Edexcel A-level Maths · Topic 5
Practice questions covering trigonometry 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 Trigonometry, each with its mark scheme and worked method. Reveal the answer when you have had a go.
1.
(3)
(Total for Question 1 is 3 marks)
2.
(3)
(Total for Question 2 is 3 marks)
3.
(3)
(Total for Question 3 is 3 marks)
4.
(4)
(Total for Question 4 is 4 marks)
5.
(4)
(Total for Question 5 is 4 marks)
6.
(4)
(Total for Question 6 is 4 marks)
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Printable question pack — 5 TrigonometryAll 96 questions across 9 specification points — 90 more than the 6 above — with the full mark scheme. Guest accessible · browser print.54 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 · Q11 · 4 marks
4 marks
Examiner report
This question involved a badge in the shape of a semicircle, less the area within a circular arc of the same radius as the semicircle, centred at one end of its diameter. Many candidates did not spot that triangle AOB was in fact an equilateral triangle which would have immediately provided the angle 3 . It was not uncommon for candidates to fail to find an appropriate angle, and these were unable to score any marks in this question. A significant proportion of the candidates found the angle 3 but it often came from considerable work on the triangle, included use of Pythagoras and trigonometry, instead of the intended, more straightforward, deduction from AOB being equilateral. Many candidates that found an appropriate angle only progressed as far as finding the area of a sector, the area of triangle AOB or the area of the semicircle, for which no credit was available unless further progress was made. Some candidates used Pythagoras to calculate the area of the right-angled triangle and occasionally missed dividing by 2. Very often candidates were able to find a correct expression for one of the segments in the circular arc for the second mark, which was also available for finding the total area of the unshaded region AOB, the latter scoring the mark much less often. To get this far, candidates needed to identify and find the area of the equilateral triangle within the cut- out section defined by the circular arc. The third mark out of the available 4 marks was much harder to obtain as this required an attempt at the required area by correctly combining their correct attempts at the required smaller areas that made up the whole shaded region. Several expressions were valid here, all of which had to arrive at the sum of a sector and the triangle in exact form for the final accuracy mark. The most common route involved subtracting their expression for the area of a segment, and since the area of the segment involved subtracting the area of the equilateral triangle, this calculation led to sign errors in final answers, losing the final accuracy mark. Candidates resorted to decimals very infrequently, although calculators could well have been used to obtain exact forms of the areas. Communication of method was, at times, very clear, but often this clarity was lacking and in combination with poor algebra/manipulation and layout of solutions this could have cost some candidates method marks. It was quite surprising how many candidates attempted this question by an attempt at integration. Of these attempts, the most common involved use of polar coordinates, presumably by further mathematics candidates. This was a demanding route, and while there were many attempts that found appropriate polar equations for the relevant circles, the limits chosen were often incorrect, while others faltered with the integration itself. As a result, many candidates attempting this method either failed to set up the problem correctly or lost their way in their processing, and very few gave a completely correct answer. Other attempts at integration, by writing e.g. 2 25 y x = − , required use of the substitution sin x = and, again, issues with selecting appropriate limits or proceeding with the following integration caused these candidates problems. Centres should encourage candidates to explore other routes to solving problems such as this but must discuss which routes are likely to be the most concise.
A badge consists of a semicircle ABCOA with centre O and diameter AC=10 cm, with A and C at the ends of the horizontal diameter and B on the semicircular arc. OB is also an arc of a circle with centre A and radius 5 cm, joining O to B. Region R is bounded by arc OB, outer semicircle arc BC and radius OC.
Edexcel 9MA0 · June 2024 · Paper 1 · Q12 · 11 marks
11 marks
Examiner report
This harmonic form question was seen to be relatively challenging, despite similar questions being seen on several previous exam papers. That said, many candidates were able to attempt all parts of the question, and many follow through marks were available to those that made errors early on in their attempt. It was pleasing that most candidates identified that this question used degrees rather than radians. It was also pleasing to see many candidates continue to work through all parts of this modelling question despite having errors or not attempting previous parts. Part (a) was, as usual, answered well, with most candidates being able to find both K and correctly. Of those who failed to score all 3 marks for this part of the question, common mistakes included using radians instead of degrees and calculating 140 arctan 480 rather than 480 arctan 140 . There were very few responses that used trigonometric functions other than tan. Responses to part (b)(i) suggested that candidates are not as confident at using the harmonic form as they are at forming it. Far too many candidates ignored their work from part (a) even if it was correct. Although given the maximum value for R, it was not uncommon to see candidates failing to use their harmonic form from part (a), but instead simultaneously setting 1500 R = , ( ) cos 30 1 t = − and ( ) sin 30 0 t = to obtain an incorrect value for A. Most of those who correctly identified that 1000 A = did then proceed to write down the full equation of the model, although many left this part with just a value for A. In part (b)(ii), some candidates struggled to calculate the value of min R as they did not recognise the link between their min R and the values of K and A. There were a few who added their values of A and K and hence did not score this B mark. Those who had found the correct value, 1000 A = , generally went on to identify the minimum value of R, but the follow through allowed those who understood the required method to gain this mark with an incorrect value for A. Most candidates were able to gain some credit for part (c) as the method mark was quite generous regarding the candidates’ interpretation of the phrasing “middle of April” in terms of the model. Approaches using a value of t condoned values in the range 3 ⩽ 𝑡 ⩽ 4.5 to be substituted into their model for R. The most common incorrect values used were 4 t = or 4.5 t = and were condoned for the method mark. However, for full marks in part (c) candidates were required to translate ‘mid-April’ into 3.5 t = and substitute this into their model. A comparison with 500 R = could then be made, and the model was seen to be reliable. Another approach requiring more work was to set 500 R = and solve for t , via ( ) cos 30 73.74 1 t + = −. This value of t could then be used to identify the month in which the model predicted R would be minimum. A common mistake seen in this approach was to incorrectly identify 3.54 t = as being ‘mid-March’ which led to mixed responses of the model being both reliable and unreliable for different candidates. Part (d) was accessible to most candidates and fully correct answers were not uncommon. The simplest and most successful approach was to set ( ) sin 30 70 1 t + = − and solve for t, then substitute this into their model for R. Some candidates incorrectly set ( ) sin 30 70 1 t + = or 0, leading to the equation 30 70 90 t + = or 180. To score full marks for this part of the question, candidates needed to find a value of t from 30 70 270 t + = , use it with sufficient accuracy when substituting into their expression for R and round their final answer to an integer. Some candidates obtained a correct decimal answer but then rounded to 3 significant figures, giving 1030 as their final answer rather than the required 1032 or 1033. Many candidates incorrectly found a value of 16 3 t = − (coming from 30 70 90 t + = − ) which should have been an indication to restart this part having recognised that time cannot be negative. A significant proportion of candidates used 90 on the right-hand side, or even 0 or 180, which were not acceptable, and these failed to score any marks. There was also a minority of candidates who thought they were being asked to find the value of F. Another minority of candidates incorrectly assumed the question was asking when the populations of foxes and rabbits were the same and wrote down equations which they quickly concluded they could not solve.
Angles in this question are in degrees. A scientist studies rabbit and fox populations in a wood for one year. Rabbits are modelled by R = A + 140cos(30t) - 480sin(30t), where t is months after the start of the year and A is constant. Foxes are modelled by F = 100 + 70sin(30t + 70).
Edexcel 9MA0 · June 2024 · Paper 1 · Q13 · 8 marks
8 marks
Examiner report
This question was very successful in distinguishing the most able mathematicians from the cohort. A significant proportion of candidates struggled, with many responses either left blank or only partially completed. The nature of a show that question ensured marks were available on both parts regardless of success in part (a) but many candidates who failed to prove the result in part (a) then went on to omit part (b) and missed out on this opportunity. However, it was not uncommon to see full marks in part (a) followed by no marks in part (b), and vice versa. Most candidates who attempted part (a) recognised the requirement to find d d x and attempted to do so, with varying levels of success. Those who differentiated successfully generally went on to substitute into the expression correctly and gained at least the first 2 marks. The flexibility in the first method mark made this accessible to most candidates who had tried at differentiation. d d 2 sin cos x a = appeared several times and benefited from the flexibility of the first method mark. The most common incorrect derivatives were 2 cos a and 2 cos a . In many cases the notation was particularly poor, although d d x a was very generously condoned provided the right-hand side was correct. Candidates struggled with the rearrangement of the integrand into the form required to successfully use the double angle formula. Many were not able to work with or simplify the square root terms successfully, and even those that did sometimes failed to recognise the need to use the double angle formula for sin 2. A minority of candidates progressed correctly to the final answer, while some of those that achieved the final expression still lost the final mark for failing to address the limits, or for missing the d or the integral sign. For those who did not achieve the first method mark this was commonly due to careless copying or premature simplification such as 2 sin a a − instead of 2 sin a a − . Those who recognised the requirement for using double angle formulae to integrate a squared trigonometric function were generally successful in part (b) of this question. Occasionally candidates struggled to deal with 2 correctly, with 2 1 cos2 sin 2 2 − = seen all too frequently. These candidates scored no marks following incorrect use of the double angle identity. The other main errors in this part were in sign errors in the double angle identity and in the integration of the trigonometric term: commonly cos4 − was integrated to 1 sin 4 4 + , and despite this still leading to the correct answer it was at the expense of the two accuracy marks. A significant proportion of candidates incorrectly attempted to increase the power and divide leading to a term in 3 sin . Some unsuccessful attempts at restarting and integrating ( ) 1 1 2 2 x a x − by parts were also seen. There was also a significant proportion of candidates who attempted integration by parts which generally gained no credit. Candidates who were successful in using integration by parts often had a succinct approach, while many reached 2 cos 2 and at this point realised the use of double angle identity would help, where it would have been more concise to use the double angle identities in the first place.
