P2.3 Quadratics

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

Examiner's comment

In part (a), most candidates showed good understanding of how to write the equation of a circle in a suitable form to extract the coordinates of the centre and the radius. There were occasional slips with the centre coordinates given as any combination of (±5, ±2). It was more common to see errors in manipulating the constants to get the correct radius. Candidates should be reminded to leave answers in the exact form, unless prompted by the question to round to a certain degree of accuracy.

Common mistakes

• complete the square (x + 5)² + (y − 2)²: centre (−5, 2), radius √28 = 2√7 exact
• sign errors in the centre
• errors with the constant for r

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

Examiner's comment

In part (b), a variety of approaches were seen, many attempting an algebraic method using the equation of the circle and discriminant rather than a geometrical approach. The geometrical approach was generally much more successful and succinct. Many candidates would have benefitted from drawing a diagram illustrating a circle and a horizontal line with equation y = k to enable them to visualise the situation. Some diagrams with y = k drawn as a diagonal line were seen. Algebraic methods were generally less successful, with errors in the algebra and errors in identifying the coefficients to apply to the discriminant. Without the context, candidates often chose the wrong range of values since the quadratic inequality obtained contained a negative 2x term. Many candidates did not provide a final answer in set notation despite identifying the correct range of values. Use of “union” instead of “intersection” was seen quite frequently.

Common mistakes

• geometric: y-coordinate of centre ± radius gives 2 − 2√7 < k < 2 + 2√7
• using the x-coordinate (5 ± 2√7) is wrong
• discriminant method error-prone (negative k² term leads to the wrong region)
• answer must be in set notation, not ∪ and not ⩽

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

Examiner's comment

Part (a) was typically answered well, with the majority of candidates attempting the substitution of 175, including reasonable explanations to secure both marks. Some responses included erroneous brackets around -175, resulting in losing the negative sign when squaring.

Common mistakes

• substitute x = 175: P = −1575 so the company makes a loss (must say it will, not might, make a loss)
• brackets round −175 lose the sign when squaring

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

Examiner's comment

Part (b) of the question was completed less successfully. A common error saw candidates replacing P with 200 000 instead of 200, leading to incorrect solutions and loss of three marks. While some candidates annotated the figure with a horizontal line at 200 to enable them to visualise that there were 2 possible solutions that gave P = 200, the word ‘highest’ in the question prompted some pupils towards an optimisation approach, differentiating and setting = 0, ignoring the minimum profit. A large number of candidates lost the final mark due to either losing the £ sign from their final answer or not giving the answer to two decimal places as is expected when working with money.

Common mistakes

• P is in thousands so set P = 200, not 200 000
• solve x² − 260x + 16 650 = 0: 114.19 < x < 145.81
• highest price £145.81 (2 d.p. with £
• £145.82 is wrong)
• differentiating to find the maximum ignores the profit condition

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

Examiner's comment

It was pleasing to see many candidates successfully completing the square in part (c). However, there were also a large number who struggled with the negative coefficient of 2x and some candidates multiplied the whole expression by -1 and many incorrectly incorporated it into the bracket resulting in various combinations of signs. Many made errors finding the constant and a large number of candidates confused this part with part (b) and attempted to complete the square on the wrong quadratic expression. There were a number of responses which did not simplify the final answer, resulting in a loss of the final mark.

Common mistakes

• complete the square with the negative x² coefficient: P = 450 − (x − 130)²
• sign errors with the −1
• must be simplified
• completing the square on the (b) quadratic

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

Examiner's comment

On the whole part (d) was quite well done. Candidates who didn’t see the connection between (c) and (d) often used differentiation instead to get the correct answer. Some marks were unfortunately lost due to lack of units or misunderstanding of the units.

Common mistakes

• read from the completed square: maximum profit £450 000 (units
• not £450) at a selling price of £130
• differentiation also works

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

Examiner's comment

This question was designed to test candidates use of trigonometric identities and solving trigonometric equations and most candidates attempted this question, with a significant number achieving at least one mark. Only a small minority of candidates did not attempt this question. Most candidates were able to recognise the use of one of the three trigonometric identities. Errors were generally in substitution where some candidates used x instead of 3x. Where candidates correctly substituted the trigonometric identity correctly they usually rearranged to the correct format, thus gaining the first two marks. A common error from this point was to miss out the square rooting and use the arcsin/arctan/arccos with the constant, thus scoring no further marks. A small number of candidates square rooted, however did then not proceed to the required format by missing out the division by 3, scoring no more than two marks. For candidates who correctly square rooted and then solved for 3x, a large majority did not achieve the final A mark as they did not consider both roots and/or change the scale for the question. It was common for candidates to achieve one or two solutions, rather than all three. Only a very small number of candidates found, additional wrong solutions within the range provided. A few gave additional solutions outside the range although this was not penalised.

Common mistakes

• use sin²3x/cos²3x = tan²3x (or sin² + cos² = 1): tan²3x = 4
• take the square root BEFORE inverse tan, then divide by 3
• x = 21.1°, 38.9°, 81.1° and no others in 0–90°
• using x instead of 3x
• forgetting to divide by 3

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

Examiner's comment

By far the most common approach to this question was to use the discriminant. Many candidates showed a good understanding of the discriminant and its effect on the roots of a quadratic equation though some candidates started to form the discriminant with the “= 0” only appearing later in the solution. The most common error seen was in an incorrect expansion of ( ) 4 5 4 q - + resulting in the loss of a negative sign and subsequently an incorrect attempt at the difference of two squares. Some candidates struggled to correctly identify a, b and c in 2 4 b ac - and some included x’s in their attempt. There were a small number of other approaches, completing the square being the next most common. There were also a large number of candidates who tried to rearrange the given equation or manipulate the given relationship between p and q with no obvious strategy. Candidates should be reminded that this was a ‘Show that’ question which requires the solution to reach the printed answer precisely, ensuring that the condition that 2 4 0 b ac - = is clear from the start. Students who find that their working is leading to an incorrect result might spend a little time checking their working, particularly when there are so few steps as here and they know the result they should be getting.

Common mistakes

• equal roots: b² − 4ac = 0 with (−3p)² − 4(5q + 4) = 0
• brackets round 3p and 5q + 4 (sign error in −4(5q + 4))
• difference of two squares 9p² − 16 = (3p + 4)(3p − 4)
• "= 0" needed from the start

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

Examiner's comment

This question was generally well answered, with many candidates demonstrating a solid understanding of the laws of logarithms. Most were able to apply the key rules, such as the power law and the subtraction law, to progress effectively through the problem. The majority of students gained the first mark for correctly applying the power law and many successfully used the subtraction rule to combine logs into a single expression and then correctly equated it to a constant. Most candidates were able to solve the resulting quadratic equation and correctly identified the roots x = 5 and x = -4 although a significant number of candidates failed to reject x = -4 and ss a result, lost the final accuracy mark. Several candidates incorrectly expanded ( ) 3 2log 1 x + as 3 3 2log 2log 1 x + or displayed other misconceptions such as ( ) ( ) ( ) ( ) 2 2 3 3 3 3 log 1 log 1 log 7 log 7 x x x x + + - + = + , showing a misunderstanding of logarithmic laws. In terms of algebra, mistakes were seen in expanding expressions such as 3(x + 7) which was sometimes incorrectly expanded to 3x + 7. A few students attempted polynomial division after correctly combining logarithmic terms, showing confusion in handling the rational expression. Generally candidates who demonstrated a secure grasp of logarithmic laws and followed a logical, structured approach were generally successful. However, persistent issues with rejecting invalid solutions and applying basic algebra correctly meant that some otherwise capable candidates lost marks unnecessarily. The question effectively discriminated between those with strong conceptual understanding and those with only procedural fluency.

Common mistakes

• power law 2log₃(x + 1) = log₃(x + 1)²
• subtraction law then 3¹ to remove logs: (x + 1)² = 3(x + 7)
• x² − x − 20 = 0
• reject x = −4, so x = 5 only
• 2log₃(x + 1) is not 2log₃x + 2log₃1
• log₃(x + 7) ≠ log₃x + log₃7

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

Examiner's comment

This question was met with mixed success. Whilst many candidates made a reasonable attempt at sketching the quadratic graph in part (a), part (b) proved more challenging, particularly in selecting the correct gradient function and providing adequate justification. Part (b) was less well answered. While some identified the correct function, many failed to justify their choice clearly or accurately. Common issues included: • no justification or vague explanations such as "it’s a cubic" (which applies to all the given options) • stating that it was correct because of the roots, but without explaining the link between the function and its derivative • attempting to integrate each option instead of recognising it as a derivative • incorrectly assuming a connection between the coefficient of 2x being 3 and the original curve being cubic Some selected the correct function but gave ambiguous or incorrect reasons, which prevented them from accessing full marks. A wide range of incorrect gradient choices were also seen. Stronger responses either correctly linked all the required aspects or ruled out the incorrect options through logical reasoning, demonstrating good understanding. Overall, this question differentiated well between candidates.

Common mistakes

• dy/dx = 3x² + 9x: positive quadratic with a root at 0 and a negative root (−3)
• "it's a cubic" or vague root comments are not a justification
• integrating each option

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

Examiner's comment

Part (b) was much more demanding with very few candidates achieving full marks and many blank responses seen. The majority of candidates failed to identify the necessary terms to set up the two required equations or made errors in expanding the brackets. The equation resulting from the constant term appeared the more accessible of the two and resulted in the first M1 at least. Where candidates were able to identify the two correct equations they generally went on to score full marks. For these more able candidates who were able to find a solution it was pleasing to see them pairing a and k correctly.

Common mistakes

• multiply (a − 2/x)(1 + 8kx + 28k²x² + …): constant a − 16k = −21, x coefficient 8ak − 56k² = −90
• substitute to a 3TQ 12k² − 28k + 15 = 0
• pair correctly: a = 3, k = 3/2 and a = −23/3, k = 5/6
• identifying the wrong terms

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

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. In part (a), the majority of candidates were able to differentiate s correctly and equate it to zero to produce a quadratic.However, a significant proportion of students did not show any method to justify their solutions from their quadratic, ignoring the instruction at the top of the question and lost two marks. A significant number of candidates did not find a derivative, choosing instead to equate s to zero. This led to t values of 0 and 3 which then had a knock on effect on the marks that could be scored in part (b).

Common mistakes

• v = ds/dt = 6t² − 24t + 18 = 0, t = 1 or t = 3
• working (e.g. factorising) must be shown, not just calculator roots
• setting s = 0 instead of v = 0 (giving t = 0 and 3)

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

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. In part (c), many found the acceleration but a number then failed to equate it to zero and so did not achieve any marks. Some did obtain t = 2 successfully but then did nothing with it. A few used the symmetry or completed the square and got to the required answer very quickly. The odd negative answer for speed was seen.

Common mistakes

• maximum speed where a = dv/dt = 12t − 24 = 0 (or by symmetry or completing the square v = 6(t − 2)² − 6): t = 2, speed 6 m s⁻¹ (must be positive)
• many found a but did not set it to 0, or stopped at t = 2

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

Examiner's comment

This question on circles was usually well answered with most learners able to score full marks. In part (b), many different strategies were seen within the responses, all duly listed in the main rubric. The most recurrent approach, usually successfully implemented, saw attempts to find the distance or the distance squared of the origin from the centre of the circle before comparing it with the radius or the radius squared and concluded. A smaller proportion of learners opted for reaching a quadratic equation, either in x or y, by setting y = 0 or x = 0 respectively, in the equation of the circle with the aim to find the intersections of the circle with the relevant axis and then check whether the x (or y) coordinates of the origin fell within these values. This approach was successfully implemented only by the most competent learners, who were also able to gain the final mark by reaching a correct conclusion. Other approaches were usually poorly implemented or outrightly incorrect as they consisted of adding the radius to either the x or y coordinate of the centre and then comparing this with the distance from the coordinate axes. A good proportion of the responses who were awarded the method mark, did not gain the final mark due to incorrect comparisons, such as 5 4 or 25 24 ≠ ≠ or incorrect final statements. The conclusion of “the origin does not lie in the circle” was often left out with some learners just writing “no”. It is important to remind learners that reasons and conclusions should always be clear and complete and that it is often useful to use the wording of the question to ensure that the question is answered. 6

Common mistakes

• distance of the centre from the origin is 5 (or d² = 25) and 5 > √24 (25 > 24) so the origin is outside
• "25 ≠ 24" is not a reason
• comparing with 24 as the radius is wrong
• must give a clear conclusion, not just "no"
• setting y = 0 and finding both roots negative also works

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

Examiner's comment

Part (b) proved to be a little more challenging with many responses not being clear enough in their explanations. By far the most popular method seen was to use the discriminant of Q(x) to show that Q(x) has no real roots. Almost every response calculated the discriminant as 7 - , then Q(x) had no real roots. A significant minority then did not go and state that as 7 0 -< on to say that “there was only one (real) root i.e. 4 -”. Learners should be reminded that in this type of question they should be referring to the question asked in their final line of working to ensure full marks are scored. The other possible methods outlined in the mark scheme were seen in a minority of cases; the quadratic formula method was well handled, however some did not complete correctly as above. A small number of learners used completing the square and usually did it correctly with some, again, not concluding correctly as above. There was only a very small number of learners who used calculus. 8

Common mistakes

• discriminant of 2x² − 5x + 4 is 25 − 32 = −7 < 0 so no real roots from the quadratic
• must conclude that −4 is the only real root (many stopped after −7 < 0)
• "−7 ≠ 0" is not a justification
• calling (x + 4) a root or −4 a factor loses the mark

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

Examiner's comment

This question tested knowledge of the intersection of lines and curves. The majority of learners found this question accessible, but some learners did not gain marks due to work particularly for part (b) appearing in part (a) and then no reference made that this should also be for part (b). For others it was for just generally not labelling the question parts and not being clear which part it belonged to, if not attempted in order. Learners should be reminded of the importance of clearly labelling and identifying each part of a question. In part (c), many responses correctly identified 6 x - as a factor and attempted algebraic division or coefficient comparison to find the corresponding quadratic factor. However, several did not show full working and instead used their calculators to find the remaining roots, thereby not gaining method and accuracy marks since the question specifically required the solution to use algebra. Some assumed that quoting the quadratic formula was sufficient and then proceeded directly to calculator-based solutions without showing the substitution steps. Some preferred to complete the square and were successful, although this approach often led to errors. Most responses recognised that only the negative solution was valid since point Q was in the third quadrant; with only a few incorrectly including both solutions. 18 A significant number of responses did not recognise that 6 x - was a factor for part (c) and instead tried to solve the cubic by factorising out an “x”, or they did not attempt it. A handful went on to find the y coordinate, which was not required, and some seemingly did not fully understand the requirement of an “exact” answer, as they gave the x coordinate as a rounded decimal. 19

Common mistakes

• (x − 6) is a factor: (x − 6)(4x² − 2x − 15)
• solve with the formula shown (not calculator roots): x = (1 − √61)/4, rejecting the positive root because Q is in the third quadrant
• exact answer

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

Examiner's comment

This modelling question required learners to construct and interpret a linear demand model, derive a profit function, solve a quadratic inequality, identify the optimal selling price, and assess the suitability of a model against real data. Overall, performance was mixed across the different parts of the question. Part (c) allowed many learners to make the most of the restart opportunity, with the majority attempting this part. Most recognised that they needed to solve the equation by setting P equal to zero and most did so by directly calculating values of x using their calculators. As this was a question involving money, responses did not gain the accuracy mark for not rounding their answers to two decimal places as required. Among those who gave a decimal answer, many did not earn the A mark either because they rounded to one decimal place or used an incorrect inequality, misinterpreting when the profit would be positive in the vicinity of their critical values, with a common incorrect answer being 13.19 < x < 51.81.

