June 2024: AS AS Paper 1: Pure Mathematics

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a very accessible question on integration to start the paper and most candidates were able to score well. Most candidates attempted to separate the fraction into two terms and achieved at least one correct index. The first term caused problems for some with errors such as 5 3 1 2 2 2 , , x x x - . A few ended up with three terms as they had added or subtracted 2 x-. A few tried to integrate the numerator and denominator or wrote 2 (2 3) x x- - but did not expand the brackets and integrated both inside and outside separately. Those who had scored the first M1 usually were able to raise the power by one on at least one term with a correct index and score the next method mark. Several made errors when processing the indices and ended up subtracting by one on at least one of the terms. A few made errors with the coefficients. The most common mistake was to forget the + c.

Common mistakes

• split into 2x^(−3/2) − 3x^(−2) before integrating (writing −3/x² is not enough)
• integrating numerator and denominator separately scores nothing
• index slips such as x^(−5/2) or subtracting 1
• forgetting + c is the most common error
• answer on one line, simplified

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question which required use of the factor theorem and solving simultaneous equations provided an opportunity to score a good number of marks. Some candidates found this question to be accessible and scored full marks or nearly full marks. Others however, encountered difficulties, particularly with part (b) and dropped marks accordingly. Some candidates lost marks because they did not pay heed to the instruction to show all stages of their working. In part (a), most candidates made progress and attempted to employ the factor theorem, as the question had instructed, and used f(4) 0 = to obtain the given answer. Some lost the second mark for not including ‘= 0’ anywhere in their proof or for algebraic slips. A few candidates attempted algebraic division and scored nothing in part (a).

Common mistakes

• use f(4) = 0 (factor theorem, as instructed)
• "= 0" must be seen
• algebraic division scores nothing
• at least one intermediate step before 10a = 32 + b

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question which required use of the factor theorem and solving simultaneous equations provided an opportunity to score a good number of marks. Some candidates found this question to be accessible and scored full marks or nearly full marks. Others however, encountered difficulties, particularly with part (b) and dropped marks accordingly. Some candidates lost marks because they did not pay heed to the instruction to show all stages of their working. For part (b), many candidates continued with the theme of the question and applied the factor theorem a second time using f(2) 0 = , obtaining a second equation in a and b. This was the easiest way to proceed. Surprisingly however, a significant number of candidates chose more difficult routes and progressed instead with entirely different methods. Many candidates had some success by expanding (2 )( 4)( 2) x k x x + - - and then comparing coefficients with the original function, obtaining further equations. Others attempted to divide the original function by ( 2) x - or sometimes by 2 6 8 x x - + , using long division. This approach was less successful because of the algebra involved and few reached a second equation. There were also a few attempts using the sum and product of the roots of a cubic equation. Some candidates attempted to use f(2) 0 = with f( ) (2 )( 4)( 2) x x k x x = + - - and made no progress in part (b). Generally, when candidates obtained further correct equations in part (b), they were then able to successfully solve them simultaneously to obtain values of a, b or k which were usually correct. This was often followed by the correct factorised expression for f(x), although a number of candidates failed to write this down and lost the final mark. A few correct answers were presented with no working, which gained no credit in part (b).

Common mistakes

• f(2) = 0 gives 8 + b = 2a
• solve with 10a = 32 + b: a = 3, b = −2, k = 3
• final answer (2x + 3)(x − 4)(x − 2) on one line
• expanding (2x + k)(x − 4)(x − 2) and comparing coefficients is longer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question which required use of the factor theorem and solving simultaneous equations provided an opportunity to score a good number of marks. Some candidates found this question to be accessible and scored full marks or nearly full marks. Others however, encountered difficulties, particularly with part (b) and dropped marks accordingly. Some candidates lost marks because they did not pay heed to the instruction to show all stages of their working. Part (c) was answered quite well, even by those who had made little or no progress in part (a) or part (b). In part (i) many identified that there were three real roots, although stating that there were two real roots was a common incorrect answer. Some also lost the mark because they listed the roots without stating how many there were. In part (ii) many realised that the largest root was 12, although 4 3 was a common incorrect answer.