Let a be a positive constant and consider the definite integral from 0 to a of sqrt(x)sqrt(a-x) dx.
Edexcel 9MA0 · June 2024 · Paper 2 · Q5 · 3 marks
3 marks
![Edexcel 9MA0 · June 2024 · Paper 2 · Q5 · 3 marks — Given that theta is small and in radians, use the small angle approximations to find an approximate numerical value of [theta tan(2theta)] / [1 - cos(3theta)].](/api/paper-crops/9ma0/2024-06/paper-2/q05/question-r1-aaf8aa81.webp)
Examiner report
This question was very accessible with many candidates scoring full marks. The most common approach was to directly apply the small angle approximations to the given expression and on the whole, this was successfully simplified to give the correct value. Using the small angle approximation 2 was most successful for the numerator, however there were a few candidates who were unable to process the multiplication. The denominator proved to be trickier to process with errors seen in the bracketing, squaring the angle, using 3cos rather thancos3or omitting the “1−”. It was very common to see errors such as 2 3 cos3 1 2 = − and 2 2 3 3 1 cos3 1 1 . 2 2 − = −− = − Once the numerator and denominator had been simplified, there was a sizeable minority who struggled with cancelling the fraction. Incorrect answers of 4 3 and 4 3 − were seen frequently, resulting from the poor use of brackets. A significant number of candidates complicated the process by introducing trigonometric identities before applying small angle approximations, presumably not realising they could substitute the appropriate approximations directly. There were many variations in the processing such as double angle formulae and compound angle formulae which all involved a lot more work and inevitably led to complicated expressions with higher order indices in the denominator. Unfortunately, only a few candidates were successful with such approaches as there were often errors in the algebra needed for the simplification or a lack of recognition that they needed to make a limiting argument.
Given that theta is small and in radians, use the small angle approximations to find an approximate numerical value of [theta tan(2theta)] / [1 - cos(3theta)].
Edexcel 9MA0 · June 2024 · Paper 2 · Q8 · 7 marks
7 marks
Examiner report
A significant number of candidates found this question challenging. In part (a) the first challenge candidates faced was combining the two terms to produce a single fraction. This was completed successfully by many candidates, either as a fraction in cosec or a fraction in sin. The second challenge was using trigonometric identities correctly. The majority used 2 2 sin cos 1 + = but the more successful candidates typically used 2 2 1 cot cosec + = . It was only a minority who managed to complete the proof successfully. It was possible to complete the proof starting either from either side of the identity, however, generally those candidates who started with the left-hand side were more successful. Few, if any, candidates gave a completely correct solution starting from the right-hand side and the majority only obtained the mark for using a correct Pythagorean identity. A small number of candidates used a “meet in the middle” approach with varying degrees of success. Notational errors were not widespread, but variables were sometimes missing, or indices misplaced resulting in the loss of the final mark. Many also lost the accuracy mark because they did not show all the necessary steps by proceeding to the given answer using sine and cosine. In part (b), candidates that spotted the connection with part (a) tended to achieve at least 3 out of the 4 marks available. Those who failed to get the first line right with the variable of 2x on both sides but just substituted 2tan sec x x or even 2tan sec on the left-hand side gained no marks in this part. Candidates took a number of different approaches to solving the correct equation. The most frequent was to obtain 2 1 tan 2 2 x = but approaches involving 2 3sin 2 1 x = or 2 3cos 2 2 x = were relatively common. There were a few candidates who used the identity 2 2tan tan 2 1 tan x x x = − to reach a quadratic in 2 tan .x These attempts were usually successful. Some candidates stated that the square root of 1 2 is 1 4 which lost them the final two marks in this part. However, those candidates that did get to one of the correct equations could normally proceed to one correct answer (which was typically x = 17.6°). It was only a minority of candidates who gave both correct answers. The final mark was regularly lost because only one solution was identified, or candidates had incorrectly found 45 degrees by incorrectly solving sec2x = 0 and interpreting this as cos2x = 0 rather than being undefined. Most candidates that achieved marks in this part of the questions could correctly identify the order of operations to solve their equations although there were some candidates who struggled with this. For the final mark, a few wrote down 17.7 instead of 17.6 and some had extra answers in the given range and therefore lost the accuracy mark. A small minority of candidates did not show appropriate working in part (b), sometimes merely writing down the two correct answers to the given equation. The question explicitly asked for all stages of a candidate’s working as the answers can be found using a calculator. Hence candidates that gave the correct answers without appropriate working were not awarded any marks in this part of the question.
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · June 2023 · Paper 1 · Q4 · 5 marks
5 marks
Examiner report
This question testing the small angle approximation for cosine and finding the equation of a straight line was accessible to many candidates with a large number scoring full marks. In part (a), some candidates formed and solved the equation but were not able to use the information that x was small to choose the correct solution. There was a very small minority of candidates who did not quote the correct small angle approximation, despite this being in the formula book. A slightly larger minority misunderstood that x being small meant that x was 0 and therefore did not include the 2x term in their solution attempt. Candidates should also pay closer attention to the question as several did not round to the required decimal places. It was also notable that some candidates at this level found it difficult to expand 2 1 1 2 2 x − when multiplying the fractions together. Part (b) was mostly well answered with the majority scoring both marks, with most candidates gaining these marks efficiently without unnecessary calculations. Most candidates found the gradient of the tangent by substituting 0 x = into the given derivative of f( ) x , however a small number of candidates took a longer approach by attempting to use the small angle approximation. Some candidates tried to substitute their answer from part (a) into the given derivative or found the gradient of the normal having originally found the correct gradient for tangent; these approaches lost both marks. A small number of candidates failed to realise the significance of the point (0,3) and therefore retained an algebraic gradient which they tried to substitute into y mx c = + . A number of responses were seen not giving their answer as an equation, just an expression 1 3 2 x + , or forgetting to substitute their value of c found, giving their equation as 1 2 y x c = + ; having been given the y intercept in the question, it was a requirement that for any marks to be scored that 3 c = .
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The curve C has equation y=f(x), x in R. Given that f'(x)=2x+(1/2)cos x, the curve has a stationary point with x-coordinate alpha, and alpha is small.
Edexcel 9MA0 · June 2023 · Paper 1 · Q8 · 10 marks
10 marks
Examiner report
Generally most candidates found this question accessible, working confidently in radians throughout, although inadvisable conversions to degrees were not uncommon with many candidates unaccountably preferring the unwieldy 2 360 r to the elegant 2 1 2 r . In part (a) a rigorous solution was expected; a very large number of candidates lost the second mark by failing to explicitly show the process of dividing the arc length by the angle (perhaps the mark most commonly lost on this question). Candidates need to appreciate that every step of working needs to be clearly demonstrated in order to gain full marks for “show that” questions and that their final result needs to be what the question asked them to show in the first place. As the question was worth 2 marks, candidates should not expect to just state 2.3 27.6 r = and for this to be sufficient. Those who chose to convert 2.3 radians to degrees not only lost time but also invariably lost the accuracy mark. Others did show the division and did not label their answer so just an expression was seen. Part (b) was typically answered well, with many candidates recognising the straight line giving a total of radians and allowing them to solve the problem. Where candidates lost marks, this was often due to a lack of brackets, e.g. − 2.3 ÷ 2 = 0.421, or an incorrect joined statement such as 0.842... 2.3 0.421 2 − = = . A small minority worked in degrees which needed much more work to secure both marks. There was a wide range of marks for part (c), although many candidates produced highly competent solutions and gained full marks. Most candidates seemed familiar with the area of a sector formula and the vast majority scored both marks for this. Candidates also seemed fully aware of the area of a triangle formula, although they did not always use a correct method to work out the length 15.7. Some candidates seemed not to realise that they needed to use the information provided, i.e. the length of the front of the stage, in order to obtain all the lengths needed for their area calculations. There were also some more complex and longer methods to find the area of a triangle which were not generally successful, but the most common error was using 12 as opposed to 15.7. Some candidates spent time working out the length of AB but then realised that this was not necessary.
A stage consists of a sector OCDEO of a circle with centre O, joined to two congruent triangles ABO and GFO. Points B, C and O are collinear with C between B and O; points O, E and F are collinear with E between O and F. Points A, O and G lie on a straight back edge AOG of length 15 m, with OA=OG. The sector angle COE is 2.3 radians and arc CDE has length 27.6 m. The two triangle angles AOB and GOF are equal. The front boundary is the path BCDEF.
