Edexcel A-level Maths · Topic 3
Practice questions covering coordinate geometry in the (x, y) plane 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 Coordinate geometry in the (x, y) plane, each with its mark scheme and worked method. Reveal the answer when you have had a go.
1.
(4)
(Total for Question 1 is 4 marks)
2.
(2)
(Total for Question 2 is 2 marks)
3.
(4)
(Total for Question 3 is 4 marks)
4.
(3)
(Total for Question 4 is 3 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 — 3 Coordinate geometry in the (x, y) planeAll 40 questions across 4 specification points — 34 more than the 6 above — with the full mark scheme. Guest accessible · browser print.27 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 · Q6 · 6 marks
6 marks
Examiner report
This question on the modulus function provided plenty of access and there were very few blank responses to any part, although many found part (c) challenging. The overwhelming majority of candidates answered part (a) correctly, demonstrating a strong understanding that the minimum y value must occur when 2 0 x − = . The answer was often written down without working. Those candidates that did not gain full marks often achieved the method mark for one correct coordinate. The majority of candidates who scored no marks found the y intercept of the modulus curve instead. In part (b), the most common approach was to remove the modulus signs and to attempt to solve two equations, usually ( ) 16 4 3 2 5 x x − = − + and ( ) 16 4 3 2 5 x x − = − − + . The more confident candidates had already established that the line would only intersect the right-hand branch of the modulus graph. The most common error in processing was thinking that 3 2 x − was equivalent to 3 6 x + , but, pleasingly, this was less common than in previous series. Some candidates kept the modulus signs in their equation throughout which made the equation difficult to manipulate. Disappointingly, a surprisingly large number of candidates rearranged their equations poorly by not performing inverse operations correctly. For those that lost the accuracy mark in part (b), the most frequent reason was because they failed to reject the solution 5 x = . Those using the diagram to aid visualisation of the problem were generally more successful in rejecting this extra “solution”. Some strong candidates demonstrated an excellent understanding of the relevant domains for each branch of the modulus graph and rejected x = 5 as they had established that an intersection with this branch required x < 2. Part (c) was found to be less accessible with few candidates scoring full marks for a correct range for k. The candidates that were the most successful in this part of the question were generally those who had adopted a visual approach in the earlier parts of the question and understood how to find the minimum and maximum value of k using their knowledge of gradients. There was more success in finding the correct lower critical value of 0.5 with the most efficient solutions using the coordinates of P with y = kx + 4, but many candidates failed to realise that there was an upper limit for k. There was a reluctance to use diagrammatic reasoning and most resorted to complicated algebra. Many set up simultaneous equations in x and k but did not appear to understand what they were attempting to achieve, failing to solve for k and therefore these attempts did not gain any marks. There were many erroneous attempts that fortuitously achieved k = 3 through an attempt to use a discriminant (despite starting with a non-quadratic equation), and these invalid approaches scored no marks on their own. A few attempted squaring both sides of the equation to form a quadratic equation and then using the discriminant, but this approach often failed due to not isolating an appropriate term (such as 2 x − ) before squaring. Relatively few candidates considered the gradient of the modulus graph and hence failed to obtain an upper bound for k, which meant many only achieved the method mark in this part. Overall, candidates' use of inequality notation was good and although set notation was not required, the vast majority of those that used it did so correctly.
Figure 1 shows a V-shaped graph with equation y = 3|x - 2| + 5. Its vertex is P and it crosses the y-axis at (0, 11). A line l has equation y = kx + 4, where k is a constant.
Edexcel 9MA0 · June 2024 · Paper 2 · Q10 · 6 marks
6 marks
Examiner report
Most candidates were successful in this question, showing that parametric differentiation was overall well understood by candidates. In part (a), many candidates successfully found the given equation of the tangent using parametric differentiation as specified in the question. By far, the most common approach seen involved finding the value of t and then substituting it into an expression for d d y x obtained by using d d d d d d y y t x t x = and then using standard coordinate geometry to find the equation of the tangent. There were a number of careless slips and algebraic errors, with 2(t + 3) = 2t + 3 being disappointingly common. Once the numerical gradient had been obtained, there were usually no issues in finding the equation of the tangent in the required form, showing at least one intermediate step as this was a given answer. Most candidates achieved the equation of the tangent via the Point-Slope form although a large number went via the Slope-Intercept form. A few candidates made coefficient and/or sign errors when manipulating their equation to obtain the given answer and unfortunately lost the final accuracy mark. Those candidates who gained few, if any, marks in this question often failed to see the need to find the value of the parameter ‘t’ for the given point P (4, 2) and were unsuccessful in using the alternative method to continue to find a value for d d y x . Some candidates used an incorrect value of ‘t’, usually t = −5 rather than t = −1, as they either did not note the restrictions for t given in the question, or did not find the t value for the y coordinate of P. A few gave two equations for each value of t found. The attempted use of the Chain Rule was not always completed accurately. Some candidates converted their gradient from the use of the Chain Rule into the gradient of the normal which resulted in an inappropriate equation. When finding the value of d d y x , some candidates substituted 4 for their value of t, confusing it with the x coordinate of point P. Some candidates did not follow the instructions stating that the question had to be answered ‘using parametric differentiation’ and, consequently, scored no marks even though they achieved the required result by obtaining the Cartesian equation of the curve. Some candidates tried to complete the question by unnecessarily converting their d d d d y t t x in terms of t to be in terms of x and/or y which often led to incorrect solutions due to algebraic and/or differentiation errors. There were many candidates who initially went wrong and restarted the question. A number of these ran out of space and incorrectly used the back pages of their answer booklet to continue this question rather than using supplementary paper. It was noted by examiners that part (b) was not attempted by a significant number of candidates. When it was attempted, it was generally not done well, with very many candidates not scoring the mark. Many achieved the correct value only to lose the mark for not stating the units. A common wrong answer was to use x = 0 in the tangent equation to give 5m. Another was to use t = −3 and then substitute this into y = 1 − t3 to give 28m, ignoring the set of values for which t was valid. Others set d 0 d y x = thinking this would give a maximum and then substituted their t value into y. Some wrongly assumed that the largest value of t for the given domain would lead to the maximum height. There was little evidence that candidates had used the sketch to help them find the maximum height.