Common mistakes

• (i) state "3", listing the roots alone scores 0
• 2 is a common wrong answer
• (ii) f(x/3) = 0 is a stretch scale factor 3 so largest root 3 × 4 = 12 (4/3 is a common wrong answer)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. In part (a), most candidates knew they needed to subtract the vectors, there were a few signs slips when doing this and quite a few candidates subtracted the wrong way round so found QP instead. There were a small but significant number who added the vectors.

Common mistakes

• PQ = q − p = −6i + 3j
• subtracting the wrong way round (QP) or adding the vectors

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. In part (b), quite a few candidates seemed unclear as to how to show 90 PQR = ; some made no attempt whilst others used the given fact within their proof often finding two angles in a triangle and showing that they were complementary, but they had assumed a right angle in order to utilise trigonometry which achieved no marks. The candidates who tended to be most successful at this were the ones who attempted to show the required property using gradients. There were a small number of signs slips and incorrect attempts at gradients but overall, this approach was usually successful. Quite a lot attempted to use Pythagoras’ theorem but there were a significant number who would only find PQ and QR and either did not attempt to find PR or had a poor attempt at finding it. It is of note that for those who successfully used Pythagoras’ theorem to complete the proof, diagrams proved useful. There were a small number who attempted the dot product or cosine rule and usually these attempts were good enough to achieve the marks. It was disappointing that the most common reasons for the M1A0 score were due to incorrect/incomplete solutions or simply a lack of conclusion.

Common mistakes

• gradients −1/2 and 2 multiply to −1
• or PQ² + QR² = PR² with PR found
• must conclude angle PQR = 90°
• assuming the right angle to use trigonometry scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall, this question proved fairly tricky with lots of candidates losing marks throughout. Part (c) proved quite tricky for a lot of candidates many of whom did not seem to have fully understood the shape with which they were working. Those who were successful usually had a good diagram identifying it as a trapezium and were able to find the area. There were a number of attempts at splitting the shape up into a rectangle and triangle or 2 triangles, but often there was a misunderstanding about which lengths were which and how they should be combined to find the area. The most common error in the rectangle and triangle approach was that the candidates failed to spot that the triangle was a right-angled triangle so could use QP as the ‘height.’ The two triangles approach would also often have the same issue. There were quite a few attempts that treated the shape either as just a rectangle or just a triangle. It was pleasing however to see that whilst a large number of candidates were unsure of how to proceed to find the area many did at least acknowledge the need to attempt the lengths of various vectors, which generally resulted in two marks being awarded.

Common mistakes

• PQRS is a trapezium (PS parallel to QR) with height PQ = 3√5
• QR = 9√5, PS = 27√5
• area = ½(9√5 + 27√5)(3√5) = 270
• a diagram helps
• treating the shape as a rectangle or parallelogram
• not spotting the right angle

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question using advanced trigonometry was generally answered well and as part (a) was a “show” question, this provided great accessibility for part (b), which candidates were typically more successful in completing. In part (a), whilst the vast majority of candidates were able to correctly state, and substitute into, the sine rule, a large proportion of them demonstrated a poor understanding of algebra and were unable to rearrange the equation to make x the subject. Many candidates, whilst arriving at the correct answer via valid algebra, used inefficient or inelegant techniques. Only a few students misquoted the sine rule e.g. misplacing the angle/opposite side pairings, using 30o and 140o instead of sin 30o and sin 140o, and writing sin x and sin( 3) x + . It was disappointing that for a “show” question where the focus is on providing a rigorous argument, that some candidates failed to work to at least 3 significant figures, instead approximating values such as sin140o too much and/or too early, which led to the answer mark being lost. Candidates should also be discouraged from working with approximated values when answering these types of questions. There was evidence of candidates resorting to using the equation solver on the calculator to find the correct value for x after having got stuck, or having made an error, whilst attempting to solve their equation, too, hence the requirement for ideally rearranging to find an expression for x before achieving the given answer.