Edexcel 9MA0 · June 2023 · Paper 1 · Q13 · 7 marks
7 marks
Examiner report
A significant number of candidates found this question very challenging, with a fair proportion of blank answers, and another significant number working with various substitutions in (a) that were quickly abandoned. Those who were able to work with both the quadratic and trigonometric models confidently were able to score highly and demonstrated a good understanding of how mathematics can be used to model real life contextual problems. In part (a), there were two common approaches that led to successfully scoring full marks. One was applying the knowledge that 2 ( 20) 0 b t − = when 60 H = which therefore meant 60 a = . This then enabled the candidate to substitute 2, 0 H t = = and find a complete equation. It was clear when candidates were able to apply this knowledge, and knowing this gave them a huge advantage in this question. Noticing that b was positive was pivotal when taking this approach. The other approach that led to successful answers involved multiplying out the brackets, and differentiating with respect tot. This enabled the candidate to find that 20 t = when d 0 d H t = , thus overcoming a shortfall in knowledge over when geometrically the maximum occurs in a negative quadratic. This second approach led to success less often, and candidates often ended up with a page of various quadratic expressions and equations that looked costly in terms of time and did not yield marks. An overwhelming majority of candidates who scored full marks in (a) successfully found H to be 2 m in part (b). Some candidates still managed to find this value with either an incorrect answer in (a) or no answer at all, by perhaps noticing that due to the symmetry of the quadratic, the height after 40 seconds was the same as at the start, which they are told was 2 m in the question. In part (c), a significant number of candidates had no real idea how to start this. Many attempts involved fruitlessly expanding cos(9 ) t + but most candidates who successfully found and did not do this. The candidates who realised that when 60 H = , cos(9 ) 1 t + = which meant 31 = had the most success. Candidates who did this typically then substituted 20 t = into cos(9 ) 1 t + = to find a value of . Candidates who found one value typically found the other. There were a number of candidates who found the two values that did not ‘find a complete equation’ as the question required. However, there were a pleasing number of fully correct solutions, and almost all candidates who found and continued to give the full model. In part (d), the most popular correct answers involved an identification of the cyclic nature of the trigonometric function, or realising the first model would become very negative. A significant number of incorrect answers involved candidates saying the original model would keep increasing over time, which showed a lack of understanding of the negative quadratic model. Others made reference to the alternative model being continuous which was too vague. Some candidates managed to score this mark without scoring any other mark on this question which was, again, pleasing to see candidates attempting later parts of questions even if they have made little progress on earlier parts.
On a roller coaster ride, passengers travel in carriages around a track and complete multiple circuits. The maximum vertical height of a carriage above the horizontal ground is 60 m, and a carriage starts a circuit at a vertical height of 2 m. The height H metres, t seconds after the carriage starts the first circuit, is modelled by H=a-b(t-20)^2, where a and b are positive constants.
Edexcel 9MA0 · June 2023 · Paper 2 · Q8 · 6 marks
6 marks
Examiner report
Part (a) of this question was generally answered very well with candidates regularly scoring full marks. Candidates seemed to be well practiced at converting to harmonic form and working in radians. The most successful candidates wrote out the full expansion of ( ) cos R − rather than trying to take a short cut. The common issues included giving a decimal solution for R or rounding the value of α to a lower degree of accuracy than was required in the question. There were a small number of candidates who worked with 2 tan 8 = and so lost the method mark and the accuracy mark. Candidates found part (b) more challenging, with many failing to link areas of mathematics, in this case trigonometry and series. Even where a candidate successfully found the sum, 9cos 36sin S x x = + , many did not relate it to part (a). Those who did recognise the form were usually able to get 4.5 " " R , usually 4.5 2 17 for part (i). Some seemed to misread the question and stated the minimum value, so lost the accuracy mark. A significant number of candidates found a correct expression in harmonic form but lost both the M1 and A1 as they kept a correct 9 17 embedded in their expression. Candidates were slightly more successful in part (ii). Most candidates who stated a correct value for the maximum, also got the value of x correct, more often than not giving it to the same accuracy as in part (a). A few benefitted from this being a follow through mark. Some candidates tried to find a subsequent value for and were often out by or 2, or attempted to rearrange 1.326 x − = . A few responses were seen where S was differentiated and set to zero to find a maximum. These candidates often lost the accuracy mark as they found an answer of 37.1. A lot of incorrect responses gave a maximum of 36 when sin 1 x = or 45 from finding 36 + 9.
The first three terms of an arithmetic sequence are cos x, cos x + sin x, and cos x + 2 sin x, where x is not an integer multiple of pi. Let S_9 be the sum of the first 9 terms as x varies.
Edexcel 9MA0 · June 2023 · Paper 2 · Q14 · 7 marks
7 marks
Examiner report
Part (a) was at least started by many but when students failed to make any progress towards the result, they tended not to attempt part (b). In part (a) the vast majority of candidates scored the first mark for using a correct trigonometric identity. The M and A marks were often both given or both lost, as candidates explored multiple manipulative steps to reach the final form for the equation. In many cases this involved expanding the brackets, using a substitution for sin 2θ, but later putting back into sin 2θ form i.e., undoing the manipulation that had been introduced. Whilst in many cases this resulted in the candidate getting there in the end, there were many examples of multiple attempts or unnecessarily long solutions. The most efficient methods did not involve multiplying out the bracket on the LHS which seemed to be the default first step for most students. A common mistake was to suddenly put the RHS equal to zero so it disappeared rather than being subtracted on the LHS. For those that were successful, there were a variety of approaches as well as variety in the number of steps required to reach the correct final form, with some methods considerably more efficient than others. Arithmetic errors were commonly seen in this question. Many successful candidates attempted to find a factor of sin 2θ on both sides of the equation before proceeding to collecting these on the same side. Another typical successful method involved writing all the terms in terms of sin θ and cos θ, collecting all terms on the same side, and then taking a factor of 2sin θ cos θ. However, many candidates who expanded tended to forget terms when expanding or factorising, leading to an expression in an incorrect format. Replacing tan θ and sin 2θ was usually handled well, as was replacing 1 + tan2 θ with sec2 θ and then 2 1 cos but many candidates didn’t follow the manipulation to the end and were unsure what to do with the sin 2θ on the right hand side, so either left it on the right hand side and lost both the M and A marks, or moved it to the left but then incorrectly dealt with it when combined with the quadratic formed on that side. Part (b) was generally done well by the candidates who managed to obtain a quadratic in part (a), with many candidates correctly solving to get 360 or 540, and then solving their quadratic. Those who found a solution to their quadratic also tended to find solutions based on that within the specified set of values. Where marks were lost, it was often by incorrectly including extra values,in particular the value of 450 degrees which had been specifically excluded in the question.
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · June 2023 · Paper 2 · Q15 · 3 marks
3 marks
Examiner report
A number of candidates lost all marks on this question by not attempting any solution. Most candidates correctly expanded the given expression and used sin2 x + cos2 x = 1 to simplify and score the first two marks. A small number incorrectly used (sin x − cos x)2 = sin2 x − cos2 x, scoring no marks. Others attempted to substitute values of 90 and 180 into the given expression instead of using algebraic manipulation. Whilst most candidates scored the first two marks, only the very best were able to score the final mark, with most candidates failing to offer a convincing reason for a contradiction. Candidates who drew a graph of sin x and cos x or sin 2x to support their understanding tended to score well. Successful candidates typically explained that 2x must be between 180 and 360, and hence sin 2x must be negative, which is a contradiction with their inequality. Those that worked with sin x cos x were generally more successful in achieving this mark than those who chose to use sin2x. However, it was clear that not all students understood what an obtuse angle was. Another typical correct conclusion involved candidates stating if x is obtuse, sin x must be positive and cos x must be negative, leading to a negative product and a contradiction. Of those who were able to offer a reason for the contradiction, some still lost the final A mark for failing to then reach a conclusion that the original statement was true or because of errors in the proof such as missing x’s or mixed variables.
A student attempts this question: Given that x is an obtuse angle, use algebra to prove by contradiction that sin x - cos x >= 1. The proof begins: Assume that sin x - cos x < 1 when x is obtuse. Therefore (sin x - cos x)^2 < 1.
Edexcel 9MA0 · June 2022 · Paper 1 · Q9 · 6 marks
6 marks
Examiner report
(Mean mark 1.9 out of 6) This vector question was set within the context of a rhombus. Surprisingly, many candidates were unable to use (or remember) the basic geometric of a rhombus and over 30% did not manage to make any progress. Part (a) required candidates to show that parallelogram PQRS was a rhombus. Although the requirement to 'show' is becoming more familiar to candidates, a significant number in this case didn't recognise the level of detail expected. So a lack of conclusion, or stating the sides were of length Ö29 without evidence of calculation, led to loss of credit. It was also necessary to conclude that the sides were equal, and not simply find the lengths and leave it to the reader to decide. A vast array of approaches were in evidence in part (b) to find the area of the rhombus. The easiest and most direct approach was to find 1 2 PR QS × . A more common method used involved finding the length of one diagonal, followed by the cosine rule to find an internal angle and then use of the formula 'ab sin C '. Unfortunately most attempts using this method only scored 3 marks as it was usual for answers to be given as 22.7 rather than the one required, Ö517. Other methods seen involved use of Heron’s formula, the “shoelace” determinant method, scalar product and vector product….. with many successfully used. Students who added a “narrative” – so explaining what they attempted to do – were in a better position to gain credit if they suffered a processing error early in their solution. This has to be considered good practice. The most common incorrect solution, seen many times over was in stating area = 29 29 29 PQ PS × = × =
PQRS is a parallelogram. The adjacent side vectors are vector PQ=2i+3j-4k and vector QR=5i-2k.