A curve C has parametric equations x = (t + 3)^2 and y = 1 - t^3, for -2 <= t <= 1. The point P(4, 2) lies on C.
Edexcel 9MA0 · June 2024 · Paper 2 · Q14 · 8 marks
8 marks
Examiner report
A common marking profile in this question was to score all three marks in (a) followed by zero marks in (b). In part (a) the most common approach was to complete the square and very often it was successful although there were a variety of incorrect answers for the centre – usually variations of ( ) 3, 7 . Quite a few knew the strategy for finding the radius although not all of them knew how to use it correctly and 91 r = was a frequently seen incorrect answer from attempting ( ) ( ) 2 2 3 9 7 49 33 0. x y − + + + + + = A surprising number of candidates evaluated 72 as something other than 49. In part (b), it was common for no marks to be scored and there were a large number of non- attempts. The most successful candidates were those who drew diagrams and looked at the problem geometrically. However, the most common approach was an algebraic one. This typically involved solving the 2 circle equations simultaneously, but the majority of such attempts were aborted before any marks were gained. There was also some poor use of algebra with things such as ( ) ( ) 2 2 3 7 25 3 7 5 x y x y − + + = −+ + = seen a surprising number of times and such attempts gained no credit. Many candidates did recognise that there would be a discriminant involved so they often arrived at a point when they thought that it would be appropriate to calculate one. However, this was usually prematurely. Despite this, it was noted that some examiners did report the occasional successful algebraic attempt. Some candidates calculated the equation of the line passing through the centres but then did not realise what they had to do with it. Most who were able to find the correct distance between the centres recognised they had to add and subtract the radius found earlier to give the limits of the solution set. Candidates who made it this far sometimes lost the final accuracy mark due to the use of incorrect notation. For candidates who got as far as the correct interval for k, the most successful were those who gave the final answer as a single set. Candidates who broke the answer into two sets, often used the incorrect symbol between them.
The circle C_1 has equation x^2 + y^2 - 6x + 14y + 33 = 0. A different circle C_2 has centre (-6, -8) and radius k, where k is a constant.
Edexcel 9MA0 · June 2024 · Paper 2 · Q15 · 12 marks
12 marks
Examiner report
This question provided a challenge to most candidates and was in general a good discriminator. In part (a), it was perhaps disappointing, although not unexpected, to see that so many candidates’ instinctive first step was to multiply out the brackets rather than realising that a simpler, quicker and more accurate method was available. In part (a), the use of the chain rule and factorisation could greatly reduce the required effort and the potential for making mistakes. Those candidates with a good understanding of the chain rule and implicit differentiation were able to use this efficient method quickly to accumulate the 5 marks available. However, a common error of omitting the 1 in (1 + d d y x ) was seen many times. Furthermore, a significant proportion of candidates who differentiated the left-hand side using the chain rule, often chose to expand the derivative, nonetheless, thus complicating the remainder of the question. Unfortunately, the majority of candidates chose to first expand ( ) 3 x y + before attempting to differentiate. Whilst this was, in many cases, completed correctly there was of course an inherent time penalty. A significant proportion of candidates made errors when expanding and displayed a lack of algebraic fluency and poor application of the binomial expansion. Common errors that were seen included missing out terms in the expansion and poor understanding of indices. Once the brackets had been expanded, there were clearly some gaps in understanding of how to successfully carry out the implicit differentiation, particularly for terms that also required application of the product rule. It was quite common to see for example 2 3x y going to d 6 d y x x and 2 3xy going to d 6 d y x . Overall, the right-hand-side was dealt with correctly although the constant term −2 was sometimes retained following differentiation and sometimes d 6 d y y x was seen. When candidates attempted to move all terms over to one side of the equation, there were sometimes sign errors or slips in copying terms. Upon completion of the process of differentiation, almost all candidates seemed to understand that the task was to gather the d d y x terms together, factorise and divide in order to isolate the d d y x , and the majority who reached this stage were able to earn credit here provided their differentiation had been sufficiently structurally correct to produce sufficient terms in d d y x . It was very rare to see a spurious d d y x term retained which has been an issue in previous sessions. In general, those candidates who were able to successfully complete part (a) found part (b) straightforward. However, the accuracy mark in this part was not available to those candidates who had not managed to achieve the correct derivative in part (a). Due to the nature of the function, setting x = 1 and y = 0, gave a ‘correct’ value of d 1 d 2 y x = in many cases even when d d y x was incorrect. As a result, candidates were often not alerted to the fact they had an incorrect d d y x . Disappointingly, candidates who did not obtain 1 2 due to errors in their differentiation, often attempted to manipulate their work to obtain the value that they could see was required from the printed answer and this often meant that the method mark was out of reach, which was a shame. Part (c) provided a challenge for many