P7.2 Differentiating standard functions

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

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

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

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

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

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

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, 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

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

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) 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 (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

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

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

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

(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

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

- 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

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 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

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) 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

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

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

Common mistakes

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

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

Examiner's comment

This 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, 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

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

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

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

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

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

(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

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 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 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 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

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

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

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

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

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

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

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

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

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

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

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

This question 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

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

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

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

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

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

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

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

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

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