Edexcel 9MA0 · June 2022 · Paper 1 · Q14 · 8 marks
8 marks
Examiner report
(Mean mark 4.0 out of 8) This question was another big discriminator. Good candidates found it familiar and accessible with some 25% scoring full marks. Strangely, for candidates who dropped marks, they seemed to be able to do either part (a) or part (b), but not both. For those candidates who attempted part (a) most recognised the need to use the compound angle expansions. Carelessness with the multiplication by 2, signs and use of the formulas was seen in a small minority of the scripts. Students who attempted to use sin (A+B) and cos (A+B) with a negative angle B, often introduced sign errors; it is clear that such a method is not to be recommended. Substituting exact numerical values was usually correct, however candidates occasionally fudged their working to get the given answer. Part (b) could be attempted without having done part (a) and many took up this opportunity. Those who paid attention to the 'hence', deducing that x = 2θ + 60, were very successful in achieving the correct answers, though quite a few only found one solution. For the many students who did not see the link between parts (a) and (b), generally encountered much more complex, longer algebraic manipulation. Such students re-started the question using the compound angle formulae with a combination of other trigonometric identities, but were often not able to reach an equation in one function, or else made numerous slips. If quadratics in tan 𝜃𝜃 were reached, they were dealt with competently. If the candidate’s efforts led to tan2θ = k, k≠ 3√3 or tan (2θ± 60) = 3√3 they were generally able to use the correct order of operations to solve their equation to find at least one solution. A fair number of correct answers were seen written down without any correct working. As the question had a warning about the use of calculator technology, these responses gained no marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · June 2022 · Paper 1 · Q15 · 10 marks
10 marks
Examiner report
(Mean mark 5.6 out of 10) This question, based on the minimum surface area of a children's toy proved to be a useful source of marks so late in the paper. In fact 25% of the cohort achieved full marks. Many candidates who could not access part (a) were able to use their skills in differentiation to find an appropriate value for the radius and prove that their value gave a minimum surface area. Unsurprisingly, part (a) was the most demanding part of this question. The key was to find an expression (using the volume of 240 cm3) for h in terms of r. Unfortunately some solutions featured the volume of a sphere, which proved a costly error, from which there was no return. Candidates produced their formula for surface area in a variety of ways – sometimes leaving it to the last line to substitute in the value for the angle, occasionally replacing 0.8 by 4/5. As with other “show” questions, sufficient detail has to be written to convince the reader that progress is genuine. Part (b) well generally well done. Many candidates scored full marks for reaching either 3 1050 or 10.2. Occasionally some candidates mistakenly wrote down 1050 and subsequently simplified their answer to 5 42 . Very few candidates who attempted this part failed to score marks. The main reason for this was usually an inability to differentiate 1680 r which usually became1680 1 or 1680ln r Most candidates were able to find the second derivative correctly in part (c). How it should be used was confused in much of the work seen. Common reasons for a loss of marks here were; • candidates solving the equation 2 2 d d S r = 0 • use of incorrect notation including 2 2 d d S r and 2 2 d d y x • incomplete solutions with calculations or conclusions missing
A solid toy is a right prism shaped like a sector of a circle. Its congruent end faces ABC and DEF are sectors of radius r cm and angle BAC=0.8 radians, with centre A on the first face. Corresponding edges AD, CF and BE are perpendicular to the sector faces and each has length h cm. Thus the three lateral faces have widths r, r and the arc length 0.8r. The volume of the toy is 240 cm^3.
Edexcel 9MA0 · June 2022 · Paper 2 · Q9 · 5 marks
5 marks
Examiner report
This was a challenging question for the majority of the candidates with very few scoring all four of the marks in part (a). For those that attempted the question, most scored at most 1 or 2 marks. These were the initial B1 mark for realising that 𝐴𝐴= 50, however many wrote that 𝐻𝐻= 50 and did not relate that 𝐻𝐻 meant 𝐴𝐴. The second B mark for finding a value for 𝑏𝑏 was very poorly answered, with many not realising a simple calculation of 180 720 was needed and attempted to use an ‘𝑅𝑅−𝛼𝛼’ method or failed to take into account transformations of the sine function, showing a lack of understanding of the properties of the sine wave. Some also tried to use d𝐻𝐻 d𝑡𝑡 with very limited success. A small minority of students achieved b = 1/4. The best answers tended to use transformation arguments, recognising that the full cycle depicted must represent a scale factor of 4 though many students attempting this approach did not appreciate the effect of the modulus in determining the value of b, obtaining instead a value of b = ½. A common incorrect method came from consideration of the wheel taking 720 seconds to complete 1 revolution leading candidates to solve the equation 720b + 1.15 = 180 or from thinking that the maximum occurs at t = 360 leading to 360b + 1.15 = 90. Of those that managed to access the question, few mixed radians with degrees and managed to give an answer as 𝜋𝜋 720 or 2𝜋𝜋 720, which scored no marks. 1010 was the most common mark profile for this question, as a significant number of students went on to find 𝛼𝛼 correctly (albeit occasionally in radians) as they were able to spot that at 𝑡𝑡= 0, 𝐻𝐻= 1. Sadly, a few that did achieve all three elements did not conclude with an equation of the model and so lost the (easy) last mark, with this final mark being scored very rarely, at less than 5% of the cohort (the second least commonly scored mark in the paper). In part (b), many candidates incorrectly referred to the fact that the passenger was 1 m above the ground and related the value of 𝑑𝑑 to this fact, for the most part because they had not appreciated that the original model already took this into account. Some responses referred to the fact that the height needed to be a positive value or could not be negative, again missing that the modulus signs in the original model accounted for this. Some responses did not refer to the context of the model in their explanation. Overall, around a fifth of candidates understood that the passenger would touch the ground without adding the constant ‘𝑑𝑑.’ There were also a few illogical answers referring to air resistance, etc. A few students also incorrectly though that ‘+ d’ was in reference to the different heights of the passengers. A small number of candidates seemed to think that the graph showed the path of the position of the passenger, referencing that the passenger “bounced off the ground” and that the ‘+ d’ had the effect of smoothing out the ride. When students were successful in this part the most common responses were that ‘+ d’ meant that: the Ferris wheel would not touch/scrape the ground, the passenger/seat cannot touch the ground or there needed to be a minimum gap between the passenger/wheel and the ground. There were many attempts at part (b) without any attempt at part (a). It was unfortunately quite common that responses to this part were very difficult to read and centres need to stress to candidates that work cannot be awarded marks if it is illegible.
The height H metres of a passenger on a Ferris wheel, t seconds after the wheel starts turning, is modelled by H = |A sin((bt + alpha) degrees)|, where A, b and alpha are constants. Over one revolution the graph is one positive arch, starting just above the t-axis and initially increasing. The maximum height is 50 m, the passenger is 1 m above the ground when the wheel starts turning, and one revolution takes 720 seconds.
Edexcel 9MA0 · June 2022 · Paper 2 · Q15 · 10 marks
10 marks
Examiner report
The majority of candidates found this question accessible and were able to score consistently throughout. Where marks were lost they were usually in parts (b) and (c). Part (a) was attempted successfully by most candidates, starting usually with 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 6𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. Candidates then successfully cross multiplied and used the appropriate identity to reach the required result. Another starting point regularly seen was 6𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡= 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቀ 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቁ 2 . Errors in the proof were rare if candidates started from a correct statement. Occasionally the “ = 0” was carelessly missing. In part (b) most candidates correctly solved the equation either by factorising or using calculator technology, but many reached the acute angle 𝜋𝜋 6 rather than the required obtuse angle of 5𝜋𝜋 6 . Others found a number of solutions but failed to identify 5𝜋𝜋 6 as the required answer. A few candidates worked in degrees and again lost the accuracy mark. Those who had obtained an angle of 𝜋𝜋 6 in (b) then usually carried this error into (c) and were in a position only to earn the method marks. Candidates using 𝜋𝜋 6 in (c) often would reach 9൫1 + √3൯, but did not seem to worry that this did not match the required answer format given in the question. There were many attempts, however, to use the correct formula for the sum to infinity with appropriate values for the first term and the common ratio. Some candidates unfortunately put effort into finding and rearranging (often incorrectly) algebraic expressions for the sum to infinity involving 𝜃𝜃, without substituting an angle in, gaining no credit, and others used 𝑎𝑎 1+𝑟𝑟 incorrectly as the formula for the sum to infinity. The most common mistake in (c) was failure to show the steps where the denominator was rationalised. Many candidates clearly typed an expression like −6√3 1−൬−√3 3 ൰ straight into their calculators to obtain 9 −9√3 losing the last two marks in the question.