candidates and was a good discriminator. It is possible that some candidates were running out of time by this part of the paper, and this seemed to lead to simple algebraic and sign errors. However, many candidates were able to make at least some progress and most recognised the need to substitute the equation from part (b), which had been provided, into the equation from part (a). Most were able to expand to obtain a cubic in x albeit with frequent errors and slips, in particular a common error of cancelling the constant terms leading to a cubic with only three terms was often seen. The alternative elimination of x to lead to a cubic in y was rarely seen. Having obtained a cubic equation, many candidates made little creditworthy further progress and the majority seemed not to have made the link to earlier parts of the question and so did not attempt to take (x − 1) out as a factor of their cubic. This was perhaps understandable when the cubic was incorrect although it seemed that even candidates with the correct cubic did not consider using the fact that x = 1 was a root. Many candidates used their calculator to obtain the three roots of their cubic and others employed dubious techniques either attempting to find a ‘discriminant’ by ignoring the 3x term or attempting to take out a factor of x whilst ignoring the constant term. Those candidates who had a three-term cubic rather than a four-term cubic were unable to progress here. The final two marks of the paper were only accessible to those who were very secure in their algebraic accuracy and knowledge as these marks were contingent on both a correct cubic and a correct quadratic factor. Most candidates who met these criteria successfully used the discriminant to demonstrate that the quadratic had no roots although some used more elaborate reasoning involving the gradient of the cubic or the minimum point of the quadratic. Unfortunately, a significant proportion of candidates who got this far, did not manage to earn both marks because their justification or conclusion was insufficient, and it was sometimes disappointing to see otherwise well-answered solutions lose the final mark because they concluded with ‘no roots’ rather than also relating this back to the question to conclude that there could be ‘no further points of intersection’.
The curve C has equation (x + y)^3 = 3x^2 - 3y - 2. The point P(1, 0) lies on C.
Edexcel 9MA0 · June 2023 · Paper 1 · Q10 · 9 marks
9 marks
Examiner report
Candidates were typically able to make good progress in part (a) with the method of completing the square being familiar to many when dealing with the equation of the circle. Part (b) was focused on using the discriminant and solving a quadratic inequality, however, the manipulation proved to be a lot for some, resulting in a wide range of marks. Again, the access was good on this question such that candidates who could not do part (a) were not restricted from attempting part (b), although those with errors from part (a) often found that this caused issues in part (b) and resulted in losing many of the marks. In general candidates obtained ( 3 , ) k k − as the centre of the circle in part (a). Common errors included (3 , ) k k − and ( 3 , ) kx ky − . If candidates were successful at completing the square, they could find the correct radius of the circle. When attempting to complete the square a common error was to subtract 2 3k instead of( ) 2 3k . Occasionally, candidates would find 2 2 10 7 r k = − but did not proceed to take the square root. Those who collected terms on the right-hand side of the equation in order to form the square of the radius made fewer sign errors than those who did not. Others tried to manipulate the equation mentally, so writing down a wrong radius with no evidence of its development (and thus no method mark). One error was to write 2 10 7 10 7 k k − = − . Sometimes there were sign slips such as writing the 7 − as +7 when rearranging the equation. In part (b), whilst a number of candidates had little idea how to proceed, the vast majority did realise what was required and attempted to produce an equation in x by substitution. Many lost their way in the algebraic complexity, or did not know what to do with the equation (sometimes trying to solve it to find a value for k). A significant number of candidates did produce a three-term quadratic for x, with coefficients in terms of k. Centres should note that the scheme required the quadratic in x to be of the form specified, and that it be set out in a way which identified the coefficients (possibly implied by later use in the discriminant). So, for example, “ (2 4) k − x” was required rather than “ 2 4 kx x − ” as this provided evidence of recognising that the coefficients were the key to solving this problem. Quite a few candidates had a constant term which was not of the required form which meant that no marks could be scored. In general candidates who found critical values for their discriminant by solving 2 4 0 b ac − = were more successful than those who expressed it as an inequality. Often the inequality was simply carried forward to a statement about k; those who identified the two critical values by solving an equation perhaps thought more carefully about where the region was. The “need” to write a single inequality was seemingly pressing for many, so 7 85 7 85 k + − was seen. A few candidates did not use exact values of k and lost the final accuracy mark. Some candidates used 'and' with their two regions, however, most recognised that this needed to be 'or', either stating this explicitly or just using a comma between their two answers. A number of capable candidates brought some interesting approaches to bear on the problem, adopting a variety of techniques from Further Maths specifications.
A circle C has equation x^2 + y^2 + 6kx - 2ky + 7 = 0, where k is a constant.