Common mistakes

• solve P = 0 and take the inside region where P > 0
• money so 2 d.p.: 13.18 < x < 51.82 (13.19 < x < 51.81 with strict signs is wrong)

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

Examiner's comment

This modelling question required learners to construct and interpret a linear demand model, derive a profit function, solve a quadratic inequality, identify the optimal selling price, and assess the suitability of a model against real data. Overall, performance was mixed across the different parts of the question. In part (d), many learners did not earn the mark due to the omission of the pound sign or failure to round their answer to two decimal places. It would be advisable for learners to pay closer attention to the word 'price' in the question, as it implies that both the correct unit and an answer given to two decimal places are required.

Common mistakes

• midpoint of the roots or x = −b/2a
• a price needs £ and 2 d.p.: £32.50

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

Examiner's comment

The final question, which involved differentiation, integration by substitution and trigonometric identities, provided challenge in dealing with the algebra but, given its layout, proved accessible. The majority of learners attempted at least part of this question with some success due to the restart opportunities available. In part (a), learners were generally able to apply the quotient rule correctly, with a smaller number attempting to apply the product rule. A few responses attempted a substitution, usually 2 1 u x = + , and this need approach often resulted in a succinct and concise solution. Sign slips were quite common, while failing to square the denominator when applying the quotient rule was also seen fairly frequently. This was probably a reflection of the fact that the denominator in the question was already squared so required a bit of extra thought and was easily missed. These errors sometimes meant that they could not make progress due to the form that they achieved. Simplifying the expression and finding the correct coordinate proved quite challenging. Responses often defaulted to unnecessary algebra, such as expanding brackets, when the expression could have been simplified more directly. The number of terms in the expression seemed to be a barrier for many learners. Of those who reached a 3-term hidden quadratic in x2, a significant number turned to the calculator for a solution despite the question stating that solutions relying on calculator technology were not acceptable. Of the responses that obtained a value of x, the vast majority were able to identify that 3 x = - was the correct value to use and then go on to find a correct, corresponding value of y. There were some, however, that found a value for x but stopped there, not attempting to find a value for y.

Common mistakes

• quotient rule with the denominator (1 + x²)⁴
• cancel a factor (1 + x²) to get 2x(x² − 3)
• solve without a calculator
• P has x = −√3 (not 0 or √3) and y = −1/8

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

Examiner's comment

This question provided some challenge and discrimination for the end of the paper. Most learners attempted this and made some progress. The attempts seen in part (c) were encouraging with many realising that the discriminant of the quadratic equation was required and for “no real roots” this must be negative. Most of these learners were able to form the correct quadratic equation in g and find the correct critical values but the correct critical region eluded some with g < 2 and g < 10 being a common error. From the correct inequalities for g it was a small step to identify that the values of g required were 4, 6 and 8 and the majority of learners who got this far were able to complete the solution and state the correct probability in the correct form.

Common mistakes

• no real roots: g² − 4(3g − 5) < 0 so (g − 2)(g − 10) < 0 and 2 < g < 10 (inside region)
• even values g = 4, 6, 8
• probability 18/420 = 3/70

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 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

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

This was a challenging question for many candidates since forces were given as vectors, but the acceleration was given as a scalar. Although most realised that the two forces needed to be combined to produce a resultant, some subtracted the components rather than adding, thererby failing to achieve the first mark. This also required the i and j terms to be collected to enable an expression for the magnitude to be found. A common starting point was to equate the vector sum of the forces to a scalar ‘ma’ term often leading to no further valid progress. Those who attempted to work in scalars sometimes equated the magnitude of their resultant to sqrt 5 (acceleration) rather than 2 sqrt 5 (ma term) and managed to secure one of the two available method marks. Where a correct unsimplified equation in c was reached, processing errors such as not squaring each side or in expanding brackets sometimes led to incorrect answers. The few who found the acceleration vector and equated the magnitude to sqrt 5 or the square of the magnitude to 5 tended to do so successfully despite the slightly harder algebra.

Common mistakes

• resultant force (2 + c)i + 2j (add, not subtract, the components)
• F = ma with magnitudes: (2 + c)² + 2² = (2√5)² = 20, or acceleration vector with magnitude √5
• c = 2 or c = −6
• equating a vector to the scalar ma
• using √5 instead of 2√5 for ma

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

Examiner's comment

( ) x In part (b), candidates generally recognised the need to set their to find values f ' 0 ( ) x is decreasing and solve their three-term quadratic to find critical of x for which f ' values. However, not all candidates observed the demand to use algebra and consequently those who used their calculator to solve the quadratic or gave an incorrect factorisation of their quadratic forfeited the first mark. Most candidates chose to simplify their quadratic before factorising and solving to find the critical values, but lack of rigour was evident in this part. Many candidates who factorised their quadratic - 2 2x jumped straight from a quadratic with a term to factorisation of the form (𝑥 ± … )(𝑥 ± … ) (possibly having used their calculator and attempting to work backwards), and consequently lost the B mark. A few used the quadratic formula and showed their working in full thus scoring the B mark, while attempts to complete the square were rarely seen. Once critical values had been obtained, candidates were required to identify the correct region and, while many candidates were successful, there were some who did not know how to proceed. It was common to see incorrect regions selected, particularly 2x - , not realising that their original quadratic had a negative coefficient of 1 4 x and therefore the outside region was required. Candidates who sketched the graph generally performed better in identifying the correct region but again, many drew a positive graph instead of a negative. There was confusion observed in the use of 'and' and commas instead of 'or', indicating weaknesses in formal mathematical language. It was notable how many candidates did not accurately identify the correct region for their x2 coefficient and critical values, often stemming from dividing their quadratic by –2 but forgetting or not knowing to change the direction of the inequality sign. Other observations that were common to see in this part of the question that were not required or caused complications included: • not recognising that the denominator would be positive due to the square • attempting to find a critical value for the denominator unnecessarily • trying to multiply the denominator and numerator and then work on the resulting quadratic • attempting to find turning points or second derivatives to identify the region, often unsuccessfully.

Common mistakes

• solve f′(x) < 0
• factorisation −2(x + 1)(x − 4) must be shown (calculator roots lose B1)
• outside region x < −1 or x > 4, not 4 < x < −1 or ∩

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

Examiner's comment

This question was answered very well by the majority of candidates with many scoring upwards of 8 marks out of the available 11. It highlighted that candidates continue to struggle with identifying the domain and range of functions, and that in most cases candidates are not aware that they should state the domain when defining a function. In part (d), some candidates were able to secure marks having found the previous parts challenging. Algebraic errors were not uncommon when equating the expressions and often the = 0 was “dropped” in favour of a stand-alone expression. Candidates generally progressed to determining the discriminant, but setting the discriminant greater than or equal to zero was often seen. Some candidates attempted to find the discriminant of a quadratic set = k which scored no marks. Some of those who attempted the correct inequalities lost the final mark due to slips when simplifying such as ( ) 36 20 4 36 20 80 k k - - == - - or failure to reverse the inequality when dividing through by –20.

Common mistakes

• 5x² − 6x + k − 4 = 0
• no real roots so b² − 4ac < 0 (not ≥ 0)
• reverse the inequality when dividing by −20
• k > 5.8

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

Examiner's comment

This question was a clear discriminator between the most secure and intermediate candidates with many making little or no progress in part (i), although many candidates made at least a respectable attempt at part (ii). The most common techniques employed in part (i) were completing the square and using the discriminant. When completing the square, candidates often stated that squared numbers are always positive and gave no indication that they could also equal 0, which was a common error the last time a similar question was asked. When finding the discriminant, candidates often obtained the answer of –4, but very little explanation followed this, and notably many seemed to think this was sufficient to conclude that the 2 k given quadratic was always positive with no reference to the coefficient of the term. It was not uncommon to see attempts analysing even and odd number scenarios, but none were seen that recovered from only considering whole numbers and extended to the entire set of real numbers. A few candidates used differentiation to find the turning point and some of these candidates calculated the second derivative to prove that their turning point was a minimum, but they usually lost the final mark as they did not write a conclusion to complete their proof. The least confident candidates simply substituted a few integer values for k into the expression and attempted to draw a conclusion from the results, scoring no marks. In part (ii), many candidates did not fully appreciate what they were being asked to do, most likely due to the unfamiliar nature of the question. As a result, some attempted algebraic manipulation of the expression, such as expanding the brackets, and were unable to progress to a solution. Other candidates believed they were being asked simply to write out the proof by adding text to the information supplied or by detailing the working for the solutions of the simultaneous equations given. However, there was a good proportion of candidates who realised that the example given was just one of many possible scenarios and proceeded to attempt to find solutions to the remaining cases. It was rare to see all of the 11 other factor pairs examined, and, in most cases, candidates only considered simultaneous equations with a positive right-hand side. The mark scheme enabled all the marks to be scored for this approach given that the simultaneous equations with negative right-hand sides could be discarded quite trivially. Often candidates only determined x or y for a couple of factors of 28 securing 1 or 2 marks. A good proportion only dealt with factor pairs ‘one way round’ and gave no justification as to why, limiting themselves to 2 marks out of the available 4. Calculators could legitimately be used to solve the simultaneous equations and candidates who did this found, almost without exception, correct values for x and y as well as saving time. Full algebraic processing of the simultaneous equations was time-consuming and prone to error. A minority of candidates realised it was possible to argue that 3𝑥+ 2𝑦 ⩾ 5 and limited themselves to examining just cases A and B from the mark scheme, but it was nice to see the strongest candidates able to distinguish themselves in this manner. Most candidates identified the reasons for rejecting factor pairs as they went along whilst a few left the rejection of the solutions to the overall conclusion, and both approaches were acceptable. Sadly, some made no comment at all on the solutions they found and were unable to score either accuracy mark as a result.

Common mistakes

• (k − 2)² + 1 with (k − 2)² ≥ 0 (not > 0)
• discriminant route also needs positive k² coefficient
• odd/even cases do not cover real k

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

Examiner's comment

In part (c), most candidates were able to solve the given quadratic equation and achieve the method mark. Overall, this was done with a calculator. Most chose the correct answer but quite a few left both answers or chose N = 45, appearing not to take into account the context of the question. They needed to realise that, when the loan was fully repaid after 36 months, the borrower would not continue to make loan repayments, so the second answer of 45 months needed to be rejected. This took significant effort for some - rechecking the sum for 36 and 45 terms and then finding the value of the 36th and 45th term and then rejecting 45 when the 45th term was negative - when thinking about the context would have been easier.

Common mistakes

• N = 36 or 45
• reject 45 in context (loan repaid after 36 months, later payments would be negative)
• leaving both values loses the A mark

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

Examiner's comment

Generally, candidates found this question accessible, with many scoring full marks or nearly full marks. In part (b)(i) most candidates were able to solve the quadratic equation, but not all identified that k = 4 3 was the only correct answer, or subsequently used this value, so lost the mark in this part of the question. Some candidates spent more time than they needed using the quadratic formula, where a factorisation of the quadratic would have been more direct. Use of the calculator equation solver was also permitted here. Candidates must learn to check their answers against any limitations given in the question. When errors did occur, they were most likely to appear in part (b)(ii), although most still answered it well. Most candidates identified the requirement of the question to add the first three terms of the sequence together, and correctly used k = 4 3 to calculate u2 =6 ( 4 3 ) -5 before finding 6 + 3 + (-1) = 8. Others first simplified the sum to obtain 6k or 6k2 + k - 4 before substituting the value of k. A common incorrect approach was to use u3 = 6k2 – 5k – 4 in the sum. Some also used k = 4 3 and k = 1 2 - to calculate the sum, without identifying which was the required answer. Alternatively, a good number of candidates attempted, correctly and incorrectly, to work out the values of u1 and u3 even though they were given in the question. It was also noticeable that a number of candidates used the quadratic expression from part (a) in order to work out values in (b)(ii), usually assuming u3 = k2 – 5k – 4. They, therefore, did not clearly understand what this equation represented. A few candidates incorrectly attempted to use the recurrence relation to find things like ( 4 3 )(1) – 5, ( 4 3 )(2) – 5 and ( 4 3 )(3) – 5, using these expressions to find their sum, and there were again occasional attempts using arithmetic or geometric series which also gained no credit. There were also a few attempts at integration. Overall, this was a question that allowed candidates to gain a good number of marks with confidence.

Common mistakes

• k = 4/3 only (k positive, reject −½)
• sum = 6 + 3 + (−1) = 8
• do not use the quadratic 6k² − 5k − 4 as u₃
• AP/GP formulae score 0

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

Examiner's comment

and candidates who achieved an incorrect quadratic model in part (a) were still able to achieve the method mark. When candidates achieved full marks in part (a) they usually went on to achieve full marks in part (c) too. It was extremely rare for H = 2.96 to be achieved and an incorrect

Common mistakes

• H = ax² + bx + c with c = 2
• use (20, 0.8) and max at x = 9 (dH/dx = 0 or −b/2a = 9)
• H = −0.03x² + 0.54x + 2
• assuming vertex form with A = 2 fails

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

Examiner's comment

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.

Common mistakes

• complete the square: centre (3, −7)
• r² = 9 + 49 − 33 = 25 so r = 5 (not √91, not ±5)

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

Examiner's comment

This question provided a challenge to most candidates and was in general a good discriminator. 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’.

Common mistakes

• substitute y = −2x + 2 into the curve: x³ − 3x² + 18x − 16 = 0 (constant term often lost)
• use (x − 1) as a known factor: x² − 2x + 16
• discriminant 4 − 64 < 0 so no other real roots
• conclude "the normal does not meet C again"
• calculator roots alone score no more

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

Examiner's comment

Whilst most students showed they had a good grasp of the concepts of correlation and (negative) quadratic probably also gave themselves lead in to part (d). Of those who did not score the mark in part (c) some simply discussed correlation rather than stating that the line had Part (d) proved challenging for most students and many simply moved on to the next question. There were many convoluted attempts involving calculus or multiplying out the bracket or substituting a value such as (1, 28). Students should be encouraged to consider the number of marks available as a guide to the amount of work required to answer the question. Those who had identified the scatter diagram as representing a parabola and saw the connection with the quadratic nature of Jane’s model could simply estimate the equation of the axis of symmetry and usually chose a suitable value of k.

Common mistakes

• k is the t value of the vertex (axis of symmetry) of the curved scatter, any k in 3 to 4.5
• 1 mark so no calculus or expanding needed

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

Examiner's comment

(b) Most candidates scored the two available marks in this part for finding the range OA. The most common method was to set y = 0 in the given equation and solve to find x. Sometimes a two-step suvat approach was adopted, generally successfully. Those who made no progress in part (a) often achieved the marks here.

Common mistakes

• set y = 0 in the path equation and solve: OA = 120 m (or double the x of the maximum, or a two-stage suvat method)

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

Examiner's comment

(c) This part was generally done well with many correct answers seen. Often a suvat approach was used to find H, the greatest height. Alternatively, many used the symmetry of the parabola and substituted x = 60 (half the range) into the given equation to find y. Some differentiated to find the maximum (or used a calculator) but occasionally gave 60 (the x value) rather than 22.5 (the y value) as their final answer. Completing the square was also employed successfully on occasion.

