Differentiation: first principles, gradients, stationary points Pure

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

Examiner's comment

Although the skills involved in part (b) were quite fundamental, the understanding of the demand of the question differentiated the candidates. Pleasingly the majority were able to differentiate correctly, substitute 2 into their derivative and set equal to 14, then solve the two equations simultaneously. On occasion those candidates who solved algebraically, as opposed to using their calculator, lost the final accuracy mark. Arithmetical errors were also seen in rearranging the second equation. A number of candidates achieved the first two marks for correctly differentiating however they then went on to set f ′(2) = 0. Other candidates achieved no marks as there was no attempt to differentiate and instead attempted f (2) = 0 or f (2) = 14 or integrated instead of differentiating. In all cases there was very minimal evidence of candidates checking that their solutions to a and b satisfied the given conditions.

Common mistakes

• differentiate f′(x) = 3ax² + 2bx + 18, then f′(2) = 14 gives 3a + b = −1
• solve with −3a + b = 5: a = −1, b = 2
• using f(2) = 14, f′(2) = 0 or integrating scores little

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

Examiner's comment

This was the first question on the paper where there was wide discrimination between the marks of responses, and many struggled with knowing how to handle a quartic function. Part (a) involved using the given information and the figure to find the set of values that satisfied f ( ) 0 x ′ and to represent this in set notation. Most identified the critical values that formed the boundaries of the regions where f ( ) 0 x ′ . In general, these critical values were used within a relevant inequality although these were often strict rather than inclusive. Some learners incorrectly interpreted the function giving incorrect inequalities or just considering the region between the two local maxima. Presenting the regions in set notation was done to varying degrees of success. Although a large proportion were able to present their solution in an acceptable form including the brackets { } and the union, ∪, there were many errors seen including intersection signs, no link between the regions, interval notation and strict inequalities. The majority used x as the variable within the inequalities but ( ) f , x ( ) f x ′ and k were all seen. Of those using a correct form of set notation, very few mentioned x was a real number although this was not specifically required on this occasion. It was clear that a significant proportion of learners did not appreciate what was required for set notation.

Common mistakes

• f is increasing (or stationary) for x ≤ −1 and 2 ≤ x ≤ 5 (inclusive inequalities)
• set notation needs { } and the union, not intersection or interval notation

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

Examiner's comment

This was the first question on the paper where there was wide discrimination between the marks of responses, and many struggled with knowing how to handle a quartic function. Part (c) involved assuming that the quartic intersects the x-axis at exactly four places with the aim to find the range of possible translation factors of the curve. Many responses found the y coordinate of the local minimum. Those who found a quartic in part (b) were often successful at this. Common errors were not appropriately using the y coordinate to describe the translation factor or not considering the minimum value. Those with positive quartic functions could not score the final mark as the figure provided already showed that this was not the case. 12

Common mistakes

• minimum at x = 2 has y = −243, so shift up by between 0 and 243: 0 < k < 243 (strict)
• needs a negative quartic from (b)

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

Examiner's comment

This question was generally well-answered, with most learners demonstrating a good grasp of the concepts. However, some common pitfalls were noted, particularly in differentiation and the inclusion of units. Part (b) involved differentiating the given model, setting this equal to zero and rearranging to the given form. This was generally well-answered, however a significant number of learners lost all the marks by not using the product rule. Accuracy marks were typically not awarded due to sign errors as a result of not using brackets correctly. Learners who wrote out the product rule separately before substitution were more likely to avoid such errors. In addition, some learners struggled with algebraic manipulation after differentiation, particularly when taking natural logarithms too early or when simplifying fractions.

Common mistakes

• product rule on te^(0.2t): dv/dt = 15 − (0.2te^(0.2t) + e^(0.2t)) = 0 (many did not use the product rule
• bracket/sign errors)
• factor e^(0.2t)(0.2t + 1) = 15 then take logs to reach the given t = 5 ln(75/(t + 5))

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

Examiner's comment

A small minority of candidates did not attempt this question at all. Some candidates attempted part (a) and then did not go on to attempt the remaining parts (generally when they were unable to complete part a)). In part (a), a common error in the first B mark was to miss out the pi for 90000 pi . Some candidates used an incorrect formula but most who wrote the correct formula, rearranged this into a different format to use later. For the second B mark, this was less well scored. Some candidates had the formula for a complete cylinder and some candidates did not write down this formula (it was often implied with later working but this resulted in the loss of both the second B mark and the A mark as this was a proof question). Where candidates had gained the first three marks, they often went on to simplify the formula and gain the final accuracy mark.

Common mistakes

• V = ½πr²l = 90 000π (π often missed)
• A = πr² (two semicircular ends) + πrl (curved surface), not a whole cylinder
• formula for A must be written down
• substitute l from V before reaching the given answer

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

Examiner's comment

A small minority of candidates did not attempt this question at all. Some candidates attempted part (a) and then did not go on to attempt the remaining parts (generally when they were unable to complete part a)). Most candidates who attempted part (b) were able to differentiate correctly to gain the first mark. A common error was to miss out the pi , thus resulting in the wrong value for r. Most candidates who attempted to differentiate often achieved the correct value for r from a cubic expression.

Common mistakes

• dA/dr = 2πr − 180 000πr⁻²
• set = 0: r³ = 90 000, r = 44.8 cm
• losing the π
• not ±44.8

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

Examiner's comment

However, a small number retrieved this mark in part (b) by correctly stating the equation at the beginning of their working. There were only a few examples of not using enough significant figures or incorrectly rounding the value of k. Misreads were fairly common on this question (usually 1200 instead of 12000 or 15000 instead of 1500) which often meant that learners were unable to proceed to a value of k as they ended up with an equation with no real solutions. In part (b), most were able to identify the need for differentiation. The execution of the method was well-attempted, although some struggled to make the necessary links for the final part of the solution. The intention was that the question would be answered in terms of A and k but many used their own model from part (a) which could also score all the marks. Nearly

Common mistakes

• dV/dt = −kAe^(−kt) then substitute Ae^(−kt) = V − 1500 to get −k(V − 1500)
• dV/dt must be seen
• working backwards by separating variables rarely scored

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

Examiner's comment

A small minority of candidates did not attempt this question at all. Some candidates attempted part (a) and then did not go on to attempt the remaining parts (generally when they were unable to complete part a)). Part (c) was less well answered and some candidates made errors with the differentiation. A small minority of candidates differentiated the 2 r - correctly but did not include the differentiation of the 2 pi r to get 2 pi . Most candidates recognised the need to substitute their value from part (b) into the second derivative to achieve a number greater than zero. A common error was to mix up parts (b) and (c). Some candidates found the first derivative (often correctly) then found the second derivative and put this equal to zero (rather than the first derivative), thus losing marks from parts (b), (c) and (d).

Common mistakes

• d²A/dr² = 2π + 360 000πr⁻³ ≈ 18.8 > 0 hence minimum (conclusion needed)
• differentiating 2πr to 2π is often missed
• setting the second derivative = 0

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

Examiner's comment

A small minority of candidates did not attempt this question at all. Some candidates attempted part (a) and then did not go on to attempt the remaining parts (generally when they were unable to complete part a)). In part (d) most candidates were able to correctly identify the need to substitute their r into the formula for A, and a good number achieved the M1 mark as a result. Fewer candidates were able to achieve the A1 mark. Some candidates multiplied or divided by 30 (or 100), but fewer recognised and used the correct conversion for centimetres squared into meters squared. Some candidates used an approximation so didn’t achieve this mark. A small number did not put their answer in the correct format for money.

Common mistakes

• A ≈ 18 928 cm² at r = 44.8
• convert cm² to m² by dividing by 100² (not 100)
• cost £56.78 (money to 2 d.p.)

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

Examiner's comment

A small minority of candidates did not attempt this question at all. Some candidates attempted part (a) and then did not go on to attempt the remaining parts (generally when they were unable to complete part a)). Part (e) was attempted by a good number of candidates. The most common correct answers related to waste and ability to buy the exact amount of material. Incorrect answers tended to relate to the width/thickness and rounding errors.

Common mistakes

• e.g. no metal is wasted, the exact amount can be bought, it can be cut accurately
• "negligible thickness" is given so scores nothing
• labour or welding costs not accepted

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 (a) was generally better attempted, with many candidates recognising the need to draw a U-shaped (positive quadratic) graph. Common errors included: • drawing a cubic curve rather than a quadratic • drawing a straight line • sketching a quadratic that did not cross the x-axis or with incorrect roots, such as crossing the positive x-axis instead of the origin and negative x-axis Some candidates left the sketch blank or attempted part (b) without completing part (a). Despite these issues, most candidates gained at least 1 mark for showing some correct features, such as a U-shape or one correct root.

Common mistakes

• gradient function is a U-shaped quadratic crossing the x-axis at the origin (minimum on the y-axis) and on the negative x-axis (maximum in quadrant 2)
• drawing a cubic or a straight line
• roots on the wrong side

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

Candidates’ performances in this exponential modelling question were varied with some able to obtain full or nearly full marks, whilst others made very little progress. Candidates who did In (d)(i) candidates were generally able to differentiate their equation of the model to obtain an expression of the correct form, earning the first mark. Many did not obtain the correct answer so were not awarded the second mark. A number of candidates incorrectly differentiated to e kt - which gained no credit in (d)(i). The first mark in (d)(i) was again at obtain expressions like aften awarded to candidates who made little progress elsewhere in this question. In (d)(ii)

Common mistakes

• dh/dt = Ake^(−kt) ≈ 0.698e^(−0.0223t)
• differentiating to −e^(−kt) or keeping the 31

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

Examiner's comment

This question using differentiation to find a minimum was generally answered well. It was usually part (a) which provided most challenge, often resulting in work which was difficult to follow. Part (b) required the use of algebraic differentiation to find a stationary value. Since the equation of the perimeter was given in the question, this task was accessible to all learners regardless of their success in (a). Differentiation was generally accurately carried out, although there were a few incorrect attempts which confused differentiation with integration and tried to introduce logs. The majority of the learners recognised the need to equate the derivative to zero to find the stationary point. Although most were successful finding a value, errors were made in the rearranging to make r the subject.

Common mistakes

• dP/dr = 2 − 480/r² = 0 gives r² = 240, r = √240 = 15.5 (not ±)
• rearrangement errors

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

Examiner's comment

This question using differentiation to find a minimum was generally answered well. It was usually part (a) which provided most challenge, often resulting in work which was difficult to follow. In part (c), learners were asked to prove by further differentiation that the value of r found gave the minimum perimeter. Most were able to make some progress and find the correct second derivative. The criteria for a minimum value was less well applied. Some responses did not use a correct inequality whilst others did not correctly evaluate their second derivative at the value of r. Incorrect second derivative notation was also not credited. 22

Common mistakes

• d²P/dr² = 960/r³ = 0.26 > 0 so minimum
• must evaluate at r = 15.5 and compare with 0, with a conclusion
• sign-change or d²y/dx² in the conclusion not credited

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

Examiner's comment

This question, involving the factor theorem and differentiation, was attempted by the vast majority of learners, with nearly all engaging with all three parts. Parts (a) and (b) were generally well-answered. Part (c), however, proved more challenging, with a significant proportion of learners relying on calculator-based methods despite the question clearly stating that these were not acceptable. In part (a), most responses correctly differentiated the function and substituted 1 4 x = - to solve for a, showing sufficient working and generally securing all four marks. The most common cause of a mark not being awarded was because any prior working to a + 4 = 0 was not shown, a step which could have been easily deduced from the given answer a = –4. Occasionally responses also found the second derivative, which was not required and sometimes introduced errors. The majority, however, followed a clear algebraic method and presented each stage of their working logically. Only a minority used the alternative verification approach, but the work was equivalent and generally well completed, with the required conclusion only occasionally omitted.

Common mistakes

• set f′(−¼) = 0 with the substitution shown (not just 4 + a = 0) before a = −4
• brackets round −¼

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

Examiner's comment

This question, involving the factor theorem and differentiation, was attempted by the vast majority of learners, with nearly all engaging with all three parts. Parts (a) and (b) were generally well-answered. Part (c), however, proved more challenging, with a significant proportion of learners relying on calculator-based methods despite the question clearly stating that these were not acceptable. Part (b) was usually answered correctly, as it required only a straightforward substitution into the original equation using the value given in part (a). Although only one mark was available, many responses showed clear working, while others simply stated the correct answer, both approaches were credited. A few responses gave the answer only as a decimal approximation, which did not gain the mark.

Common mistakes

• exact value 4735/768
• a decimal approximation scores 0

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

Examiner's comment

This question, involving the factor theorem and differentiation, was attempted by the vast majority of learners, with nearly all engaging with all three parts. Parts (a) and (b) were generally well-answered. Part (c), however, proved more challenging, with a significant proportion of learners relying on calculator-based methods despite the question clearly stating that these were not acceptable. Part (c) was the most challenging part of the question but was well supported by a clear diagram that helped to improve learners’ ability to solve the problem. A good range of marks were seen for this part of the question, and it differentiated well between the varying understanding of learners across the cohort. While many learners correctly identified that the number of real solutions to the equation f(x) = k depended on the location of k relative to the turning points of the curve, a substantial number relied on calculator methods to find the critical values of k, rather than using written methods as instructed. There were many responses where the required quadratic of x2 – 4 (or maybe 4x2 – 16) was never seen, the cubic was either followed directly by x = 1 4 - , 2, –2, or (4x + 1)(x – 2)(x + 2). These responses scored M0A0 and could only receive limited credit under the special case. Those who did attempt the algebra often reached the correct values but struggled with set notation. A common error when writing the final answer was using the union of two intervals instead of an intersection, while some used inclusive inequalities instead of strict inequalities and others made no attempt to use set notation. Despite this, many understood the underlying concept and were able to identify the correct range for k, even if it was not expressed with the required detail.