Edexcel 9MA0 · June 2023 · Paper 2 · Q9 · 7 marks
7 marks
Examiner report
Most candidates found part (a) challenging but it was usually attempted. There were not many blank responses. Of the four methods listed on the mark scheme, one common successful method was to substitute t in terms of x into y = 6ln(t + 3) via “completing the square” (way 1). When substituting into y = 6ln(t + 3), if a correct t equation was created, then this usually led to full marks. Another common approach was to rearrange t in terms of y, usually successfully as 6 e 3 y t = − and substitute into x = t2 + 6t 16 (way 3). When using this method some candidates had difficulty squaring 6 e y meaning they were unable to earn full marks. Some candidates incorrectly set 25 5 x x + = + thus also losing the accuracy mark. A significant number of candidates factorised x = t2 + 6t 16 to give x = (t 2)(t + 8) and then wrongly thought x = t 2. This then often led to y = 6ln(x + 5) which scored no marks in this part. Candidates were generally more successful in part (b) than part (a). Many candidates found the y intercept by substituting t = 2 into the parametric equation for y rather than substituting x = 0 into the Cartesian equation. The first method mark was more challenging. Some candidates attempted the derivative from the Cartesian equation and they usually reached either ( ) 3 x A + or 3 x whilst others attempted to use the chain rule approach. Some candidates obtained the correct derivative but failed to substitute x = 0. If a candidate achieved the first method mark, they usually went on to achieve the dependent method mark for attempting to find the equation of the tangent. There were very few cases where the candidate used the negative reciprocal of their gradient. The most successful way candidates achieved full marks was differentiating their Cartesian equation. Those that differentiated parametrically were often caught out by arithmetic and algebraic errors or substituted t = 0 instead of t = 2 and failed to achieve the gradient correctly. For the final answer, some candidates had failed to realise the form of equation the question was asking for so left their answer as “y =” or they didn’t have 𝑎, 𝑏 and 𝑐 as integers. Many candidates who had incorrectly answered (a), were able to go on to achieve full marks in (b) by using parametric differentiation to work out the gradient. There were very few blank responses for this part of the question.
The curve C has parametric equations x = t^2 + 6t - 16 and y = 6 ln(t + 3), where t > -3.
Edexcel 9MA0 · June 2022 · Paper 1 · Q3 · 5 marks
5 marks
Examiner report
(Mean mark 3.5 out of 5) This question on the geometry of a circle had standard bookwork in part (a) followed by problem solving in part (b). Success in part (b) was usually achieved by candidates who drew a diagram. In part (a) most candidates scored all three marks. Some candidates made sign errors when completing the square but still obtained some version of (± 5, ± 8). Many candidates were still able to get the correct radius following this slip. A significant minority of candidates obtained a centre of (5, - 4) seemingly from square rooting the 16 rather than halving. In part (b), there was full follow through on the centre and radius, which allowed many candidates to score full marks despite errors in part (a). Candidates who drew a diagram worked out that they need to find the distance of the centre from the origin and add on the radius. One lengthy method involved finding the intersection of the circle with the line through O and C, but this was unlikely to give the requested exact answer, and some candidates gave up at the point where the quadratic had to be solved as it had unpleasant coefficients. A common error was based on simply doubling the radius, either as 26 or finding the length of the vector (10, -16) as 2 Ö89.
The circle C has equation x^2+y^2-10x+16y=80.
Edexcel 9MA0 · June 2022 · Paper 1 · Q16 · 9 marks
9 marks
Examiner report
(Mean mark 1.9 out of 9) The final question on the paper proved to be a challenge for many candidates and there were several blank responses seen. It is possible that this was due to time management. Some candidates attempted unsuccessfully to change to a Cartesian equation, suggesting that possibly parametric integration had not been covered. In part (a), candidates who recognised the need to use parametric integration were usually able to differentiate correctly to find d d x t , multiply this by y and then use the double angle formulae to find an acceptable expression for d d x y t . It was common for candidates to score the first two marks. The next two marks were more elusive, and it was common to see candidates perform a lot of onerous manipulation without making any significant progress. Those who were most likely to be successful were the ones who had an expression in terms of 2 sin 2t and then spotting the need to use the double angle formula to form a term in cos 4t. For many candidates the mark for the upper limit was all that was achieved in part (a). However, some failed to achieve it as they left it as 4 (value for x not t) or 45 (degrees). It was good to see students who were unsuccessful at part (a) still attempting part (b), and in some cases scoring full marks. Most candidates were able to integrate the first two terms correctly, but a significant proportion were unable to integrate the third term to get an expression in 3 sin .t The majority of correct solutions came from the use of the reverse chain rule for the final term. Those who attempted to use integration by substitution or by parts were usually unsuccessful. Candidates who had successfully completed the integration often went on to score full marks.
The curve C has parametric equations x=8sin^2(t) and y=2sin(2t)+3sin t for 0<=t<=pi/2. The region R is bounded by C, the x-axis and the vertical line x=4; it is the area under the first part of C from the origin to x=4.
