2018 · Paper 2 · Q14 · 10 marks
10 marks
![Edexcel 9MA0 · June 2018 · Paper 2 · Q14 · 10 marks — A population of mice is modelled by N = 900/[3 + 7e^(-0.25t)], where N is the number of mice t months after the start of the study and t >= 0.](/api/paper-crops/9ma0/2018-06/paper-2/q14/question-r1-1803160d.webp)
Examiner report
This was the most demanding question on the paper. Parts (a) and (d) were accessible to many students. Most students made some creditable progress in part (b), but many students found it difficult to access the demanding material in part (c). In part (a), most students evaluated or deduced that there were 90 mice at the start of the study. In part (b), most students used either the chain rule or the quotient rule in their attempt to find an expression for d . d N t There were many incorrect attempts at the chain rule such as 0.25 0.25 2 d 6300e (3 7e ) d t t N t or 0.25 2 d 1575(3 7e ) d t N t while the most common incorrect attempt at the quotient rule was 0.25 0.25 0.25 2 d (3 7e )(1) 1575e . d (3 7e ) t t t N t Although some impressive algebra was seen in manipulating a correct 0.25 0.25 2 d 1575e d (3 7e ) t t N t to achieve d (300 ) d 1200 N N N t , many students were unsuccessful in their attempts at eliminating t or the 0.25 e t term. Some students substituted 0.25 900 3 7e t N into d (300 ) d 1200 N N N t and used algebra to give 0.25 0.25 2 d 1575e d (3 7e ) t t N t , and provided some indication that their proof was complete. Some students differentiated N by using implicit differentiation while other students attempted to rearrange the given equation for N to make t the subject before finding an expression for d . d t N A few students were successful in applying an alternative method of solving the differential equation d (300 ) d 1200 N N N t to give 0.25 900 . 3 7e t N In part (c), most students misinterpreted the question and instead applied a method to find the time T for the maximum value of N, rather than the time T at which the rate of growth d d N t was a maximum. Many students, therefore, used 300 N and manipulated 0.25 900 300 3 7e T to give 0.25 e 0. T A few students differentiated d , d N t with most writing 2 2 d , d N t which was condoned, and solved the equation 1 (300 2 ) 1200 N to give 150. N It was rare to see students using the symmetry of the quadratic function (300 ) 1200 N N to deduce that 150. N Although some students gave their answer to part (c) as 150, T most students who progressed this far solved the equation 0.25 900 150 3 7e T to give the correct value for T. In part (d), many students deduced 300 P with only a few students deducing 299, P which was also accepted.
A population of mice is modelled by N = 900/[3 + 7e^(-0.25t)], where N is the number of mice t months after the start of the study and t >= 0.
Edexcel 9MA0 · sample · Paper 1 · Q6 · 7 marks
7 marks
A company plans to extract oil from an oil field. The daily volume V, in barrels, depends on the time t years after drilling starts. The model assumes V=16000 at t=0, V=9000 at t=4, and that V decreases over time. Model A is the straight line through (0,16000) and (4,9000). Model B is a decreasing exponential curve through the same two points.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 1 · Q12 · 9 marks
9 marks
In a controlled experiment, N is the number of microbes in a culture T days after counting starts. It is expected that N=aT^b, where a and b are constants. A plot of log_10(N) against log_10(T) has crosses approximately at (0,1.9), (0.28,2.4), (0.60,3.2), (0.70,3.35), (1.00,4.15) and (1.20,4.50). Its straight line of best fit has vertical intercept approximately 1.8 and gradient approximately 2.3; the displayed data cover 0<=log_10(T)<=1.2 and approximately 1.8<=log_10(N)<=4.5.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q4 · 5 marks
5 marks
The functions f and g are defined by f(x) = e^x for x real, and g(x) = 3ln(x) for x > 0.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q5 · 4 marks
4 marks
The mass m grams of a radioactive substance, t years after observation begins, is modelled by m = 25e^(-0.05t).
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q16 · 12 marks
12 marks
A meerkat population is modelled by dP/dt = (1/22)P(11 - 2P), for t >= 0 and 0 < P < 5.5, where P is the population in thousands and t is the time in years since the study began. Initially there were 1000 meerkats, so P = 1 when t = 0.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q5 · 10 marks
10 marks
Let f(x) = x^3 + ax^2 - ax + 48, where a is a constant. It is given that f(-6) = 0.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q3 · 4 marks
4 marks
A cup of hot tea is placed on a table. At time t minutes, its temperature theta degrees C is modelled by theta = 25 + Ae^(-0.03t), where A is a constant. The temperature is 75 degrees C when the cup is placed on the table.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q7 · 12 marks
12 marks
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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Everything here stays free. If exponentials and logarithms is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.