Common mistakes

• substitute x = 60 (half the range) or complete the square or use dy/dx = 0: H = 22.5
• giving 60 (the x value) as the answer
• 45/2 not accepted

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

Examiner's comment

This question was a friendly and well-attempted start to the paper. Most candidates were able to score a high proportion of the marks available here. In part (b), a significant minority of candidates were unclear on the correct way to identify the correct critical values and, as a result, some attempted to solve 2 2 d 0 d y x = in pursuit of a ‘critical value’. Perhaps because the question had asked where y was decreasing, an occasional misconception led candidates to find y values corresponding to the x values resulting from d 0 d y x = followed by an attempt to use these y coordinates to define a range rather than using the values of x. Nonetheless, most candidates recognised the need to set d 0 d y x = to find the turning points of the curve and most were able to solve the three-term-quadratic set equal to zero. Commonly, the use of calculators was apparent. But many others took an algebraic approach with sometimes mixed results. Factorisation was the most successful of these, with errors mainly occurring with the signs within the linear factors. The quadratic formula was sometimes inaccurately applied and omissions of the – sign in front of ‘b’ in the numerator, or the use of ‘2ac’ in the denominator, caused issues and the resulting loss of the M mark. Completing the square was seen but it seemed to be a fairly unreliable approach for the candidates that chose to use it. A significant minority of candidates did not write down their critical values clearly, instead confining them to being labelled only on a graph of the d d y x curve. A significant number of candidates who had found critical values appeared to believe that stating them was sufficient to define the required range, or perhaps they were unable to make further progress in stating the range and so stopped short by simply stating the two values of x. Those candidates who did realise, however, that an interval needed more careful definition, usually did so correctly. This was most often achieved using standard inequality notation. The most successful candidates were those who drew a sketch of the quadratic to accompany their determination of the range for x, but some candidates determined 2 2 d d y x as a means to identifying the correct region. A minority of candidates lost a mark for inconsistent use of inequalities by mixing, for example, and in their statement of the range. Only a relatively small proportion of candidates defined the outside region; a small number of whom wrote non-sensical inequalities such as 1 4 2 x - . Answers such as 1 , 4 2 x x - were also not uncommon. The use of set notation tended to be more problematic for those that used it as there was sometimes some confusion between the union and the intersection symbols.

Common mistakes

• set dy/dx < 0 (not d²y/dx² = 0)
• critical values −1/2 and 4 from (2x + 1)(x − 4)
• choose the inside region −1/2 < x < 4
• stating just the two values is not a range
• use x values, not y values
• do not mix strict and non-strict inequalities or confuse union and intersection

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

Examiner's comment

This question was generally not particularly well answered, especially as it appeared early in the paper, with the majority scoring two or fewer marks out of the available four marks. The majority of candidates identified that a quadratic equation needed to be solved to progress, although some struggled to reach a suitable equation. It was notable that significant proportion of candidates ignored the question’s requirement to show all stages of their work, directly solving their quadratic in u with their calculator. They immediately, therefore, forfeited the first two marks which required candidates to demonstrate that they can solve a quadratic without reliance on a calculator. Those who did show their method and attempted to factorise, generally did so successfully and reached solutions in terms of u. Another common error, however, was then to forget to square their values of u so they did not reach values of x. Furthermore, candidates very often also did not reject the invalid 25 4 x = , which resulted from 5 2 x = - , so that the majority of candidates did not score the final mark. While factorising was by far the most common approach to solve the quadratic, other approaches included use of completing the square or the quadratic formula, and a very few squared the original equation and attempted to solve 36x2 – 289x + 400 = 0.

Common mistakes

• form 6u² + 7u − 20 = 0 and show a method (factorise, formula or completing the square), not calculator roots
• square u = 4/3 to get x = 16/9
• reject u = −5/2 (√x cannot be negative), so x = 25/4 must be discarded

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

Examiner's comment

Most candidates recognised the need and were able to solve the equation 2 4 3 23 x + = to determine the upper limit, 5 x = , although some candidates did not use the diagram to discard the 5 x = - and ended up integrating between the incorrect limits. It was also very common to see the limits 3 and 23 being used, with these candidates unclear about the area that is found through an integration strategy. The most successful strategy for finding the area under the curve was through finding ( ) 5 2 0 4 3 d x x + and then subtracting that from the area of the rectangle. However, there were plenty of candidates that failed to subtract the area under the curve 29 5 3 from the rectangle, and these candidates generally scored 3 marks out of 5. Those candidates that elected to find the difference between the curve function and the line y = 23 and then integrated their expression were more likely to have slips in their working: some failed to subtract correctly, resulting in ( ) 5 2 0 23 4 3 d x x - + ; while others subtracted the wrong way round, leading to a negative answer that some failed to recover to a positive answer. However, in both cases, once the integrand was determined, the majority of candidates using this strategy were able to achieve method mark for being able to perform integration of at least a term in the integrand. There were very few candidates that tried to integrate the function bounded by the y-axis and those that did were generally unsuccessful: often their rearrangement to make x the subject was poorly done and in these cases they often lacked the experience of Year 2 integration to be able to integrate function of the form ( ) n ax b + , where n is a fraction.

Common mistakes

• upper limit x = √5 from 4x² + 3 = 23 (reject −√5)
• limits 3 and 23 are y values, not x limits
• area = rectangle 23√5 minus ∫(4x² + 3)dx, or ∫(20 − 4x²)dx
• forgetting to subtract from the rectangle
• k = 40/3

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

Examiner's comment

Part (a) of this question was well answered as candidates appeared to be able to complete the square to determine the coordinates of the centre as being ( ) 3, 5 - .

Common mistakes

• complete the square: (x − 3)² + (y + 5)² = 34 − k
• centre (3, −5), sign errors common

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

Examiner's comment

Very few candidates achieved full marks on this question. The vast majority of candidates were able to find the linear equation successfully, with a significant proportion making no further attempt at the question. While the majority did recognise the form of the quadratic equation using the two x-intercepts, most simply stated 2 6 y x x = - or equivalent and did not use the coordinates on the graph to find the coefficient of the 2x term. There were a lot of candidates who did not understand what the final part of the question required as they were trying to find intercepts (often assuming that the region needs to be defined as 3 10 x - ) or find the area between the curves. Very few candidates seemed to understand what the regions were and how to use the inequalities to determine the region. Incorrect use of notation was also seen here with R used instead of y linking the two equations.

Common mistakes

• line l: y = 2x + 60
• curve y = ax(x − 6) and use (10, 80) to get a = 2 (not just y = x² − 6x)
• region 2x(x − 6) ⩽ y ⩽ 2x + 60 in y, not R or f(x)
• do not mix < and ⩽
• finding area or x-intercepts is not what is asked

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

Examiner's comment

This question was an effective discriminator between candidates. There was evidence of log misconceptions in a substantial proportion of candidates’ work and there were a number of completely blank scripts seen. Often though, even weaker candidates were able to pick up the first mark for evidence of a log law correctly applied, albeit amidst otherwise erroneous work which limited them to the first mark only. Most commonly this was earned in isolation either for use of the power law ( ) ( ) 2 5 5 2log 3 2 log 3 2 x x - → - or for 2 2 5 → when both sides of the equation were raised to base 5 in an attempt to remove logs. It was not uncommon, however, to see the right-hand side become 25 rather than 52. Another common error involved a misapplication of the log law for subtraction whereby log log A B - was believed to be equal to log log A B with the log terms on numerator and denominator subsequently ‘cancelled’. In contrast, however, there were a large number of candidates who were well-prepared for this question and were able to demonstrate clear understanding and apply their log knowledge accurately to gain the first three marks. Following a successful removal of logs, most were able to rearrange their equation into a solvable form and gain the final method mark for solving the resulting quadratic equation. There was a surprising proportion of candidates who chose to attempt to simplify the algebraic fraction 2 9 12 4 x x x - + , often correctly, to 4 9 12 x x - + . Of these, a good proportion managed to recover, but many candidates simply used an equation solver on their calculator to arrive at the two solutions, and thus forfeited the final two marks. Similar errors to those observed in question 1 were also apparent here in the solving of a quadratic equation and factorisation proved again to be the most successful algebraic method. The final mark was a particularly good discriminator as only a minority of stronger candidates checked the viability of each root in the original equation and so most candidates failed to reject the 1 9 x = root. It was again disappointing to see the number of attempts that jumped straight to 1 9 x = without showing sufficient log work to achieve an equation without logs. Candidates must be clear that using calculators to solve equations must be treated with caution and that they risk losing a substantial number of marks.

Common mistakes

• 2log₅(3x − 2) = log₅(3x − 2)²
• 2 = log₅25 (not 2⁵)
• subtraction law gives a quotient, not log A / log B
• (3x − 2)² = 25x leads to 9x² − 37x + 4 = 0
• reject x = 1/9 (3x − 2 < 0), x = 4 only
• full log work must be shown, not calculator

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

Examiner's comment

This question involving the use of trigonometric identities and solving a trigonometric equation was generally very well answered with the exception of part (c). There were very few blank scripts. Part (a) was generally well answered. The majority of candidates scored the first mark for correctly identifying sin tan cos x x x = . Two common errors were to either use sin 1 cos x x = - , or to include the number 4 in the denominator 4sin i.e., 4tan 4cos x x x = . Some candidates made errors in their notation (which was only penalised if it occurred more than once) with a few using 2 sin x . A small number started with the answer to try to work backwards, but this approach rarely scored full marks.

Common mistakes

• tan x = sin x / cos x (not 4 tan x = 4 sin x / 4 cos x)
• multiply by cos x and use cos²x = 1 − sin²x
• show all steps and write "= 0" in the final line

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

Examiner's comment

This question involving the use of trigonometric identities and solving a trigonometric equation was generally very well answered with the exception of part (c). There were very few blank scripts. Part (b) was attempted by the majority of candidates with some good success. Calculators will solve this equation directly, and as a result there were some candidates (although a minority) who were unable to work out the correct answer using a clear written method combining the quadratic formula and using the inverse candidates were able to solve the equation and find at least one value of x (usually awrt 42.6). Generally, where candidates understood the requirements of the question, both angles were found to the required accuracy. Some candidates stated additional incorrect answers inside the given range and so lost the final accuracy mark. It was surprisingly common to see candidates rounding their values from the quadratic equation (i.e., to 0.7) which resulted in inaccurate angles and as a result they lost the final accuracy mark.

Common mistakes

• sin x = (−2 + √29)/5 ≈ 0.677 (reject the root below −1)
• x = 42.6° and 137.4°
• rounding sin x to 0.7 loses accuracy
• extra solutions in the range lose the final mark
• show the quadratic being solved

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

Examiner's comment

This question began with the familiar calculator warning which was unfortunately ignored by a significant proportion of candidates. As mentioned previously, candidates must be clear on what is allowed when these warnings are stated, or they risk losing a number or marks. In part (b), the vast majority of candidates achieved the cubic equation required by equating the two given equations and rearranging to set equal to 0, but a significant proportion ignored the guidance at the top of the question and proceeded to solve the question directly on their calculator, resulting in a maximum mark of 1 out of 5. For those that proceeded further, many used inspection to factorise the cubic using the given factor ( ) 1 x - , while others attempted algebraic division, generally very successfully, achieving the correct quadratic 2 4 6 x x - -. Occasionally, the rearrangements that candidates used resulted in a negative cubic, but despite making the division more challenging it was pleasing that candidates were still generally able to complete this part successfully. However, having arrived at a suitable quadratic, another significant proportion relied on their calculator to solve the resulting equation (set equal to 0) and forfeited the final two marks. For those that showed full working, many assumed that 0 k and chose the incorrect solution, despite often writing 0 k as their justification for rejecting (the correct) 2 10 - . As a result, there were very few candidates who achieved full marks for this question.

Common mistakes

• equate to get x³ − 5x² − 2x + 6 = 0
• (x − 1) is a factor from (a): divide to get x² − 4x − 6
• solve exactly by completing the square or formula, not calculator
• k < 0 so k = 2 − √10 (2 + √10 is positive)

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

Examiner's comment

final part. Since the shape of the graph used in part (d) was the same shape as the graph used

Common mistakes

• acceleration 1 m s⁻² means Sam reaches X after X seconds
• area ½X² + X(27.5 − X) = 100 gives a quadratic in X only
• X = 3.92 (3 s.f., choose the root less than 27.5)
• common errors 22.5 instead of 27.5, one equation in two unknowns

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

Examiner's comment

This question was accessible and generally well answered, with a large number of candidates gaining full marks with concise solutions. Some, however, did not pay heed to the instruction to show all stages of their working. In part (b)(i), most candidates had no problems solving the cubic, getting the values 0, 3 4 and 2 - for a. However, many did not take note of the fact that a was positive, and that the question asked for the value of a, so lost the mark for not rejecting 𝑎= 0 and 𝑎= -2. There was confusion as to the reasons for rejecting values. Some thought that a had to be an integer so rejected 3 4 . Some chose 2 - as their solution giving spurious reasons such as “a is a real number.” Of those who did not identify 3 4 a = as the only valid answer, many recovered by opting to use only 3 4 a = in part (b)(ii). Part (b)(ii) was well done by those who showed every step of their working, but some failed to reject the values of a that were not needed. Once the cubic was obtained by substituting 3 4 a = and setting f( ) 3 x = , those who solved the cubic without factorising to 2 (4 5 10) x x x + - lost the final A mark. Candidates were asked to show all stages of their working and not proceeding as far as the quadratic factor was penalised. Many gave the exact solutions followed by rounded decimal solutions, or even just the decimal solutions, so were not paying heed to the question asking for exact solutions. There were some who were unable to solve the cubic or the quadratic, having taken out a factor x outside the brackets. These were mainly cases where they tried to complete the square or their value for a meant that they could make no further progress. Some did not collect terms on one side with 0 on the other side, and having taken out a factor of x, solved their quadratic ignoring the constant on the other side. The solution 𝑥= 0 was omitted by many candidates due to division by x instead of using factorisation. Others rejected this particular solution, if it was found, possibly confusing the restriction that a was positive rather than x. Those who did reach the correct quadratic factor were usually successful in finding the two exact roots for the A1 mark. A common incorrect approach was to find f(3) rather than setting f( ) 3 x = . This approach gained no credit.

Common mistakes

• must reject a = 0 and a = −2 (a is positive)
• set f(x) = 3, not find f(3)
• take out the factor x to see the quadratic factor (dividing by x loses x = 0)
• exact roots, not decimals

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

Examiner's comment

This question testing the small angle approximation for cosine and finding the equation of a straight line was accessible to many candidates with a large number scoring full marks. In part (a), some candidates formed and solved the equation but were not able to use the information that x was small to choose the correct solution. There was a very small minority of candidates who did not quote the correct small angle approximation, despite this being in the formula book. A slightly larger minority misunderstood that x being small meant that x was 0 and therefore did not include the 2x term in their solution attempt. Candidates should also pay closer attention to the question as several did not round to the required decimal places. It was also notable that some candidates at this level found it difficult to expand 2 1 1 2 2 x - when multiplying the fractions together.

Common mistakes

• substitute cos x ≈ 1 − x²/2 into f′(x) = 0 and keep the 2x term (x small is not x = 0)
• choose the small root only
• 3 d.p.

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

Examiner's comment

This was a more demanding than expected question on geometric sequences with a number of candidates getting confused with the value of k and the subsequent values of a and r required for the question. There was still access to later parts as the quadratic in k was given such that part (b) and part (c) could be attempted without part (a). In part (b)(i) candidates were told that the given sequence converges. Candidates were required to find the value of k and to give a reason for their answer. Solving from calculator, or without working, was allowed and the majority of candidates were able to correctly solve the quadratic reaching the terms 20 k = and 2 3 k = . However, many found giving a reason particularly challenging. Of those who did score both marks, listing values in the sequence was a common explanation, as was finding the value of r and using the fact that 1 r . When attempting to reason via a calculation of the common ratio, there were often errors. A common response was to assume k was equal to r and then selecting 2 3 k = as the final value. Some did not select a final value, keeping 20 and 2 3 . Some candidates who had chosen 20 k = gave comments such as “because it is an integer/whole number” or “because it is a constant”. A significant number chose 20 k = giving a reason “because the sequence is convergent” which they had already been told and did not demonstrate an understanding of what the term means. Part (b)(ii) proved to be very challenging for many candidates. It was quite often the case that candidates tried to use 2 3 r = and just found the value 6 a = when 2 3 k = . It was very frequent that those who had correctly identified 20 k = in part (b)(i) did not show their working to find a and r and lost all marks in (b)(ii) as the sign error in substituting in 3 4 - meant that it was unclear whether the correct value for r had been found, or if the correct sum to infinity formula was being used. Some candidates having found 256 7 correctly then proceeded to give a rounded decimal as their answer which was condoned. Those who found 6 a = and 5 3 r = were still able to score the B mark for a correct pair of values for a and r, even though they could not score the mark for using the sum to infinity formula with an invalid value for r.