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The first three terms of a geometric series are 12cos(theta), 5 + 2sin(theta), and 6tan(theta).
Edexcel 9MA0 · October 2021 · Paper 1 · Q10 · 8 marks
8 marks
Examiner report
This was found to be another discriminating question. Those who were familiar with the topic made good progress yet there were many other attempts that scored no marks at all. Part(a) There were a significant number of fully correct solutions for this part, although some candidates did not achieve the stage where both the numerator and denominators of their fraction were factorised correctly with a common factor of ( sinθ+cosθ ). Another difficulty encountered by some, was how to use the different forms of the identity for cos2θ in order to deal with, and remove, the presence of both “1’s ” in the fraction. Very few notational errors were seen, which was a pleasing improvement. Part (b) A significant number of candidates used the result of (a) correctly and started their solution with tan2x=3sin2x. However, it was also the case that tanx =3 sin2x was seen several times as well. Of those candidates who started correctly with tan2x = 3sin2x, the significant majority simply cancelled out the resulting factor of sin2x and in so doing neglected to consider the possibility that sin2x=0 and in so doing the solution x = 90° was omitted from the fully correct list of three answers. Many candidates however, did state that cos2x =⅓ and went on to get 35.3° and 144.7° and so only one mark was lost.
Show all stages of working; solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · October 2021 · Paper 2 · Q4 · 3 marks
3 marks
Examiner report
The first method mark in this question was often gained for sight of but the second mark proved more challenging, with errors seen in the correct application of the small-angle approximation. Some candidates missed out the square completely and just wrote . Many of those who did attempt to expand made mistakes with fractions and powers, with 2 2 3 1 4 θ θ − + occasionally seen. Candidates who realised they could use and then use the approximation fared better. In contrast, those using the double angle formula performed less well, with many fraction and sign errors seen with this approach. It was not uncommon to see candidates who followed the main method ‘multiply everything by 4’ at the end leading to an incorrect final expression of , forfeiting the final A mark.
Given that theta is small and measured in radians, use the small angle approximations to show that 4sin(theta/2) + 3cos^2(theta) is approximately a + btheta + ctheta^2, where a, b and c are integers to be found.
Edexcel 9MA0 · October 2021 · Paper 2 · Q6 · 5 marks
5 marks
Examiner report
Performance on this question was variable and approximately 25% of candidates failed to score any marks. (a) The majority of candidates were successful with this part. Common errors were stating that angle AOB was or , sometimes following through to gain marks in (b) and (c). (b) Most candidates followed the correct strategy for finding the area of the logo by finding the sum of the areas of sectors OCD, OAB and OEF. Common errors were omitting the brackets round the 2r when finding the area of the sector OCD and algebraic errors when simplifying their expression for the sum of the two smaller sectors OAB and OEF. A small number of candidates showed insufficient working when simplifying their area of the sum of the two smaller sectors to arrive at the given result. (c) Again, most candidates attempted this part well. A common errors was not to include 4r in the expression for the perimeter of the logo and there were some algebraic errors when simplifying the expression for the sum of the perimeters of sectors OAB and OEF.
The boundary OABCDEFO is a logo made from three sectors with common centre O. Points O, B and C are collinear, points O, E and D are collinear, and A, O and F form a straight line. OAB and OFE are congruent sectors of radius r. ODC is a sector of radius 2r, with angle COD = theta radians. Thus OA = OB = OE = OF = r, OC = OD = 2r, and the boundary consists of OA, arc AB, BC, arc CD, DE, arc EF and FO.
Edexcel 9MA0 · October 2021 · Paper 2 · Q9 · 3 marks
3 marks
Examiner report
This question was fairly well answered by a number of candidates, who understood the need to use the sum to infinity formula, but a sizeable majority made little or no progress, not realising for example, that the sum started from n = 2 and not n =1 – hence they often did not get the first mark for a correct first term and/or common ratio. 9/16 was seen fairly often, but not always attributed to the first term, with some candidates thinking this was the value of the sum. One candidate misread the infinity symbol as an 8 and applied the sum to 8 terms correctly, but was obviously not able to score any marks. A number of students found the sum to infinity, using n = 1 to infinity and then subtracting the first term. This was done well by candidates who went down this route. Unfortunately, many students achieved no marks on this part due to a wide variety of errors at the beginning. Calculating cos (360) on the calculator in radians instead of degrees was seen. Some tried to split the summation incorrectly into ∑ ቀ 3 4ቁ 𝑛𝑛 ∞ 2 × ∑ cos 180𝑛𝑛 ∞ 1 and find the sum to infinity of the trigonometric terms separately. A few wrote ( 3 4)2cos (360) but did not evaluate the cos(360). There were some very good solutions but in many responses there was often a lack of clear structure to their proof.
Show that the infinite series from n = 2 to infinity of (3/4)^n cos(180n degrees) has sum 9/28.
Edexcel 9MA0 · October 2021 · Paper 2 · Q15 · 11 marks
11 marks
Examiner report
This question discriminated well between candidates that got this far, but it did appear that many candidates had simply run out of time, perhaps indicative of a generally lower understanding than in a June series because of it being a resit paper following the pandemic. Part (a) was answered very accurately by the vast majority of candidates, recognising the need to use the compound angle formulae for cos(𝜃𝜃+ 𝛼𝛼). Two common errors occurred, namely tan 𝛼𝛼= − 1 2, following inaccurate equating of the coefficients of cos 𝜃𝜃 and sin 𝜃𝜃, and an approximate value for 𝑅𝑅, which usually arose from candidates using their value of 𝛼𝛼 to find 𝑅𝑅 from either sin 𝛼𝛼= 1 𝑅𝑅 or cos 𝛼𝛼= 2 𝑅𝑅. These candidates should be encouraged to find 𝑅𝑅 using Pythagoras. Part (b)(i) was usually well attempted by those who had answered part (a), however, some candidates did not recognise the need to double 𝑅𝑅 and arrived at 3 + 𝑅𝑅 rather than 3 + 2𝑅𝑅 instead. Others simply assumed that the maximum was 3, 5 or 7, from a poor understanding of combining two trigonometric functions and the purpose of using the compound angle formulae to combine these two functions. Part (b)(ii) proved to cause problems for many candidates, with the majority incorrectly assuming that 0.5𝑡𝑡+ 𝛼𝛼= 0, and proceeded to ignore that this equation results in a negative value of 𝑡𝑡 (usually −0.9) by simply making this value of 𝑡𝑡 positive. They had failed to recognise that the first maximum would occur when 0.5𝑡𝑡+ 𝛼𝛼= 2𝜋𝜋. Perhaps the use of a sketch or graphical calculators would have helped these candidates. An alternative approach to part (b) required candidates to differentiate 𝐻𝐻, either in its original form or having used part (a) and solve the equation d𝐻𝐻 d𝑡𝑡= 0. This approach was more complicated and resulted in needing to solve tan(0.5𝑡𝑡) = − 1 2 and find the third solution, in order to avoid finding the maximum when 𝑡𝑡< 0 or the first minimum. Generally, candidates taking this approach were unsuccessful in scoring the accuracy mark, although the method mark was scored reasonably frequently. Part (c) was another challenging part to this question and many candidates made no attempt. A surprising proportion of candidates attempted to solve the equation cos(0.5𝑡𝑡+ 𝛼𝛼) = −1, apparently in an attempt to find the minimum, but did not appear to understand how to use this to solve the problem. It was uncommon to see this method result in a correct value for 𝑇𝑇. More commonly, candidates were successful in finding a value for 𝑡𝑡 that satisfied 𝐻𝐻= 0, but incorrectly interpreted this as the solution to the problem. Instead, they were required to find a second value of 𝑡𝑡 and find the difference between these times. Again, there were a number of efficient and concise solutions to the problem that demonstrated an excellent understanding of what was required. The final mark in this question, in part (d), was rarely scored, and demonstrated that candidates are generally not well prepared to answer questions about refining models, as required in AO3.5c. Many candidates did attempt this part, often despite making little or no progress with the earlier parts of the question. Most concentrated on having an extra constant, wanted the average water level to be used, or simply restated the question by saying that the varying water level needed to be accounted for, without suggesting how this might be done. Others knew that they needed to adjust the model, but suggested exponential, linear or quadratic terms to do this, which would not have been appropriate. Again, there were some excellent and detailed explanations of how a trigonometric function could be included to taken into account the varying water level .
A water wheel rotates about a fixed axis through its centre C. Point P is at the end of a paddle. The water level is horizontal and is initially assumed constant. The vertical height H metres of P above the water level, t seconds after the wheel starts rotating, is modelled by H = 3 + 4cos(0.5t) - 2sin(0.5t).