Common mistakes

• sine rule x/sin 30° = (x + 3)/sin 140° with sides opposite the correct angles
• must rearrange to x(2 sin 140° − 1) = 3 or an expression for x before 10.5
• verification by substituting 10.5 scores 0
• poor rearranging algebra

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question using advanced trigonometry was generally answered well and as part (a) was a “show” question, this provided great accessibility for part (b), which candidates were typically more successful in completing. Candidates were more confident when using the cosine rule and so part (b) resulted in many achieving the correct answer. This may reflect the fact that they simply needed to substitute in the given correct value for x into the formula and then do the numeric processing i.e. no algebraic work was required. The vast majority of candidates realised that the most efficient solution used triangle ABD with an angle of 150o. Those that chose to use triangle ABC tended to arrive at the correct answer, although they probably took longer to do so, as there was an additional side length to calculate. A number of candidates did not substitute side lengths into the cosine rule, instead working with x and 3 x + . They then attempted to manipulate the resulting algebraic expressions. Some failed to realise that the value of x was given; others noticed this at the last minute, after considerable algebra. This was another example where, had candidates read the question carefully and realised the requirement was to give the answer to 3 significant figures, then the need to work algebraically was no longer necessary as an exact answer would not need be required.

Common mistakes

• cosine rule in triangle ABD with angle ADB = 150° (180° − 30°), AD = 8.5, BD = 13.5
• AB = 21.3 cm
• using 30° instead of 150°
• triangle ABC route is longer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested candidates’ ability to sketch graphs and to be able to divide polynomials. Part (a) required the sketching of a reciprocal graph. A significant proportion of candidates did not even attempt to sketch the graph; those that did often produced a good positive reciprocal graph with both branches, scoring the first mark. The second mark proved harder with many graphs being in the wrong place, with an extra asymptote and the x intercept not labelled. The y- axis with a dashed line next to it also resulted in losing the second mark as candidates were not aware that this was implying that their asymptote was not the y-axis. Part (a)(ii) required both the y = x = 3 0 equations of both asymptotes: was common but was not seen as often.

Common mistakes

• y = 6/x translated up 3: both branches, crossing the x-axis only at (−2, 0)
• asymptotes y = 3 and x = 0 (the y-axis, not a separate dashed line)
• x ≠ 0 is not an equation

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested candidates’ ability to sketch graphs and to be able to divide polynomials. In part (b), candidates were required to equate both curves and rearrange to obtain a given answer. This was very well answered with the most common error being slips on the indices and not equating the final answer to zero. Those who were unsure what was being asked just proceeded to solve the cubic on their calculator.

Common mistakes

• equate 6/x + 3 = 3x² − 4x − 10 and multiply through by x
• at least one intermediate line
• "= 0" needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested candidates’ ability to sketch graphs and to be able to divide polynomials. Many candidates did not heed the warning in part (c) that “solutions relying on calculator technology are not acceptable” and this cost them marks. Overall, this question had a wide spread of marks. Part (c) proved more difficult, and this is where the calculator warning was ignored. Many candidates simply stated the three roots with no working which scored no marks. Others tried to work backwards and failed to appreciate the requirement to use the given information. It is not acceptable to just use the calculator to solve to find the three roots, then above write the factorised version and then multiply out above this to achieve the cubic equation as there is not sufficient evidence that the given information had been used. If it was not factorised into ( ) x + or evidence that another factor had been found, 3 2 (quadratic) demonstrated by an algebraic technique such as algebraic division, or using the factor theorem then full marks could not be awarded. Just factorising by inspection using one of the other factors which have mysteriously appeared was not accepted and these approaches should be discouraged. Those that realised that algebraic division was 2 x + required scored well, a significant number of candidates choose to divide by rather 3 x + than the expected 3 . The resultant quadratic was then factorised to give the 2 additional points of intersection.

Common mistakes

• use the given root −2/3: divide by (3x + 2) to get x² − 2x − 3, then factorise
• x = −1 and x = 3
• three linear factors written down from a calculator score 0
• quadratic factor needed first

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a very accessible question and candidates clearly felt comfortable with this topic. In particular, parts (a) and (b) were well answered, although part (c) did provide much more challenge and discriminated candidates well. Part (a) was generally done well with most candidates able to form a correct equation and proceed to find a. Those who used 5 8 a = - usually forgot to write a conclusion. There were a few who expanded and achieved 2 2 1 12 66 ax a x + + but then did not equate the terms or substitute in a and did not score any marks.