Common mistakes

• use the factor theorem on f′(x): (4x + 1)(x² − 4) must be seen, not roots from a calculator
• lower minimum f(−2) = −5
• four solutions between the lower minimum and the local maximum: {k : −5 < k < 4735/768}
• intersection not union
• strict inequalities
• must use set notation

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

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

exact (surd) answer was required so those who only stated a decimal approximation did not sqrt 322 -162 thereby not gaining both marks. gain the accuracy mark. A few did not account properly for the negative sign and evaluated In the third part the majority of learners recognised the requirement to differentiate to find the velocity vector and most did so successfully. Occasionally i and j were omitted entirely resulting in an expression which was no longer a vector and the loss of the method mark. Attempts at suvat or integration were seen but only rarely. Virtually all gave their answer in terms of i, j and t as required. In part (d) those learners who achieved the marks in part (c) generally achieved the first two 2 was not here for differentiating to find the acceleration vector. Occasionally the power of 1 handled correctly but differentiation skills were mostly sound. The final two marks proved more challenging and many either could not relate the gradient to their coefficients in i and j or simply stopped after finding the acceleration vector. For those who did attempt a solution, some recognised the need to use the ratio of their i and j coefficients but the majority did not Other errors seen included ignoring i and j and setting the expression equal to ±3 or ± 1 identify and use the correct perpendicular gradient. Some just equated i and j coefficients. 3. Of those who achieved the correct equation in T, a few solved it incorrectly as a result of errors 1 -. A very small number of learners used the scalar product successfully. in handling T 2

Common mistakes

• differentiate r: v = 6t^(1/2) i − 2t j
• i and j must stay in the answer

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

Examiner's comment

In part (d) those learners who achieved the marks in part (c) generally achieved the first two 2 was not here for differentiating to find the acceleration vector. Occasionally the power of 1 handled correctly but differentiation skills were mostly sound. The final two marks proved more challenging and many either could not relate the gradient to their coefficients in i and j or simply stopped after finding the acceleration vector. For those who did attempt a solution, some recognised the need to use the ratio of their i and j coefficients but the majority did not Other errors seen included ignoring i and j and setting the expression equal to ±3 or ± 1 identify and use the correct perpendicular gradient. Some just equated i and j coefficients. 3. Of those who achieved the correct equation in T, a few solved it incorrectly as a result of errors 1 -. A very small number of learners used the scalar product successfully. in handling T 2

Common mistakes

• a = 3t^(−1/2) i − 2j
• direction perpendicular to y = x/3 has gradient −3, so (j component)/(i component) = −3
• wrong: equating i and j coefficients or setting the expression to ±3 or ±1/3
• handling T^(−1/2)
• T = 81/4

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

Examiner's comment

Many candidates were successful across all parts of this question demonstrating a good understanding of the Newton Raphson method, but this question did start to discriminate between the least confident and those who were more careful in their processing. The differentiation in part (b) was generally well attempted with many candidates achieving both marks. The first mark was awarded for recognising that 1 tan 2 x differentiated to 2 1 sec 2 x which many candidates were able to do successfully.

Common mistakes

• tan(½x) → ½sec²(½x)
• the x differentiates to 1 (often lost)
• f′ not f⁻¹
• quotient rule via sin/cos is error-prone

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

Examiner's comment

This question was very similar to previous differentiation from first principles questions and many candidates scored full marks, being much better prepared than in previous series. Most candidates used ( ) ( ) f f x h x h + - to find the gradient of the chord, although a small number of candidates attempted to use ( ) ( ) f f x x h - + but failed to change the denominator to .h - Misunderstanding the definition of the derivative led to some candidates losing some or all of the marks, by writing, for example ( ) ( ) f ^2 f x ^2 x h h + - or ( ) ( ) f ^2 f x^2 x h h + - . Slips when expanding ( ) 2 x h + were condoned for the method mark but the 2x - was required. Most candidates were able to correctly expand and proceed to 2 2xh h h + or 2x h + to score the first two marks. Common errors included incorrectly simplifying the fraction, failure to cancel h correctly or an incomplete limiting argument (e.g. stopping at 0 lim2 h x → ), each of which led to the loss of the final accuracy mark. Notation was generally good, and most candidates included a reference to 0 h → somewhere in their working. A significant minority of candidates failed to include d d y x = or equivalent anywhere in their answer despite being asked to show d 2 d y x x = in the question, thereby losing the final mark.

Common mistakes

• gradient of chord ((x + h)² − x²)/h with the −x² kept
• cancel h to 2x + h
• limit h → 0 must be written and must end "= 2x"
• dy/dx = must appear

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

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

Common mistakes

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

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

Examiner's comment

Part (b) was usually answered correctly with many candidates scoring full marks. As the instruction was to “show that”, candidates were required to present all their working and it was not acceptable to work back from the given answer. Most candidates found a x = correct derivative and clearly substituted into this. It was common to see both 4 ( ) - = - = + y y m x x and y used to find the equation of the tangent. The former mx c 1 1 c = provided a more efficient solution with the latter required clear working to find 48 = + which then needed to be substituted back into A small number of candidates . y mx c = - + y 12 x 48 failed to get to the required form, leaving their answer in the form and hence lost the final mark. Others jumped directly to the given answer without evidence of a correct method for the equation of the straight-line and these candidates failed to score the method mark. The final part of the question proved more challenging and differentiated between the

Common mistakes

• dy/dx = 8 − (5/2)x^(3/2) = −12 at x = 4
• y − 0 = −12(x − 4)
• must reach 12x + y = 48 exactly with working shown

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

This question was very accessible with the vast majority obtaining full marks on all three parts. In part (a), a few candidates integrated the expression and hence obtained no marks as they did not have a linear equation to solve in part (b), but this was a rarity. One of the more common reasons to lose a mark was the inclusion of a spurious “+ c” with one or both derivatives. A few candidates attempted the differentiation but made mistakes with the coefficients (usually resulting from dividing by rather than multiplying by the index). Another error seen was to “simplify” the second derivative from 24x - 14 to 12x - 7.

Common mistakes

• multiply by the index (not divide)
• no "+ c" on a derivative
• simplify (no x¹ or x⁰)
• integrating instead scores 0
• second derivative 24x − 14

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

Examiner's comment

This question was very accessible with the vast majority obtaining full marks on all three parts. In part (b) candidates could access the first mark for setting their linear second derivative equal to 0 and solving for x. There were a few candidates who lost the accuracy mark by giving a value of 24 14 instead of 14 24 and there were a surprising number of candidates who went from 24x = 14 to 1 2 x = .

Common mistakes

• set the second derivative 24x − 14 = 0
• exact x = 7/12 (not 0.583 rounded)
• 14/24 inverted to 24/14
• 24x = 14 does not give x = ½

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

Examiner's comment

In part (a) candidates struggled with the differentiation of the exponential function and were much more confident in dealing with the natural logarithm. For the exponential function, the most successful candidates often wrote an intermediate step such as and then used the chain rule. Common incorrect responses seen in (a)(i) were 2 4 1 e x A -(A was often 4 or 8 when this was seen), ( ) 2 2 4 1 4 1 e x x - - (with or without brackets), 2 2 4 1 4 e x x -and expressions involving 2 4e . x Common incorrect responses seen in (a)(ii) were 1 8x and 1 . x In (a)(ii) a few candidates wrote 8 8ln ln x x = first and achieved the correct alternative . It was disappointing to see a sizeable number of candidates confused between the “dash” notation for differentiation and inverses. So, it was fairly common to see attempts at the inverse of each function, resulting in no marks for parts (a) or (b).

Common mistakes

• chain rule: f′(x) = 8x e^(4x² − 1) (not (4x² − 1)e^(…) or 8e^(…))
• g′(x) = 8/x
• f′ is the derivative, not the inverse

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

This question provided a challenge to most candidates and was in general a good discriminator. In general, those candidates who were able to successfully complete part (a) found part (b) straightforward. However, the accuracy mark in this part was not available to those candidates who had not managed to achieve the correct derivative in part (a). Due to the nature of the function, setting x = 1 and y = 0, gave a ‘correct’ value of d 1 d 2 y x = in many cases even when d d y x was incorrect. As a result, candidates were often not alerted to the fact they had an incorrect d d y x . Disappointingly, candidates who did not obtain 1 2 due to errors in their differentiation, often attempted to manipulate their work to obtain the value that they could see was required from the printed answer and this often meant that the method mark was out of reach, which was a shame.

Common mistakes

• substitute (1, 0): dy/dx = ½, normal gradient −2
• y = −2x + 2
• incorrect derivatives may fortuitously give ½ so A needs a correct (a)

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

Examiner's comment

In part (a), most differentiated the given expression for v with respect to t to get to an expression for their acceleration. Candidates usually handled the powers correctly but there were the occasional errors in the coefficients or an error in the substitution of t = 4. Some tried to use suvat formulae to find the acceleration and a few integrated rather than differentiated.

Common mistakes

• a = dv/dt = 2 − 7/(2√t), so a = 0.25 m s⁻² at t = 4
• integrating instead of differentiating
• suvat cannot be used (acceleration not constant)

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

Examiner's comment

(b) The majority of candidates realised the need to differentiate the given expression for r, with c = 3 substituted, to obtain the velocity, and most of these differentiated correctly. They then substituted t = 4 and used Pythagoras to find the speed, giving their answer either as an exact value or an appropriate decimal equivalent. A minority correctly found the velocity but then stopped and lost two marks. A few candidates delayed substituting c = 3 until the end and a few delayed putting t = 4 which made the working a lot more demanding. About half worked with column vectors instead of using i - j form. Common errors included calculation slips when evaluating the speed or when substituting t = 4, misreading the expression such as losing the negative sign in the j-component and losing the i and j and not recovering to use components. A few candidates tried to use a suvat approach not realising that this was not a constant acceleration situation. (c) Many of those who differentiated in (b), successfully differentiated their velocity vector to obtain the acceleration vector. However, a significant number of candidates did not know how to proceed from there and the second stage of the process proved to be a discriminator. It was only the stronger candidates who realised that their acceleration vector had to be a multiple of the given vector. Those who did make this link generally were able to complete the question successfully. A number of unsuccessful candidates just set the i-component of the acceleration equal to -1 (or more rarely to -27). Most of k -- those who compared a to ( 27 ) i j were able to reach an equation in t only which they then tried to solve. Fewer candidates used a ratio approach but a significant minority of these used the reciprocal of the correct ratio but were still able to score the method mark. Having reached a correct equation in t or T only, some then struggled to obtain the correct answer. There were a few candidates who tried to use a suvat approach for this part not realising that this was not a constant acceleration situation and scored nothing.

Common mistakes

• differentiate r (powers of t decrease by 1): v = (3/2)t^(−1/2)i − (3/4)tj
• put t = 4 in both components then Pythagoras: speed = 3√17/4 ≈ 3.09 m s⁻¹
• stopping at the velocity vector
• suvat invalid (acceleration not constant)

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

Examiner's comment

(c) Many of those who differentiated in (b), successfully differentiated their velocity vector to obtain the acceleration vector. However, a significant number of candidates did not know how to proceed from there and the second stage of the process proved to be a discriminator. It was only the stronger candidates who realised that their acceleration vector had to be a multiple of the given vector. Those who did make this link generally were able to complete the question successfully. A number of unsuccessful candidates just set the i-component of the acceleration equal to -1 (or more rarely to -27). Most of k -- those who compared a to ( 27 ) i j were able to reach an equation in t only which they then tried to solve. Fewer candidates used a ratio approach but a significant minority of these used the reciprocal of the correct ratio but were still able to score the method mark. Having reached a correct equation in t or T only, some then struggled to obtain the correct answer. There were a few candidates who tried to use a suvat approach for this part not realising that this was not a constant acceleration situation and scored nothing.

Common mistakes

• a = −(3/4)t^(−3/2)i − (3/4)j must be a multiple of (−i − 27j): use the ratio of components (not the reciprocal)
• setting the i component equal to −1 is wrong
• T = 9

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 (a), most candidates scored both marks for a correct differentiation. Occasionally a persisting ‘ 5 + ’ or an introduced ‘ c + ’ led to the loss of a mark here. The most common error, however, arose in the coefficients of 2x or x: with the fractional coefficients causing something of a challenge to some candidates. It was rare that attempts to integrate were seen.

Common mistakes

• fractional coefficients: (2/3)x³ gives 2x², (7/2)x² gives 7x
• the constant 5 must go (a persisting + 5 or an added + c loses the mark)

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 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 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. Part (b) was mostly well answered with the majority scoring both marks, with most candidates gaining these marks efficiently without unnecessary calculations. Most candidates found the gradient of the tangent by substituting 0 x = into the given derivative of f( ) x , however a small number of candidates took a longer approach by attempting to use the small angle approximation. Some candidates tried to substitute their answer from part (a) into the given derivative or found the gradient of the normal having originally found the correct gradient for tangent; these approaches lost both marks. A small number of candidates failed to realise the significance of the point (0,3) and therefore retained an algebraic gradient which they tried to substitute into y mx c = + . A number of responses were seen not giving their answer as an equation, just an expression 1 3 2 x + , or forgetting to substitute their value of c found, giving their equation as 1 2 y x c = + ; having been given the y intercept in the question, it was a requirement that for any marks to be scored that 3 c = .