Common mistakes

• k = 20 because the modulus of r (¾) is less than 1 (k = ⅔ gives r = 5/3)
• S∞ = a/(1 − r) with a = 64, r = −¾

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

Examiner's comment

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.

Common mistakes

• complete the square to get centre (−3k, k) and radius √(10k² − 7)
• not ±√ and not halved

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

Examiner's comment

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 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.

Common mistakes

• substitute y = 2x − 1 into the circle
• b² − 4ac > 0
• critical values 7 ± √85
• outside region with "or" or ∪, not "and"

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

Examiner's comment

A significant number of candidates found this question very challenging, with a fair proportion of blank answers, and another significant number working with various substitutions in (a) that were quickly abandoned. Those who were able to work with both the quadratic and trigonometric models confidently were able to score highly and demonstrated a good understanding of how mathematics can be used to model real life contextual problems. In part (a), there were two common approaches that led to successfully scoring full marks. One was applying the knowledge that 2 ( 20) 0 b t - = when 60 H = which therefore meant 60 a = . This then enabled the candidate to substitute 2, 0 H t = = and find a complete equation. It was clear when candidates were able to apply this knowledge, and knowing this gave them a huge advantage in this question. Noticing that b was positive was pivotal when taking this approach. The other approach that led to successful answers involved multiplying out the brackets, and differentiating with respect tot. This enabled the candidate to find that 20 t = when d 0 d H t = , thus overcoming a shortfall in knowledge over when geometrically the maximum occurs in a negative quadratic. This second approach led to success less often, and candidates often ended up with a page of various quadratic expressions and equations that looked costly in terms of time and did not yield marks.

Common mistakes

• a = 60 (maximum height)
• use H = 2 at t = 0 to find b = 0.145
• complete equation in H and t

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

Examiner's comment

There were very few non-attempts or zero scores here and this question proved to be a good discriminator with a spread of marks achieved by candidates. Most students managed to score at 5 - In (b) part (i), almost all candidates were able to correctly state the values ‘6’ and ‘ ’, it was 3 clear that in many cases candidates were making use of their calculators to solve the quadratic equation which was acceptable. Part (ii) however, was answered less reliably. Although most 5 x = - candidates had some understanding of the domain of a log function and knew that was 3 not a valid solution, many were unable to provide reasoning that was precise enough for the mark here. It was common to see, for example, “a log can’t be negative”, which was not considered sufficiently clear. Others stated simply ‘because it is negative’ which was also insufficient. A range of responses which engaged specifically with the domain of a log function needing to be positive were considered acceptable such as; ‘log of a negative value is undefined/impossible/gives a math error’.

Common mistakes

• roots 6 and −5/3
• −5/3 is rejected because log₂(−5/3) is not real/undefined (log of a negative number)
• "logs can't be negative" or "because it is negative" is not enough

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

Examiner's comment

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.

Common mistakes

• complete the square: x = (t + 3)² − 25 so t + 3 = √(x + 25), y = 6ln√(x + 25) = 3ln(x + 25), A = 3, B = 25
• common error x = (t − 2)(t + 8) so "x = t − 2" giving 6ln(x + 5) scores 0

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

Examiner's comment

Part (b) was generally done well by the candidates who managed to obtain a quadratic in part (a), with many candidates correctly solving to get 360 or 540, and then solving their quadratic. Those who found a solution to their quadratic also tended to find solutions based on that within the specified set of values. Where marks were lost, it was often by incorrectly including extra values,in particular the value of 450 degrees which had been specifically excluded in the question.

Common mistakes

• sin2x = 0 gives 360° and 540° (450° is excluded: tan x undefined)
• the quadratic factorises as (23cos x + 15)(cos x − 1), cos x = −15/23 gives 491°
• answers 360°, 491°, 540° only, including 450° is the common error

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

Examiner's comment

The majority of candidates who attempted the second part used a correct method, setting the i and j components of the velocity equal to each other or both equal to a constant which was then eliminated. A few also successfully used trial and error. However, a significant number set both i and j coefficients equal to 1, solved one or two separate equations and scored nothing.

Common mistakes

• equate the i and j components (t² − 3t + 7 = 2t² − 3), t = 2 only
• setting each component equal to 1 scores nothing

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

Examiner's comment

by not going on to find the magnitude of their velocity vector. Parts (b) and (c), however, proved to be significantly more challenging. A few candidates wrote down a correct vector equation but were unable to make any further progress without focussing on separate components. Those who used integration to find r were often more successful, because the initial position of A was the integration constant. Of those who used r = ut + 0.5at2, many forgot to include the initial position of A and/or made a sign error and lost marks in both (b) and (c). In part (b), occasionally the quadratic equation was solved correctly but the wrong value of T was selected and in part (c) there was often a

Common mistakes

• r = r_A + ut + ½at² must include the initial position 44i − 10j
• equate i-components for a quadratic in T
• choose T = 10 since T > 5

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

Examiner's comment

the final answer was stated exactly as printed in order to achieve all the marks. In part (c) most candidates realised they should solve the given quadratic in tan alpha to give two possible angles of projection. Consideration of the situation should have indicated that the larger of the two angles would lead to the greatest possible height. Although many recognised that the greatest height is reached when the vertical component of velocity is zero, some chose to use the smaller 45o angle. Some found the greatest heights for both angles but then failed to select the relevant one. Other errors included using the time to greatest height as half the answer in part (a) or choosing sin alpha = 1 to give the maximum height in

Common mistakes

• tan α = 1 or 3: the larger angle gives the greatest height
• vertical velocity zero: 0 = (28 sin α)² − 2gH, H = 36 m
• not using half the time from (a) or sin α = 1

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

Examiner's comment

This question proved to be a successful question for many candidates with many being able to score most of the marks. In part (c), the majority of candidates understood the idea of using the discriminant to prove the quadratic part had no solutions. Most were able to correctly evaluate the discriminant correctly although some achieved a value of 41 - instead. With many responses, they often lacked a convincing conclusion clearly stating that as the quadratic had no solutions then f( ) 0 x = has only one solution. Many just stated that the quadratic had no real roots or just showed the discriminant was negative and so only one root. Some incorrectly stated (x+3) as the real root. A few candidates chose to use the quadratic formula and show two imaginary roots and one real root. These candidates were often successful, but some just embedded the values into the quadratic formula and stated “math error” for example which was not sufficient explanation.

Common mistakes

• discriminant (−1)² − 4(2)(5) = −39 < 0 so the quadratic factor has no real roots, so f(x) = 0 has only one real root
• both statements needed (−39 < 0 alone is not enough)
• −41 is a common slip
• "math error" is not an explanation

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

Examiner's comment

5 Many candidates did not attempt part (a)(i) or tackled it poorly. However, the majority of these candidates were still able to go on and score some marks in part (a)(ii) and part (b) and this resulted in a significant proportion of candidates achieving all 3 of the final marks. In (a)(i), the correct statement of the cosine rule usually produced a full solution, even though several responses clearly showed amendments to once flawed working to be able to reach the given equation. Here, some candidates attempted to fudge their working to correct their work but were unsuccessful in correcting each line fully. The common errors seen were the failure to change signs when removing brackets after a minus sign and candidates multiplying the entire RHS of the equation by 1 2 when applying cos60 rather than just the 2 ( 7) x x - - . The absence of " 0" = was rare but still occurred.

Common mistakes

• cosine rule with BC = 3x + 10 opposite the 60° angle
• cos 60° = ½ multiplies only the 2(x + 2)(7x) term, not the whole right-hand side
• sign errors removing brackets after a minus
• "= 0" needed

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

Examiner's comment

In (a)(ii), the 16 17 x = - was not always rejected, although almost all of these candidates used 3 x = in part (b) and were able to gain credit. Very few candidates used both values in part (b).

Common mistakes

• x = 3 only: reject x = −16/17 (lengths must be positive)

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

Examiner's comment

This question often enabled candidates to score marks in the first part, however, part (b) proved to be very good at discriminating between the most able candidates. In part (a), there were very few non attempts. Unfortunately, some did not know how to complete the square correctly and used 10 and 8, thus losing the marks. The majority of candidates achieved full marks, however or two marks because of a slip, usually in finding the radius.

Common mistakes

• complete the square: (x − 5)² + (y − 4)² = 9
• centre (5, 4), radius 3 (not using 10 and 8)

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

Examiner's comment

There were a number of completely blank responses to both parts of this question. This was either due to timing on the paper or possibly due to this topic not having been covered in sufficient depth due to disruption to learning. In part (b) it was pleasing to see candidates making an attempt although there were quite a few with nothing at all correct. These often had incorrect or irrelevant identities, resulting in equations which could not be solved for sin 2x, meaning that they could not access the last method mark. Those who did achieve 2 sin 2 3 x = (or -1) usually solved correctly to get, at least, one value of x, usually x = 20.9 and very often the second value too. Most who got to 11 this stage usually rejected any other values and scored full marks. Often, they had cancelled the cos 2x earlier in their working, so they did not have to reject the extra values at the end. Usually this should be avoided, although on this occasion the domain meant that this did not affect their final answer. It was pleasing to see that candidates had not tried to answer the question entirely by calculator by just stating the angles and the use of radians rather than degrees was not typically seen. Many did not recognise the relevance of part (a) and began this part from scratch, often successfully, even though they had not shown any relevant working for part (a).

Common mistakes

• multiply by cos 2x and use cos²2x = 1 − sin²2x (or use (a) with θ = 2x): 3sin²2x + sin 2x − 2 = 0
• sin 2x = 2/3 (reject −1 and x = 45° as cos 2x ≠ 0)
• find 2x then halve: 20.9°, 69.1°

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

Examiner's comment

Candidates who had not read the question and used g as 9.8 or 9.81 were few and far between. common approach in part (b) was to write down a quadratic equation in t and to solve it using the quadratic formula. There were occasional sign errors in the equation and some were either unable to deal with the quadratic or misquoted the formula. Nevertheless, a significant number did successfully find the two values of t and gave the correct one as their answer. The alternative approach of ‘up’ and ‘down’ separately was seen, but often only one of the times was calculated correctly. The other most common method being use of v = u + at but a significant number made a sign error in their equation and 0.2 was a very common wrong answer. In part (c), correct refinements seemed to be in the majority but only just. A more

Common mistakes

• e.g. 10 = −8 + 10T or 1.8 = −8T + 5T² gives T = 1.8
• sign errors and misquoted quadratic formula
• reject the second root (two answers is A0)
• if split up/down, both times needed

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

Examiner's comment

Part (a) was generally well answered by candidates with accurate and efficient methods frequently shown. The majority knew to differentiate to obtain an expression for velocity and were able to do so correctly and differentiating again to find the acceleration was almost always done correctly also. Some didn't set their velocity equal to zero and instead put their acceleration equal to zero, commonly scoring 3 out of the 6 marks available in this part. A number of candidates integrated instead of differentiating but this was rare. A tiny minority attempted to use suvat formulae in this variable acceleration question. Part (b) was much more challenging and most candidates didn't see the connection with the work they'd done in part (a) and what this meant for the direction of travel and total distance travelled. The majority of the successful candidates showed three separate sets of working for each of the three sections and then knew they needed to add the modulus of each section. This was often supported by sketching a graph. Integrating the velocity between 0 and 4 or using the given equation and a single calculation between 0 and 4 was the most common error leading to an answer of 16/3. A significant number of candidates did find s2, s3 and s4 (and often s1 as well) but then just added these values together so just scored the first method mark in this part.

Common mistakes

• v = ds/dt = t² − 5t + 6
• at rest means v = 0 (not a = 0), t = 2 or 3
• a = 2t − 5 gives 1 and −1
• integrating or suvat score 0

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

Examiner's comment

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.

Common mistakes

• complete the square on x and y: (x − 5)² + (y + 8)² = 169
• centre (5, −8), radius 13 (not ±13)
• centre (5, −4) from rooting 16 instead of halving

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

Examiner's comment

This was the third modelling question on the paper, this time based around an exponential model. This should have been expected given the Advance Information and candidates performed well on parts (a) and (b). Part (c) was more of a challenge with only higher achieving students scoring all 4 marks. Part (c) was the most demanding part of the question with many candidates just picking up the first mark for equating the two expressions. Having equated the two expressions, the significance of the negative exponential term seemed to be poorly understood. Very common errors included • changing the 0.05 e t - term to 0.05 e t and then solving • applying 'ln' to each term in the equation to form a linear equation in t • slips in coefficients with 220 becoming 200 A number of candidates reverted to numerical methods or the use of their calculator to solve the equation rather than solve by algebraic methods. The strongest candidates recognised the quadratic equation in 0.05 e t and were able then to solve it correctly, many after making a substitution 0.05 =e t x .

Common mistakes

• multiply by e^(0.05t): 220(e^(0.05t))² + 35e^(0.05t) − 800 = 0
• e^(0.05t) = 1.829 (reject negative)
• T = 20 ln 1.829 = 12.08
• taking ln of each term or changing e^(−0.05t) to e^(0.05t)
• calculator-only answer is SC

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

Examiner's comment

This question was based around the intersection of a cubic and a quadratic equation. It proved to be very accessible with many candidates able to make progress in both parts. In fact 30% of all candidates scored full marks on this question. In part (b) most candidates made some progress with weaker students scoring only the first mark for setting up the cubic equation. The majority were then able to divide by either 1 2 x - or ( ) 2 1 x - to form a quadratic factor and then equation which could then be solved to find the positive root. Marks lost were usually as a result of • slips in establishing the cubic equation or else errors in the division • not giving the exact answer of 4 33 -+ but instead its decimal equivalent • using a calculator to solve the cubic equation 2 3 2 15 42 17 0 x x x + - + =

Common mistakes

• 2x³ + 15x² − 42x + 17 = 0
• divide by (2x − 1) to get x² + 8x − 17
• x = −4 + √33 only (x > 0, reject −4 − √33)
• exact not decimal
• calculator cubic solving scores little

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

Examiner's comment

This question on arithmetic sequences and series proved to be very discriminating. The proof of the sum of an arithmetic sequence is standard, but was not answered well by many candidates. Most candidates however were able to then made progress with the rest of the question, having been given the required formula. Good candidates often did well in this question however, with 17% of the cohort achieving full marks. In part(i) those who had learned the standard proof for the sum of the first n terms of an arithmetic series often gained full marks. It was however, unusual to see a well set out, fully correct proof. Candidates often failed to write a correct expression for Sn at the start of their proof, or sometimes having an incorrect first or last term or extra terms (most often a+nd) within their sum. Other errors included not displaying a sufficient number of terms, not writing S= or Sn =, and using commas instead of + sign between the terms. In part(ii) most realised that they could make headway even if they hadn’t attempted part (i). In (ii) (a) many of the candidates used a=20, d = -0.8 and Sn = 64 in the sum formula to score the M mark. Poor notation such as (n-1)-0.8 was frequently seen, but candidates usually recovered from this. Following a correct equation most were able to proceed correctly to the given quadratic for the accuracy mark. A common error was where candidates used +0.8 as the common difference rather than -0.8 Part (ii) (b) was generally answered without issues. Almost all candidates managed to get B1 for getting the two values of n, either by factorising or by using a calculator. Part (ii)(c) was not answered very well with most candidates identifying 10 as the valid solution but not adequately justifying this choice. The reasons provided were often too vague or not related to why 10 was better than 16. More tended to score the mark by explaining that at 16 weeks he would be saving a negative amount rather than using the fact that 10 was the smaller number or that he would have had enough by 10 weeks so would not continue saving until 16 weeks.

