P3.1 Straight lines

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

Examiner's comment

Part (a) was generally answered well by the majority of candidates. Most were able to correctly calculate the gradient and went on to correctly form the equation of the line. The most commonly used method was the point-gradient form ( ) 1 1 y y m x x - = - which was usually applied successfully to arrive at the correct equation 4 12 y x = + . Others used the form y = mx + c, substituting coordinates to find c. Common errors included: • an incorrect gradient method of 2 1 2 1 x x y y - - • sign errors when substituting coordinates • slips when expanding brackets or rearranging • finding the equation of a perpendicular line Despite these errors, most candidates achieved full marks for this part.

Common mistakes

• gradient (22 − 0)/(2.5 − (−3)) = 4, then y − 0 = 4(x + 3)
• y = 4x + 12
• gradient upside down (x over y)
• sign slips
• finding a perpendicular line

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

Examiner's comment

The performance in part (b) was much more varied. Whilst many candidates who correctly completed part (a) were able to identify and write the inequalities describing the bounded region, e.g. 2 12 y x < + and 𝑦𝑦⩾2𝑥𝑥2 + 5𝑥𝑥+ 3, a significant number misunderstood the question and presumably seeing the shaded area, attempted to find the area between the line and the curve. Of those who interpreted the question correctly, as in previous series, there were a large number of candidates who defined the region using R instead of expressing inequalities in terms of y. This usually resulted in the loss of both marks. Of those who did attempt inequalities, common errors included • writing inequalities the wrong way round or using inconsistent inequality symbols • using set notation incorrectly • defining the region purely in terms of x e.g. 5 3 2 x -< < This part of the question highlighted a significant gap in understanding how to describe regions using inequalities.

Common mistakes

• y < 4x + 12 (dashed line, strict) and y ≥ 2x² + 5x − 3 (solid curve)
• inequalities in terms of y, not R
• "or" or ∪ loses the second mark
• finding the area instead
• −3 < x < 5/2 alone scores nothing

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

Examiner's comment

For part (b) many students misinterpreted the information in the first bullet point and used the incorrect equation 3.5 = a + 500b together with the correct one of 20 = a + 1700b. Those who had interpreted the information correctly generally used it in conjunction with the second bullet point to obtain 16.5 = a + 2200b as their equation together with 20 = a + 1700b ∓3.5 although a few of those using this approach used 1200b instead of 2200b. 500 were more successful although some of Those who used the gradient approach with b = these did not pick up that the gradient was negative. There were some students who had the gradient upside down.

Common mistakes

• gradient b = −3.5/500 = −0.007 (negative, not upside down)
• 3.5 = a + 500b is wrong (common misreading of the first bullet)
• alternatively 16.5 = a + 2200b with 20 = a + 1700b

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

Examiner's comment

In part (a), it appeared that many students did not recognise the sum given as that used in the Common errors were sqrt 394 and 394 original definition of variance/standard deviation and so were unable to attempt an answer. 25 although other wrong solutions involving extensive calculations were also seen. For part (b) many students misinterpreted the information in the first bullet point and used the incorrect equation 3.5 = a + 500b together with the correct one of 20 = a + 1700b. Those who had interpreted the information correctly generally used it in conjunction with the second bullet point to obtain 16.5 = a + 2200b as their equation together with 20 = a + 1700b ∓3.5 although a few of those using this approach used 1200b instead of 2200b. 500 were more successful although some of Those who used the gradient approach with b = these did not pick up that the gradient was negative. There were some students who had the gradient upside down.

Common mistakes

• substitute their b into 20 = a + 1700b to get a = 31.9
• 1200b used instead of 2200b in the simultaneous-equation approach

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

Examiner's comment

This modelling question required learners to construct and interpret a linear demand model, derive a profit function, solve a quadratic inequality, identify the optimal selling price, and assess the suitability of a model against real data. Overall, performance was mixed across the different parts of the question. Part (a) proved to be the most accessible part of the question and was generally well handled demonstrating a sound understanding of linear modelling. The majority were able to formulate an appropriate equation for the model in the form y = mx + c, correctly identifying the constants m and c. Only a small number of responses encountered difficulties, primarily due to confusion between x and y coordinates when calculating the gradient.

Common mistakes

• gradient (300 − 1500)/(50 − 30) = −60, not with x and y swapped
• y = 3300 − 60x

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

Examiner's comment

This question, assessing understanding of modulus functions, composite functions and the range, was accessible to almost all learners. Generally, responses scored well on the first two parts but did not find a simple strategy for the last part. Learners found part (c) difficult, and many scored few marks as a result. This part is not dissimilar to questions from previous series but the idea of there being no solutions clearly caused greater difficulties for some. Those who were most successful approached the question thinking graphically, drawing on the given graph a line from the origin to the vertex, finding the gradient, then realising the gradient was between this value and –4 (inclusive of –4). Often the strict inequality for –4 was given and did not gain the final mark, but on the whole this approach was quick and successful. Other responses showed one of the more difficult routes to finding the upper limit for k and invariably had processing errors at some stage. Finally, many responses considered that a discriminant had to be involved because the question mentioned there being “no solutions”, however for most this ended up with them trying to find the discriminant of expressions that were not quadratic.