Common mistakes

• gradient is f′(0) from the given derivative, not from the answer to (a) or the normal
• answer must be an equation y = ½x + 3

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

Examiner's comment

This question on exponential functions caused problems for the majority in part (b), as they failed to recognise that differentiation was required. However, generally parts (a) and (c) were answered well. In part (b), the majority of candidates failed to recognise that the 15.3 cm per year meant that they were required to differentiate to obtain a rate of change, and attempted instead to substitute t = 4 into the expression for h. This answer resembled 15.3 cm, so many incorrectly assumed that they had answered the question, even though the subsequent conversion from metres to cm made no sense. Some attempted to divide this value by 4, showing some appreciation for the units given in the question, but lacking an appreciation that the information given provided an instantaneous rate of change and not an average rate over the four years. Without an attempt at differentiation no marks could be scored in (b) but those candidates who did differentiate generally successfully answered this part of the question.

Common mistakes

• rate of growth means dh/dt = 0.34e^(−0.2t), not h at t = 4 (1.536 m) or h/4
• 0.153 m per year = 15.3 cm per year

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

Examiner's comment

This differentiation from first principles question was similar to the one that appeared in 2018 and candidates demonstrated a much greater confidence with the way in which they should approach this topic. The limits were provided in the question and the accessibility was evident as it was very rare to see a completely blank script, which is not always true for a differentiating from first principles question. Most candidates were able to obtain the first 3 marks of this question, with very few sign errors seen when using the addition formula. Most correctly started with the fraction for the gradient of the chord f( ) f( ) x h x h + - as given in the formula booklet. Occasionally all a candidate would attempt was the addition formula in terms of A and B, but more often than not it was in the correct format and correct. Scoring the last two marks was not as common, but still achieved more often than not. There were a variety of layouts used for the candidates to justify the replacement of terms with 0 and 1. If candidates lost marks here, it was often because they failed to separate their terms in h correctly. Some candidates introduced extra h’s, forming expressions likecos sin xh hh . At other times, candidates simply tried to cancel terms incorrectly or stopped. Notation was generally good with limiting arguments correct. The main mistakes that resulted in lost marks were either failing to isolate the required expressions and jumping to the answer or trying to replace each of the 3 terms separately, failing to spot the common factor of sin x in two of them. There were also several candidates incorrectly splitting up the product of each term resulting in an extra h in the denominators. Incorrect notation resulted in the loss of the final mark, in particular the limit notation was often seen right through to the end, in the final answer, even after the limits had been used. A small number of candidates proceeded by using the small angle formula which was generally done well; where errors were seen it was generally not cancelling correctly. One surprising thing to note was the number of candidates who felt it acceptable to not clearly identify the numerator of the fraction by using very small fraction lines, resulting in their solution looking like only one term was being divided by h. Many candidates failed to write d cos d y x x = at the end, despite this being what they had been asked to show. If f ( )x had not been acceptable as an answer, then many candidates would have lost the final A mark. Only a very small handful of candidates used small angle approximations.

Common mistakes

• gradient of chord (sin(x + h) − sin x)/h
• compound angle formula
• isolate (cos h − 1)/h and (sin h)/h
• h → 0 must be seen

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

Examiner's comment

This question brought together a variety of topics and was appropriately placed at the end of the paper. This was very good at discriminating between candidates at the higher end, but it was also pleasing to see a good number of weaker candidates who were still able to attempt later parts such as part (c) and part (d) to a pleasing degree of success. In part (a) a significant proportion of candidates understood that they needed to use the product rule on 7 ex x and most candidates who did, applied it correctly. A common error was just to write 7ex or 7 ex x . Most candidates were unable to achieve the B1 mark, with most not applying the chain rule and so missing terms or making errors with powers. Candidates usually used their expressions correctly with the quotient rule, but few candidates scored the final method mark due to not having the correct expressions. Most candidates chose to use the quotient rule method, but those who used product rule instead, and initially rewrote the denominator as ( ) 1 3 2 e 2 x - - , were generally successful in achieving the first four marks. Some, however, omitted brackets in their expression which lost the final two marks. The final A1 mark was achieved by very few candidates as most did not complete the rearrangement process. Some candidates adjusted their answer by using the given answer in part (b) to work out what the values of A and B should be. Whilst candidates should not typically use later information in earlier parts, this was condoned, as in some cases it would have been too difficult to determine whether this had been done so the benefit of the doubt was given. To achieve the correct form in part (a) from correct differentiation deserved full credit, regardless of the route to the final form required. It should be noted, however, that there were a few who integrated instead of differentiating.

Common mistakes

• product rule on 7xeˣ, chain rule on √(e³ˣ − 2), then quotient rule
• A = −4, B = −4

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

Examiner's comment

This question brought together a variety of topics and was appropriately placed at the end of the paper. This was very good at discriminating between candidates at the higher end, but it was also pleasing to see a good number of weaker candidates who were still able to attempt later parts such as part (c) and part (d) to a pleasing degree of success. In part (b) very few candidates attempted a solution having not completed a solution to part (a). Some determined candidates attempted a solution using A and B and so were able to achieve the method mark. Some candidates attempted a solution by using the given solution to work out what the values of A and B should be and then attempted a solution with those values. A minority of candidates who attempted a solution did not show all the steps required for a show that question.

Common mistakes

• set the numerator = 0, collect the x terms and factorise

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

Examiner's comment

Generally, the candidates that made a full attempt at this question did very well by achieving full marks or at least 4 out of 6. However, many failed to use the information in the second bullet point fully (that the curve has a stationary point at ( ) 4,3 ). The majority of the candidates achieved the first mark by understanding that the stationary point at ( ) 4,3 meant that the first derivative must be equal to zero at 4 x = , arriving at 16 2 0 a b + + = . Again, the majority went on to perform the integration of the gradient function ( ) 4 d x a x b x + + to achieve ( ) 3 2 2 2 f 2 3 x x ax bx c = + + + with the majority able to determine that 5 c = -. A common error was to achieve 3 2 3 2 ax following integration, although those who had written the unsimplified 3 2 3 2 ax were still able to access the intermediate accuracy mark. Common problems were for candidates to misinterpret the information about the stationary point, either using ( ) f 4 3 = or ( ) f 4 0 = , or to miss out a constant of integration, which often resulted in concluding that b was 5 -. It was fairly common to see candidates substitute a rearranged 16 2 0 a b + + = into their integrated expression (usually for b) and this typically did not hinder further progress, although it did result in some sign errors that meant their final expression ( ) f x was incorrect.

Common mistakes

• stationary at (4, 3) means f′(4) = 0 (16 + 2a + b = 0) and f(4) = 3, not f′(4) = 3
• integrate a√x to (2/3)ax^(3/2)
• + c with c = −5 (b = −5 is wrong)
• a = −15, b = 14
• f(x) = 2x² − 10x^(3/2) + 14x − 5

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

Examiner's comment

This question provided an accessible start to the paper for the vast majority of candidates, most of whom picked up at least some of the marks available. In part (a) most candidates were able to differentiate f (x) twice and were able to earn both marks. If candidates did lose marks it was mainly due to arithmetic or processing errors such as differentiating 3x2 to 5x. The most common error was a slip in the final coefficient of x.

Common mistakes

• differentiate twice: f'(x) = 3x² + 4x − 8, f''(x) = 6x + 4
• slips in the final coefficient (e.g. 3x² → 5x)
• do not then simplify 6x + 4 to 3x + 2

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

Examiner's comment

This question provided an accessible start to the paper for the vast majority of candidates, most of whom picked up at least some of the marks available. In (b) part (i), most candidates were able to make a successful attempt to solve f ′′(x) = 0 for x. The mark in this part was sometimes lost due to sign errors when rearranging. It was surprisingly common to see 6x + 4 = 0 leading to 2 3 x = . A significant number of candidates did not simplify their answer, but this was condoned as were values which followed correctly from an incorrect second derivative of the correct form. Part (ii) proved to more of a challenge and many responses highlighted a lack of understanding or awareness of the condition for a function to be concave. A number of blank responses were seen here. Many who attempted this part of the question deduced, or guessed, that it was necessary to consider the sign of the second derivative but were often confused about which sign was required. Others attempted incorrect calculations setting f ′(x) = 0 and solving the resulting quadratic in x. Occasionally, candidates attempted sketches of f (x) and/or f ′(x) in order to determine the range of f (x) which was not required. Others attempted to demonstrate a sign change in attempt to prove that the solution from part (i) was an inflection point. For those employing the correct method, both strict and non-strict inequalities were accepted.

Common mistakes

• (i) x = −2/3 (sign slips give 2/3)
• (ii) concave where f''(x) ≤ 0, so x ≤ −2/3 (< condoned)
• confusion over which sign of f'' means concave
• setting f'(x) = 0 instead
• giving both inequalities without choosing scores 0

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

Examiner's comment

Those scoring full marks in this question were in a significant minority of candidates. Part (b) had mixed outcomes for candidates. Interpretation of the given initial rate of cooling was problematic for many candidates. There was a general lack of appreciation that the rate of cooling was a continuous variable which represented d . d H t - Those candidates not recognising this and using a difference method (similar to those introduced at GCSE) between t = 0 and t = 1 were restricted to the initial B mark in part (a). The most common attempt at a response for these candidates was setting H(1) = 85 - 7.5 and working from there. Of those who recognised the need to first find d d H t and then substitute t = 0, many incorrectly used d d H t = 7.5 rather than the correct d d H t = -7.5 to express a negative increase in temperature. This led to the loss of the last mark. Another error was to forget to differentiate the + 30 to zero leaving a constant in their d d H t , only scoring at most the second method mark for substituting t = 0 and their A into their d d H t . Some differentiated incorrectly to obtain -55Bte-Bt. Several students missed out on the final mark despite correct work by not writing out the final equation. It is important that centres make candidates aware that when a question asks for the complete equation for a model that it is given as such and not just the values of any required constants.

Common mistakes

• rate of cooling is dH/dt = −55Be^(−Bt) = −7.5 at t = 0 (sign matters), B = 3/22 ≈ 0.136
• give the complete equation H = 55e^(−0.136t) + 30
• using H(1) = 85 − 7.5 (difference method) is wrong
• derivative of +30 is 0

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

Examiner's comment

Part (a) of this question was accessible for the majority of candidates. Most understood that the derivative at a stationary point is zero and were able to achieve both the method mark and the accuracy mark, substituting x = 3 into the given derivative and setting it equal to zero. Occasionally the accuracy mark was lost due to insufficient working following substitution e.g. 2(3)3 – 9(3)2 +5(3) + k =0 followed by k = 12 thus omitting a required intermediate step. In a ‘show that’ question candidates need to ensure that they have completely justified the given answer. Occasionally students showed the substitution of x = 3 into the derivative, processed the powers and simplified to obtain k – 12 and then jumped to k = 12. These responses lacked the appreciation that the correct answer had been achieved because the gradient equals 0 at x = 3. There were some rare instances of candidates incorrectly substituting x = 3 and d 10 d y x = - to find the value of k. Some candidates automatically integrated the given derivative but then realised that the differentiated expression was required in order to answer part (a) which they then successfully did. Other candidates integrated the expression and then substituted in the value for k = 12 to achieve an answer for part (b). Occasionally they then used the answer to part (b) to find k = 12, so were unable to access the marks available for this part of the question. A rarely seen alternative was to use algebraic division to divide the given polynomial by the linear factor (x – 3) to achieve a remainder of 12 + k, and then equate that remainder to zero. This approach was met with varying degrees of success and students were often unable to complete the algebraic division successfully.

Common mistakes

• turning point so dy/dx = 0 at x = 3: 54 − 81 + 15 + k = 0
• show the evaluated powers before k = 12
• do not substitute dy/dx = −10

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

Examiner's comment

In part (b), candidates generally understood that they were required to integrate the expression and the straightforward integration meant the first method mark was commonly achieved. Most realised that they then needed to find the value of their constant of integration using x = 3 and y = 10 to determine the point where the curve crossed the y-axis and achieved the second method mark. The majority of students were successful in achieving full marks in this part, with most stating their answer in coordinate form (0, -28) though some candidates left their answer as c = 28, not making the link with the y intercept explicit. Very few candidates made numerical slips in their processing to find the value of their constant of integration. Where mistakes were made, a common reason for not scoring the second method mark was due to substituting in a value of 0 instead of 10 for y in the integrated expression. Where the constant of integration had been omitted candidates went directly to substituting x = 0. Less commonly, some candidates differentiated rather than integrated and tried to find the roots of the second derivative rather than the y intercept.

Common mistakes

• integrate with + c, use (3, −10) to find c = −28, answer (0, −28) as a point
• substituting y = 0 instead of −10
• c = −28 alone not linked to the y-axis
• circular arguments avoiding (a)

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

Examiner's comment

denominator) which led to sign errors in part (b). Others made slips in the rearrangement which lost the final mark. In some cases, there were issues with clarity of notation for the final statement of d y x in terms of the positioning of the minus sign and candidates should be advised d to take care when drawing the vinculum, particularly when the starting term of the numerator is negative to make sure the sign is in the correct place. In part (b), the majority of candidates were able to use the correct approach to find the equation of the normal and candidates were often able to earn at least some of the marks here even if they had been unsuccessful in part (a). The least successful approach for calculation of the gradient

Common mistakes

• substitute (−2, 5) first: tangent gradient −11/13, normal gradient 13/11
• 13x − 11y + 81 = 0 with integers and "= 0"
• finding the tangent instead
• rearranging to dx/dy before substituting is error-prone

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

Examiner's comment

Candidates were generally more successful in part (b) than part (a). Many candidates found the y intercept by substituting t = 2 into the parametric equation for y rather than substituting x = 0 into the Cartesian equation. The first method mark was more challenging. Some candidates attempted the derivative from the Cartesian equation and they usually reached either ( ) 3 x A + or 3 x whilst others attempted to use the chain rule approach. Some candidates obtained the correct derivative but failed to substitute x = 0. If a candidate achieved the first method mark, they usually went on to achieve the dependent method mark for attempting to find the equation of the tangent. There were very few cases where the candidate used the negative reciprocal of their gradient. The most successful way candidates achieved full marks was differentiating their Cartesian equation. Those that differentiated parametrically were often caught out by arithmetic and algebraic errors or substituted t = 0 instead of t = 2 and failed to achieve the gradient correctly. For the final answer, some candidates had failed to realise the form of equation the question was asking for so left their answer as “y =” or they didn’t have 𝑎, 𝑏 and 𝑐 as integers. Many candidates who had incorrectly answered (a), were able to go on to achieve full marks in (b) by using parametric differentiation to work out the gradient. There were very few blank responses for this part of the question.