Edexcel 9MA0 · June 2022 · Paper 2 · Q16 · 12 marks
12 marks
Examiner report
This question proved to be a good differentiator between candidates. It provided a challenge for many candidates but equally there were a number of accessible marks here. Most candidates attempted part (a), many attempted part (b) and at least partially attempted part (c), usually attempting to find the lower limit for k. There were, however, very few completely correct answers to this question and a substantial minority made no attempt at any part, perhaps having run out of time by the time they reached the last question on the paper. Part (a) was mostly approached correctly with a good number scoring full marks. Candidates generally noticed the instruction to use parametric differentiation. Whilst d𝑥𝑥 d𝑡𝑡 was most often correct, d𝑦𝑦 d𝑡𝑡 proved a challenge for many who struggled to differentiate sec2𝑡𝑡. Many did not seem to know that these functions were standard derivatives presented in the formula booklet and instead there were some long-winded and overly complex attempts to convert to sine or cosine functions and such approaches often led to errors. However, even some of the most inefficient approaches managed to persevere through lengthy algebra to obtain a correct d𝑦𝑦 d𝑥𝑥 eventually. Almost all who attempted d𝑥𝑥 d𝑡𝑡 and d𝑦𝑦 d𝑡𝑡 knew and attempted d𝑦𝑦 d𝑥𝑥= d𝑦𝑦 d𝑡𝑡 ÷ d𝑥𝑥 d𝑡𝑡. Those who made errors when determining d𝑦𝑦 d𝑥𝑥 often exhibited an over-reliance on trying to match the printed result and thus ‘adjusted’ their d𝑦𝑦 d𝑥𝑥 in an attempt to obtain the required value for the normal gradient. Often this meant the loss of a method mark. Candidates would be better advised to utilise a correct method rather than incorrectly manipulating their answers to match the result they are attempting to achieve and caution should always be advised when working backwards from a given answer. Nonetheless, application of negative reciprocals and equations of lines was generally successfully attempted, and candidates were clearly confident in these areas. There were a few attempts at a Cartesian approach in part (a) which gained zero marks as the question had specified the use of parametric differentiation. In part (b), there were a multitude of approaches seen and candidates were often at least partially successful. Most candidates recognised the need to begin with a trigonometric identity, usually sec2𝑡𝑡= 1 + tan2𝑡𝑡 and rearranged the expressions for x and y to obtain expression for sec2𝑡𝑡 and tan2𝑡𝑡 before substituting into the identity to obtain an equation in 𝑥𝑥 and 𝑦𝑦 which could in turn be rearranged to the required form. However, sometimes candidates took a far lengthier approach, beginning with e.g. sin2𝑡𝑡+ cos2𝑡𝑡= 1 and using 𝑥𝑥 and 𝑦𝑦 to find expressions for sin𝑡𝑡 and cos𝑡𝑡. An alternative and popular, though inelegant, method was to start from the target Cartesian equation and substitute in the trigonometric forms for 𝑥𝑥 and 𝑦𝑦 before rearranging to obtain an identity. Provided a conclusion was given, this approach was credited with both marks. Occasionally candidates used the equation for 𝑥𝑥 to obtain an expression for 𝑡𝑡 in terms of arctangent before substituting into the expression for 𝑦𝑦 which, if correctly done, was an acceptable approach. Several students just substituted the coordinates of points such as P into the given Cartesian equation, stating that they satisfied the equation. Such attempts did not constitute a proof and so gained no marks. Part (c) provided a particular challenge and usually only the lower limit was attempted. Most candidates appeared to completely miss the need to find an upper limit. Of those who did manage to find both limits, only a small minority of candidates got to the final solution with correct inequality interpretation. It is worth noting that candidates achieving full marks for this part very often had clear construction lines on their graph that enabled them to see the correct approach. Most often though, candidates gained the method mark for the lower limit, 43 8 , for identifying that two intersections were required and set up a correct inequality for the discriminant for the quadratic equation obtained by equating the equations of the line and the curve. Common errors in manipulation, particularly of the coefficient of ½ from the quadratic meant that a number of candidates did not earn the accuracy mark for 43/8. Many candidates stopped after obtaining a lower limit for k, but others explored the parameter limits at 𝑡𝑡= − 𝜋𝜋 4 and 𝑡𝑡 = 𝜋𝜋 3 to identify 𝑘𝑘 values of 13 2 and 23 2 −√3 for the line passing through the associated points. Very few, however, recognised 13 2 as an upper limit and a number of candidates assumed incorrectly that 13 2 < 𝑘𝑘 < 23 2 −√3. A minority of candidates incorrectly thought that setting d𝑦𝑦 d𝑥𝑥 = 0 or d𝑦𝑦 d𝑥𝑥 =− 1 2 would provide limiting criterion for k values. In summary this was an interesting question which gave rise to a wide variety of responses in all three parts. There were instances of both succinct and ingenious approaches seen, particularly in part (c) and it was pleasing to see tenacity in some response where candidates had inadvertently opted for less efficient approaches but had persevered nonetheless to obtain the required results.
The curve C has parametric equations x = 2tan(t) + 1 and y = 2sec^2(t) + 3, where -pi/4 <= t <= pi/3. The line l is the normal to C at the point P where t = pi/4.
Edexcel 9MA0 · October 2021 · Paper 1 · Q7 · 9 marks
9 marks
Examiner report
The first few marks in this question were very accessible but the final part of the question proved to be a useful discriminator especially around grade C and below. Part(a) The majority of candidates achieved full marks in this part. Sign errors were responsible for most errors, although failure to halve was also seen with either or both of (x ±10)2 and (y ± 4)2 being seen. Having achieved (x-5)2+(y+2)2 = 18, it rare to see the centre or radius not found correctly. Part (b) This part was found to be discriminating with complete solutions seen from only the most competent of candidates. The most common approach was to substitute y = 3x + k into the circle equation to find a quadratic in x (with coefficients in terms of k) for the points of intersection. Errors mostly arose when (3x + k +2)2 was expanded incorrectly or the 2x + 6kx terms not being gathered together. Many solutions then stopped after the substitution and expansion step, as candidates did not seem to appreciate what to do with an equation of this complexity in two variables. Application of b 2 = 4ac to find equal roots was used on many occasions but errors frequently occurred. Only a very few solutions involved attempts via b2 - 4ac > 0. It was not always obvious to see how a student had solved their quadratic (formula not quoted and values for a, b and c not seen) but a limited number of correct solutions were seen. Using a calculator quadratic solver was frequent and acceptable. Several attempts started by using implicit differentiation on the circle equation. Although this is a possible method, it did not often go beyond using dy/dx = 3 to get x+3y+1= 0. Another method seen several times was to find the equation of the normal ( i.e. the radius ) to try to make progress towards finding values of k, but this method also seldom went further than the equation of the line.