Common mistakes

• ¹²C₁ × a = −15/2 so 12a = −15/2
• binomial coefficient must be evaluated
• conclusion needed if verifying with a = −5/8

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a very accessible question and candidates clearly felt comfortable with this topic. In particular, parts (a) and (b) were well answered, although part (c) did provide much more challenge and discriminated candidates well. In part (b), many candidates achieved the correct value for k. The most common mistake was to not square the a and ended up with 164 4 k = - . A few who just gave the answer as a decimal had rounded and lost the A mark.

Common mistakes

• k = ¹²C₂ × a² = 66 × 25/64 = 825/32
• forgetting to square a
• rounding a decimal

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a very accessible question and candidates clearly felt comfortable with this topic. In particular, parts (a) and (b) were well answered, although part (c) did provide much more challenge and discriminated candidates well. In part (c), a minority of candidates achieved both marks. A common mistake was to solve 5 17 1 8 16 x - = to find an appropriate value of x but to stop there and not substitute it into their expansion. A few substituted it back into 12 5 1 8 x - and ended up just evaluating 12 17 16 . For those who did substitute their x into the expansion, several had made errors when solving the equation and had an incorrect value of x. Again a few made rounding errors and lost the accuracy mark. Some candidates also found the next term in 3x which was not required and took up additional time, however, successful responses were credited with both marks.

Common mistakes

• solve 1 − (5/8)x = 17/16 to get x = −0.1 and substitute into 1 − (15/2)x + kx²
• awrt 2.0078
• stopping after finding x
• calculator value 2.0699 with no method scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested the understanding of logarithmic graphs and exponential growth. This was successful in differentiating between candidates, although surprisingly

Common mistakes

• log₁₀P = 3.3 − 0.2x so a = 10^3.3 ≈ 1995 and b = 10^(−0.2) ≈ 0.6310 (4 s.f.)
• gradient slips
• using 3.3 instead of 10^3.3 for a
• writing y for log₁₀P

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested the understanding of logarithmic graphs and exponential growth. This was successful in differentiating between candidates, although surprisingly candidates still struggle with this relationship. In addition, part (b) provided challenging for most who were unable to interpret in context. = x P ab Many candidates lacked confidence in relating the exponential form of the model to the log P against x. Whilst some very poor log work was seen, some straight-line graph of 10 candidates also made fundamental errors whilst forming the equation of a straight line, for - 6 0 = - 5 example, an incorrect method for calculating the gradient, e.g. - 2.1 3.3 log P when writing down the equation of the straight line. Many candidates used y instead of 10 log 3.3 within their Some candidates appeared confused as to whether they should use 3.3 or 10 equation, whilst others also showed a lack of robust understanding in manipulating indices after achieving a correct equation for P. A significant minority of candidates who achieved the correct values for both a and b failed to then write down the complete equation required. In part (b), the vast majority of candidates were unable to correctly explain the meaning of “ab”. ab = = 1005 0.6310 1259 Some thought that they simply had to calculate its value i.e . Many others assumed they needed to give the meanings of each of a and b separately, as on previous papers.

Common mistakes

• ab is the concentration (parts per million) 1 km from the chimney
• "amount of smoke" or "when x = 1" is not in context

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Marks awarded in this question were generally quite varied. Often candidates made progress in (c) only, by attempting some integration with little correct work seen in (a) or (b). Each part of this question presented a given answer, and candidates who failed to observe the warning in the question about showing all stages of their working, or relying entirely on calculator technology, often lost marks. In part (a) candidates who identified how to proceed attempted two approaches. Some differentiated the equation of the given curve and set the answer equal to 2 - to obtain 2 3 14 2 x - = -. Others subtracted the equations of the curve and the line to obtain 3 12 16 x x - + and used the derivative of this to obtain 2 3 12 0 x - = . A common error was attempting to solve 2 3 14 0 x - = instead. Candidates then proceeded either to solve their equation to obtain 2 x = , or substituted 2 x = into their equation to verify the solution. Some candidates lost the final mark in part (a) for reaching 2 x = and failing to select 2 x = . Other candidates often failed to make any progress in part (a) at all. These often attempted to substitute 2 x = into both the equations for the curve and the line or tried to solve 3 12 16 0 x x - + = . Neither of these approaches gained credit because of the requirement of the question for use of calculus.