Common mistakes

• P at x = 0, y = 3ln25 = 6ln5 (t = 2, not t = 0)
• gradient 3/25 from the Cartesian form or dy/dt ÷ dx/dt at t = 2
• 25y − 3x = 150ln5
• answer must be in the form ax + by = c ln5

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

Examiner's comment

- but there were a few who rearranged their v = 0 equation and in part (b) and then obtained 8 - not understanding then differentiated and they got 2t+2 as their derivative. Many stopped at 8 that magnitude had to be positive. A few tried to do some sort of “magnitude” calculation using Pythagoras and so either sqrt (82 + 82) or even sqrt (82 + 32) was sometimes seen. In part (c),

Common mistakes

• a = dv/dt = −2t − 2, at t = 3 a = −8
• magnitude 8 must be positive
• Pythagoras-style "magnitude" calculations are wrong

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

Examiner's comment

Part (a) was generally a well answered question. Some found the velocity vector at t = 0 and went no further, either forgetting, or not being aware that they needed to use Pythagoras’ Theorem to find the speed. A small number of candidates scored the M mark only, if they had an incorrect velocity vector but used Pythagoras’ Theorem correctly. 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

• differentiate each component of v
• answer must be a vector (2t − 3)i + 4tj

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

Examiner's comment

There was a very mixed response to this question on a cubic function with only one in eight candidates scoring all 6 marks. Part (a) was for finding the set of values of x for which f ' (x) < 0. Candidates generally knew what to do and most scored this B mark. Marks were lost when candidates gave the inequality as or else 2 < . ′ f ( ) 6 2 6 x x <

Common mistakes

• f′(x) < 0 where the curve is decreasing: 2 < x < 6 (strict inequalities)

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

Examiner's comment

Generally, candidates scored well in the first two parts of this question, by factorising the cubic expression and drawing a graph. The final part, however, where the candidates were required to find a range, was very poorly answered, with few gaining any marks and very few reaching the correct final answer. Part (c) proved to be very challenging. Candidates were unable to identify the need to differentiate with many of them treating the equation as if it were quadratic and attempting to find the discriminant. Candidates who recognised the need to differentiate generally went on to find the x coordinates successfully, however, it was common for these candidates to write an inequality in terms of x rather than moving forward to find the associated y coordinates. It appears many candidates are not truly confident in the specific meaning of the terms “domain” and “range”. Of the relatively few candidates who did correctly calculate the y coordinate of the turning points, some still lost the final mark by failing to use correct set notation. A few responses were seen where decimal approximations to 6 3 had clearly been obtained from calculator technology alone and these gained no credit.

Common mistakes

• three intersections means k lies between the turning values: dy/dx = 9 − 3x² = 0, x = ±√3, y = ±6√3
• using a discriminant is wrong (cubic)
• answer in k, not x
• set notation {k : −6√3 < k < 6√3} (union is wrong)
• calculator decimals score nothing

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

Examiner's comment

This was the second modelling question on the paper. The given equation modelled the speed of a car as it travelled between a pair of traffic lights. Many weaker candidates missed out part (b) and scored marks in the more accessible parts (a) and (c). The first two marks in part (b) were often achieved by those who attempted it. Using the product rule to find d v t and then setting this = 0 gave them one form of the given equation. The final two marks were only achieved d by the best of students. In reality, all that was required was to make t the subject of the formula, but it was beyond most.

Common mistakes

• product rule: dv/dt = −0.4 ln(t + 1) + (10 − 0.4t)/(t + 1)
• set = 0 and make t appear once: t = (25 − ln(t + 1))/(1 + ln(t + 1))
• the rearrangement defeated most

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

Examiner's comment

This question was reasonably well attempted. Logs continue to be an area of challenge for many candidates but despite this, a significant proportion were able to earn a good number of marks here. Part (c) of this question was a good differentiator and was a challenge to many candidates. For a significant proportion of candidates, “rate” did not signal the need for rate of change 8 and hence differentiation. It was common to see candidates simply substituting in t = 2 into their equation for P to obtain P = 1.32. Of those candidates who did attempt to find d d P t many seemed to struggle with the chain rule. Some candidates obtained results of the form Ate-0.5t, whilst others retained the constant term 0.8 when differentiating. Others began by substituting in t = 2 and then proceeded from e-2 to e-1 or ‘differentiated’ e-1 to get e-2. It was common for candidates who were able to progress to a correct value to give the final answer of -0.258 rather than 0.258 but this was condoned.

Common mistakes

• rate means dP/dt = −0.7e^(−0.5t) (the constant k disappears
• not Ate^(−0.5t))
• substituting t = 2 into P is wrong
• 0.258 kg/cm² per minute

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

Examiner's comment

This question was, on the whole, well-attempted and provided some differentiation between candidates. In part (a), the majority of candidates were able to correctly work through the steps required to derive the equation of the tangent to curve C. The vast majority were able to successfully and accurately differentiate the equation of the curve and to evaluate it at x = 4 to find the gradient. Most were then able to determine the y coordinate of P (although some candidates appeared to use the equation of the tangent to find this value rather than the equation of the curve). Most were confident and well prepared to correctly find the equation of the tangent, commonly using 1 1 ( ) y y m x x - = - but also often y mx c = + equally successfully. A small number of candidates incorrectly worked with the gradient of the normal rather than the gradient of the tangent and there was sometimes confusion between the value of the gradient and the value of the y coordinate at P: 13 6 and13 3 respectively. There were a small number of 9 candidates who worked from the equation of the tangent and who appeared to be attempting to verify this equation rather than deriving it.

Common mistakes

• dy/dx = (2/3)x − x^(−½), gradient 13/6 at x = 4
• y = 13/3 from the curve (not from the line)
• tangent not normal
• derive the equation rather than verify it

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

Examiner's comment

10 This was one of the most challenging questions on the paper for candidates, although it was a familiar structure to this type of problem and there was access to later parts as part (a) was a “show” question. Most candidates achieved at least a method mark in (b) for attempting to differentiate. A significant number of candidates set the second derivative equal to 0 and solved, instead of the first derivative.

Common mistakes

• dC/dr = 0.26πr − 28.4/r²
• set the first derivative (not the second) to 0
• r = ∛(1420/13π) = 3.26

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

Examiner's comment

10 This was one of the most challenging questions on the paper for candidates, although it was a familiar structure to this type of problem and there was access to later parts as part (a) was a “show” question. In part (c), it was common for many of those who had understood the demands in part (b), to progress and find the second derivative. A lot were unsure what to do with it at this point, however, with a significant number equating it to zero and solving (sometimes when they had already solved correctly for 3.26 in (b)). Of those who achieved a positive answer in (b), most were able to substitute into the model for M1 here.

Common mistakes

• d²C/dr² = 0.26π + 56.8/r³ > 0 for r > 0 (≈ 2.45), hence minimum
• setting the second derivative to 0 is wrong

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

Examiner's comment

10 This was one of the most challenging questions on the paper for candidates, although it was a familiar structure to this type of problem and there was access to later parts as part (a) was a “show” question. It is clear that many candidates had not read the question carefully or checked their answer, as a lot of candidates gave the answer £13 instead of 13p in part (d), which fortunately for them was condoned on this occasion. Candidates could be encouraged to sense check their modelling questions more as there were a variety of clear unfeasible answers here, including negative answers, and many hundreds of pounds. Many also did not read that their answer should be rounded to the nearest integer, and lost marks unnecessarily.

Common mistakes

• substitute r = 3.26 into C (not into dC/dr)
• C = 13 pence to the nearest integer (£13 condoned)
• sense-check the size of the answer

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

Examiner's comment

This question, based on the minimum surface area of a children's toy proved to be a useful source of marks so late in the paper. In fact 25% of the cohort achieved full marks. Many Part (b) well generally well done. Many candidates scored full marks for reaching either 3 1050 or 10.2. Occasionally some candidates mistakenly wrote down 1050 and subsequently simplified their answer to 5 42 . Very few candidates who attempted this part failed to score marks. The main reason for this was usually an inability to differentiate 1680 r which usually became1680 or 1680ln r 1 Most candidates were able to find the second derivative correctly in part (c). How it should be

Common mistakes

• dS/dr = 1.6r − 1680/r² = 0 gives r³ = 1050, r = 10.2
• differentiating 1680/r to 1680 ln r

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

Examiner's comment

This question, based on the minimum surface area of a children's toy proved to be a useful source of marks so late in the paper. In fact 25% of the cohort achieved full marks. Many Most candidates were able to find the second derivative correctly in part (c). How it should be used was confused in much of the work seen. Common reasons for a loss of marks here were; 2 d S • candidates solving the equation = 0 2 d r 2 2 d d S y and • use of incorrect notation including 2 2 d d r x • incomplete solutions with calculations or conclusions missing

Common mistakes

• d²S/dr² = 1.6 + 3360/r³ = 5 > 0 at r = 10.2 so minimum
• solving d²S/dr² = 0
• notation d²y/dx²
• conclusion needed

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

Examiner's comment

The final question on the paper proved to be a challenge for many candidates and there were several blank responses seen. It is possible that this was due to time management. Some candidates attempted unsuccessfully to change to a Cartesian equation, suggesting that possibly parametric integration had not been covered. In part (a), candidates who recognised the need to use parametric integration were usually able to differentiate correctly to find d d x t , multiply this by y and then use the double angle formulae to find an acceptable expression for d d x y t . It was common for candidates to score the first two marks. The next two marks were more elusive, and it was common to see candidates perform a lot of onerous manipulation without making any significant progress. Those who were most likely to be successful were the ones who had an expression in terms of 2 sin 2t and then spotting the need to use the double angle formula to form a term in cos 4t. For many candidates the mark for the upper limit was all that was achieved in part (a). However, some failed to achieve it as they left it as 4 (value for x not t) or 45 (degrees).

Common mistakes

• area = integral of y dx/dt dt with dx/dt = 16 sin t cos t
• sin 2t = 2 sin t cos t
• cos 4t = 1 − 2sin²2t
• x = 4 gives sin²t = ½ so a = π/4
• changing to Cartesian fails

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

Examiner's comment

This question on differentiation from first principles saw good scoring and the method was widely recalled. A small number of slips with the algebra were seen and some replaced f (x + h) with 2x2 + h or (2x + h)2 or 2(x2 + h) instead of 2(x + h)2. There was the occasional poor squaring and mishandling of the “2”s e.g. expanding 2(x + h)2 to get 2x2 + 2xh + h2 was surprisingly common. Some leeway was afforded to the students with notation here although use of h = 0 instead of h 0 was penalised if no indication of the limiting process was seen. A small number of responses did not have an “f′(x) =” or equivalent and so forfeited the final mark.

Common mistakes

• gradient of chord (2(x + h)² − 2x²)/h
• expand 2(x + h)² correctly (not 2x² + 2xh + h²)
• reach 4x + 2h and show h → 0 (not h = 0)
• dy/dx = or f′(x) = must appear

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

Examiner's comment

Part (a) was generally well done with many candidates scoring 3 out of 4 but very few scoring 4 out of 4. Candidates generally understood that the question required an initial differentiation and this was done correctly on the whole. However, incorrect attempts at differentiation included: • 16cosቀ 1 2 𝑥𝑥ቁ-3 • 8cosቀ 1 2 𝑥𝑥ቁ-3 • -4cos ቀ 1 2 𝑥𝑥ቁ-3 Candidates also understood that they had to set their derivative to zero which required a combination of manipulating the algebra, applying arccos, and doubling their answer. The difficulty encountered in this part was the recognition that it was the 3rd solution which was required and most gave either the 1st or 2nd turning point. As a diagram was given in the question, candidates should use this to help them understand what is required. Some students who gave the 1st turning point also checked by further differentiation that their point was a maximum. A few candidates unnecessarily found the corresponding y value for their x value. In establishing their value of x, it was disappointing to see all of the following: • candidates halving instead of doubling • using degrees • arriving at a correct answer but leaving it as 14.01 instead of 14.0

Common mistakes

• f′(x) = 4cos(½x) − 3 (not 8cos or 16cos)
• cos(½x) = ¾, double (not halve) the angle
• use the sketch to pick the third turning point x = 14.0 (not 1.45 or 11.1)
• radians
• 14.0 not 14 or 14.01

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

Examiner's comment

Part (c) was answered well with many candidates scoring full marks. Amongst those who didn’t score full marks, many did not heed the demand in the question to “Show your method” and just quoted their calculator’s solution without any explanation whilst others just wrote down a formula but did not show an application with values embedded. Others made rounding mistakes such as giving a final answer of 4.805 or 4.81. Premature rounding of f (5) and f′(5) also led to a final answer of 4.81. A significant number of candidates erroneously used degrees but could score the method mark if appropriate working was shown. Some candidates were able to “recover” minimal working due to the accuracy of their answer 4.804...., and gained the two marks. Some candidates also evaluated f (5) incorrectly even though it was given in the question. It was disappointing to see students misquoting The Newton-Raphson formula, typically with f(x) and f′(x) the wrong way round, even though the formula was given in the formula book.