The circle C has equation x^2+y^2-10x+4y+11=0. The line l has equation y=3x+k, where k is a constant.
Edexcel 9MA0 · October 2021 · Paper 1 · Q13 · 3 marks
3 marks
Examiner report
A solution gaining all three marks was relatively rare to see, and there were many very poor demonstrations of algebraic manipulation seen. Many candidates managed to score the first M mark but usually were unable to make further progress. The attempts were split between candidates who attempted to substitute into the LHS of the Cartesian equation to give an equation/expression in t and those who attempted to eliminate t, usually via t 2 = (5-x)/(x-1), to give an equation involving x and y only. The errors then ranged from incorrect squaring of a bracket to incorrect manipulation of fractions with very few unable to complete the proof
A curve C has parametric equations x=(t^2+5)/(t^2+1) and y=4t/(t^2+1), for real t.
Edexcel 9MA0 · October 2021 · Paper 2 · Q8 · 9 marks
9 marks
Examiner report
Many candidates gained some credit in this question but few gained full marks. Part (a) was relatively well answered. Most candidates identified that implicit differentiation was required and made good progress by correctly differentiating the given equation. When mistakes happened at this stage, they usually appeared in the differentiation of rather than in the differentiation of or . There were a few candidates, however, who clearly were not able to employ implicit differentiation and who gained no marks. Once a differentiated expression had been obtained, many were able to rearrange correctly to make the subject, obtaining the required answer of . A common mistake, however, was to make sign errors in the rearranging process and incorrect answers like and were surprisingly common following correct differentiation. In (b) most candidates gained some marks but few used a fully correct method to achieve the correct answer of and . Most were able to rearrange the equation of the normal, and identified that its gradient was . They then identified the gradient of the tangent as and used their result from (a) at and to attempt to form a correct equation in and such as . A number of candidates incorrectly equated the numerators and denominators to form the equations and and solved these to fortuitously arrive at the correct solution, losing most marks in (b). Another incorrect approach was to form the equation of the normal of the form and to rearrange to the form followed by 'comparing coefficients' with (an incorrect method). Many candidates successfully substituted and into the equation of C to get . This step was missed, however, by quite a few candidates, who could not then gain full marks for the question.
The curve C has equation px^3 + qxy + 3y^2 = 26, where p and q are constants.
Edexcel 9MA0 · October 2021 · Paper 2 · Q13 · 6 marks
6 marks
Examiner report
This question involving differentiation of parametric equations was found to be challenging by the majority of students, particularly the differentiation of 𝑦𝑦= cosec3 𝜃𝜃 and part (b). The majority of candidates knew what was expected in part (a) and proceeded to attempt to use the chain rule to find d𝑦𝑦 d𝑥𝑥, however they found differentiating 𝑦𝑦= cosec3 𝜃𝜃 challenging. Some did not check in the formula book that the derivative of cosec θ was negative, or omitted the constant 3, and this cost 3 of the 6 marks available in the whole question. For those that attempted to write 𝑦𝑦= sin−3 𝜃𝜃 and use either the chain rule or the quotient rule success was mixed. Other common incorrect derivatives included d𝑦𝑦 d𝜃𝜃= (cosec3 𝜃𝜃cot3 𝜃𝜃) and variants of this. Differentiating 𝑥𝑥= sin 2𝜃𝜃 proved less troublesome, with most doing so correctly. There was a proportion of candidates that rewrote sin 2𝜃𝜃 as 2 sin 𝜃𝜃cos 𝜃𝜃 and used the product rule, creating more work but ultimately doing so successfully in most cases. The method mark was scored by the majority of candidates as a result, although there were a few candidates who tried to write 𝑦𝑦 in terms of 𝑥𝑥 and differentiate, with no marks being scored for this approach. A number of candidates, having found a correct derivative, proceeded to attempt to tidy this up, unfortunately using time that could have been spent elsewhere in the paper. Part (b) proved more challenging for many candidates, with some unable to make a start as they were unable to process 8 = cosec3 𝜃𝜃 to arrive at either a value for sin 𝜃𝜃 or 𝜃𝜃. Of those that did arrive at a value for sin 𝜃𝜃 or 𝜃𝜃, many were able to work in exact form using their value to find an exact value for their 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑, others were unable to proceed or attempted to give a decimal answer. 24√3 and 8√3 were common incorrect final answers. There were, however, some excellent and concise responses to this question.
The curve C has parametric equations x = sin(2theta) and y = cosec^3(theta), where 0 < theta < pi/2.
Edexcel 9MA0 · October 2020 · Paper 1 · Q7 · 5 marks
5 marks
Examiner report
A very significant majority of the candidates did not seem to know how to define a region using inequalities. It was common to see a response where the area of region R was attempted using integration. It was only a small minority of candidates who made a correct inequality statement at the end, even with the allowance of follow through on their quadratic curve and their line equations. It was often the case that this final mark was lost due to a candidate using the letter R rather than y in their inequality. Finding the equations of the boundaries of the region is essential when defining the inequalities, so candidates who went on to find an area were mostly still trying to do this and if successful would have only lost the final mark. A very small minority of candidates found the area below the line by using a trapezium area formula and so did not manage to gain the marks for the line equation. Finding the line equation correctly was achieved by a majority of candidates, although a slip with the line gradient did occur occasionally even if the diagram showed a convincingly positive gradient. Finding the equation of the quadratic curve was found to be more difficult, with a common error being f(x)=(x+2)2+13. Occasionally a candidate spotted that this was not consistent with the curve going through (0,25) and tried to adjust their answer, but not often in a correct way. Attempts at using f (x)=a(x+2)2+13 were often successful at finding a, whilst attempts at using f (x)=ax2+bx+c were much less often correct.