Edexcel 9MA0 · October 2020 · Paper 1 · Q6 · 7 marks
7 marks
Examiner report
This was the second modelling question on the paper. Part (a) was generally well done, with few errors on giving the exact value of R or the angle in radians to the required degree of accuracy. A very small number of candidates gave R as a decimal or the angle in degrees. Part (b) was also well done with most of those candidates giving the exact answer for the temperature. Occasionally candidates obtained an answer of 8° C from 5 + 1 + 2, which they obtained from taking the maximum values of sin and cos. Part (c) was more challenging. The correct answer could be obtained fairly quickly by solving 1.107 3 12 2 t + − = π π but often the 1.107 and/or the 3 were omitted. Stating the time in an acceptable format also proved surprisingly difficult.
The temperature theta degrees C inside a room is modelled by theta=5+sin(pi*t/12-3)+2cos(pi*t/12-3), for 0<=t<24, where t is hours after midnight.
Edexcel 9MA0 · October 2020 · Paper 1 · Q11 · 8 marks
8 marks
Examiner report
This proved to be a challenging question with very few candidates achieving full marks overall, and with many more difficulties encountered in (b) rather than (a). (a) Candidates used a variety of approaches. The most common started by using algebra to find the coordinates of the points where the circles meet. This was achieved correctly by a majority of those who did the question this way. Then the required angle was found either by trigonometry in a right angled triangle and doubling an angle, or by the cosine rule in triangle AOB. Errors in this approach included processing x2+y2=100 into x+y=10. Although a minority of candidates’ answers stopped after finding coordinates for A or B, there were a significant number who went on to score the remaining 2 marks. However, some did not seem to realise that length OA=10 as it is a radius of C1, and had to use the distance formula to find this length. Another approach that was seen, but not so frequently was to use knowledge of circle equations in finding the radii and centres, to establish the lengths of the sides of triangle OAX, with X being the centre of circle C2. Using the cosine rule then led to half of the required angle. When used, this method often gained all four marks. (See relevant diagram below) (b) It was very common to see just the first mark scored in this part. This was due to many candidates wrongly assuming that the given angle AOB=0.635 was the same size as the angle AXB at the centre of circle C2. For the minority of candidates who tried a correct method to find the size of angle AXB, or half of it, most of these achieved the correct values of either 1.03 or 0.516. Even for the minority of candidates who found one of these angles correctly, getting the final answer correct proved a challenge and only a very small minority managed this. Note that it was possible to do part (b) even if (a) was not attempted. However, it seemed that failure to find a way to do (a), led in nearly every case to no valid attempt at (b).
Circle C1 has equation x^2+y^2=100 and circle C2 has equation (x-15)^2+y^2=40. Their centres are O=(0,0) and X=(15,0), and they intersect at A above the x-axis and B below it. The shaded region is the union of the two circular discs, so its outer boundary consists of the major arc AB of C1 and the major arc AB of C2.
Edexcel 9MA0 · October 2020 · Paper 1 · Q12 · 8 marks
8 marks
Examiner report
Part (a) was generally answered well and there were remarkably few solutions containing missing variables. On this occasion, candidates who started with the RHS fared better as those who worked from the LHS occasionally missed the required step between 𝑐𝑐𝑐𝑐𝑐𝑐2𝜃𝜃/sin𝜃𝜃 and cos𝜃𝜃cot𝜃𝜃, while those who worked from both sides would often get confused or, if they progressed correctly they simply failed to make the necessary conclusion. Part (b), however was very often left completely blank. Many who attempted it failed to see the connection with (a) and got lost in complicated trig expansions which led nowhere. Of those who did use part (a) to answer part (b), the most common solution was simply 𝑥𝑥=25 after cos 𝑥𝑥 had been cancelled, thus losing the solution given by cos 𝑥𝑥 =0. Only a small minority of the most able found all 3 solutions.
Show all stages of working; solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · October 2020 · Paper 2 · Q10 · 8 marks
8 marks
Examiner report
Many students were able to get started in part (a) with correct use of cos(2A + A) alongside the addition formula. Candidates who managed this, regularly went on to apply the double angle formulae correctly to achieve an expression in cosA only. Often circuitous routes were taken, most notably with the use of cos2A = cos2A - sin2A and then making a second substitution for sin2A (this sometimes resulted in bracketing and/or sign errors). A small number of candidates added a proof for the double angle formulae from the addition formulae even though this was unnecessary. A high proportion of students who managed to get the first mark went on to get at least 3 if not 4 marks in this part of the question. It was very much an all or nothing question with many only getting the first mark. A significant number presented their work in a way that was sometimes difficult to follow even if they did manage to score full marks. In part (b), it was common for candidates to achieve no marks for part (a) and then full marks for this part. Most of the students gained the first method mark and managed to produce an equation in cosx only. Mistakes were made with signs and brackets. In particular, candidates failed to write brackets around 4cos3x – 3cosx and hence would obtain an incorrect quadratic equation to solve. Several candidates ‘solved’ the cubic/quadratic equation using the polynomial equation function on their calculator and did not show any evidence of their working. Some incorrect equations led to complex solutions but most of these candidates still did not realise that they had made an error. A large number of candidates divided by cosx to simplify their expression but they failed to recognise this lost a solution and therefore were not awarded the final mark. A small number of candidates failed to gain any marks on this part as they replaced the right hand side with cosx but left the left hand side with cos3x and then proceeded to treat this as a cubic/quadratic. When candidates had identified the correct equation and solutions they were very good at finding all the solutions in the range. A large number who did get solutions did not include -90° or included 180° and therefore lost the final mark.
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · June 2019 · Paper 1 · Q2 · 5 marks
5 marks
Examiner report
(Mean Mark 2.5 out of 5) This question proved to be more challenging for a number of candidates although many were still able to score full marks. In part (a) it was important for candidates to pay attention to the scale of the graph. Most candidates who did draw a straight-line graph were able to make the link between one point of intersection and one real root to the equation, however, a significant minority of these were unable to construct the correct straight-line graph with enough accuracy to gain both marks for this part of the question. In part (b) almost all candidates were able to write down the small angle approximation for cosx even if they were unable to use it in order to answer the question. Most candidates did substitute correctly into the equation in order to gain a quadratic equation, although some did contain a mixture of variables. Sign errors sometimes occurred at this stage and the weakest candidates were unable to deal correctly with resulting fractions in order to simplify the equation. Many candidates appear to be using calculators in order to solve quadratics and in this question, this usually resulted in full marks. The final mark was lost by some candidates who correctly solved the quadratic equation but gave both roots as their final answer.
The equation cos(x)-2x-1/2=0 can be interpreted by comparing the curve y=cos(x), with x measured in radians, with a suitable straight line.
Edexcel 9MA0 · June 2019 · Paper 1 · Q6 · 8 marks
8 marks
Examiner report
(Mean Mark 4.4 out of 8) This was generally a well attempted question with most candidates able to score over half of the marks. In part (a) the correct identities were well known and efficiently used in order to obtain an equation in one function. Almost all candidates who reached this stage were then able to proceed to find at least one correct angle. It was clear from the graphs and diagrams sketched by most candidates that they had been taught to look for all angles in the required range and many were able to give all four of the angles needed. Only a minority of candidates identified 0, 180 as solutions to the equations. Most candidates divided through by sin or cos and failed to identify any of these values. In part (b) a significant number used 2 50 x rather than 25 x .Of those who used one of their solutions to part (a), the majority then proceeded to the correct answer, although a significant number did not give the smallest positive solution to the equation. This was usually due to the fact that they used their smallest positive solution to part (a) (i.e. 18.4) in the equation, not realising that they needed to use – 18.4.
Solutions based entirely on graphical or numerical methods are not acceptable.
Edexcel 9MA0 · June 2019 · Paper 1 · Q14 · 7 marks
7 marks
Examiner report
(Mean Mark 2.4 out of 7) A number of candidates did not attempt this question. Whether this was due to lack of understanding or time constraints is unknown. Part (a) was generally well done with a surprising proportion using implicit differentiation. A good number who found d d y x correctly failed to substitute in y =0 to find the value of d d y x at the origin. Candidates who tried to expand sin2y using the double angle formulae and then differentiating using the product rule often went wrong. The small angle approximation in (b) part (i) was also well handled. Most settled for the form 8 x y although some rearranged to y = . Marks were generally lost when sin 2y was replaced by 2 instead of 2y. Many candidates failed to see the link between (a) and (b)(i) even though they had both answers correct. They were unable to see that the gradients were the same or else failed to explain themselves coherently. Only a small percentage of candidates were able to secure full marks for part (c) and it was not uncommon to see the whole question left unanswered. This type of question was common on Core 3 and usually well answered.
The curve C is defined by x=4sin(2y), with -pi/4<y<pi/4, and passes through the origin O.
Edexcel 9MA0 · June 2019 · Paper 2 · Q3 · 3 marks
3 marks
Examiner report
Q03 proved to be the most accessible question on this paper with many candidates obtaining full marks. In Q03(a), many candidates explained that the angle 40 should have been converted to its equivalent in radians in order for it to be applied to the formula 2 1 . 2 r Some candidates’ explanations lacked sufficient detail, e.g. they stated ‘the angle is in degrees’ but failed to mention that the angle used in the formula should have been expressed in radians. A few candidates did not give a reason in Q03(a), but correctly calculated the area of sector AOB in Q3(b). Q03(b) was also well-answered. Most candidates converted 40 to radians and applied a correct 2 9 to the given formula 2 1 . 2 r Some candidates applied a correct formula 2 360 r with 40 . Errors included converting 40 to either 40 360 or 9 ; 2 applying 2 1 2 (5 ) ; 2 9 and applying incorrect formulae such as 2 1 sin 2 r or 2 1 . 2 r Most candidates gave the correct answer in exact form as 25 , 9 and some gave an answer 8.73 which was rounded to 3 significant figures, while others gave both exact and rounded answers. Most candidates gave the units 2 cm in their answer to Q03(b). On this occasion, a lack of units in their final answer was condoned.