Common mistakes

• x₁ = 5 − f(5)/f′(5) with values shown
• f(x) and f′(x) the right way round
• radians
• awrt 4.80 (not 4.8, 4.805 or 4.81 from premature rounding)

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

Examiner's comment

Part (a) was generally well answered. The quotient rule was usually applied correctly for the first two marks and, given the quadratic factor was given, most candidates were able to arrive at the correct expression for g(𝑥𝑥). Some used the product rule instead and usually gained the first two marks although it was a little more difficult to get the final answer and some algebraic errors were made. Although most candidates gained the M mark in (a) their brackets were very often incorrect or missing. This was usually recovered in later steps allowing full marks to be obtained. When differentiating 4x2 + k, some candidates left k in the answer, or differentiated to 8x + 1, gaining no marks. Occasionally the terms on the numerator were reversed and these attempts gained no marks in (a) as in incorrect method was implied. Some missed the 3 when differentiating e3x using the chain rule. Most candidates understood there was no need to expand the denominator. Where they did so, they generally did it correctly and no marks were lost. This was a “Show that” question so “f′(x) =” or “ d d y x = ” needed to be seen and was sometimes missing, losing the final A mark.

Common mistakes

• quotient rule numerator the right way round
• d/dx e³ˣ = 3e³ˣ, d/dx(4x² + k) = 8x
• g(x) = e³ˣ/(4x² + k)²
• f′(x) = must be seen

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

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

and went on to use Pythagoras correctly to find an exact answer or an answer correct to the i and j and then finding 12-6 sqrt 2 rather than apply Pythagoras. A small number 2sf or better. Common mistakes were to stop once the velocity had been found, losing subtracted the squared values rather than adding in their Pythagoras. In the second part, most candidates identified that this was variable acceleration and differentiated correctly to find the acceleration vector. Common mistakes were losing either or both i and j and not recovering, integrating instead of differentiating and, despite the question specifying the answer was to be given in i j form, final answers given in column vector form were occasionally seen. Many candidates integrated to achieve the first two marks in part (c),

Common mistakes

• variable acceleration so differentiate v (not integrate, not suvat)
• keep both i and j
• answer 6ti − 3t^(−1/2)j in i, j form, not a column vector

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

Examiner's comment

Aspects of this question proved to be more demanding than expected, but overall good marks were scored by candidates with most achieving over half marks in this question. Part(a) Almost all candidates identified the need to differentiate and set their expression equal to zero. Although the derivative was often accurately found – the chain rule aspect of the log expression was omitted by some so that the expression 4x-4 was not seen as the numerator of the required fraction. Some numerators seen included 2x-2, 2x- 4 or most often 1. Manipulating their equation into the desired form was also an issue for some, with sign errors causing problems, whilst others could not properly convert their equation involving fractions into one not involving fractions.

Common mistakes

• chain rule on ln(2x² − 4x + 5) gives numerator 4x − 4 (not 1 or 2x − 2)
• set f′(x) = 0 and multiply through by the quadratic
• sign errors

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

Examiner's comment

This question was generally well attempted, although it was rare for candidates to score full marks. In part (a), candidates knew that they needed to differentiate, and most had read the question carefully to realise that they also needed to substitute 2 x = into their d d y x , enabling them to proceed often to the correct answer. 5

Common mistakes

• dy/dx = 6x then substitute x = 2: gradient 12

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

Examiner's comment

This question was generally well attempted, although it was rare for candidates to score full marks. Part (b) was completed to a high degree of success. Candidates are clearly more comfortable with the idea of finding the gradient of a chord in terms of h and most were able to proceed to the correct answer. Some made slips in their multiplying out of the brackets and there was the occasional response where the candidate had differentiated again and substituted in 2 h + which could not score.

Common mistakes

• gradient of chord (3(2 + h)² − 2 − 10)/((2 + h) − 2) = (12h + 3h²)/h = 12 + 3h
• denominator must be h
• differentiating again is wrong

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

Examiner's comment

This question was generally well attempted, although it was rare for candidates to score full marks. In part (c) candidates still struggled with bringing the concept of the gradient of a chord tending to the gradient at a point on a curve it was extremely rare to see a correct response, although this could be due to candidates having made an error in (a) or (b) so they were unable to see clearly what the link was between them anyway.

Common mistakes

• as h → 0, 12 + 3h → 12: the gradient of the chord PQ tends to the gradient of the tangent at P
• correct responses were extremely rare

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

Examiner's comment

This was the first significant modelling question on this paper. As with other series, candidates find aspects of modelling difficult. The barriers to success here were parts (b) and (c). In part (b) many students used an incorrect method to find the rate of increase in the number of bacteria. Some substituted t =8 into their expression for N, which just found the number of bacteria after 8 hours; others then went on to divide this number by 8 or even 1000 to try and find the rate of increase. Those who realised that the expression for N needed to be differentiated to find the rate of increase often scored at least the method mark.

Common mistakes

• rate of increase means dN/dt = 1000k e^(kt) at t = 8: 420 (2 s.f.)
• substituting t = 8 into N, or dividing N by 8, is wrong

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 question proved equally challenging and very few fully correct solutions were seen. The modal mark achieved here was 2, scored by over 40% of the candidates, most often for a fully correct attempt at dy/dx. Most solutions seen used the quotient rule although incorrect application of the rule often meant that no marks could be awarded. The most common errors were subtracting the wrong way round on the numerator, or else adding the two expressions on the numerator. Differentiation of sqrt x suprisingly caused a few problems as well. Product rule expansions seldom got past the first expression. The ability to manipulate the ensuing algebraic expression was often lacking, with fractional terms within a fraction, the main cause for failure. Alternative methods which might have simplified the algebra were seen, but only rarely. Only a handful spotted that a difference of two squares method applied to the numerator would have made this a very simple problem.

Common mistakes

• quotient rule numerator the right way round
• multiply through by √x to simplify to 1/(2√x), A = 2
• spotting x − 4 = (√x + 2)(√x − 2) gives y = √x − 2 directly
• dy/dx must be seen

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

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. In part (a) candidates struggled with the information being presented in reverse order. As a result, candidates typically formed the equation in a and b but then got stuck as they could not see how they would find a. It was those who differentiated and equated to 3 - who were able to fully complete part (a) correctly. It was noted that a large number did not seem to be that confident with turning points and derivatives and it may have been that this had not been covered which prevented candidates from doing this question particularly well.

Common mistakes

• use the gradient first: f′(2) = 12a + 60 − 39 = −3 gives a = −2 with an intermediate step
• then (2, 10) on C gives b = 44
• candidates who started with the point were stuck with two unknowns

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

This question was accessible to the majority of students however part (b) proved to be a good discriminator with only the most able gaining all 7 marks. The majority of students gained the first three marks of this questions by correctly differentiating the expression twice. A few arithmetic errors were seen occasionally, although if this was the case then students would only lose 1 out of the three marks. Part (b) differentiated out the students to a greater extend. Most students could identify the correct method to find the stationary point, although there were a disappointing number of students who failed to follow this up with a concluding statement or preamble. A number of students got confused between parts (i) and (ii) and tried to use the second derivative to identify stationary points. The most common method to identify the stationary point was using the first derivative and substituting the value x = 1, although there were a significant number who chose to deduce the value x = 1 by factorising their cubic expression. Both approaches tended to lead to a positive outcome, and full marks. Part (ii) was much more mixed in response. A significant number of students gained no marks on this part as they assumed that when both first and second derivative equal zero then the point is a point of inflection. Of those students who successfully answered this question there were an equal number of students who considered the first derivative as those that considered the second derivative. Most of these students considered an appropriate intervals to gain full marks here, however a significant number of students considered inappropriate intervals such as 0 and 2 or 1.5 and so were unable to gain the accuracy mark.

Common mistakes

• dy/dx = 20x³ − 72x² + 84x − 32
• second derivative 60x² − 144x + 84

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

Examiner's comment

This question was accessible to the majority of students however part (b) proved to be a good discriminator with only the most able gaining all 7 marks. The majority of students gained the first three marks of this questions by correctly differentiating the expression twice. A few arithmetic errors were seen occasionally, although if this was the case then students would only lose 1 out of the three marks. Part (b) differentiated out the students to a greater extend. Most students could identify the correct method to find the stationary point, although there were a disappointing number of students who failed to follow this up with a concluding statement or preamble. A number of students got confused between parts (i) and (ii) and tried to use the second derivative to identify stationary points. The most common method to identify the stationary point was using the first derivative and substituting the value x = 1, although there were a significant number who chose to deduce the value x = 1 by factorising their cubic expression. Both approaches tended to lead to a positive outcome, and full marks. Part (ii) was much more mixed in response. A significant number of students gained no marks on this part as they assumed that when both first and second derivative equal zero then the point is a point of inflection. Of those students who successfully answered this question there were an equal number of students who considered the first derivative as those that considered the second derivative. Most of these students considered an appropriate intervals to gain full marks here, however a significant number of students considered inappropriate intervals such as 0 and 2 or 1.5 and so were unable to gain the accuracy mark.

Common mistakes

• substitute x = 1 into dy/dx and conclude
• f″(1) = 0 alone is not enough: check the sign of f″ either side (interval within x < 1.4) or a third derivative
• reasoned conclusion needed

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

Examiner's comment

Part (a) was generally well answered, with most students able to differentiate correctly and understand the need to use the first derivate to find the gradient of the tangent line. A few tried to use the diagram to obtain the gradient, even after differentiating correctly so were only able to gain the first B mark, but they were in a minority. The vast majority of candidates who correctly differentiated (a few integrated, hence lost the marks), were able to go on to find the correct equation of the line, often using y = mx + c rather than 𝑦𝑦-𝑦𝑦1 = 𝑚𝑚(𝑥𝑥-𝑥𝑥1). A small minority displayed poor algebraic skills when rearranging their linear equation to find c, and a few used a gradient of 1 2 instead of 2.

Common mistakes

• gradient from dy/dx at x = 5 (= 2), not from the diagram
• y = 2x − 23

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

Examiner's comment

Many candidates gained some credit in this question but few gained full marks. In (b) most candidates gained some marks but few used a fully correct method to achieve the correct answer of and . Most were able to rearrange the equation of the normal, and identified that its gradient was . They then identified the gradient of the tangent as and used their result from (a) at and to attempt to form a correct equation in and such as . A number of candidates incorrectly equated the numerators and denominators to form the equations and and solved these to fortuitously arrive at the correct solution, losing most marks in (b). Another incorrect approach was to form the equation of the normal of the form and to rearrange to the form followed by 'comparing coefficients' with (an incorrect method). Many candidates successfully substituted and into the equation of C to get . This step was missed, however, by quite a few candidates, who could not then gain full marks for the question.

Common mistakes

• substitute P into C for p − 4q = 22
• normal gradient −19/26 so tangent gradient 26/19
• equating numerators and denominators separately or comparing coefficients with the normal scores nothing
• p = 2, q = −5

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

Examiner's comment

This question involving differentiation of parametric equations was found to be challenging by the majority of students, particularly the differentiation of 𝑦𝑦= cosec3 𝜃𝜃 and part (b). The majority of candidates knew what was expected in part (a) and proceeded to attempt to use the chain rule to find d𝑦𝑦 d𝑥𝑥, however they found differentiating 𝑦𝑦= cosec3 𝜃𝜃 challenging. Some did not check in the formula book that the derivative of cosec theta was negative, or omitted the constant 3, and this cost 3 of the 6 marks available in the whole question. For those that attempted to write 𝑦𝑦= sin-3 𝜃𝜃 and use either the chain rule or the quotient rule success was mixed. Other common incorrect derivatives included d𝑦𝑦 d𝜃𝜃= (cosec3 𝜃𝜃cot3 𝜃𝜃) and variants of this. Differentiating 𝑥𝑥= sin 2𝜃𝜃 proved less troublesome, with most doing so correctly. There was a proportion of candidates that rewrote sin 2𝜃𝜃 as 2 sin 𝜃𝜃cos 𝜃𝜃 and used the product rule, creating more work but ultimately doing so successfully in most cases. The method mark was scored by the majority of candidates as a result, although there were a few candidates who tried to write 𝑦𝑦 in terms of 𝑥𝑥 and differentiate, with no marks being scored for this approach. A number of candidates, having found a correct derivative, proceeded to attempt to tidy this up, unfortunately using time that could have been spent elsewhere in the paper.

Common mistakes

• dy/dθ = −3cosec³θ cotθ (chain rule
• derivative of cosec is negative
• keep the 3)
• dx/dθ = 2cos 2θ

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

Examiner's comment

(a) Most candidates knew that they needed to differentiate but integration was also seen once. 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).

Common mistakes

• differentiate v to get a = 10 − 2t
• integrating is wrong
• setting a = 0 is not required

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

Examiner's comment

Part (a) was well answered by most candidates who realised that they needed to differentiate although a few candidates did omit the vector notation and therefore achieved zero marks. Part (b) proved to be the least popular part of this question.

Common mistakes

• a = dv/dt: differentiate both components (powers decrease by 1)
• keep the vector notation (M0 if i, j disappear)
• answer (3/2)t^(−½) i − 2j

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

Examiner's comment

This question was generally answered well with most candidates scoring full marks. Most candidates were confident in differentiating the cubic correctly and substituting in 2 x = to find the gradient, although a few incorrectly set their gradient equal to 0 and attempted to solve to find x. The majority of candidates were then able to use their gradient and the given point to correctly find the equation of the tangent, although a few candidates incorrectly thought that they needed to use the negative reciprocal of their gradient instead to find the equation of the normal at P.