Common mistakes

• set dy/dx = 3x² − 14 equal to the gradient −2 (or differentiate curve − line and set = 0)
• solving 3x² − 14 = 0 is a common error
• equating curve and line is not calculus (M0)
• must reject x = −2

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Marks awarded in this question were generally quite varied. Often candidates made progress in (c) only, by attempting some integration with little correct work seen in (a) or (b). Each part of this question presented a given answer, and candidates who failed to observe the warning in the question about showing all stages of their working, or relying entirely on calculator technology, often lost marks. Those who attempted part (b) employed many varied methods. The easiest approach was to simply to verify that both the line and the curve passed through the point ( 4, 15) - showing sufficient working. Many though, subtracted the equations of the line and the curve again to obtain 3 12 16 y x x = - + . Candidates then either verified that 4 x = - was a solution of 3 12 16 0 x x - + = , or attempted to solve this equation. Those who reached 4 x = - by factorising via 2 ( 4)( 2) 0 x x + - = were in a position to score both marks, but a number relied on their calculator equation solver, and could only be awarded the first mark in part (b). Candidates often lost the final mark for not writing a full conclusion. To score full marks, candidates needed to reach 4 x = - showing sufficient working using a valid method, and also indicate that they had completed the proof.

Common mistakes

• substitute x = −4 into both equations to get y = 15 and conclude
• or factorise x³ − 12x + 16 = (x − 2)²(x + 4)
• stating (−4, 15) alone is insufficient

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Marks awarded in this question were generally quite varied. Often candidates made progress in (c) only, by attempting some integration with little correct work seen in (a) or (b). Each part of this question presented a given answer, and candidates who failed to observe the warning in the question about showing all stages of their working, or relying entirely on calculator technology, often lost marks. In part (c) candidates generally knew that they needed to proceed with integration, and often the first two marks were scored for correctly integrating even if they made no further progress. Successful candidates usually attempted to evaluate 2 3 4 14 23 d x x x - - + and then subtracted 2 4 2 7 d x x -- + . Some used the area of a trapezium to give 1 (3 15) 6 2 + instead of integrating the equation of the line. Alternatively, some attempted to evaluate 2 3 4 12 16 d x x x - - + reaching the required area of 108 in one step. Candidates were required to make their method clear, including showing the use of the limits 4 - and 2. Some failed to do so, clearly relying on calculator technology to evaluate the integrals. Others failed to show the complete correct strategy for finding the area and often incorrectly identified the trapezium as a triangle. Also, correct notation was required and appropriate use of the integral sign and the ‘dx’ on either side.

Common mistakes

• ∫ from −4 to 2 of (x³ − 14x + 23) dx minus the trapezium ½(15 + 3) × 6 = 54, or ∫(x³ − 12x + 16) dx
• correct limits 2 and −4
• integral sign and dx needed
• calculator integration scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question, testing the topic of logarithms, was poorly answered with the majority of candidates simply unable to engage with parts (b) and (c). Some candidates did not appear to understand what was being asked of them. This portrayed both a lack of understanding of logarithms, and also a lack of practice on similar past paper questions. Part (a) however, was answered relatively well, with many candidates reaching the correct answer of 2p. Errors included, for example, candidates writing log 256 2log 16 a a = without going onto write this in terms of p. Many appeared to think they had fully answered the question because it is given that log 16 a p = .

Common mistakes

• log 256 = 2 log 16 = 2p
• stopping at 2 log₂16

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question, testing the topic of logarithms, was poorly answered with the majority of candidates simply unable to engage with parts (b) and (c). Some candidates did not appear to understand what was being asked of them. This portrayed both a lack of understanding of logarithms, and also a lack of practice on similar past paper questions. Similarly, in part (b), many candidates wrote log 100 log 25 log 4 a a a = + but never proceeded further than this. These candidates lost the A mark, seemingly thinking they had given a complete correct answer.