Common mistakes

• n = 10 or 16

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

Examiner's comment

Candidates found part (b) more challenging. Although most knew to set f′(x) to zero for stationary points, there were many marks lost due to insufficient rigour in the presentation of method when looking for one or more roots. Generally candidates understood that it was the quadratic expression that needed to be used and not any part of g(x). Overall, most realised that the number of solutions for a quadratic was determined by the discriminant although a significant number did not. The majority attempted b2 – 4ac with the appropriate values but candidates often set 𝑏𝑏2 -4𝑎𝑎𝑎𝑎= 0 rather than 𝑏𝑏2 -4𝑎𝑎𝑎𝑎 >= 0. The B mark was usually gained by the use of the discriminant > or >= 0, it was rarely gained by making an appropriate statement. Some stated g(𝑥𝑥) ≠0 and a few g(𝑥𝑥) > 0 which may have been required if full justification had been required. Most candidates gained the M mark for calculating the value of the discriminant using the correct values for a, b and c. Other methods included completing the square to find the least value of the quadratic and then considering the y coordinate. Some differentiated the quadratic to find the minimum turning point and again considered the y value. These alternative methods were often incomplete and unsuccessful in most cases seen. Few realized the range had a lower limit, but this was condoned for the A mark as the k was defined as positive in the question. A common error was to give the final inequality as 4 9 k < .

Common mistakes

• stationary points need 12x² − 8x + 3k = 0 (g(x) ≠ 0)
• b² − 4ac ≥ 0 not = 0: 64 − 144k ≥ 0 gives 0 < k ≤ 4/9 (not k < 4/9)

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

Examiner's comment

The majority of candidates found this question accessible and were able to score consistently throughout. Where marks were lost they were usually in parts (b) and (c). 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 6𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 Part (a) was attempted successfully by most candidates, starting usually with 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. Candidates then successfully cross multiplied and used the was 6𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡= 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቀ 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቁ 2 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 appropriate identity to reach the required result. Another starting point regularly seen . Errors in the proof were rare if candidates started from a correct statement. Occasionally the “ = 0” was carelessly missing. 𝜋𝜋 In part (b) most candidates correctly solved the equation either by factorising or using 6 rather than the required 5𝜋𝜋 5𝜋𝜋 calculator technology, but many reached the acute angle 6 . Others found a number of solutions but failed to identify 6 as the obtuse angle of 𝜋𝜋 required answer. A few candidates worked in degrees and again lost the accuracy mark. 6 in (b) then usually carried this error into (c) and 𝜋𝜋 Those who had obtained an angle of 6 in (c) often would reach 9൫1 + sqrt 3൯, but did not seem to worry that this did not match the required answer were in a position only to earn the method marks. Candidates using format given in the question. There were many attempts, however, to use the correct formula for the sum to infinity with appropriate values for the first term and the (often incorrectly) algebraic expressions for the sum to infinity involving 𝜃𝜃, without common ratio. Some candidates unfortunately put effort into finding and rearranging 𝑎𝑎 1+𝑟𝑟 incorrectly as the formula substituting an angle in, gaining no credit, and others used for the sum to infinity. The most common mistake in (c) was failure to show the steps 3 ൰ straight into their calculators to obtain 9 -9 sqrt 3 losing the last two marks in -6 sqrt 3 where the denominator was rationalised. Many candidates clearly typed an expression 1-൬- sqrt 3 like the question.

Common mistakes

• sinθ = ½ (25/2 impossible)
• θ obtuse so 5π/6, not π/6 or 150°

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

Examiner's comment

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 (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 - sqrt 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 - sqrt 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.

Common mistakes

• lower limit: tangency, discriminant of x² − x + (11 − 2k) = 0 gives k = 43/8
• upper limit from the end point t = −π/4, (−1, 7): k = 13/2
• 43/8 < k ≤ 13/2 (most missed the upper limit
• draw lines on the sketch)

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

Examiner's comment

This was a short opening question to the paper, which enabled nearly all candidates to score at least one mark. However, several candidates did not find their critical values correctly, sometimes equating to 20 rather than rearranging and equating to 0 or making errors when factorising. Most, however, were able to find 5 and -4. Typically, most realised that they needed to select the outside region, although there were a few responses with incorrect inequalities. It was very rare for candidates to correctly give their answer using set notation, and very few being able to score full marks.

Common mistakes

• rearrange to x² − x − 20 > 0 before finding critical values 5 and −4 (do not equate to 20)
• outside region
• set notation {x : x < −4} ∪ {x : x > 5} needed for A1 (very rarely seen)
• {x < −4, x > 5} not accepted

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

Examiner's comment

This was a question which was completed to a high degree of accuracy by a pleasing number of candidates. There seemed to be more notice taken by many candidates regarding showing all stages of their working. In part (a) it was common for candidates to score full marks. Typically, candidates factorised out the x and then they proceeded to correctly factorise the resulting quadratic proceeding to the correct solutions. It was rare for a candidate to divide by x and hence lose the solution of 0.

Common mistakes

• take out the factor x: x(3x + 1)(x − 6) = 0
• dividing by x loses x = 0
• x = 0, −1/3, 6 and no extras

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

Examiner's comment

This was a question which was completed to a high degree of accuracy by a pleasing number of candidates. There seemed to be more notice taken by many candidates regarding showing all stages of their working. Part (b) proved to be a good discriminator. Some candidates tackled this part well recognising the link from (a) and showing sufficient working of using their solutions to proceed to correct ones in (b). However, a significant number came unstuck with this part and tried to multiply out the brackets and start again. Some also tried to set 2 ( 2) y - equal to their solution, but then multiplied out the brackets and solve using the formula or factorising a new quadratic, rather than just square rooting and adding 2. It was rare for candidates to just use their calculator which could not score. Some squared rather than square rooted, and some forgot the negative solution from square rooting. Others thought you could square root a negative.

Common mistakes

• spot the structure: (y − 2)² = 0, −1/3, 6 from (a)
• reject the negative value (no real root)
• y = 2, 2 ± √6 (remember the ± and do not square instead of square root)
• multiplying out the brackets and starting again wastes time

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

Examiner's comment

Candidates were usually able to make good progress on the first half of this question on solving trigonometric equations, but they tended to run into problems halfway through. It was rare for full marks to be scored, although most were able to score around half of the available marks. In part (i), nearly all candidates were able to proceed to the correct quadratic in sin , although the occasional candidate tried to incorrectly substitute sin into the right- = - 1 cos hand side of the equation. Most candidates were able to apply the quadratic formula and proceed to finding all three angles. Some found additional angles, which even if they are outside the range, prevent full marks from being scored. Most candidates seemed to opt for sketching trigonometric graphs to find the other angles as opposed to using the CAST diagram. Part (ii)(a) was poorly answered, however. Most candidates failed to score either mark, although those who did were usually able to make some reference to cancelling out sin x . Just stating that not all the angles had been found was insufficient explanation as to the errors made so candidates should be encouraged to provide greater detail if they are able to do so. Part (ii)(b) was attempted by most candidates, with many able to equate to the correct angle, rearrange and proceed to the correct answer. There were a number who equated to several angles, but either did not make their final answer clear, or did not rearrange to find the required angle. 8

Common mistakes

• use cos²θ = 1 − sin²θ to get 5sin²θ + 6sin θ − 5 = 0
• sin θ = (−3 + √34)/5 only (other root < −1)
• θ = 34.5°, 145.5°, 394.5° in 0 < θ ≤ 450°
• extra values outside the range lose the final mark

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

Examiner's comment

This question was one of the most successful on the paper. It tested aspects such as completing the square, simple integration to find an area and finding a turning point which candidates seemed comfortable finding in nearly all cases. Part (a) was typically correctly answered. Most opted to complete the square rather than equate coefficients and it was only the minority who made slips with the 3 - coefficient for a. Where errors occurred, it was usually in multiplying out to get the coefficient for c, but they were still able to score 2 marks.

Common mistakes

• take out the factor −3: −3(x − 2)² + 20
• slips with the −3 when finding c

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

Examiner's comment

This question was one of the most successful on the paper. It tested aspects such as completing the square, simple integration to find an area and finding a turning point which candidates seemed comfortable finding in nearly all cases. Part (b) required candidates to find the turning point. Whilst the intention was for candidates to use their part (a) answer, and follow through was allowed on this, it was much more common for candidates to differentiate the quadratic and find the turning point. Therefore, nearly all candidates score both marks on this part as most seemed more confident using calculus and proceeded to the correct answer anyway.

Common mistakes

• maximum at (−b, c) = (2, 20) from the completed square (differentiation also works)

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

Examiner's comment

This question was one of the most successful on the paper. It tested aspects such as completing the square, simple integration to find an area and finding a turning point which candidates seemed comfortable finding in nearly all cases. Part (c) was typically completed correctly. Candidates usually opted to integrate the curve and find the area between 0 x = and 2 x = and then subtracted this area from 40. Those candidates who tried to form a single integral to solve the problem were the ones were errors typically occurred. Some candidates equated the line 20 y = to the curve and rearranged to one side and integrated this expression. This method can work, although some responses did not appear that the candidates were that sure as to why they had achieved a negative value for the integral if they collected terms on the wrong side. In addition, errors in collecting terms resulted in an incorrect integrated expression and, ultimately, an incorrect final answer.

Common mistakes

• R = rectangle 2 × 20 − ∫₀² (−3x² + 12x + 8) dx = 40 − 32 = 8
• use the limits 0 and 2
• sign errors if integrating 20 − f(x) as a single integral

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

Examiner's comment

9 This question provided a good range of marks between the candidates, although nearly all candidates struggled to proceed to the final answer in part (c). There were also several blank responses on this question which may have been down to candidates either running out of time or just getting overwhelmed towards the end of a paper. Part (c) was either not attempted at all or candidates made an attempted but abandoned due to the expressions formed appearing so complicated that it seemed as though they were convinced they had gone wrong. Usually, candidates opted to substitute their part (a) answer into their part (b) equation of the circle. To score the first mark candidates had to multiply out the brackets and collect terms together to form a 3TQ where k was in both the b and c coefficients of the quadratic. Some candidates stopped before getting this far, although it was pleasing that some continued to persevere and pick up the first mark. Those who continued, realised the discriminant needed to be used, but disappointingly once they had their expression also decided to stop even if they had a quadratic in k which they could solve. Therefore only 1 or no marks was scored in nearly all responses. The easiest way was the vector approach, which was seen, but far too frequently, and even then, it was only a few candidates who were able to score full marks on this question.

Common mistakes

• the other tangent touches at the opposite end of the diameter through P: (13/2, 11/2), then k = 10/3
• via simultaneous equations the quadratic in x has b and c depending on k and b² − 4ac = 0 (k = 0 is rejected)
• candidates often abandoned the messy algebra

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

Examiner's comment

The last question on the paper was attempted by most candidates, although there were a significant number of blank responses. Again, this was possibly either due to time, or more likely due to candidates just feeling overwhelmed as the question was the longest on the paper. Part (b) also demonstrated the lack of confidence with using calculus in relation to turning points. Quite often they were able to differentiate to get a quadratic in x, however they seemed unsure where to go from there and often just state that there were no solutions or that it would not factorise. It was typical for candidates to only score 1 mark in this part. 10

Common mistakes

• f′(x) = −6x² + 30x − 39
• discriminant 30² − 4(−6)(−39) = −36 < 0 so f′(x) ≠ 0 and there are no stationary points
• a conclusion with reason is needed
• "does not factorise" is not enough

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

Examiner's comment

The last question on the paper was attempted by most candidates, although there were a significant number of blank responses. Again, this was possibly either due to time, or more likely due to candidates just feeling overwhelmed as the question was the longest on the paper. Most candidates did not attempt part (d), whilst some were able to find that 20 x = they failed to state the coordinates so hardly any candidates scored any marks on this part.

Common mistakes

• Q(x) has no real roots so the only x-intercept of y = f(x) is (4, 0)
• stretch scale factor 5 parallel to the x-axis gives (20, 0)
• y-intercept unchanged (0, 44)
• give coordinates, not just x = 20

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

Examiner's comment

Some went on to put a = 0, solving to get t = 5 which they went on to use in part (b) but this subsequent working was ignored in part (a). (b) The majority used v = 0 when t = 6, found a value of k and obtained a quadratic. Unfortunately, many made sign errors and their expression had +24 instead of -24. If they went on to solve their incorrect quadratic using a correct method, these candidates scored two of the four marks available. There were several completely correct answers.

Common mistakes

• instantaneously at rest means v = 0
• v = 0 at t = 6 gives k = 24
• sign errors (+24 instead of −24)
• solve the quadratic for t = 4

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

Examiner's comment

This focussed on the ability of a student to complete the square and proved to be as accessible as the first question. The vast majority of responses seen were fully correct. In part (a) most candidates were able to score full marks. The form (x-2)2 + .... was almost always achieved, although occasionally an incorrect (x+2)2 +…. was offered. The most common error was having +9 instead of +1.

Common mistakes

• (x − 2)² + 1
• +9 instead of +1 is the common error
• (x + 2)² is wrong

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

Examiner's comment

This focussed on the ability of a student to complete the square and proved to be as accessible as the first question. The vast majority of responses seen were fully correct. In response to part (b) both P and Q were often stated correctly with some candidates able to score the follow through mark for their Q from their value of ‘b’ from part (a). Occasionally a mark was lost if they had P =5 instead of y =5 when x=0. Very few sketches were seen, which if correctly labelled would have also scored full marks in (b). It was relatively common in (b)(ii) to see differentiation used, rather than the completed square expression.

Common mistakes

• P = (0, 5), not P = 5
• minimum Q = (2, 1) read from the completed square (−a, b), no need to differentiate

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

Examiner's comment

This proved to be a little more demanding than the first two questions but over 40% of candidates still scored all 6 marks. Errors seen by candidates are summarised below. In part (b) candidates usually found both roots of the given quadratic either by factorisation or by use of the quadratic formula, with a small number by completing the square. A significant minority failed to gain the A mark here as they did not state that 6 was an integer or that 40/3 was not an integer, with some excluding 40/3 on the grounds that it was irrational.

Common mistakes

• roots 6 and 40/3
• k = 6 because k must be an integer (40/3 is not an integer
• it is not irrational)

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

Examiner's comment

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.

Common mistakes

• complete the square: (x − 5)² + (y + 2)² = 18
• centre (5, −2)
• r = 3√2, not ±3√2 or √18
• halving errors

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

Examiner's comment

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 (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.

Common mistakes

• substitute y = 3x + k into the circle to get 10x² + (6k + 2)x + k² + 4k + 11 = 0
• tangent means b² − 4ac = 0
• k = −17 ± 6√5
• alternatives via the radius gradient −1/3 or implicit differentiation rarely completed

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

Examiner's comment

This was the second modelling question on this paper. It proved to be a significant discriminator between candidates, with a significant minority making little or no progress. It certainly seems to be a topic that students find hard to access. Part (a) The most common approach which led to either partial or complete success was to start with H=ax2+bx+c and use (0,3) to find that c =3. Many candidates then used (120,27) to find an equation connecting a and b, but sometimes no more progress was made. Of those who did make further progress, the most common approach was to use differentiation and the knowledge that the gradient was zero when x=90 to get a second equation in a and b. Candidates using this approach often went on to get (a) fully correct and then also to gain most if not all of the marks in (b) as well. It was only rarely that there was a misunderstanding about H=0 rather than H=3 when x=0. There were other methods used in (a) and any of these could have led to success if applied correctly. For example, using the fact that the quadratic has a turning point when x =90 and so H = a(x-90)2+c was used correctly by some candidates. Alternatively, using additional implied points such as (60,27) or (180,3) due to the symmetry of quadratic curves, was used appropriately by candidates.

Common mistakes

• H = ax² + bx + 3 (H = 3 when x = 0, not 0)
• use H = 27 at x = 120 and dH/dx = 0 at x = 90 (or vertex form a(x − 90)² + c, or symmetry points (60, 27), (180, 3))
• H = −x²/300 + 3x/5 + 3

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

Examiner's comment

This was the second modelling question on this paper. It proved to be a significant discriminator between candidates, with a significant minority making little or no progress. It certainly seems to be a topic that students find hard to access. Part (b) Both parts were done well by candidates who had achieved a correct answer to (a). Part (b) (i) There were only a very few who omitted the units here. A wrong answer to (a) would have had made it difficult to gain the accuracy mark here. Part (b)(ii) A wrong answer to (a) could have scored a method mark here if the candidate’s quadratic had been solved correctly.