Common mistakes

• dy/dx = 6x² − 4 then substitute x = 2 (show it: 20 alone could be from a calculator)
• do not set dy/dx = 0
• tangent not normal (no negative reciprocal)
• answer y = 20x − 27 in the form asked

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

• product rule with d/dx e^(−2x) = −2e^(−2x)
• f′(x) must appear
• no bracketing errors on the way to the given answer

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

Unsurprisingly candidates found part (c) more challenging, especially (c) (ii) where the significance of x=0 was rarely understood. It was however a good discriminator for stronger candidates.

Common mistakes

• (i) g(x) ≥ −8e² (double the minimum)
• (ii) h has domain x ≥ 0: lower bound h(0) = 2f(0) − 3 = −19, upper bound 16e⁻⁴ − 3 from the maximum
• significance of x = 0 rarely seen
• exact values and correct range notation

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

Examiner's comment

The final question was not attempted in a significant number of cases. This may have been due to the demand of the question for the majority of candidates. Of those candidates who did attempt this question, it was rare for full marks to be achieved. In part (a), candidates struggled with getting started which then cost them dearly. They seemed to struggle with forming an equation without being given one, despite it being a cubic equation, which then meant they lacked the confidence to pursue any further. Common errors were not appreciating that there was no constant term which meant they produced simultaneous equations with too many unknowns. Others knew they needed to differentiate g(x) and sub in 2 x = but then they equated it to 9 and not 0. Clearly having lots of information provided at the same time and having to select which is appropriate for each stage of their working the majority of candidates found difficult. The lack of understanding 12 surrounding coefficients may have led to not being able to set up the cubic with the coefficients for 3x and x being the same. Some even assigned numerical values rather than unknowns. There were some candidates who were able to form two simultaneous equations, however, and most of these successfully found a and b. They were asked to find the function, so it was essential that they did state their answer at the end to score full marks in this part.

Common mistakes

• g(x) = ax³ + bx² + ax (no constant: through the origin)
• use g(2) = 9 and g′(2) = 0 (not 9)
• solve 10a + 4b = 9, 13a + 4b = 0
• state g(x) = −3x³ + (39/4)x² − 3x

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

Examiner's comment

The final question was not attempted in a significant number of cases. This may have been due to the demand of the question for the majority of candidates. Of those candidates who did attempt this question, it was rare for full marks to be achieved. In part (b), if candidates were able to come up with a cubic for part (a), they usually appreciated the need to differentiate to find the second derivative and substitute in 2 x = . Their knowledge on second derivatives meant they knew it would need to be negative to prove it was a maximum. However, there were a number of candidates who had incorrect derivatives or evaluated incorrectly and still stated they had a maximum, even though their answer would have been positive. This, for some, should have prompted further investigation into their earlier work. Some candidates still substituted in a value of x either side of 2 x = into either g(x) or g ( )x . This was far more working to score both marks, so candidates should be encouraged with this type of question to prove the nature of the turning points using second derivatives.

Common mistakes

• g″(2) = −33/2 < 0 so maximum
• state the sign
• checking either side takes longer

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

Examiner's comment

Although fully correct solutions to this question were very rare, many candidates were able to score marks from both parts. In (b) candidates often scored the M1 for an unsimplified second derivative or a correct structure with a sign incorrect. Where the first A1 was scored in (b), it was usually 2 2 d d y y 4 27 awarded for a correct simplified , or solving to get 0 x = x = 2 2 d d x 2 d y Very few candidates seemed aware that they needed to show a change in sign of 2 d x 4 27 either side of to show a point of inflection. x =

Common mistakes

• quotient rule: d²y/dx² = 54(x⁴ − 27)/(x⁴ + 81)²
• d²y/dx² = 0 alone is not enough: show the sign change either side of x = ⁴√27 and conclude

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

Examiner's comment

In part (a), most candidates used a correct method to differentiate, the majority using the quotient rule. A few applied the rule the wrong way round (uv’ – vu’)/v2. Candidates who initially wrote the quotient as 2𝑥𝑥 3 2 + 1 2 𝑥𝑥 1 2 were often more successful with the differentiation. A few candidates forgot about the 4lnx term but most dealt with this correctly. Errors from missing brackets or when simplifying their differentiated expression, in particular dealing with the powers, meant that many did not gain full marks in this part.

Common mistakes

• split into 2x^(3/2) + ½x^(1/2) before differentiating (or quotient rule the right way round)
• ln x differentiates to 1/x
• no missing brackets in a show that

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

Examiner's comment

In part (b), although almost all candidates made dy/dx = 0 and reached 12x2 + x - 16 sqrt x = 0 the majority then failed to make progress and did not gain any marks. Several made no attempt to rearrange this equation and others made x2, x or sqrt x the subject but did not make the key step of dividing by sqrt x first.

Common mistakes

• set dy/dx = 0, then divide by the square root of x first (key step usually missed)
• make 12x^(3/2) the subject before raising to the power 2/3

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

Examiner's comment

Many candidates left the entire question or parts of this question completely blank and clearly found it difficult to access the content being assessed here, perhaps due to a lack of familiarity with integration using parametric equations. Many tried to establish the Cartesian equation to answer various parts. That said, there were many candidates who provided clear and concise solutions to each part. In part (a)(i), most candidates recalled the formula for integrating parametric equations and knew to find 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑. Awareness of the double angle formula for sin2t was good. The main problem caused for these candidates was a lack of “dt” throughout their work, and only included this in the final line, losing the final accuracy mark. There was a significant proportion of candidates who attempt ed some differentiation using the chain rule: 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑= 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 x 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 , scoring no marks. In part (b) many candidates who had found part (a) difficult gave up and lost the opportunity of scoring some easier marks. Many candidates who scored no marks elsewhere were still able to access at least the first two marks in part (b) by setting sin2t = 4.2 5 and solving, frequently getting at least one correct angle; but too many of these found their second value of t by subtracting their first value from pi instead of calculating pi – 𝑠𝑠𝑠𝑠𝑠𝑠-1( 0.84) before dividing by 2. The following mark was for finding two values of x, but many simply substituted their first value in to find x, thinking that this gave them the width of the walkway. Too many marks were lost by approximating too soon and hence not having values with the required accuracy. A few lost the last mark because they omitted the units.

Common mistakes

• area = integral of y (dx/dt) dt
• sin 2t = 2sin t cos t must be seen explicitly
• keep dt throughout
• dy/dx by the chain rule is the wrong method

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

Examiner's comment

Part (b) was attempted by most candidates although it was not uncommon to see attempts to find the inverse of g(𝑥𝑥) instead of differentiating. The differentiation in part (b) was however correctly carried out by many candidates, with the majority using the quotient rule. Those who stated the correct formula and the individual components of u, v, u’ and v’ were often more successful. A few used ቀ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑ቁ 2 instead of 𝑣𝑣2 in the denominator, while others made slips on expanding the bracket 3(𝑙𝑙𝑙𝑙(𝑥𝑥) - 2). Some attempted the product rule, usually with less success. Many candidates did not attempt to simplify the numerator of g′(x). Often those that did made slips and it was frustrating to see candidates losing marks due to missing brackets and sign errors when expanding. It was common to see numerators incorrectly simplified to –13 𝑥𝑥 or 5 𝑥𝑥 although many did manage the correct simplification to 1 𝑥𝑥. Some candidates unnecessarily expanded the denominator, sometimes incorrectly writing ln (𝑥𝑥2) instead of (ln 𝑥𝑥)2. Candidates who managed to correctly simplify their expression for g′(x) argued correctly why g′(x) > 0 for all 𝑥𝑥 in the domain, but some simply stated this with no reason or gave an incomplete argument of “𝑥𝑥> 0 so g′(𝑥𝑥) > 0” and lost the final mark in this part.

Common mistakes

• quotient rule with v² in the denominator
• numerator simplifies to 1/x
• conclude g′(x) > 0 because x > 0 AND (ln x − 2)² > 0

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

Examiner's comment

In part (a), a significant minority misunderstood the question and solved v = 0 instead of a = 0. The majority, however, did multiply out and differentiate correctly to find a before then equating it to zero. A few errors in multiplying out the brackets were seen and a significant number had changed the sign of v which then had consequences in part (b).The majority, for part (b), realised that the particle the answer. A few, however, used the t value that had been found in part(a) and lost the method mark.

Common mistakes

• multiply out v = 14 + 3t − 2t² and differentiate
• stops accelerating means a = 0 (not v = 0), t = 3/4

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

Nearly all students who were correct in part (a) achieved the mark for part (b), including many who had failed to cancel the (x+1). Indeed, there were many cases where a correct answer to part (b) was seen following incorrect work in (a), presumably due to the use of a calculator to plot and inspect the graph.

Common mistakes

• dy/dx < 0 when (x + 1)³ < 0 so x < −1

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

Examiner's comment

Almost all candidates were able to access some of the marks available in this question with many gaining full marks in (a). However, (b) was done with much less confidence with many candidates making little or no attempt. (a) The majority of candidates achieved the first two marks (for the 6x), with very few integrating instead of differentiating. Common errors with the 24/x term included rewriting the term as +24x or 24x1/2 before differentiating. A few wrote the term correctly as 24x-1 but differentiated to -24.

Common mistakes

• write 24/x as 24x⁻¹ (not 24x or 24x^½)
• derivative −24x⁻², not −24

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

Examiner's comment

Almost all candidates were able to access some of the marks available in this question with many gaining full marks in (a). However, (b) was done with much less confidence with many candidates making little or no attempt. (b) Some candidates seemed unfamiliar with what was meant by an increasing function and did not attempt to use dy/dx > 0. Common incorrect attempts included finding d2y/dx2 or treating the original function as a quadratic with attempts to complete the square or find the discriminant. Of those who constructed the correct inequality, a number lacked the algebraic manipulation skills to deal with inequalities with both positive and negative powers of x. Candidates should ensure they provide answers in the form requested by the question. It was very common to see final answers given as rounded decimals only, when an exact answer was required.

Common mistakes

• increasing means dy/dx > 0 (not d²y/dx² or the discriminant)
• 6x − 24/x² > 0 gives x³ > 4
• exact answer x > ∛4, not a rounded decimal

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) and (b) more accessible to students compared to part (c) and (d). In part (a) most students were successful at using the product rule and setting dy/dx equal to 0 (though the latter was implied rather than being explicit in many cases). Missing the -0.25 factor when differentiating the e(-0.25x) was fairly common, but the differentiation was mostly correctly carried out. A few students failed to use the product rule correctly, instead multiplying the derivatives of each term, but these were in the minority. Having attempt the derivative the majority did set the result equal to zero (possibly implied) and proceeded to cancel the exponentials. Reasoning for the cancellation was usually not given, but in this instance there was no requirement for such reasoning. However it might be well noted that such explanations may be required under the new specification. Many students lost the final accuracy mark as they did not state or show division by cos x. Students need to be clear that in a ‘show that’ question all steps need to be shown.

Common mistakes

• product rule
• missing the −0.25 from differentiating e^(−0.25x)
• set f′(x) = 0, cancel the exponential and show division by cos x

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) and (b) more accessible to students compared to part (c) and (d). There was less success in part (c). Attempts at this part were varied, with a sizeable proportion not making the connection with part (a) at all and attempting to find the value of H(t) at pi /2 or some other value, or failing to substitute into H(t) at all. Many solved tan x = 4 to find the acute angle but did not then look for the further solution which was needed for the height between the first and second bounces and so achieved only the first mark. Also, use of degrees instead of radians was a common error, but this was allowed the first method mark. The magnitude of their resulting answer should have alerted them to the fact that something had gone very wrong.

Common mistakes

• need the second solution of tan t = 4, t = 4.47 (not 1.33)
• radians not degrees
• H ≈ 3.18 m

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

A number of candidates did not attempt this question. Whether this was due to lack of understanding or time constraints is unknown. Part (a) was generally well done with a surprising proportion using implicit differentiation. A good number who found d d y x correctly failed to substitute in y =0 to find the value of d d y x at the origin. Candidates who tried to expand sin2y using the double angle formulae and then differentiating using the product rule often went wrong.

Common mistakes

• dx/dy = 8cos 2y then invert
• substitute y = 0 to get 1/8 (often omitted)

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

Examiner's comment

A number of candidates did not attempt this question. Whether this was due to lack of understanding or time constraints is unknown. The small angle approximation in (b) part (i) was also well handled. Most settled for the form 8 x y although some rearranged to y = . Marks were generally lost when sin 2y was replaced by 2 instead of 2y. Many candidates failed to see the link between (a) and (b)(i) even though they had both answers correct. They were unable to see that the gradients were the same or else failed to explain themselves coherently.