Common mistakes

• log 100 = log 4 + log 25 = ½p + q
• stopping at log 4 + log 25 without writing log 4 in terms of p

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question, testing the topic of logarithms, was poorly answered with the majority of candidates simply unable to engage with parts (b) and (c). Some candidates did not appear to understand what was being asked of them. This portrayed both a lack of understanding of logarithms, and also a lack of practice on similar past paper questions. It was rare in part (c) for candidates to make any progress, having already struggled on part (b). The lack of the use of brackets often caused issues, too. It was rare for candidates to progress to the impressive, simplified solution of 2 2 1 4 p q - . The general understanding of the laws of logarithms was very poor. Examples of conceptual misunderstandings were numerous. E.g. ( ) 2 2 log 256 log 16 a a p = = , log 16 a p = so log 256 16 a p = , log 25 4 4 4 a q q = = , log (80 3.2) log 256 2 a a p = = . There were, of course, some excellent solutions, too.

Common mistakes

• log 80 = log 16 + log 5 and log 3.2 = log 16 − log 5, with log 5 = ½q
• answer (p + ½q)(p − ½q) = p² − ¼q²
• very poorly answered
• missing brackets
• false laws such as log(80 × 3.2)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question required problem solving skills and focused on coordinate geometry and circles. Part (a) required candidates to obtain a given equation. The most common route for this was to work out the gradient between P and Q and to set this equal to the negative reciprocal of the gradient of the tangent. Candidates attempting this route scored well. The other common route was to find two equations for radius using the coordinates of P and Q and the gradient of 1 2 . In general, candidates who had a strategy for the question did well on this part with just bracketing and sign errors being the major problem. Unfortunately, many did not appreciate the relevance of the gradient of the tangent of l would enable them to find the gradient of the normal to the curve at Q (hence the equation of the line between P and Q) and only typically scored one mark for being able to form an equation using the coordinates for Q with the given gradient of 2 - and not knowing where to go from there.

Common mistakes

• tangent gradient −2 so radius PQ has gradient ½
• (k² − 2k − k − 8)/(3 + 1) = ½
• "= 0" needed
• many did not link the tangent to the gradient of PQ

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question required problem solving skills and focused on coordinate geometry and circles. Part (b) required candidates to find the equation of the circle. Most candidates found both values of k but ignored the information in the question that k was a positive constant. Those who did appreciate this and used 5 k = went on with good success to find the coordinates of P, giving them the centre of the circle and Q which allowed them to calculate the radius and complete the question. This part was more often successfully answered than the “show” part (a), which showed good accessibility, as well as good exam technique by many candidates to still attempt a later part, even if they were unable to successfully answer the earlier part.

Common mistakes

• k = 5 only (k positive, reject −2)
• centre (−1, 13)
• r² = PQ² = 4² + 2² = 20
• (x + 1)² + (y − 13)² = 20

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was a modelling one relating to the price per gram of two different metals. It enabled candidates to demonstrate understanding of working with exponential functions and differentiating them. In part (a), a significant proportion of candidates scored both marks, achieving the correct answer of 240. A few candidates halved instead of doubled and got 60 and a few left p as 120. A few made errors when evaluating 0 20e or 0 40e thinking this was equal to 1 (or making a careless slip).

Common mistakes

• V_A at t = 0 is 100 + 20 = 120, so p = 240
• halving (60) or leaving 120
• e⁰ = 1

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was a modelling one relating to the price per gram of two different metals. It enabled candidates to demonstrate understanding of working with exponential functions and differentiating them. Part (b) was generally poorly answered, and a number of candidates made no attempt to answer this part. A common error was to set A B V V = . For those that found d d A V t and d d B V t some did not put them equal to each other, or they set one or both equal to 0 instead, but the most common mistake was to keep the negative and write 0.04 0.02 0.8e 4.8e t t - = - thus forfeiting the last three marks. The minority who did set d d A V t d d B V t = - were usually able to use a correct method to solve the equation. Some candidates did appreciate the issue of trying to take logs of a negative was not possible, but rather than go back to the original equation, they opted to either ignore it or scribble out the negative at some point of the solution which was not acceptable to score these final marks.