Common mistakes

• maximum at x = 90: H = 30 m (units needed)
• set H = 0 and reject the negative root: 185 m

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

Examiner's comment

acceleration formulae. Part (d) allowed those who had made little progress on the first 3 parts, to score some marks as it only required the use of the given velocity vector to get started. For those who attempted it, the use of Pythagoras’ theorem on the velocity components was very accessible and a number of candidates solved the quadratic to obtain t = 4. Sign errors here were common, however, with many ending up with a wrong t value. Some candidates then scored a further two M marks for substituting their value of t into their position vector and then applying Pythagoras to find the distance. The question required an exact distance and so only surd answers earned the final mark.

Common mistakes

• speed 10 means 9t + 4t² = 100, t = 4 (sign errors common)
• substitute into r: 14i − 16j
• distance is the magnitude of r
• exact answer 2√113 (√452), decimals lose the final mark

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

Examiner's comment

The binomial expansion seemed a welcome relief with many candidates achieving full marks in part (a). The usual problems with exponents crept in but fortunately there were no ‘negatives raised to powers’ to contend with. The simple equation in part (b) was not answered quite as well but coefficients were followed through from part (a) allowing some marks if they multiplied the correct expression by 3. In part (b), those who got an answer for part (a) would usually achieve at least the first mark in part (b). Some candidates were unable to form a correct equation for the coefficients and simply set them equal to each other without multiplying the linear term by 3. There was a small number of cases where the wrong side was multiplied by 3. It was disappointing that a large number are still uncomfortable with the term “coefficient” as there were a lot of scripts that kept the x and x3 in the equation and even tried to solve to find x. Others removed the x and x3 later on, but this should be discouraged as a general approach.

Common mistakes

• 120k³ = 3 × 10k (multiply the x coefficient by 3, not the x³ one)
• coefficients do not include x
• cancel k (k ≠ 0) to get k = ±½

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

Examiner's comment

7 The first part of this question was generally answered well with many candidates scoring full marks, but the second part proved to be more challenging for a disappointing number of them. In part (b), some candidates struggled with identifying the given equation as a hidden three term quadratic, with a few trying unsuccessfully to simplify by multiplying every term by k, for example. Some candidates tried to square both sides but were mostly let down in this approach by mistakes in expanding out the brackets or simply squaring individual terms. Those candidates who were successful often proceeded by setting y k = , for example, and proceeding correctly to two values, which they then squared, but a number did not realise that they needed to discard the solution of 3 k = -. Some candidates confused themselves by setting k k = to produce a three term quadratic in k but then failed to realise they had not found the actual value for k.

Common mistakes

• quadratic in √k: (3√k − 4)(√k + 3) = 0
• reject √k = −3 so k = 9 is not a solution
• square to get k = 16/9 (not √k)

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

Examiner's comment

This question was attempted by most candidates with many being able to secure nearly all the marks in part (c). It was rare, however, for full marks to be scored overall, with most struggling with the transformations in part (b). Part (c) proved to be high scoring for many candidates, however, some did not recall the identities correctly and made little progress. In particular, candidates made errors with rearranging 2 2 sin cos 1 x x + = or missing the square from either or both terms. The given interval was unusual, although it was pleasing to see many of the candidates who proceeded to find a value for were able to find the one that was in the required interval. 9

Common mistakes

• multiply by cos θ: 3cos²θ = 8 sin θ, then cos²θ = 1 − sin²θ
• 3sin²θ + 8sin θ − 3 = 0 gives sin θ = ⅓
• only 520.5° lies in 450 to 720

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

Examiner's comment

Candidates were generally successful in answering the first part of this question, but found the second part more challenging, with only a few gaining full marks. In part (i), most candidates were confident in completing the square to find the coordinates of the centre of the circle, and many could then successfully find the gradient of the line from the centre to the given point. There were still a number of candidates who made sign errors when finding the gradient, however. Some candidates forgot to then find the negative reciprocal of this gradient in order to find the tangent, and a few did not give their final answer in the requested form, but this was generally answered well. Candidates should be reminded to check the form that the answer should be given in to ensure that they score marks that they are more than capable of achieving. In part (ii), again most candidates were confident in completing the square and finding the radius of this circle in terms of k, but a number then stated this radius needed to be less than 6, rather than saying it needed to be less than 4. There were a large number who made errors + 52 k just rearranging the equation resulting in the radius as . Only a very small number of k candidates picked up the final mark for realising that, in addition to , we also required 36 k in order for the circle to actually exist. 52

Common mistakes

• complete the square: centre (−9, 1)
• radius gradient 3/2, tangent gradient −2/3 (perpendicular)
• sign errors
• answer 2x + 3y − 11 = 0 with integers

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

Examiner's comment

Candidates were generally successful in answering the first part of this question, but found the second part more challenging, with only a few gaining full marks. In part (i), most candidates were confident in completing the square to find the coordinates of the centre of the circle, and many could then successfully find the gradient of the line from the centre to the given point. There were still a number of candidates who made sign errors when finding the gradient, however. Some candidates forgot to then find the negative reciprocal of this gradient in order to find the tangent, and a few did not give their final answer in the requested form, but this was generally answered well. Candidates should be reminded to check the form that the answer should be given in to ensure that they score marks that they are more than capable of achieving. In part (ii), again most candidates were confident in completing the square and finding the radius of this circle in terms of k, but a number then stated this radius needed to be less than 6, rather than saying it needed to be less than 4. There were a large number who made errors + 52 k just rearranging the equation resulting in the radius as . Only a very small number of k candidates picked up the final mark for realising that, in addition to , we also required 36 k in order for the circle to actually exist. 52

Common mistakes

• centre (4, −6), radius √(52 − k)
• in the 4th quadrant needs radius < 4 (distance from the centre to the y-axis, not 6) so k > 36
• circle exists needs 52 − k > 0
• 36 < k < 52

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

Examiner's comment

answer. In part (c), a significant minority failed to understand that s = 1.2 was required with many using 15.6. Those using one equation often made a sign error. Calculators were usually used to solve the resulting quadratic equation. The other method which seemed to produce fewer errors was to find the time to the top and the time down and then add. In part (d), very few correct graphs seen. Many

Common mistakes

• displacement is 1.2 m below A, not 15.6 m
• 1.2 = −5t + 5t² gives 5t² − 5t − 1.2 = 0, t = 1.2 s
• sign errors in the one-equation method

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

Examiner's comment

had consequences in part (b).The majority, for part (b), realised that the particle would change its direction of motion when v = 0 and solved this equation to give t = 3.5. They were then able to integrate v correctly and use their t value to obtain the answer. A few, however, used the t value that had been found in part(a) and

Common mistakes

• changes direction when v = 0, t = 3.5 (not the t from (a))
• integrate v (suvat scores M0)
• s = 931/24 = 38.8 m

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

Examiner's comment

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.

Common mistakes

• l is y = 6x + 25
• C from the vertex form y = a(x + 2)² + 13 through (0, 25), y = 3(x + 2)² + 13
• region 3(x + 2)² + 13 < y < 6x + 25 (use y, not R or C
• ∩ not ∪)
• many found the area by integration instead

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

Examiner's comment

Parts (a) and (b) of question 9 were standard bookwork involving the product rule of differentiation using an exponential function. Well rehearsed candidates were very successful here scoring the majority of marks. The diagram should have helped in determining the maximum and minimum points of the function.

Common mistakes

• f′(x) = 0 when 2 + x − x² = 0 (e^(−2x) is never 0), x = −1 and 2
• exact points (−1, −4e²) and (2, 8e⁻⁴)
• extra point from x = 0 penalised

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

Examiner's comment

This question on sequences proved very discriminating, with weaker candidates scoring only the first mark. As with many questions on this paper however, there were some excellent and well written responses. Part (c) could be done independently to (a) and (b), and many candidates took advantage of this. It was not uncommon to see the terms 2, 4, 1 - - in (c) where only the first mark in (a) had been scored. Methods of finding the sum to 80 terms generally revolved around the calculation ( ) ( ) ( ) 26 2 4 1) 2 4 x + -+ - + + - . Incorrect attempts involved uses of both arithmetic or geometric sum formula or ones involving an average of the terms.

Common mistakes

• k = −2 (k ≠ 1) gives repeating terms 2, −4, −1
• 26 full cycles plus 2 + (−4) = −80
• arithmetic or geometric sum formulas do not apply

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

Examiner's comment

The first part of part (b) was straightforward and gave an opportunity to re-enter the question for those who had come unstuck in part (a). Indeed, for some candidates this was the only mark (𝑥𝑥-1)(𝑥𝑥-8) or gave ‘solutions’: (𝑥𝑥-1) and (𝑥𝑥-8). Only a very small number of candidates attained in this question. It was surprising that a number of candidates left their ‘solution’ as incorrectly obtained 𝑥𝑥= -1 or 𝑥𝑥= -8. Most candidates were able to identify 𝑥𝑥= 8 as the value which was not a solution. However, in The second part of part (b) was more discriminating and a careful explanation was required here. log(4 -8) is undefined. Too many candidates simply stated ‘logs can’t be negative’ which was order to gain credit, it was also necessary for candidates to make specific reference to the fact that incorrect and quite different to ‘you can’t take logs of a negative number’. Candidates should be reminded that reasoning should be precise and specific.

Common mistakes

• roots 1 and 8 (not factors or −1, −8)
• reject 8 because log(4 − 8) = log(−4) cannot be found
• "logs can't be negative" is not accepted

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

Examiner's comment

In part (b), it was common for candidates to achieve no marks for part (a) and then full marks for this part. Most of the students gained the first method mark and managed to produce an equation in cosx only. Mistakes were made with signs and brackets. In particular, candidates failed to write brackets around 4cos3x – 3cosx and hence would obtain an incorrect quadratic equation to solve. Several candidates ‘solved’ the cubic/quadratic equation using the polynomial equation function on their calculator and did not show any evidence of their working. Some incorrect equations led to complex solutions but most of these candidates still did not realise that they had made an error. A large number of candidates divided by cosx to simplify their expression but they failed to recognise this lost a solution and therefore were not awarded the final mark. A small number of candidates failed to gain any marks on this part as they replaced the right hand side with cosx but left the left hand side with cos3x and then proceeded to treat this as a cubic/quadratic. When candidates had identified the correct equation and solutions they were very good at finding all the solutions in the range. A large number who did get solutions did not include -90° or included 180° and therefore lost the final mark.

Common mistakes

• brackets round 4cos³x − 3cos x
• factor out cos x, do not divide by it (loses cos x = 0)
• solutions −90°, 0°, 90°, 139°
• include −90°, exclude 180°

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

Examiner's comment

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.

Common mistakes

• tangent so b² − 4ac = 0
• simplifies to r² − 18r + 36 = 0
• r = 9 ± 3√5 as simplified surds

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

Examiner's comment

A number of the responses were completely blank suggesting perhaps that candidates had run out In part (b), the first method mark was gained by the majority of students. Some didn’t divide by “a” or wrongly cancelled r10 and r5 in their equation, so lost the first accuracy mark. However, many of the students who gained the first two marks continued to gain the last two marks by dealing with the quadratic and achieving the correct solution for “r”. A few candidates used (n - 1) in the formula rather than n and a number of candidates got confused with n and r, replacing r = 10 and 5 in the denominator of each fraction. Only around 50% realised that they had ended up with a quadratic equation in r5 and candidates who did not cancel the "a" out early often ended up with an overcomplicated equation which they could not solve. Very few did not eliminate r = 1 if they found it, although a few solved the quadratic and then forgot to cube root the answer and so had r = 3.

Common mistakes

• cancel a and (1 − r): 1 − r¹⁰ = 4(1 − r⁵)
• quadratic in r⁵
• reject r = 1
• r = fifth root of 3 (not r = 3)
• S₁₀ = 4S₅, not S₅ = 4S₁₀

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

Examiner's comment

In the first part, the majority of candidates used v = u + at with t = 2 to obtain a correct constant, in this case, the initial velocity. In part (b), most used r = ut + 1 velocity. Some chose to integrate, often successfully, although some failed to include a 2 at2 to find a displacement vector. Again, some chose to integrate, mostly correctly. Having equated their displacement expression to λi – 4.5j, some did not know how to proceed. A few tried to use magnitudes and reduce the problem to scalars and some had mixed up velocity and displacement. Those who realised that equating j components gave an equation in T generally solved the resulting quadratic and found the correct answer. Part (c) required the marks in part (b) went on to gain the two marks here. A method mark was available for those carrying through a wrong value for T provided that it had been found from a valid method in part (b).

Common mistakes

• r = ut + ½at² (not with u = 0)
• equate the j components only: −4.5 = 2T − (5/2)T² gives T = 1.8
• do not take magnitudes or mix up velocity and displacement

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

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• perpendicular to i means the i component of v is zero: t − 2t² + 36 = 0 gives t = 4.5
• equating the j component to zero (or i to j) scores nothing
• use velocity, not displacement

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

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• differentiate r to get v = (2t − 1)i + 3j
• speed is the magnitude: (2t − 1)² + 3² = 5² gives t = 2.5
• equating the vector to 5 or dropping i and j is wrong
• not √(x² − y²)

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

Examiner's comment

(i) Many candidates approached this question well and were able to manipulate the indices correctly to achieve a = 0.25. Several candidates who derived a fractional index of x on one side of the equation were unable to successfully ‘undo’ the index e.g. x3/2 = k often became x = sqrt (k3). Few managed to deduce a =0 due to cancelling a factor of a instead of factorising it out. Many candidates squared both sides of the given equation to have integer indices, though they frequently did not square the coefficients as well. Depending on how they processed their equation was key to whether they got both solutions. Those who factorised or used calculator methods to solve the quartic equation were more likely to get both solutions. A significant number of candidates displayed very poor algebra skills when attempting to isolate a in this part and for the most part the same candidates also tackled (ii) poorly. (ii) Many candidates were able to spot the quadratic equation in b2 which then led to correct factorising and the elimination of the imaginary solutions. Reasons for losing marks were: 2 2 giving only , instead of the full solution 3i including the imaginary solutions so losing the final mark. solving the equation using their calculator, a method forbidden by the question.

Common mistakes

• quadratic in b²: (b² + 9)(b² − 2) = 0
• reject b² = −9 (real solutions only, no ±3i)
• give both ±√2
• calculator-only solutions score nothing

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

Examiner's comment

In general, parts (a) and (b) were well done with part (c) causing some issues. (c) Candidates who identified the link between the discriminant and the fact that l was a tangent, generally secured full marks. Those who attempted the alternative method of completing the square were often defeated by the algebra. Attempts using differentiation tended not to gain any credit.

Common mistakes

• tangent means repeated root: b² − 4ac = 0 for 2x² − 4x + k² = 0, so 16 − 8k² = 0
• k = ±√2 (both)
• differentiation attempts rarely worked

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

Examiner's comment

Most were able to attempt this question, but lack of units cost a significant minority a mark. (a) All but the weakest candidates produced a correct value.

Common mistakes

• n = 1 gives 117 tonnes
• units required

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

Examiner's comment

Most were able to attempt this question, but lack of units cost a significant minority a mark. (b) Again, the majority of candidates found this value. However, many did not appreciate the form of the given equation and embarked on differentiation, thus expending much effort for a single mark.

Common mistakes

• completed-square form gives maximum 1200 tonnes when n = 20
• no need to differentiate

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

Examiner's comment

Most were able to attempt this question, but lack of units cost a significant minority a mark. (c) Most candidates could not distinguish between ‘in 2023’ and ‘up to the end of 2023’ and offered 432 as their final answer (from n = 4). Those who had some understanding of the requirement of the question were about equally split between the correct T5 – T4 (= 93) and T4 – T3 (=99).