A sector AOB has centre O, radius 5 cm and angle AOB = 40 degrees. A student writes: Area of sector = (1/2)r^2 theta = (1/2)(5^2)(40) = 500 cm^2.
Edexcel 9MA0 · June 2019 · Paper 2 · Q4 · 6 marks
6 marks
Examiner report
Q04 was well attempted by both medium ability and higher ability candidates. Lower ability candidates struggled to make progress, with most of them scoring no more than one mark. There were two common methods that were used by candidates. The first method, covered by Way 1 in the mark scheme, was the substitution of the parametric equations of 1 C into the Cartesian equation of 2 C to give an equation in t only; the trigonometric identity 2 2 sin cos 1 t t being used to obtain an equation in 2 sin t (or 2 cos t ) only, and a value for sint (or cos )t found. The second method, covered by Way 2 in the mark scheme, was the application of the trigonometric identity 2 2 sin cos 1 t t to the parametric equations of 1 C to give 2 2 1, 10 4 2 x y the Cartesian equation of 2 C being used to obtain an equation in 2x (or 2) y only, and a value for x (or y) found. A few candidates used a correct method of progressing from 2 2 (10cos ) (4 2 sin ) 66 t t to either 2 tan 1 t or cos2 0. t In all these methods, substituting back yielded their coordinates for S, with most candidates realising that 0 Sx and 0. Sy Some candidates incorrectly stated S as (5 2 , 4). In the first method, successful candidates used either 7 4 t or , 4 t and then applied symmetry to find the correct coordinates for S as either (5 2 , 4) or (7.07, 4). An area of concern centred around arithmetical slips or errors in elementary algebra. Examples of the former included 2 (4 2) becoming 8 or 64; 2 1 sin 2 t becoming 1 sin ; 4 t and 2 2 1 100 32 x y becoming 2 2 32 100 1 or 100 or 32. x y Examples of the latter included the invalid methods of 2 2 66 y x becoming 66 y x and 2 2 sin 1 100 x t becoming sin 1 . 10 x t Candidates who made no creditable progress included those who differentiated the parametric equations for 1 , C and those who obtained, e.g. 4 2 sin arccos 10 x y leading to 2 2 32sin arccos 66. 10 x x
The curve C1 has parametric equations x = 10cos(t), y = 4sqrt(2)sin(t), for 0 <= t < 2pi. It meets the circle C2, x^2 + y^2 = 66, at four distinct points. One intersection point S lies in the fourth quadrant.
Edexcel 9MA0 · June 2019 · Paper 2 · Q12 · 7 marks
7 marks
Examiner report
Q12(a) discriminated well between the higher ability candidates and it was possible for candidates of all abilities to gain access to Q12(b). In Q12(a), many candidates struggled to apply a complete strategy which would help them to make significant progress in proceeding from cos3 sin3 , sin cos via an intermediate stage of cos(3 ) , sin cos 2 2 cos2 (cos sin ) sin cos or cos2 , sin cos to the given result 2cot2 . Candidates who used Way 1, as described in the mark scheme, gained some credit for rationalising cos3 sin3 sin cos to give cos3 cos sin3 sin . sin cos Of those candidates who correctly applied 1 sin cos sin 2 2 to the denominator, only a few realised that the numerator could be written as cos2 . A common error in using Way 1 was to simplify cos3 sin3 sin cos to give 2 2 cos 3 sin 3 sin cos or cos4 sin 4 . sin cos Way 2, as described in the mark scheme (and variations of this), was the most popular approach among candidates who completed the proof successfully. Candidates only started to access the marks for Way 2 when they combined their fractions by rationalising the denominator. Errors using Way 2 included the incorrect expansion of cos(2 ) or sin(2 ) , bracketing errors, manipulation errors or sign errors. A few candidates started their proof from 2cot2 . They usually progressed as far as writing either 2cos2 2cot 2 sin 2 or 2 2(1 tan ) 2cot 2 , 2tan but could make no creditable progress until they applied a Way 3 method as described in the mark scheme. Some candidates spent a considerable amount of time attempting to write cos3 sin3 sin cos in terms of to give e.g. 3 3 4cos 3cos 3sin 4sin . sin cos candidates only started to access the marks for this method when they combined their fractions by rationalising the denominator. In Q13(b), most candidates used the result given in Q13(a) to deduce the equation 2cot 2 4. Most of these then attempted to solve 1 tan 2 , 2 and some obtained the correct answer 103.3 . Common errors included attempting to solve tan2 2 and finding more than one value for . A few candidates obtained 103.3 by solving the equation 2 2(1 tan ) 4. 2tan
Angles are measured in degrees. Let theta != 90n degrees for integer n.
Edexcel 9MA0 · June 2018 · Paper 1 · Q1 · 3 marks
3 marks
Examiner report
This question was accessible to most students and many gave fully correct solutions. Almost all students were able to score a mark for sin3 3 but there were a number who simplified the denominator to 6 instead of 2 6 resulting in a final solution of 4 . 3 Replacing the cos 4𝜃 proved more problematic with common incorrect attempts on the numerator including 2 1 4 1 2 and more frequently 2 4 1 1 2 where the 4was not squared leading to an incorrect answer of 1/3. Other common incorrect attempts involved simplifying 1 cos4 to 2 8 rather than 2 8 leading to the incorrect answer of - 4/3. Even in cases where both substitutions were performed correctly many careless errors were made when simplifying. There were a small number of students who tried to expand and simplify the expression using the double angle formula for cos4, but in most cases this resulted in an incorrect expression which did not lead to the correct answer.
The angle theta is small and is measured in radians.
Edexcel 9MA0 · June 2018 · Paper 1 · Q3 · 4 marks
4 marks
Examiner report
This question was well answered by the vast majority of students. The formulae for arc length and area of a sector were well known and applied, and although the 4 was occasionally seen on the incorrect side of the equation, this was rare. Most students worked in radians and used simultaneous equations to find the value of before proceeding to find an exact (positive) value for r. Algebraic and numerical slips, however, when manipulating equations were not an uncommon reason for losing marks. If method errors were made in this question, it was usually to miss the half from the sector area or to confuse the area of a sector with that of a triangle.
A sector AOB has centre O, radius r cm and angle AOB=theta radians. Its area is 11 cm^2. The perimeter of the sector is four times the length of arc AB.
Edexcel 9MA0 · June 2018 · Paper 1 · Q5 · 5 marks
5 marks
Examiner report
The majority of students knew to use the quotient rule in this question, although the slips on signs and bracketing in the terms of the numerator were commonplace. We would encourage all students to state the formula being used, with their expressions for ‘u’ ‘v’ ‘du’ and ‘dv’, to ensure that their method is made clear to examiners. Getting a correct derivative into the required form using trigonometric identities proved to be more challenging than the differentiation in this question. Students recognised the need to use the Pythagorean identity for sine and cosine to simplify their numerator and denominator but were less confident when applying the double angle formula for sine, and it was not uncommon for a final answer to be left as 3 2 2sin 2 , or for there to be errors in an attempt to write this fraction in the required form.
Given y=(3sin(theta))/(2sin(theta)+2cos(theta)), where -pi/4<theta<3pi/4:
Edexcel 9MA0 · June 2018 · Paper 1 · Q8 · 5 marks
5 marks
Examiner report
This question involved modelling the depth of water in a harbour using a trigonometric function. The mark in part (a) was usually scored, although some students did forget to include the units for the answer of 4.48 metres. Other errors seen were as a result of substituting incorrect values of t , usually 0 t , into the equation for D or prematurely rounding their answer to 4.5 metres. Most students also started part (b) correctly and were able to proceed to the intermediate point of sin30 0.6 t without much difficulty. The final part of the question, however, proved to be much more difficult, with many students not finding a value of t greater than 8.5, giving 𝑡= 7.2 then a time of 7: 14𝑎𝑚. Of those who correctly found 𝑡= 10.77 a large number failed to give their final answer as a time, with 10 hours and 46 minutes being common. A number of successful students were noteworthy in terms of the clear presentation and communication of their solutions, often producing sketch graphs of the sine curve to aid their understanding.
The depth of water D metres in a harbour is modelled by D=5+2sin(30t degrees), for 0<=t<24, where t is the number of hours after midnight. A boat enters at 6:30 am, takes 2 hours to load, and requires a depth of at least 3.8 m before it can leave.