Common mistakes

• rate of increase of A equals rate of decrease of B: 0.8e^(0.04T) = 4.8e^(−0.02T) (drop the negative sign)
• e^(0.06T) = 6 so T = 29.9 months
• setting V_A = V_B or derivatives = 0
• logs of a negative

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

coming unstuck on part (a) without an attempt at later parts. This was a shame particularly as later parts of the question provided a good source of marks for many candidates. In part (a) attempts to find expressions for the surface area and perimeter of the given shape were often successful, although errors and slips were not uncommon. A correct expression for the perimeter was perhaps more often successfully achieved than a correct expression for the area. A relatively common error was in setting x as the diameter of the sector rather than the radius. A significant number of candidates found expressions in terms of r and could earn no marks until r was replaced by x. A minority of candidates equated the perimeter, rather than the area, to 100, but the vast majority made an attempt at the key step of rearranging their equation to find an expression for y in terms of x. Some candidates were more unusual in their choice of substitution: rather than finding an expression for y or 2y, they used effective and efficient substitutions for functions such as 100 x . Once the critical substitution was achieved, most candidates went on to successfully obtain the required result, although occasionally marks were lost due to arithmetical and/or sign errors, or for the omission of ‘P =.’ y using their value of x and the relationship derived in part (a), to determine whether y was greater than the required lower limit of 1. This approach was usually the most successful. Slightly more circuitous routes were also seen which involved employing the given expression for P to determine the minimum perimeter and then an additional y = and either x = 1 expression for P in terms of x and y, used with (to find P) or 10 P = (to find x). Often this approach fell slightly short of being complete as a final 40 comparison of x with the value 10, or a comparison of P with the value 40 was required to justify whether the swimming pool dimensions were suitable. Some candidates tried to draw conclusions but gave insufficient supporting statements such as ‘10.8 is close to 10, so suitable’, or ‘10.8 > 0, so suitable’ rather than the relevant ‘10.8 > 10, so suitable’. Other candidates were unable to gain marks here for irrelevant statements unsupported by calculation such as ‘suitable, although the pool might be a bit small.’ Others were x = confused about what value of x they should use and erroneously substituted (the 0.4 x = result value of the second derivative at ) into either their expression for y or their 10 expression for P.

Common mistakes

• area 2xy + πx²/4 = 100 gives y = (400 − πx²)/8x
• perimeter 2x + 4y + 2πx/4
• substitute to reach P = 2x + 200/x
• x is the radius, not the diameter
• equating perimeter to 100

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) required differentiation to determine the position of a stationary point and the vast majority of candidates were well versed in this process; the majority earning full marks in this part. There were a few errors arising from errors in differentiation for example in the sign or the coefficient of the 200 x term. Other errors arose from poor rearrangement of the equation d P x = - x , resulting in , or in the inclusion of the 0.1 d x = - x = root or stating a spurious . 10 0

Common mistakes

• dP/dx = 2 − 200x^(−2) = 0 gives x = 10 only (reject −10)
• sign/coefficient errors on the 200/x term

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (c), differentiation to obtain the second derivative was commonly successfully achieved. However, it was surprising that a significant minority of candidates felt that determining the second derivative was sufficient, neither evaluating it for their value of x, nor reasoning why it was positive for all positive x. Others lost the final mark in this part for a lack of conclusion. A number of approaches to part (d) were observed. Most commonly, candidates evaluated

Common mistakes

• d²P/dx² = 400x^(−3) = 0.4 > 0 at x = 10 so minimum
• a conclusion is needed
• just finding the second derivative is not enough
• checking the gradient either side scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A number of approaches to part (d) were observed. Most commonly, candidates evaluated y using their value of x and the relationship derived in part (a), to determine whether y