A quadratic curve C, y=f(x), meets a straight line l at (-2,13) and (0,25). The point (-2,13) is the minimum turning point of C. The finite region R lies above C and below l between the intersection points.
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 2 · Q12 · 11 marks
11 marks
Examiner report
Many candidates left the entire question or parts of this question completely blank and clearly found it difficult to access the content being assessed here, perhaps due to a lack of familiarity with integration using parametric equations. Many tried to establish the Cartesian equation to answer various parts. That said, there were many candidates who provided clear and concise solutions to each part. In part (a)(i), most candidates recalled the formula for integrating parametric equations and knew to find 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑. Awareness of the double angle formula for sin2t was good. The main problem caused for these candidates was a lack of “dt” throughout their work, and only included this in the final line, losing the final accuracy mark. There was a significant proportion of candidates who attempt ed some differentiation using the chain rule: 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑= 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 x 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 , scoring no marks. In part (a)(ii), many knew that the integral was a multiple of cos3 t but had trouble getting to the printed answer of 20 with correct steps. As the answer was given, it was a requirement to see the limits calculated separately as 0 – ( -20 ) = 20. Integration by parts was a common approach for unsuccessful candidates, while others attempted to use a substitution, often not using the necessary u = cosx. Those candidates with an understanding of the integration of f’(x)[f(x)]n were generally very successful here. Some lost the penultimate accuracy mark usually due to a sign error. In part (b) many candidates who had found part (a) difficult gave up and lost the opportunity of scoring some easier marks. Many candidates who scored no marks elsewhere were still able to access at least the first two marks in part (b) by setting sin2t = 4.2 5 and solving, frequently getting at least one correct angle; but too many of these found their second value of t by subtracting their first value from π instead of calculating π – 𝑠𝑠𝑠𝑠𝑠𝑠−1( 0.84) before dividing by 2. The following mark was for finding two values of x, but many simply substituted their first value in to find x, thinking that this gave them the width of the walkway. Too many marks were lost by approximating too soon and hence not having values with the required accuracy. A few lost the last mark because they omitted the units.
A curve C is defined parametrically by x = 6sin(t), y = 5sin(2t), for 0 <= t <= pi/2. The region R lies between C and the x-axis. The same curve models the cross-section of a dam. A horizontal walkway at height y = 4.2 meets C at M and N; its width is the horizontal distance MN.
Edexcel 9MA0 · October 2020 · Paper 2 · Q14 · 7 marks
7 marks
Examiner report
Part (a) presented some thinking challenges for many candidates, in particular spotting that a = b = r in their equation for the circle. About half of the candidates were not able to deduce the correct equation of the circle. The most commonly seen incorrect equations were 𝑥𝑥2 + 𝑦𝑦2 = 𝑟𝑟2 , (𝑥𝑥+ 𝑟𝑟)2 + (𝑦𝑦+ 𝑟𝑟)2 = 𝑟𝑟2 and (𝑥𝑥 ± 𝑎𝑎)2 + (𝑦𝑦 ± 𝑏𝑏)2 = 𝑟𝑟2. Some of these candidates managed to gain the method mark for using 𝑦𝑦= 12 – 2𝑥𝑥, but a disappointingly high number didn’t progress. Relatively few candidates who initially used 𝑎𝑎 and 𝑏𝑏, correctly replaced them with 𝑟𝑟 to attain at least one mark. Some candidates tried to use a sketch to help them. This did allow some to identify the centre as (𝑟𝑟, 𝑟𝑟), but many put the circle in the wrong quadrant and proceeded to use (𝑟𝑟, −𝑟𝑟) or (−𝑟𝑟, 𝑟𝑟). The majority of candidates who gained the B1 and M1 expanded their expressions correctly to give the correct equation and attain the A1, although the algebra involved in squaring the 3-term expression (𝑟𝑟2 −24𝑟𝑟+ 144)was not always carried out correctly. Candidates were generally more successful when they expanded (𝑦𝑦−𝑟𝑟)2 first before substituting 𝑦𝑦= 12 − 2𝑥𝑥. Part (b) proved to be more accessible to most candidates. Many recognised the need to use the discriminant and equate to zero, though poor algebraic manipulation often cost them the accuracy marks. Almost all candidates used a calculator to solve the quadratic, though there were examples of the quadratic formula being used, mainly correctly. A small number resorted to calculus to find a gradient, but very few successfully attempted to differentiate their equation and set the gradient to -2. A few tried to use the idea of gradient or even equate the equation with 12 – 2𝑥𝑥.
A circle of radius r lies in the first quadrant and touches both coordinate axes. The line 2x + y = 12 intersects the circle.