Common mistakes

• mass mined in 2023 is T(5) − T(4) = 525 − 432 = 93 tonnes, not T(4) = 432 or T(4) − T(3)

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

Examiner's comment

Most were able to attempt this question, but lack of units cost a significant minority a mark. (d) Whilst some candidates realised the model could not work after 20 years (or 40), most could not explain why. The second mark was rarely awarded as candidates failed to appreciate that it was the total mass mined (T) that could not decrease. Many responses commented on the context of the question rather than the model and went on to discuss the logistics of mining and how the tin might or might not run out over time.

Common mistakes

• model valid only for n ≤ 20 because the total mass mined cannot decrease after the maximum
• comment on the model (total T), not on mining logistics

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

Examiner's comment

(a) This was generally answered successfully with candidates able to complete the square and set the equation in the correct format. Usual numerical errors when manipulating equation led to sign errors and a wrong value for r 2.

Common mistakes

• complete the square: (x − 2)² + (y + 4)² = 28
• centre (2, −4)
• radius √28 = 2√7 (sign slips with r²)

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

Examiner's comment

As a general point, it is worth noting that a good examination technique is to look at the mark allocation in a question. Some candidates submitted virtually no work in part (b) where there were 4 marks, yet produced half a page of work in (c) or (d) which had an allocation of just 2 marks. The (b) As in part (a), a ‘show that’ question always needs a concluding statement. Most candidates were very adept in algebraic division and did so clearly and accurately gaining the first two marks. The majority went on to factorise the 3TQ successfully. What too many candidates failed to do at this stage was to refer back to f(x), write it out in its complete factorised form of , and then state that because there was a repeated factor, the equation had only two roots. Confusion between roots and factors cost some candidates the final as a root. Quite a few candidates found b2– 4ac for the mark as they incorrectly referred to quadratic expression, stated that as it was > 0 and therefore two roots, clearly thinking that they had answered the question. There was also clear evidence of use of calculators to find the roots followed by an attempt to work backwards to obtain the factors; this was unacceptable. those merely explained what they had done in (b).

Common mistakes

• quadratic factor 2x² − 5x − 12 = (2x + 3)(x − 4)
• write f(x) = (x − 4)²(2x + 3) and explain 4 is a repeated root, so roots 4 and −3/2 only
• b² − 4ac > 0 of the quadratic is not an answer
• no calculator roots

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

Examiner's comment

f 4 4 2 3 x x x x f 0 x 4 x 0 , 2 This was an open ended question, types of which are expected in the new specification. Almost all candidates were able to achieve some credit in this question. Poor notation (usually the omission of dx from the integral) caused many to forfeit the final mark. Some candidates seemed unaware that it was unacceptable to rely entirely on a calculator and that they needed to show detailed working in this “show that” question. Many candidates attempted to integrate (usually correctly) but with no other work of any value, some using 0 and k as limits and some even showing the substitution of k and 0 and subtracting. But without a value of k found via differentiation, only the second B mark was available to them. Other candidates thought they could find k by finding the roots of the original cubic despite it being clear from the diagram that the origin was the only root. Of those who realised differentiation was required, most arrived at x = 5/3, 4. Some of these candidates spotted from the graph that k = 4, but many wasted time, either by finding the corresponding y- coordinates (and using the x value corresponding to the lower of these) or using the second differential (where d2y/dx2 > 0) to determine which was k. A few candidates having found the second differential, equated this to zero and erroneously used the resulting value as k. Occasional mistakes with limits included using 5/3 instead of 0 as the bottom limit and subtracting the wrong way round (or even adding) after substituting. Some candidates forfeited the final two marks for failing to show intermediate working.

Common mistakes

• find k by dy/dx = 0: 6x² − 34x + 40 = 0 gives x = 5/3 or 4, choose k = 4 (the minimum, not a root of the cubic)
• integrate from 0 to 4 showing substituted values
• dx and correct notation needed for the given answer

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

Examiner's comment

In the first part, the majority of candidates knew that finding acceleration involved differentiating the given expression for the velocity and, apart from the occasional slip, this was generally done well. However, a fair number then found either t when a = 0 or a when t = 0 without realising that, to find the relevant value of t, it was necessary to solve the quadratic equation obtained by setting v = 0 (when the particle is at instantaneous rest) . Many of those with the correct working lost the final mark for giving a negative answer for the ‘magnitude’ of the acceleration. Most candidates knew they had to integrate for part (b) which they did mostly successfully and many

Common mistakes

• instantaneous rest means v = 0: solve 12 + 4t − t² = 0 to t = 6 (reject −2)
• a = dv/dt = 4 − 2t = −8
• magnitude 8 m s⁻² (positive)
• not a when t = 0 or t when a = 0

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

Examiner's comment

This question proved to be more challenging for a number of candidates although many were still able to score full marks. In part (b) almost all candidates were able to write down the small angle approximation for cosx even if they were unable to use it in order to answer the question. Most candidates did substitute correctly into the equation in order to gain a quadratic equation, although some did contain a mixture of variables. Sign errors sometimes occurred at this stage and the weakest candidates were unable to deal correctly with resulting fractions in order to simplify the equation. Many candidates appear to be using calculators in order to solve quadratics and in this question, this usually resulted in full marks. The final mark was lost by some candidates who correctly solved the quadratic equation but gave both roots as their final answer.

Common mistakes

• cos x ≈ 1 − x²/2
• x² + 4x − 1 = 0
• give only the small root −2 + √5 ≈ 0.236, not both roots

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

Examiner's comment

This question on completing the square and the quadratic function was accessible to all, especially the first 6 Part (a) was very straightforward and many students were able to write down the answer without any difficulty. x 2 2( 1) 8 Although errors were rare, one common incorrect answer was . Part (b) was equally straightforward, even for candidates who did not achieve the correct result in part (a). 2 ( ) a x b c Many used the form or their graphical calculators to produce accurate and well drawn curves. Marks were lost here when candidates drew V shaped curves or had incorrectly placed turning points. As stated earlier, part (c) was a lot more demanding. Rather strangely part (i) was found more accessible than

Common mistakes

• 2(x + 1)² + 7
• common error 2(x + 1)² + 8

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

Examiner's comment

This question on completing the square and the quadratic function was accessible to all, especially the first 6 Part (b) was equally straightforward, even for candidates who did not achieve the correct result in part (a). 2 ( ) a x b c Many used the form or their graphical calculators to produce accurate and well drawn curves. Marks were lost here when candidates drew V shaped curves or had incorrectly placed turning points. As stated earlier, part (c) was a lot more demanding. Rather strangely part (i) was found more accessible than

Common mistakes

• U shape not V
• y-intercept (0, 9)
• minimum (−1, 7) in the second quadrant

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

Examiner's comment

This question on completing the square and the quadratic function was accessible to all, especially the first 6 marks. Part (c) was more demanding but there were some excellent responses here. As stated earlier, part (c) was a lot more demanding. Rather strangely part (i) was found more accessible than part (ii). Many candidates sketched the new curve and compared the positions of the turning points noting that had ''moved'' to 1,3 . Disappointingly a rather sizeable majority could only describe this ''movement'' 1,7 21 h( ) x as a ''transformation'' rather than ''translation'' . Fewer students could use the form to 2 2( 1) 7 x 21 x y 0 find the range of the function. Many resorted to substituting producing a range with as one of 9 the limits. Common incorrect answers for the range of h scoring one mark were h( ) 3 x or 0 f ( ) 3 x .

Common mistakes

• (i) needs the word translate and the vector (2, −4), not just "transformation"
• (ii) h(x) = 21/(2(x + 1)² + 7) so 0 < h(x) ≤ 3
• common errors h(x) ≤ 3 or using 9 as a limit

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

Examiner's comment

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

Common mistakes

• substitute the parametric equations into x² + y² = 66 and use sin²t + cos²t = 1 (or convert C₁ to x²/100 + y²/32 = 1)
• (4√2)² = 32 not 8
• S in the 4th quadrant is (5√2, −4), not (−4, 5√2)

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

Examiner's comment

Q06 discriminated well between candidates of all abilities, with only a few candidates gaining full marks. There were some marks in Q06(b) and Q06(d) which were accessible to the majority of candidates. x In Q06(a), many candidates struggled with the split domain for g( )x (i.e. for 2, and for 2 g( ) ( 2) 1 x x 2, g( ) 4 7) x x x and only a minority gained both marks. Many candidates found a correct g(0) 5, but a significant number used the incorrect 2 gg(0) (5 2) 1 10. method of substituting ‘5’ to give Only a minority used a correct method and found gg(0) 4(5) 3 13. Occasionally candidates found an algebraic expression for gg( ), x although this was often incorrect. Very few obtained a x x x 2 2 correct answer of 13 by substituting into or 0 4(( 2) + 1) 7 4( 2) 3. x In Q06(b), many candidates attempted to solve the inequality g( ) 28 by finding the critical values for x. Most solved the linear equation to find a correct critical value 35 , x x 2 3 3. and many solved the quadratic equation to find A minority 4 x x 2 3 3 2 3 3 rejected and deduced that was the second critical value. Only a few candidates used the critical values and the diagram in Q6 to write down the correct 35 35 , x 2 3 3 . x x solution Some candidates, however, incorrectly gave 4 4 35 x 2 3 3 2 3 3 x x or gave as their final answer. A few candidates 4 made no creditable progress in Q06(b) by using a method of equating both parts of 2 ( 2) 1 4 7. x x g( )x and attempting to solve the equation Although Q6(c) required a comment for both functions, f and g, some candidates only wrote a comment about one of the two functions. A significant number of candidates believed that g did not have an inverse due to its being defined in ‘two parts’. Candidates who were successful in Q6(c) usually gave a reason such as, ‘g is a many- one function and h is a one-one function’. A few correctly stated that the inverse of g is one-many, which is not a function, whereas the inverse of h is one-one, which is a function. A few candidates incorrectly described g as a ‘one-many function’. 1 h . In Q06(d), only a few candidates applied the method of finding x by applying 2 As this method was less prone to error, most candidates who used it scored full marks. 1 h ( ), x Most candidates applied a complete method of finding their inverse followed 1 1 h ( ) . x by using their inverse to form and solve the equation Many candidates 2 x who found a correct answer lost the final mark in Q06(d). This was because 7.25 1 h ( ) 2 1 x x they used the incorrect inverse (instead of the correct 1 ) which led to their ‘solving’ the incorrect x 1 h ( ) 2 1 x x 1 (which has 2 x 7.25. no solutions) to give A few candidates in Q06(d) incorrectly believed that 1 h ( )x referred to the reciprocal of h, or even to the first derivative of h with respect to x.

Common mistakes

• solve each branch = 28
• reject 2 + 3√3 (not in x ≤ 2)
• answer x < 2 − 3√3 or x > 35/4 (not "and" or ∩)
• do not equate the two branches

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

Examiner's comment

Q06 discriminated well between candidates of all abilities, with only a few candidates gaining full marks. There were some marks in Q06(b) and Q06(d) which were accessible to the majority of candidates. x In Q06(a), many candidates struggled with the split domain for g( )x (i.e. for 2, and for 2 g( ) ( 2) 1 x x 2, g( ) 4 7) x x x and only a minority gained both marks. Many candidates found a correct g(0) 5, but a significant number used the incorrect 2 gg(0) (5 2) 1 10. method of substituting ‘5’ to give Only a minority used a correct method and found gg(0) 4(5) 3 13. Occasionally candidates found an algebraic expression for gg( ), x although this was often incorrect. Very few obtained a x x x 2 2 correct answer of 13 by substituting into or 0 4(( 2) + 1) 7 4( 2) 3. x In Q06(b), many candidates attempted to solve the inequality g( ) 28 by finding the critical values for x. Most solved the linear equation to find a correct critical value 35 , x x 2 3 3. and many solved the quadratic equation to find A minority 4 x x 2 3 3 2 3 3 rejected and deduced that was the second critical value. Only a few candidates used the critical values and the diagram in Q6 to write down the correct 35 35 , x 2 3 3 . x x solution Some candidates, however, incorrectly gave 4 4 35 x 2 3 3 2 3 3 x x or gave as their final answer. A few candidates 4 made no creditable progress in Q06(b) by using a method of equating both parts of 2 ( 2) 1 4 7. x x g( )x and attempting to solve the equation Although Q6(c) required a comment for both functions, f and g, some candidates only wrote a comment about one of the two functions. A significant number of candidates believed that g did not have an inverse due to its being defined in ‘two parts’. Candidates who were successful in Q6(c) usually gave a reason such as, ‘g is a many- one function and h is a one-one function’. A few correctly stated that the inverse of g is one-many, which is not a function, whereas the inverse of h is one-one, which is a function. A few candidates incorrectly described g as a ‘one-many function’. 1 h . In Q06(d), only a few candidates applied the method of finding x by applying 2 As this method was less prone to error, most candidates who used it scored full marks. 1 h ( ), x Most candidates applied a complete method of finding their inverse followed 1 1 h ( ) . x by using their inverse to form and solve the equation Many candidates 2 x who found a correct answer lost the final mark in Q06(d). This was because 7.25 1 h ( ) 2 1 x x they used the incorrect inverse (instead of the correct 1 ) which led to their ‘solving’ the incorrect x 1 h ( ) 2 1 x x 1 (which has 2 x 7.25. no solutions) to give A few candidates in Q06(d) incorrectly believed that 1 h ( )x referred to the reciprocal of h, or even to the first derivative of h with respect to x.

Common mistakes

• quickest: x = h(−½) = 7.25
• if finding h⁻¹ it must be 2 − √(x − 1) (domain of h is x ≤ 2), using 2 + √(x − 1) loses the final mark
• h⁻¹ is not the reciprocal or derivative

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

Examiner's comment

Most students gained marks in at least one part of this question. However, the rigorous approach required to achieve full marks was generally lacking. In part (i), most attempted to complete the square but a common error was to state that a square number must be positive, with no reference to the possibility of it being zero. Those who found the discriminant rarely scored more than the first mark as they did not realise the need to consider the shape of the curve. A small number attempted differentiation and whilst they stated (4,1) was a minimum it was very unusual to have this confirmed (e.g. using the second derivative). It was rare to see a fully graphical approach but those who tried usually gained two marks for a correct sketch with the correct minimum shown. A few students tried to give an explanation in words or used a few numerical examples. In part (ii), most students used a numerical approach and the majority concluded that the statement is ‘sometimes true’. It was common, however, to show working for a negative value only, for which the statement is not true. Some attempted a true and a false value, as required, but failed to conclude that the statement is ‘sometimes true’. Many students unsuccessfully tried to write general statements referring to positive and negative values being used, but only scored marks if they produced a counter-example. Of those who attempted an algebraic approach, many reached 6x + 9 > 0 but tended to stop there and did not gain the final mark for a correct conclusion.

Common mistakes

• complete the square (x − 4)² + 1
• reason must be (x − 4)² ≥ 0 so expression ≥ 1 (saying a square is always positive is incomplete)
• discriminant method also needs the U shape
• a conclusion is required

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

Examiner's comment

Success on this question was partially determined by students realising that P was measured in thousands of pounds. Additionally, students should be aware that units are necessary in modelling questions and in particular the answer to (c)(i) is £100 000 and not 100, £100 or 100 000 as commonly seen. (a) Most students gained at least one mark on this part. The easiest method is to substitute x =15 into the given equation and explain that it isn’t a sensible selling price as the company would be making a loss of £125 000. The most common errors in this part were: (1) substituting in P = 0 and finding x, (2) stating that the company would not make any money or “may/could” make a loss.

Common mistakes

• substitute x = 15: P = −125, i.e. a loss of £125 000
• must say it would make a loss (not "may" make a loss)
• setting P = 0 is the wrong approach

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

Examiner's comment

Success on this question was partially determined by students realising that P was measured in thousands of pounds. Additionally, students should be aware that units are necessary in modelling questions and in particular the answer to (c)(i) is £100 000 and not 100, £100 or 100 000 as commonly seen. (b) Many students attempted this part. The most common error was to use 80,000 rather than 80. This error resulted in a score of 000. Students who did use 80 usually attempted to solve the quadratic, but many by expanding and factorising rather than by using the form given. Some students only found one solution, rather than the two required. Very few students gained the final accuracy mark, with most students incorrectly stating £7.21 as the final value, not taking into account the need to round up. A small number of students used a trial and improvement method, which was often more successful in achieving the correct answer. Others used their graphical calculators to gain the correct answer with limited working.