Common mistakes

• y = (400 − 100π)/80 = 1.07 m ≥ 1 so suitable
• "1.07 > 0" is wrong comparison

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The trigonometric equation, again, demonstrated how candidates are usually successful in solving these types of equations, however, they often struggle with the manipulation to show a given result. Part (a) held one of the highest demands in the paper and showed a clear differentiation between the most capable mathematicians and those who were familiar with trigonometric identities but struggled to apply and manipulate them. Overall, this proved to be difficult for most candidates with quite a few gaining one or two marks but very few were able to gain all four marks. Multiple attempts were seen which showed good resilience but further evidenced the complexity of the question. Most candidates were able to gain the first M mark which usually was for using sin tan cos = , (even if this followed from incorrect work which was quite often) quite a few candidates attempted to divide by cos as they seemed to spot it would give them a tan in their working, however not many recognised the need to divide by 2 cos and those that did would often forget to divide the 4 on the right hand side by it as well; they also tended to replace 2 2 7sin cos with 2 3tan . Of the candidates who did manage to pick up the first two marks, many could not reach a 3TQ in tan and thus the required answer. The successful approaches would often start with using 2 2 sin cos 1 + = on the right-hand side and proceed from there or work backwards from the given answer.

Common mistakes

• divide every term by cos²θ and use tanθ = sinθ/cosθ
• 4/cos²θ needs sin²θ + cos²θ = 1 (or sec²θ = 1 + tan²θ)
• dividing only some terms
• final answer in θ with correct notation

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The trigonometric equation, again, demonstrated how candidates are usually successful in solving these types of equations, however, they often struggle with the manipulation to show a given result. Part (b) was much more accessible due to the given answer in (a) not being a “show” which required constants to be found. Candidates achieved much more success with this part with most able to find the required roots and use these to find at least two of the required angles. There were occasional sign errors and a very small number attempted to solve the original equation from (a) without success. There were quite a few candidates who did not find all the required angles. This would often be because they would reject one of the solutions, 2 3 - or the angle it gave being negative seemed to be a reason that candidates ruled it out. The other root, 2, would often be ruled out presumably because of a misconception that the range of all trigonometric graphs was |𝑦| ⩽1. There were a small number who used an incorrect method to find other values, such as adding 90, or subtracting from 180 from their angles. Quite a lot of candidates did manage to find all the required angles, however.

Common mistakes

• roots tanθ = 2 and −2/3
• four answers 63.4°, 146.3°, 243.4°, 326.3°
• rejecting the negative root or missing the third/fourth quadrant solutions

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The trigonometric equation, again, demonstrated how candidates are usually successful in solving these types of equations, however, they often struggle with the manipulation to show a given result. Part (c) was usually poorly attempted with the vast majority not evidencing an understanding of graph transformations and just adding 63.4 to 720 or multiplying 243.4 by 4 as this would give an angle in the required range.

Common mistakes

• 4α = θ + 360n: 4α in (2880°, 4320°) gives 4α = 2880 + 63.4, α = 735.9°
• common errors adding 63.4 to 720 or 4 × 243.4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be a challenge for the majority of candidates and full marks were not common. The instruction to ‘use algebra’ was lost on a number of candidates who employed logical approaches or attempted to test an array of integer values for n. Perhaps a casualty of being the last question on the paper, there were also a number of blank responses seen. For those employing a logical approach, some credit was given when algebra was used to factorise the quadratic followed by a consideration of the parity of the factors. This was often left incomplete however, and so rarely scored the two marks out of four for a complete attempt via this approach. The majority of candidates who attempted the question did grasp the need to consider the general form for odd and even integers and made a reasonable attempt. Although some candidates incorrectly used ‘ 1 k + ’ or considered even integers twice e.g. ‘ 2k ’ and ‘ 2 2 k + ’. Unfortunately, marks were very commonly, yet surprisingly, lost due to basic mistakes in simple algebraic manipulation. These errors occurred in both the expansion and the factorisation of brackets and included errors such as 5(2 1) 10 1 k k + = + or 2 2 4 10 2( 5 ) k k k k + = + . It was surprisingly common for candidates to use n in their general form i.e. ‘ 2 n n = ’ which led to the loss of the final mark, especially since this has been in previous mark schemes that this would not garner full marks. A large proportion of candidates failed to give an overall conclusion to their proof which was also penalised with the final mark. A small minority of candidates employed proof by induction and, when seen, this was usually confidently executed.

Common mistakes

• consider n = 2k and n = 2k + 1: 4k² + 10k = 2(2k² + 5k) and 4k² + 14k + 6 = 2(2k² + 7k + 3)
• give a reason each is even and a final conclusion
• testing values or pure logic (odd × even) without algebra scores little