Common mistakes

• sin 2y ≈ 2y (not 2θ) gives x = 8y
• the gradient of x = 8y is 1/8, the same as (a)
• "8 and 1/8 are reciprocals" alone not enough

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

Examiner's comment

Q02 proved to be challenging, with a significant number of candidates struggling to realise that the length of the runway could be estimated by the area under the speed- time curve generated from the table of values given in the question. In Q02(a), only a minority of candidates estimated the length of the runway by applying the trapezium rule, which was considered the most appropriate method, with many of these giving the correct answer of 415m. While the table of values clearly shows an b a h n 5, h interval width the application of the formula with instead of 6 n 25. h n led some candidates to use an incorrect strip width Other candidates 5 6 made calculation errors, bracketing errors or transcription errors. Some candidates used the equations of motion with constant acceleration (i.e. the suvat equations) to estimate the length. Many of the candidates who used this method applied the calculation 2 42 (25) to obtain an estimate of 550m. A few used v u at 2 1 2 s ut at to find the acceleration, followed by to estimate the length. Some 2 u candidates used in their suvat calculations, in contradiction to the table of values 0 A few candidates used a method which was 1 2 ms . which indicated an initial speed of equivalent to the trapezium rule, of applying suvat to each of the 5 time-intervals, to give the correct answer of 415m. Other methods seen included: summing up the area of rectangles which enclosed the area above the curve for each of the 5 time intervals; summing up areas of rectangles which enclosed the area below the curve for each of the 5 time intervals; and applying (average speed) (total time). The most common methods which were deemed incorrect included 25 42 1050; 1 (25)(42) (speed)(time) (0)(2) (5)(5) (10)(10) ... (25)(42) 2005; 525; 2 or a 6 rectangle approach (instead of 5 rectangles). As for Q02(b), while many candidates who used a trapezium rule or suvat method in Q02(a) identified their estimate of the length as an overestimate, some found difficulty in articulating a reason for this. Those who were most successful drew a diagram which showed clearly the extra area generated by the trapezium rule in relation to the curve. Some candidates sketched the speed-time curve and explained that the curve was convex or that the acceleration was continually increasing. Only a few candidates who used a rectangle method were able to give a correct reason. Some candidates answered Q02(b) by finding an alternative estimate and comparing it with their estimate in Q02(a). For example, some found 415 in Q02(a) and 550 in Q02(b), and then concluded that ‘415 must be an underestimate’. Other candidates, who also received no credit, gave physical explanations about friction, air resistance or the shape of the aircraft, while others stated that ‘the jet accelerated smoothly’.

Common mistakes

• distance = area under the speed-time graph
• trapezium rule with h = 5 (not 25/6, 6 strips): 415 m
• suvat with constant acceleration (550) earns the method mark only
• 25 × 42 or a sum of speed × time scores nothing

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

Examiner's comment

Q02 proved to be challenging, with a significant number of candidates struggling to realise that the length of the runway could be estimated by the area under the speed- time curve generated from the table of values given in the question. In Q02(a), only a minority of candidates estimated the length of the runway by applying the trapezium rule, which was considered the most appropriate method, with many of these giving the correct answer of 415m. While the table of values clearly shows an b a h n 5, h interval width the application of the formula with instead of 6 n 25. h n led some candidates to use an incorrect strip width Other candidates 5 6 made calculation errors, bracketing errors or transcription errors. Some candidates used the equations of motion with constant acceleration (i.e. the suvat equations) to estimate the length. Many of the candidates who used this method applied the calculation 2 42 (25) to obtain an estimate of 550m. A few used v u at 2 1 2 s ut at to find the acceleration, followed by to estimate the length. Some 2 u candidates used in their suvat calculations, in contradiction to the table of values 0 A few candidates used a method which was 1 2 ms . which indicated an initial speed of equivalent to the trapezium rule, of applying suvat to each of the 5 time-intervals, to give the correct answer of 415m. Other methods seen included: summing up the area of rectangles which enclosed the area above the curve for each of the 5 time intervals; summing up areas of rectangles which enclosed the area below the curve for each of the 5 time intervals; and applying (average speed) (total time). The most common methods which were deemed incorrect included 25 42 1050; 1 (25)(42) (speed)(time) (0)(2) (5)(5) (10)(10) ... (25)(42) 2005; 525; 2 or a 6 rectangle approach (instead of 5 rectangles). As for Q02(b), while many candidates who used a trapezium rule or suvat method in Q02(a) identified their estimate of the length as an overestimate, some found difficulty in articulating a reason for this. Those who were most successful drew a diagram which showed clearly the extra area generated by the trapezium rule in relation to the curve. Some candidates sketched the speed-time curve and explained that the curve was convex or that the acceleration was continually increasing. Only a few candidates who used a rectangle method were able to give a correct reason. Some candidates answered Q02(b) by finding an alternative estimate and comparing it with their estimate in Q02(a). For example, some found 415 in Q02(a) and 550 in Q02(b), and then concluded that ‘415 must be an underestimate’. Other candidates, who also received no credit, gave physical explanations about friction, air resistance or the shape of the aircraft, while others stated that ‘the jet accelerated smoothly’.

Common mistakes

• overestimate: the curve is convex (acceleration increasing) so the tops of the trapezia lie above it
• a diagram showing the extra area is best
• comparing with another estimate, friction/air resistance or "accelerated smoothly" scores nothing

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

Examiner's comment

In general, Q11 discriminated well between candidates of all abilities. Q11(a) was demanding with most candidates scoring no more than one mark. Q11(b), Q11(c) and Q11(d) proved accessible to candidates of all abilities. In Q11(a), most candidates took logs, as advised by the question, and many obtained either ln ln y x x or log log . y x x At this point a significant number of candidates made no further creditable progress. Some candidates used a complete method of applying the product rule on x and implicit differentiation to obtain a correct ln x Most of these candidates then applied d y 1 d y x 1 y x e . x 0 to obtain Some 1 ln . x d d x log ’ method to obtain candidates, who received partial credit, used a ‘ after 0.1, 10 1 d y x finding A few candidates used logarithms without reference to a 1 log . x 10 d y x base. Those who proceeded to obtain received partial credit and those who 0.1 1 e x obtained were allowed full marks. There were some alternative methods employed by a few candidates in Q11(a). A few d y y x ln e x x x ln x x y x candidates rewrote as and applied to give a (1 ln )e 0 x d 1 e . x Others differentiated both sides of the equation ln ln y x x correct with 1 e . x respect to y and usually proceeded to find A very few candidates applied the ln y , x quotient rule to the equation but were rarely successful with this method. ln x xx x x xx or 2 1.6. In Q11(b), most candidates evaluated either at both and 1.5 Many candidates compared their values with either 2 or 0 (depending on whether they xx xx or 2 ). Although many of these concluded that 1.5 1.6, evaluated only a few made any reference to the curve being continuous in this interval. A few candidates, x who made no creditable progress, used values of x outside the given range, e.g. 1.4 x x 1.6. and Very few candidates used values within the range e.g. and 1.51 x 1.59, which meant that they were able to earn both marks. x 1.673, 4x In Q11(c), most candidates found a correct although a few did not state 4 , x x and 3x as part of their correct to 3 decimal places. Some wrote down the iterates 1 2 4x with no intermediate work. solution, while others stated In Q11(d), some candidates gave correct descriptions such as ‘sequence oscillates between 1 and 2’ or ‘the sequence is periodic with period 2’. Other descriptions such as ‘alternates between 1 and 2’, ‘keep getting 1 and 2’ or ‘fluctuates between 1 and 2’ were condoned. Some credit was given for partial explanations such as ‘sequence is non-convergent’, ‘sequence oscillates’ or ‘sequence is divergent’.

Common mistakes

• ln y = x ln x, implicit differentiation with product rule: (1/y)dy/dx = 1 + ln x
• set = 0 so x = e⁻¹ (0.368)
• using log₁₀ to get x = 0.1 max 3 marks
• no marks without working

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

Examiner's comment

Q13 discriminated well between the medium and higher ability candidates. Lower ability candidates, however, struggled to gain any creditable progress. It was obvious that some candidates had been well prepared for this ‘optimisation’ question and the quality of their responses reflected this. candidates who attempted this question generally found Q13(b) and Q13(c) accessible but some struggled to complete the proof in Q13(a). In a few cases, candidates who scored zero marks in Q13(a) went on to achieve full marks in the remainder of Q13 by using the given result for the surface area. There were, however, many instances where candidates who struggled with Q13(a) then gave up and did not attempt the remainder of the question. In Q13(a), successful candidates followed a correct strategy of forming an equation for the volume of the tank, rearranging their equation to give an expression for h in terms of r and substituting their expression for h into their formula for the surface area of the tank. Some candidates did not halve the given sphere formulae so that they could use these for the hemispherical shell. This led to errors in both their volume equation and surface area expression. Other candidates did not include the area of the circular base in their surface area expression. These errors led to an incorrect volume equation 4 2 3 2 and incorrect surface area expressions and 6 r h r 5 2 A r rh 3 2 In other cases, bracketing errors, manipulation errors, or using 2 2 . A r rh incorrect formulae for the curved surface area of a cylinder, prevented candidates from achieving the given result. In Q13(b), many candidates who attempted this part applied a complete method to find their value of r for which d A r r 0. Most successful candidates found metres, 1.05 d 3 18 r with a few giving an exact answer metres, which was condoned. Some 5 candidates differentiated A incorrectly to give d 10 A 12ln , r r while others used d 3 r 12 10 0. r incorrect algebra when solving the correct equation A few 2 3 r 2 d A 0. candidates used the incorrect method of finding a value of r for which 2 d r In Q13(c), most candidates who attempted this part correctly substituted their value of r, found from solving d A 12 5 r 2 0, , A r into the model with equation with d 3 r 2 17m . A few candidates, who many obtaining a correct minimum surface area of 2 d A 0, made no creditable progress, substituted their value for r, found from solving 2 d r into the model for A.

Common mistakes

• 6 = πr²h + ⅔πr³ (halve the sphere formulae)
• A = πr² + 2πrh + 2πr² (include the circular base)
• substitute h to reach the given A

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

Examiner's comment

Q13 discriminated well between the medium and higher ability candidates. Lower ability candidates, however, struggled to gain any creditable progress. It was obvious that some candidates had been well prepared for this ‘optimisation’ question and the quality of their responses reflected this. candidates who attempted this question generally found Q13(b) and Q13(c) accessible but some struggled to complete the proof in Q13(a). In a few cases, candidates who scored zero marks in Q13(a) went on to achieve full marks in the remainder of Q13 by using the given result for the surface area. There were, however, many instances where candidates who struggled with Q13(a) then gave up and did not attempt the remainder of the question. In Q13(a), successful candidates followed a correct strategy of forming an equation for the volume of the tank, rearranging their equation to give an expression for h in terms of r and substituting their expression for h into their formula for the surface area of the tank. Some candidates did not halve the given sphere formulae so that they could use these for the hemispherical shell. This led to errors in both their volume equation and surface area expression. Other candidates did not include the area of the circular base in their surface area expression. These errors led to an incorrect volume equation 4 2 3 2 and incorrect surface area expressions and 6 r h r 5 2 A r rh 3 2 In other cases, bracketing errors, manipulation errors, or using 2 2 . A r rh incorrect formulae for the curved surface area of a cylinder, prevented candidates from achieving the given result. In Q13(b), many candidates who attempted this part applied a complete method to find their value of r for which d A r r 0. Most successful candidates found metres, 1.05 d 3 18 r with a few giving an exact answer metres, which was condoned. Some 5 candidates differentiated A incorrectly to give d 10 A 12ln , r r while others used d 3 r 12 10 0. r incorrect algebra when solving the correct equation A few 2 3 r 2 d A 0. candidates used the incorrect method of finding a value of r for which 2 d r In Q13(c), most candidates who attempted this part correctly substituted their value of r, found from solving d A 12 5 r 2 0, , A r into the model with equation with d 3 r 2 17m . A few candidates, who many obtaining a correct minimum surface area of 2 d A 0, made no creditable progress, substituted their value for r, found from solving 2 d r into the model for A.

Common mistakes

• dA/dr = −12r⁻² + 10πr/3 = 0 gives r³ = 18/(5π), r = 1.05 m
• not d²A/dr² = 0
• 12/r does not integrate/differentiate to 12 ln r

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

Examiner's comment

Q13 discriminated well between the medium and higher ability candidates. Lower ability candidates, however, struggled to gain any creditable progress. It was obvious that some candidates had been well prepared for this ‘optimisation’ question and the quality of their responses reflected this. candidates who attempted this question generally found Q13(b) and Q13(c) accessible but some struggled to complete the proof in Q13(a). In a few cases, candidates who scored zero marks in Q13(a) went on to achieve full marks in the remainder of Q13 by using the given result for the surface area. There were, however, many instances where candidates who struggled with Q13(a) then gave up and did not attempt the remainder of the question. In Q13(a), successful candidates followed a correct strategy of forming an equation for the volume of the tank, rearranging their equation to give an expression for h in terms of r and substituting their expression for h into their formula for the surface area of the tank. Some candidates did not halve the given sphere formulae so that they could use these for the hemispherical shell. This led to errors in both their volume equation and surface area expression. Other candidates did not include the area of the circular base in their surface area expression. These errors led to an incorrect volume equation 4 2 3 2 and incorrect surface area expressions and 6 r h r 5 2 A r rh 3 2 In other cases, bracketing errors, manipulation errors, or using 2 2 . A r rh incorrect formulae for the curved surface area of a cylinder, prevented candidates from achieving the given result. In Q13(b), many candidates who attempted this part applied a complete method to find their value of r for which d A r r 0. Most successful candidates found metres, 1.05 d 3 18 r with a few giving an exact answer metres, which was condoned. Some 5 candidates differentiated A incorrectly to give d 10 A 12ln , r r while others used d 3 r 12 10 0. r incorrect algebra when solving the correct equation A few 2 3 r 2 d A 0. candidates used the incorrect method of finding a value of r for which 2 d r In Q13(c), most candidates who attempted this part correctly substituted their value of r, found from solving d A 12 5 r 2 0, , A r into the model with equation with d 3 r 2 17m . A few candidates, who many obtaining a correct minimum surface area of 2 d A 0, made no creditable progress, substituted their value for r, found from solving 2 d r into the model for A.