Edexcel 9MA0 · June 2018 · Paper 2 · Q6 · 6 marks
6 marks
Examiner report
This was a well-answered question with part (b) and part (c) providing most of the discrimination. In part (a), nearly all students found f (2) 0 and many used this result to deduce that ( 2) x was a factor of f( ). x The method of long division was the most popular and successful method to find the quadratic factor of f( ). x Those who used the more time-efficient method of comparing coefficients were slightly more prone to making sign or manipulation errors. Most students arrived at the correct answer 2 f( ) ( 2)( 3 2 5), x x x x with some writing the equivalent 2 f( ) (2 )(3 2 5). x x x x There were a few students who found both correct factors but did not express their final answer as a product of two algebraic factors. In part (b), many students used part (a) to write down 2 4 2 ( 2)( 3 2 5) 0 y y y and most deduced the two real solutions 2. y Surprisingly a few students solved 2 2 y to give 2. y A significant number of students stated, without proof, that 2 ( 3 2 5) x x or 4 2 ( 3 2 5) y y where 2, x y had no real solutions. Some students applied ‘ 2 2 4 (2) 4( 3)( 5) 56 0 b ac ’ with an appropriate conclusion to show that 4 2 ( 3 2 5) 0 y y had no real solutions. Common errors included just stating ‘ 2 4 0 b ac ’ with no reference to a discriminant calculation or the omission of ‘ 0 ’ after a correct discriminant calculation. Only a minority of students were able to bring the two strands of the proof together to show that the given equation had only two real solutions. In part (c), some students correctly deduced that the given equation had 3 real solutions for 7 10 , with a few students incorrectly deducing that there were 6 real solutions. Most students misunderstood the requirement of the question by finding 23.1, 26.2, 29.4, with those students who did not go on to state that there were 3 real solutions losing the mark for this part.
Let f(x) = -3x^3 + 8x^2 - 9x + 10 for real x.
Edexcel 9MA0 · June 2018 · Paper 2 · Q7 · 9 marks
9 marks
Examiner report
Part (i) was accessible to most students, whereas some students struggled to gain access to part (ii). In part (i), many students replaced secx with 1 cos x and arrived at 4sin cos 1. x x A few students replaced 4sin cos x x with sin 4 ,x but most used the identity sin2 2sin cos x x x to give the correct 1 sin 2 . 2 x Many students obtained both correct exact solutions in radians, with only a few omitting the second solution 5 . 12 Some students made no attempt at part (ii) while a significant number of other students did not employ a correct strategy for solving 5sin 5cos 2. The majority of students, however, attempted one of several correct strategies. The most popular strategy was to square both sides of the given equation and apply the identities 2 2 cos sin 1 and sin2 2sin cos to achieve 21 sin 2 . 25 A second strategy was to square both sides of 5sin 2 5cos and apply the identity 2 2 cos sin 1 to achieve a quadratic equation in either cos or sin . In both of these strategies, extra (incorrect) solutions were generated as a result of squaring both sides of their equation. Only a few students rejected these extra solutions to leave the correct solutions 61.4 , 208.6 . A third strategy was to rewrite the given equation as 50 sin( 45 ) 2 or even 50 cos( 45 ) 2, which were usually solved to give 61.4 , 208.6 . A few students applied other correct strategies which led to solving quadratic equations in either tan or cot .
Solutions based entirely on graphical or numerical methods are not acceptable.
Edexcel 9MA0 · June 2018 · Paper 2 · Q12 · 9 marks
9 marks
Examiner report
This question discriminated well between students of all abilities with part (a) more accessible to students than part (b). In part (a), most students provided a proof which progressed from one side of the identity to the other by using methodical steps and correct trigonometric identities. There was no preference seen for starting their proof on a particular side of the identity. Some students manipulated both sides of the identity to give a common expression 2 2sin , with many providing some indication that their proof was complete. Part (b) proved to be discriminating, with some students not realising that they could use the result from part (a). Some students substituted 1 cos2x with tan sin2 x x and then proceeded to cancel tan sin2 x xfrom both sides, while other students substituted tan sin2 x x with 1 cos2x and then proceeded to cancel 1 cos2x from both sides. These two strategies together with applying the identity 2 2 sec 1 tan x x usually led to a 3-term quadratic in tan .x Most students used a correct method, usually by factorising, to solve their quadratic equation in tan x but only a few stated all three correct solutions. As a result of cancelling either tan sin2 x x or 1 cos2x from both sides of their equation, many students missed out the correct solution 0. x Those students who factorised out these terms were usually more successful in finding 0. x Some students did not give their other two solutions in the required form of the exact 4 x and 1.326 x to 3 decimal places.
Angles are measured in radians where no degree symbol is shown.
Edexcel 9MA0 · sample · Paper 1 · Q2 · 5 marks
5 marks
The shape ABCDOA consists of a sector COD of a circle with centre O joined to the shaded sector AOB of a different circle, also with centre O. Points A, O and D lie on a straight line in that order, and points O, B and C lie on a ray in that order. The arc length CD is 3 cm, angle COD is 0.4 radians, and AD=12 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 1 · Q7 · 5 marks
5 marks
Triangle ABC has vector AB=2i+3j+k and vector BC=i-9j+3k.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 1 · Q9 · 5 marks
5 marks
Let theta be such that theta!=npi/2 for integer n.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 1 · Q10 · 5 marks
5 marks
![Edexcel 9MA0 · sample · Paper 1 · Q10 · 5 marks — The angle theta is measured in radians. You may assume the formula for sin(A+B) and that, as h tends to 0, sin(h)/h tends to 1 and [cos(h)-1]/h tends to 0.](/api/paper-crops/9ma0/sample/paper-1/q10/question-r1-21f9186b.webp)
The angle theta is measured in radians. You may assume the formula for sin(A+B) and that, as h tends to 0, sin(h)/h tends to 1 and [cos(h)-1]/h tends to 0.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 1 · Q13 · 13 marks
13 marks
The curve C has parametric equations x=2cos(t) and y=sqrt(3)cos(2t), where 0<=t<=pi. Point P lies on C where t=2pi/3, and the line l is the normal to C at P.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 1 · Q15 · 8 marks
8 marks
![Edexcel 9MA0 · sample · Paper 1 · Q15 · 8 marks — For 0<=x<=pi, f(x)=4sin(2x)/e^[sqrt(2)x-1]. The curve y=f(x) has a maximum turning point P followed by a minimum turning point Q.](/api/paper-crops/9ma0/sample/paper-1/q15/question-r1-4aa0e2f8.webp)
For 0<=x<=pi, f(x)=4sin(2x)/e^[sqrt(2)x-1]. The curve y=f(x) has a maximum turning point P followed by a minimum turning point Q.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q2 · 3 marks
3 marks
Some A level students were asked to solve cos(theta) = 2sin(theta) for -90 degrees < theta < 90 degrees. Student A divided and wrote tan(theta) = 2, giving theta = 63.4 degrees. Student B squared both sides, used cos^2(theta) = 1 - sin^2(theta), obtained sin^2(theta) = 1/5, and gave theta = +/-26.6 degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q12 · 8 marks
8 marks
Angles are measured in degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q13 · 9 marks
9 marks
A Ferris wheel is modelled by H = a - 10cos(80t degrees) + 3sin(80t degrees), where H metres is the height of a passenger t minutes after the wheel starts. At t = 0 the passenger is 1 m above the ground. Solutions based entirely on graphical or numerical methods are not acceptable.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q12 · 7 marks
7 marks
Consider the definite integral from theta = 0 to theta = pi/2 of sin(2theta)/(1 + cos(theta)) dtheta.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q13 · 11 marks
11 marks
Tom models the water depth D metres at Southview harbour on 18 October 2017 by D = 6 + 2sin(4pi t/25) - 1.5cos(4pi t/25), for 0 <= t <= 24, where t is hours after 00:00. Jolene proposes H = 6 + 2sin(4pi x/25) - 1.5cos(4pi x/25), for 0 <= x <= 24, where x is hours after 00:00 on 19 October 2017.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q14 · 5 marks
5 marks
A curve C has parametric equations x = 4cos(t + pi/6), y = 2sin(t), for 0 < t <= 2pi.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q1 · 4 marks
4 marks
A circle has centre O. Points A and B lie on its circumference. The area of the major sector AOB is 135 cm^2 and the reflex angle AOB is 4.8 radians.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q2 · 5 marks
5 marks
Given that theta is small, use the small angle approximation for cos(theta). Adele then tests the resulting approximation using theta = 5 degrees. Her calculator gives 1 + 4cos(5 degrees) + 3cos^2(5 degrees) = 7.962, but she substitutes theta = 5 into 8 - 5theta^2 and obtains -117, so concludes that the approximation is not true for 5 degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q12 · 7 marks
7 marks
A curve C is given by sin(x) + cos(y) = 0.5, where -pi/2 <= x < 3pi/2 and -pi < y < pi. A point P lies on C and the tangent to C at P is parallel to the x-axis.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q13 · 10 marks
10 marks
Angles in this question are measured in degrees where the degree symbol is shown.
Pearson does not publish a direct link to this question paper.
Question extracts are shown for study alongside a link to the original Pearson paper. Reference-only rows carry series, paper, question number and marks.
Everything here stays free. If trigonometry is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.