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 · Q7 · 7 marks
7 marks
Examiner report
Q07 discriminated well between the medium and higher ability candidates. Lower ability candidates struggled to make much progress with this modelling question and most of them scored no more than one mark. In Q07(a), most candidates wrote down a correct equation for the model in the form , y kx c with constants k and c. A minority struggled to give a suitable form of the linear equation, with some omitting the fixed cost constant, c. Some candidates gave incorrect answers, such as ; y x equations involving exponential models; or differential equations. Q07(b) proved challenging for those candidates who struggled to grasp the concept ‘profit sales cost ’, with many of them incorrectly assuming that the model from Q07(a) was for the profit made. Many of these candidates used (800, 500) and either (300, 80) or (300, 80) to form two linear equations, with some erroneously finding a gradient of 0.84. Those candidates who understood the link between profit and cost usually found the total cost to the factory of £1100 and £680 for 800 and 300 bars of soap respectively. Most used (800, 1100) and (300, 680) to form two linear equations and proceeded to find the gradient, 0.84, and y-intercept, 428. A large proportion concluded by stating the given equation 0.84 428. y x A few candidates, however, did not complete their analysis with a suitable conclusion, and so lost the final mark. Some candidates used (800, 1100) and (300, 680)to verify the model 0.84 428. y x Many of these were successful but often failed to give a suitable conclusion stating that the given model was true. Q07(c) gave a significant number of candidates their only creditable access to Q07. Some candidates gave a correct interpretation such as 0.84 represented ‘the cost of making an extra bar of soap’. Incorrect explanations included that 0.84 represented ‘the selling price of a bar of soap’ or ‘the profit per bar of soap made’. There was a mixed response to Q07(d). The most popular approach was to form and solve a correct 2 0.84 428 n n or 2 0.84 428, n n which mostly led to the correct answer 369. n A final answer left as an inequality such as 368 n or 369, n without reference to the statement that 369 was the number of bars required, was not sufficient for the final mark. Some candidates used trials of both 368 n and 369 n in a complete trial and error method leading to a correct conclusion of 369. n
A small factory makes bars of soap. On any day, its total cost y pounds for making x bars is modelled as a fixed cost plus a cost proportional to x. Bars sell for 2 pounds each. When 800 bars are made and sold the profit is 500 pounds; when 300 bars are made and sold the loss is 80 pounds.
Edexcel 9MA0 · June 2018 · Paper 1 · Q6 · 10 marks
10 marks
Examiner report
Part (a) was a ''show that'' question in which students used the idea of perpendicular gradients to find the equation of line PA. It was very straightforward and well done with almost all students scoring the 3 marks. Part (b) required the students to find the equation of the circle C using the information given. Again most students attempted this with ease, finding firstly the coordinates of point P, and then the radius PA, before writing down the equation of the circle. Errors witnessed tended to be arithmetic, although some students complicated the question by attempting to solve 2 2 2 7 5 x y r and 2 1 y x simultaneously in an attempt to find r. There were numerous ways to find the value of k in part (c). One of the best involved using vector geometry to find the coordinates of the point N on C, where 2 y x k meets C. Using the fact that PA AN enabled students to find N = (11,3) and then k by substituting this point into 2 . y x k Again simultaneous equation methods were common, some more successful than other. The most common correct attempt involved solving 2 y x k with 2 2 7 5 20 x y and then finding where the resulting equation had one root. Care was required with the algebra, especially the squaring of terms, as well as the application of the discriminant 2 4 0. b ac This proved to be too great a challenge for many.
A circle C has centre A=(7,5). The line l, with equation y=2x+1, is tangent to C at P.
Edexcel 9MA0 · June 2018 · Paper 1 · Q14 · 10 marks
10 marks
Examiner report
The final question on the paper was a challenge to all. Part (a) was straightforward with most recognising the need to use the identity 2 cos2 1 2sin t t in some form to prove the required result. This was one of the more successful ''show that'' parts of the paper. Only stronger students, however, were able to understand and explain why the curve C did not include all points of the curve 2 6 3 . y x Many drew the whole of the parabola in part (b) scoring one out of the three marks. Those who managed to sketch the correct part of the parabola rarely were able to explain why the domain (and range) were restricted. Stating that ''because 0 2 t '' was not enough to explain why for instance 1 5 x . The best students were able to explain that as 1 sin 1 t then 3 2sint would have a maximum value of 3 2 5 and a minimum value of 3 2 1. Part (c) was also demanding and better students annotated their graphs to help their understanding of the question (See Figure 1). Most however used the discriminant condition 2 4 0 b ac for two roots and were able to proceed to two values. Arithmetic errors were quite common using this method. Most students who proceeded to the answer were well versed in writing their answer using set notation. Figure 1
A curve C has parametric equations x=3+2sin(t), y=4+2cos(2t), for 0<=t<2pi.
Edexcel 9MA0 · sample · Paper 1 · Q3 · 4 marks
4 marks
A circle C has equation x^2+y^2-4x+10y=k, where k is a constant.
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 2 · Q8 · 7 marks
7 marks
ABCD is a rectangle. A lies on the y-axis, while B and D lie on the x-axis. The side AB lies on the line 5y + 2x = 10, and AD is perpendicular to AB through A.
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 · Q9 · 9 marks
9 marks
A circle with centre A(3, -1) passes through P(-9, 8) and Q(15, -10). A point R also lies on the circle.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q10 · 9 marks
9 marks
A curve C has parametric equations x = ln(t + 2), y = 1/(t + 1), for t > -2/3. A finite region R is bounded by C, the line x = ln(2), the x-axis and the line x = ln(4).
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 coordinate geometry in the x y plane is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.