Common mistakes

• substitute r = 1.046 (from dA/dr = 0) into A: 17 m² to nearest integer

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

In the first part, the majority of candidates attempted to differentiate the given vector expression for velocity to find the acceleration as required. This was completed mostly successfully with just the occasional slip. Occasionally the expression was integrated or the fact that it was a vector just ignored with the coefficients of i and j being added; however, such instances were relatively rare. Those who completed the differentiation generally substituted t = 4 to obtain the acceleration and many correct answers were seen. Some went on to calculate the magnitude which was not actually required here but generally the marks had already been achieved for a correct vector. Some candidates worked in column vectors throughout and some in i and j components; both are equally acceptable. In part (b) integration in part (a).

Common mistakes

• a = dv/dt: differentiate both components (powers decrease by 1)
• do not integrate or add the i and j coefficients
• answer 6i − 15j as a vector (finding the magnitude is not needed)

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

Examiner's comment

Part (a) of this question allowed the demonstration of good skills in differentiation and was well answered by the majority of students. A small minority made careless errors on the indices, for example writing x as 2 x , leading to the loss of both marks as their derivative was not of the required form. Others dealt with the index correctly, but made arithmetic errors on the coefficient when differentiating the term. The second derivative was almost always either correct, or a correct follow through, from the students first derivative.

Common mistakes

• write 24√x as 24x^½ before differentiating
• dy/dx = 2x − 2 − 12x^(−½)
• second derivative 2 + 6x^(−3/2)
• index and coefficient slips

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

Examiner's comment

In part (b) a significant number of students were seen attempting to solve the equation dy/dx = 0 although most quickly realised that they could substitute x = 4 into dy/dx to obtain 0. To be awarded the final mark of part (b), students were required to interpret their value of dy/dx at x=4 and conclude that the point was a stationary point of the curve. Many students omitted this final step thus failing to fully answer the question.

Common mistakes

• substitute x = 4 into dy/dx (no need to solve dy/dx = 0)
• must show dy/dx = 0 and state "hence stationary point" – conclusion often omitted

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

Examiner's comment

For part (c) there was a little more variation. Most students were aware of the need to substitute 4 x into their second derivative and evaluate its sign, although some failed to do this and instead considered the sign of either y, dy/dx and/or d2y/dx2 either side of 4 x (usually at 4.1 and 3.9). A follow through mark enabled students with an incorrect negative or fractional index from part (a) to achieve both marks in the last part of the question. A significant proportion of students seemed unable to interpret what they had proven from their calculations, and many seemed to think they were looking for an inflection point, or else determining that the curve was either convex or concave. As with part (b), the most common source of lost marks was therefore a lack of correct reasoning before making a conclusion.

Common mistakes

• substitute x = 4 into the second derivative: 2.75 > 0 hence minimum
• need calculation, reason and conclusion
• not a point of inflection or convexity question

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

Examiner's comment

This proved to be a challenge and a large number of students did not recognise that it was a differentiation problem until asked to find the second derivative in (b). In (a), the common error was to set C = 0 and solve the resulting quadratic equation. A few tried to start with 1500/v + 2v/11 =0. Some used trial and improvement to find the minimum cost, even arriving at v = 91 and C=93; this achieved no marks as the question did state that ‘Solutions based entirely on graphical or numerical methods are not acceptable’. Many of those who did differentiate struggled with the negative index. In (a)(ii), students usually recognised the need to substitute their value of v into the original equation. Some obtained unrealistic values from (i) and did not consider their answer in the context of the question. (e.g. v = 0.06 km/h). It was not uncommon to find this part omitted even by good students; perhaps they had not read the question carefully enough.

Common mistakes

• differentiate: dC/dv = −1500/v² + 2/11
• set = 0: v² = 8250, v = 90.8 km/h
• minimum cost awrt £93
• setting C = 0 is wrong
• trial and improvement scores 0
• errors with the negative index

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

Examiner's comment

This proved to be a challenge and a large number of students did not recognise that it was a differentiation problem until asked to find the second derivative in (b). In (b), it was disappointing to see so many students successfully find the second derivative but gain no marks as they failed to find a value for it at their v or establish that it would be positive.

Common mistakes

• d²C/dv² = 3000/v³
• must evaluate it (≈ +0.004) or argue its sign at their v and conclude > 0 so minimum
• finding d²C/dv² without a value or sign scores nothing
• correct notation needed for the A mark

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

Examiner's comment

This proved to be a challenge and a large number of students did not recognise that it was a differentiation problem until asked to find the second derivative in (b). Part (c) was sometimes the only mark collected in this question by less able students. Some simply re- stated the model’s assumption of a constant speed without reference to this not being possible. Others gave vague or irrelevant statements such as “there might be different routes”.

Common mistakes

• the lorry could not travel at a constant speed over the whole journey (traffic, stops)
• just restating the constant-speed assumption, or vague comments such as different routes, score 0

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 north is where dy/dx = 0: solve y = x with the curve and take the positive solution
• "put x = 0 or y = 0" is wrong
• rarely scored

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

Examiner's comment

This is a new topic on the specification. There were some very good solutions, showing full understanding of limits with all necessary steps shown. However, several students did not understand what was meant by ‘prove from first principles’ and simply stated and applied the general rule: if y=xn, dy/dx= nxn-1 or wrote comments such as ‘multiply by the power and subtract one from it’. Those who knew what was required usually gained the first two marks but poor algebra often cost the third (and hence the fourth) mark. The final mark was frequently lost by failing to link the gradient of the chord and the gradient of the curve in some way.

Common mistakes

• consider ((x + h)³ − x³)/h
• expand (x + h)³ correctly
• gradient of chord 3x² + 3xh + h²
• state h → 0 and link the gradient of the chord to the derivative
• quoting dy/dx = nxⁿ⁻¹ scores 0

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

Examiner's comment

Some students were clearly running out of time when they reached this question, as there were many unfinished answers and some blanks. However, there were a great many accurate, concise and well written solutions. Few attempts at the line minus the curve approach were seen. This was an unstructured question in which the students were required to find their own route through the problem. There were two discrete elements to the solution, one via differentiation and one via integration that were required to be tied together to show the given answer. A good number completed the differentiation part correctly. There were slips in the differentiation of 2 x , but generally the method of finding the equation of a normal was well known. An equal number completed the integration part finding the area under the curve between 2 and 4. Similar slips were made on the integration of 2 x . Tying up these two techniques into finding the area of R was less well done. Standard errors seen here involved: (1) attempting the area of a triangle between 2 and 16 (2) integrating under the curve from 2 to 16 (3) finding the equation of the tangent to the curve at P.

Common mistakes

• dy/dx = −64/x³ + 3 gives gradient 2 at P, so the normal has gradient −½ and meets the x-axis at x = 16
• area under the curve from 2 to 4 = 10
• triangle ½ × 12 × 6 = 36
• total 46
• slips differentiating and integrating 32/x²
• many ran out of time

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

Examiner's comment

This was a well-answered question with part (a) and part (c) providing most of the discrimination. In part (a), most students substituted a defined 3 2 f( ) 2 1 n n n x x x and a correct 2 f ( ) 6 2 n n n x x x into the Newton-Raphson formula 1 f( ) . f ( ) n n n n x x x x Many students used correct algebra to arrive at the given result. Some students moved directly from 3 2 1 2 2 1 6 2 n n n n n n x x x x x x to 3 2 1 2 4 1 6 2 n n n n n x x x x x without showing the correct intermediate step of rationalising the denominator. There were bracketing errors, sign errors and the error of omitting ‘ 1 nx ’ to write only the right-hand side of the Newton-Raphson formula.

Common mistakes

• f′(x) = 6x² + 2x
• common-denominator intermediate step must be shown before the given answer
• keep the x_n term
• bracket/sign errors

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

Examiner's comment

In part (c), only a few students gave an acceptable explanation that detailed why the Newton- Raphson method did not work for this question. Some of these students stated ‘because there is a stationary point at 0, x then the method could not work’ while others explained that a tangent to the curve at 0 x would never meet the x-axis. The explanation, ‘not being able to divide by zero in the formula’, by itself, was not sufficient to gain credit in part (c).

Common mistakes

• stationary point at x = 0 so the tangent is horizontal and never meets the x-axis
• "cannot divide by zero" or "f′(0) = 0" alone scores B0

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

Examiner's comment

In part (a), many students realised they had to find the velocity to determine when the particle is at instantaneous rest and attempted to differentiate the given expression for x. Some failed to multiply out the brackets first and made no valid progress. Nevertheless, a fair number derived the relevant cubic and solved it successfully to find the 3 values for t. On occasions, t was cancelled out from the equation and the solution t = 0 was lost.

Common mistakes

• multiply out first: x = ½(t⁴ − 2t³ + t²), v = 2t³ − 3t² + t
• v = 0 gives t(2t − 1)(t − 1) = 0 so t = 0, ½, 1
• cancelling t loses t = 0

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

Examiner's comment

This was a well-answered question considering that this was the first time that differentiation from first principles has been tested on the new specification. Most students wrote down a correct expression for the gradient of the chord joining the points ( , cos ) and ( , cos( )) h h for a small value h. Many students simplified their gradient expression to give a correct sin cos 1 sin cos h h h h and applied the limiting arguments given in the question to deduce that d cos sin . d Some students lost the final mark for not applying limiting arguments correctly. Common errors included: stating that sin cos 1 1, 0 h h h h ; not using a limiting argument; or applying the limiting argument inconsistently. Other students lost the final mark for the use of inconsistent notation in their proof. These students, who usually referred to both x and in their proof, wrote incorrect statements such as d cos sin dx or f( ) cos . x

Common mistakes

• gradient of chord (cos(θ + h) − cos θ)/h
• compound angle formula
• isolate (sin h)/h and (cos h − 1)/h
• limit h → 0 must be applied correctly and notation kept consistent (θ not x)

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

Examiner's comment

In part (b), a minority of students struggled to make any creditable progress. These students usually evaluated f ( )x for increasing x-values and concluded that the corresponding values for f ( )x were continually decreasing. The majority used the chain rule to differentiate f( ). x A common error at this stage was for students to differentiate 1 2(1 2 )x to give 2 2(1 2 ) , x while other students tried to apply an incorrect method of showing f ( ) 0. x Only a few of the students who obtained a correct 2 2 f ( ) 4( 3) 4(1 2 ) x x x provided an acceptable explanation detailing why f ( ) 0. x

Common mistakes

• use the partial fractions: f(x) = 3 + 4(x − 3)⁻¹ − 2(1 − 2x)⁻¹
• chain rule gives −4(1 − 2x)⁻² (common error: factor −2 missed)
• explain both squares positive so f′(x) < 0
• evaluating f at increasing x scores nothing

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

Examiner's comment

This question, despite being unstructured, proved accessible, with many students scoring at least 7 out of the 10 marks available. Those students who annotated the diagram were generally more successful in formulating a correct strategy to find the area of the region R. Most students differentiated ln y x x by using the product rule to give the correct d 1 ln . d y x x A few students found the correct d ln d y x x x x and simplified this incorrectly to give d ln . d y x x Many students used a complete method to find the x-value where the normal to C at (e, e) cuts through the x-axis, with the majority finding the correct 3e. x Most students integrated ln y x x correctly by using integration by parts to give 2 2 1 1 ln . 2 4 x x x Many students applied the correct limits of e and 1 to obtain a correct 2 1 1 e 4 4 with some making a bracketing error when applying the limit of 1 to obtain the incorrect 2 1 1 e . 4 4 The majority of students attempted to find the area of the triangle by using the formula 1 (base)(height). 2 While some students applied the formula correctly to give 1 (3e e)(e), 2 there was a considerable number that applied a base length of 3e by writing 1 (3e)(e). 2 A few students used the equation of the normal to find the area of the triangle by applying the correct 3e e 1 3 e d , 2 2 x x but not many evaluated this to arrive at the correct answer of 2e . Several incorrect strategies were seen such as trying to evaluate e 0 ln d , x x x 3e 1 ln d x x x or even 3e 1 1 3 ln e d . 2 2 x x x x Only a minority of students applied the calculation 2 2 1 1 e e 4 4 to achieve the correct 2 5 1 e . 4 4

Common mistakes

• product rule dy/dx = 1 + ln x
• normal gradient −½ meets x-axis at 3e
• ∫ x ln x dx by parts = ½x² ln x − ¼x² with limits 1 and e (bracketing error at x = 1)
• triangle base 3e − e not 3e
• area = 5e²/4 + 1/4

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 (b), most students used either the chain rule or the quotient rule in their attempt to find an expression for d . d N t There were many incorrect attempts at the chain rule such as 0.25 0.25 2 d 6300e (3 7e ) d t t N t or 0.25 2 d 1575(3 7e ) d t N t while the most common incorrect attempt at the quotient rule was 0.25 0.25 0.25 2 d (3 7e )(1) 1575e . d (3 7e ) t t t N t Although some impressive algebra was seen in manipulating a correct 0.25 0.25 2 d 1575e d (3 7e ) t t N t to achieve d (300 ) d 1200 N N N t , many students were unsuccessful in their attempts at eliminating t or the 0.25 e t term. Some students substituted 0.25 900 3 7e t N into d (300 ) d 1200 N N N t and used algebra to give 0.25 0.25 2 d 1575e d (3 7e ) t t N t , and provided some indication that their proof was complete. Some students differentiated N by using implicit differentiation while other students attempted to rearrange the given equation for N to make t the subject before finding an expression for d . d t N A few students were successful in applying an alternative method of solving the differential equation d (300 ) d 1200 N N N t to give 0.25 900 . 3 7e t N

Common mistakes

• chain rule on 900(3 + 7e^(−0.25t))⁻¹ (common errors in the chain or quotient rule)
• eliminate e^(−0.25t) using N
• alternatively substitute N into RHS and show LHS = RHS with a conclusion

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