Common mistakes

• P is in thousands, so use P > 80 not 80 000
• (x − 9)² < 3.2 gives 9 − √3.2 < x < 9 + √3.2 (use the completed-square form rather than expanding)
• round up to £7.22 (£7.21 is not acceptable)
• find both values

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

Examiner's comment

Success on this question was partially determined by students realising that P was measured in thousands of pounds. Additionally, students should be aware that units are necessary in modelling questions and in particular the answer to (c)(i) is £100 000 and not 100, £100 or 100 000 as commonly seen. (c) Most students attempted this part but were unable to use the given equation to just state the answers. Some students used differentiation to find the maximum. Common errors were omitting units, giving the answers in the wrong order (with no labelling) and giving the answer for the maximum profit as £100 instead of £100,000.

Common mistakes

• read from the completed-square form: maximum profit £100 000 (with units
• not 100 or £100) at a selling price of £9
• differentiation is unnecessary
• label (i) and (ii)

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

Examiner's comment

Overall this question was well done, with many students taking advantage of being able to attempt (b) without completing (a). The few who restarted unnecessarily in (b) lost valuable time in the examination. In (b) students generally managed to solve the quadratic, often using a calculator to find the two roots. The most common error was to give the signs of the roots the wrong way around, even after correct factorisation in some cases. Almost all found solutions for 3x but many failed to divide by 3. A substantial number of students who went on to find values for x missed a solution (often the 80o) or gave extra solutions outside the range e.g. 103.9°

Common mistakes

• solve 6c² − c − 2 = 0: cos 3x = 2/3 or −1/2 (signs of the roots often reversed)
• find 3x in 0 ≤ 3x < 270° then divide by 3
• x = 16.1°, 40°, 80° only (80° often missed
• 103.9° is outside the range)

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

Examiner's comment

Part (a) was very accessible, part (b) less so. Hence in part (a) there were a large number of students who scored full marks. There were however occasional sign errors witnessed when the students completed the square.

Common mistakes

• complete the square: (x − 3)² + (y + 5)² = 25
• centre (3, −5), radius 5
• sign errors when completing the square

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

Examiner's comment

Part (a) was very accessible, part (b) less so. In (b), the first method mark was scored by the majority of students who showed their substitution of y = k x . Those who substituted into the equation given in the question rather than their (a) answer often coped better with the less challenging algebra. Collecting terms to reach a correct three term quadratic was a significant hurdle for many, although the most common error was to write kx2 for (kx)2. Most students correctly quoted b2 – 4ac before using it. Having done this they were usually able to factorise their expression to obtain k = 0 and another critical value. Having done the bulk of the work, many spoiled an otherwise perfect solution by stating the inside region.

Common mistakes

• substitute y = kx: (1 + k²)x² + (10k − 6)x + 9 = 0 (write (kx)², not kx²)
• b² − 4ac > 0 gives critical values 0 and 15/8
• outside region k < 0 or k > 15/8 (not the inside region and not joined with "and")

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

Examiner's comment

again we expect answers to be given to 3 significant figures. The final part required students to either form a list of values for Y , or even Y – X, or solve an inequality to realise that they needed P(X …2). Some were able to do this and this proved to be a good discriminator on this section. SECTION B: MECHANICS Introduction A significant proportion of the students were poorly prepared and showed little understanding of basic mechanics and displayed poor algebraic/numerical skills. There were a number of blank or incomplete responses to question 9 but it was difficult to determine whether this was due to a lack of time or a lack of knoeledge. Question 7 was by far the most successfully answered with just under a third of students scoring 6 out of 7 but the last two questions worked well as discriminator questions for the more able students. Just under 30% of the students scored nothing on question 8 and about a quarter failed to score on question 9. In calculations the numerical value of g which should be used is 9.8, unless otherwise stated. Final answers should then be given usually to 3 significant figures (unless otherwise stated) – more accurate answers will be penalised, including fractions but exact multiples of g are usually accepted. If there is a printed answer to show then students need to ensure that they show sufficient detail in their working to warrant being awarded all of the marks available. In all cases, as stated on the front of the question paper, students should show sufficient working to make their methods clear to the examiner and correct answers without working may not score all, or indeed, any of the marks available. If a student runs out of space in which to give his/her answer than he/she is advised to use a supplementary sheet.

Common mistakes

• 12/X − X ≤ 4 gives X² + 4X − 12 ≥ 0, so X ≥ 2 since X > 0 (or tabulate Y − X = 11, 4, 1, −1)
• P(X ≥ 2) = 0.85
• a good discriminator

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

Examiner's comment

This proved to be a reasonably accessible starter to the Mechanics section with a modal mark of 4 out of 4, achieved by just over a third of the students. There were various possible approaches to this question which involved a ball moving up and down freely under gravity. Those who attempted a solution using one suvat equation generally included appropriate terms, but sign errors were very common. If the resulting quadratic equation was incorrect an explicit method (such as factorising or using the quadratic formula) was required for the second method mark. Some split the motion into two or three separate stages with a fair degree of success. Again, however, sign errors (showing a lack of understanding of how to incorporate directions) were often seen. Those who just found the time to the greatest height or to the same horizontal level received no credit, and use of g = 9.8 m s-2 rather than 10 m s-2 as specified in the question was penalised as an accuracy error.

Common mistakes

• one suvat for the whole flight with up positive: −2 = 9t − 5t², so 5t² − 9t − 2 = 0 and T = 2 only (reject −0.2)
• sign errors are common
• use g = 10 as given (9.8 is penalised)
• time to greatest height alone scores 0

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

Examiner's comment

distance. Part (c) also proved to be a challenge with few students scoring either of the two available marks. Those who did write the expression as a perfect square (achieving the first mark) often claimed that therefore it had to be positive ignoring the possibility of zero. Common incorrect responses included: ‘time cannot be negative’ and ‘t 2’ is positive and any number times by a positive number is always positive’. A handful of students used their values from (b) to describe the motion in detail and thereby deduce the fact that the particle never moved on the negative x-axis.

Common mistakes

• x = ½t²(t − 1)², half a perfect square so x ≥ 0 (zero is possible, so not "always positive")
• "time cannot be negative" is not a reason

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

Examiner's comment

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 meets C. 2 y x k Using the fact that PA enabled students to find N = (11,3) and then k by substituting AN this point into Again simultaneous equation methods were common, some more 2 . y x k successful than other. The most common correct attempt involved solving with 2 y x k 2 2 and then finding where the resulting equation had one root. Care was 7 5 20 x y required with the algebra, especially the squaring of terms, as well as the application of the 2 discriminant This proved to be too great a challenge for many. 4 0. b ac

Common mistakes

• best route: other tangent point N = A + PA vector = (11, 3), then k = −19
• discriminant route 5x² + (4k − 34)x + k² − 10k + 54 = 0 with b² − 4ac = 0 has algebra/squaring errors
• reject k = 1

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

Examiner's comment

Question 9(a) was well answered. Students generally demonstrated a good understanding of implicit differentiation as well as the chain and product rules. Most reached the given result scoring all 4 marks. In question 9(b) however, many made the numerator of the fraction equal zero (i.e. y = x) instead of the denominator (x = 3y). This kind of error usually resulted in only two of the five marks being scored. There were however many full and accurate solutions, with very few students unable to pick out the coordinates of the point P from the two solutions to their simultaneous equations. Only a small number of students gained the B1 mark in question 9(c). They simply needed to state that the point furthest north can be found by substituting y= x into the equation of the curve and picking out the positive solution. Explanations tended to be incorrect or x y 0 incomplete, with many stating that it could be found by putting either or 0.

Common mistakes

• furthest west/east means vertical tangent: denominator 3y − x = 0, not numerator (y = x is a common error)
• solve with the curve, x = ±5√3
• choose P = (−5√3, −(5/3)√3), exact

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

Examiner's comment

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 then 3 2sint students were able to explain that as 1 sin 1 t would have a maximum and a minimum value of 3 2 value of 3 2 5 1.

Common mistakes

• sketch only the arc from (1, 2) to (5, 2) with maximum (3, 6), not the whole parabola
• reason: −1 ≤ sin t ≤ 1 so 1 ≤ x ≤ 5 ("0 ≤ t < 2π" alone is not enough)

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

Examiner's comment

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 for two roots and were able to proceed to two values. Arithmetic errors were 4 0 b ac quite common using this method. Most students who proceeded to the answer were well versed in writing their answer using set notation. Figure 1

Common mistakes

• lower value k = 7 from the end point (5, 2)
• x² − 7x + (k + 3) = 0 with b² − 4ac = 0 gives k = 37/4
• answer {k : 7 ≤ k < 37/4} in set notation
• annotate the sketch

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

Examiner's comment

This was a well-answered question with most students scoring at least 4 of the 5 marks available. In part (b), most students found the difference between OC and , OA and applied Pythagoras’ Theorem to 4 AC to give the correct 2 2 2 2 ( 2) (2) (2) (4) . a The resulting quadratic equation was generally solved correctly but not always efficiently. Too often, students expanded to form a 3-term quadratic rather than spotting that their quadratic was already in a ‘completed square’ form. Despite the question stating that 0, a some students gave both solutions 2 2 2 a as their final answer, while others mistakenly rejected 2 2 2 a to leave 2 2 2. a

Common mistakes

• AC = OC − OA
• (a − 2)² + 2² + 2² = 16 is already completed-square form, (a − 2)² = 8
• reject 2 + 2√2 because a < 0, answer a = 2 − 2√2

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

Examiner's comment

This was a well-answered question with part (b) and part (c) providing most of the discrimination. In part (b), many students used part (a) to write down 2 4 2 ( 2)( 3 2 5) 0 y y y and most deduced the two real solutions 2. y Surprisingly a few students solved 2 2 y to give 2. y A significant number of students stated, without proof, that 2 ( 3 2 5) x x or 4 2 ( 3 2 5) y y where 2, x y had no real solutions. Some students applied ‘ 2 2 4 (2) 4( 3)( 5) 56 0 b ac ’ with an appropriate conclusion to show that 4 2 ( 3 2 5) 0 y y had no real solutions. Common errors included just stating ‘ 2 4 0 b ac ’ with no reference to a discriminant calculation or the omission of ‘ 0 ’ after a correct discriminant calculation. Only a minority of students were able to bring the two strands of the proof together to show that the given equation had only two real solutions.

Common mistakes

• replace x by y²: (y² − 2)(−3y⁴ + 2y² − 5) = 0
• discriminant 4 − 60 = −56 < 0 must be calculated and compared with 0
• y² = 2 gives exactly y = ±√2
• bring both strands together

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

Examiner's comment

This question discriminated well between the average and more able students. There was a significant minority of students, however, that made no creditable progress. In part (a), a minority of students struggled to write down a suitable form of the quadratic model. Some students, who tried to model the height of the rugby ball as a negative quadratic cutting the x-axis at 0 x and 40, x wrote down either ( 40) H x x or (40 ), H x x but struggled, due to the lack of a constant, to fit the condition (20, 12) to their model. Those who wrote (40 ) H Ax x usually applied (20, 12) to their model to give the correct 3 (40 ). 100 H x x Other students who adopted a model of the form 2 H ax bx c were usually successful in using the conditions (0, 0), (40, 0) and (20, 12) to give the correct 2 0.03 1.2 . H x x Only a few students used the model 2 12 ( 20) H x with condition (40, 0) to find a correct 2 3 12 ( 20) . 100 H x A few other elaborate methods were seen, including one where (40, 0) and , 12 2 b a were applied to 2 , H ax bx which usually resulted in a correct 2 0.03 1.2 . H x x

Common mistakes

• H = Ax(40 − x) needs a constant A (fit (20, 12) to get A = 3/100)
• or H = 12 − λ(x − 20)², or H = −0.03x² + 1.2x

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

Examiner's comment

This question discriminated well between the average and more able students. There was a significant minority of students, however, that made no creditable progress. In part (b), many students substituted 3 H into their quadratic model and most applied a correct method for solving their resulting 3-term quadratic equation or their quadratic equation of the form 2 ( 20) , 0. x k k Many students found the largest solution in the range (0, 40) and gave their answer, usually to 3 significant figures, in metres.

Common mistakes

• set H = 3: x² − 40x + 100 = 0 or (x − 20)² = 300
• choose the larger root 20 + √300 = 37.3 m

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

Examiner's comment

This question discriminated well between students of all abilities with part (a) more accessible to students than part (b). Part (b) proved to be discriminating, with some students not realising that they could use the result from part (a). Some students substituted 1 cos2x with tan sin2 x x and then proceeded to cancel tan sin2 x xfrom both sides, while other students substituted tan sin2 x x with 1 cos2x and then proceeded to cancel 1 cos2x from both sides. These two strategies together with applying the identity 2 2 sec 1 tan x x usually led to a 3-term quadratic in tan .x Most students used a correct method, usually by factorising, to solve their quadratic equation in tan x but only a few stated all three correct solutions. As a result of cancelling either tan sin2 x x or 1 cos2x from both sides of their equation, many students missed out the correct solution 0. x Those students who factorised out these terms were usually more successful in finding 0. x Some students did not give their other two solutions in the required form of the exact 4 x and 1.326 x to 3 decimal places.

Common mistakes

• replace (1 − cos 2x) by tan x sin 2x (or vice versa)
• factorise, do not cancel, so that x = 0 is kept
• sec²x = 1 + tan²x gives tan²x − 3tan x − 4 = 0
• x = −π/4 (exact) and 1.326

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

Examiner's comment

This was the most demanding question on the paper. Parts (a) and (d) were accessible to many students. Most students made some creditable progress in part (b), but many students found it difficult to access the demanding material in part (c). In part (c), most students misinterpreted the question and instead applied a method to find the time T for the maximum value of N, rather than the time T at which the rate of growth d d N t was a maximum. Many students, therefore, used 300 N and manipulated 0.25 900 300 3 7e T to give 0.25 e 0. T A few students differentiated d , d N t with most writing 2 2 d , d N t which was condoned, and solved the equation 1 (300 2 ) 1200 N to give 150. N It was rare to see students using the symmetry of the quadratic function (300 ) 1200 N N to deduce that 150. N Although some students gave their answer to part (c) as 150, T most students who progressed this far solved the equation 0.25 900 150 3 7e T to give the correct value for T.

Common mistakes

• maximise dN/dt, not N (common error using N = 300)
• N(300 − N)/1200 is maximum at N = 150 by symmetry or d²N/dt² = 0
• e^(−0.25T) = 3/7 so T = 4 ln(7/3) ≈ 3.4 months

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

Examiner's comment

(b), most started with the valid strategy of using horizontal and vertical equations (s = ut + 1/2at2). Not all included the appropriate correct distance (-1.25) and there were occasional sign errors. By substituting for t and then for U (using the expression from (a)) it was possible to obtain a quadratic equation in tan alpha . Not all attempts were successful, with errors in simplification leading to equations with a variety of trig functions or else not quadratics. Alternatively, rather than substituting for t and U in the original equation, a quadratic in t could be found by an initial substitution for Usin alpha . The resulting value of t could then be used in ’20 = Utcos alpha ’ to find alpha . This method was seen successfully employed on occasion. Similarly, the motion could be split into two stages and the times to and from the highest points found. This approach was only very rarely seen. Some candidates made several attempts at this part of the question; it is worth remembering that if none are crossed out only the last most complete solution will be credited. Reference to air resistance was by far the most common correct response about the limitations of the model in part (c). Other valid comments relating to spin, dimensions of the ball and wind effects were

Common mistakes

• 20 = Ut cos α and −1.25 = Ut sin α − ½gt² (vertical displacement −1.25, not ±2 or ±3)
• substitute for t and U² to get quadratic 400tan²α − 80tan α − 5 = 0 (or in t), tan α = 1/4, α = 14°
• sign errors and simplification errors common