Common mistakes

• lines y = kx through O: parallel to the left branch at k = −4 (no solutions, included) up to the line through the vertex (3, −5), k = −5/3
• −4 ≤ k < −5/3 (strict −4 loses the final mark)
• discriminant of a linear equation is invalid

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

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

This question on the modulus function provided plenty of access and there were very few blank responses to any part, although many found part (c) challenging. Part (c) was found to be less accessible with few candidates scoring full marks for a correct range for k. The candidates that were the most successful in this part of the question were generally those who had adopted a visual approach in the earlier parts of the question and understood how to find the minimum and maximum value of k using their knowledge of gradients. There was more success in finding the correct lower critical value of 0.5 with the most efficient solutions using the coordinates of P with y = kx + 4, but many candidates failed to realise that there was an upper limit for k. There was a reluctance to use diagrammatic reasoning and most resorted to complicated algebra. Many set up simultaneous equations in x and k but did not appear to understand what they were attempting to achieve, failing to solve for k and therefore these attempts did not gain any marks. There were many erroneous attempts that fortuitously achieved k = 3 through an attempt to use a discriminant (despite starting with a non-quadratic equation), and these invalid approaches scored no marks on their own. A few attempted squaring both sides of the equation to form a quadratic equation and then using the discriminant, but this approach often failed due to not isolating an appropriate term (such as 2 x - ) before squaring. Relatively few candidates considered the gradient of the modulus graph and hence failed to obtain an upper bound for k, which meant many only achieved the method mark in this part. Overall, candidates' use of inequality notation was good and although set notation was not required, the vast majority of those that used it did so correctly.

Common mistakes

• line through (0, 4): critical gradients through P (k = ½) and parallel to the right branch (k = 3)
• many miss the upper limit
• ½ < k < 3 joined with "and"/∩, not "or"
• discriminant on a non-quadratic is invalid

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 (c). The final part of the question proved more challenging and differentiated between the candidates, with many producing concise and fully correct answers. Most candidates realised it required the use of integration and often scored the B mark for integrating the curve. Many candidates failed to note that the question required an exact answer and use of algebraic integration. This meant that using a calculator to provide a decimal approximation was insufficient for the final method mark. The quality of notation in this part of the question was very varied with the dx often missing from integrals and spurious integral signs included throughout working. 1 = The majority of successful candidates used or integration to find the area of A bh 2 the triangle and then subtracted the area under the curve. However, a significant proportion of candidates made the question much more difficult than necessary, using a mixture of Pythagoras and trigonometry to find the area of the triangle-this often led to errors or premature rounding. Other common errors when finding the area of the triangle included assuming the triangle was isosceles, and therefore that the intersection x = point was at , or splitting the triangle into two right-angled triangles and assuming 2 they both had the same area. Candidates attempting to use the alternative method using lines – curve were more likely to make errors with signs, brackets or limits. They were also required to correctly 2 2.4 7 ( ) cancel the terms to find an exact answer and so most often were unable to 2 7 score the final method and accuracy marks. The most common invalid method was attempting to combine the lines and curve before integrating using the limits 0 and 4, and this scored no marks unless the point of intersection had been found.

Common mistakes

• l1 and l2 meet at (2.4, 19.2)
• triangle 38.4 minus ∫₀⁴ (8x − x^(5/2)) dx = 192/7
• exact 384/35 (calculator decimal loses marks)
• triangle is not isosceles
• missing dx

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

Examiner's comment

Most candidates were successful in this question, showing that parametric differentiation was overall well understood by candidates. In part (a), many candidates successfully found the given equation of the tangent using parametric differentiation as specified in the question. By far, the most common approach seen involved finding the value of t and then substituting it into an expression for d d y x obtained by using d d d d d d y y t x t x = and then using standard coordinate geometry to find the equation of the tangent. There were a number of careless slips and algebraic errors, with 2(t + 3) = 2t + 3 being disappointingly common. Once the numerical gradient had been obtained, there were usually no issues in finding the equation of the tangent in the required form, showing at least one intermediate step as this was a given answer. Most candidates achieved the equation of the tangent via the Point-Slope form although a large number went via the Slope-Intercept form. A few candidates made coefficient and/or sign errors when manipulating their equation to obtain the given answer and unfortunately lost the final accuracy mark. Those candidates who gained few, if any, marks in this question often failed to see the need to find the value of the parameter ‘t’ for the given point P (4, 2) and were unsuccessful in using the alternative method to continue to find a value for d d y x . Some candidates used an incorrect value of ‘t’, usually t = -5 rather than t = -1, as they either did not note the restrictions for t given in the question, or did not find the t value for the y coordinate of P. A few gave two equations for each value of t found. The attempted use of the Chain Rule was not always completed accurately. Some candidates converted their gradient from the use of the Chain Rule into the gradient of the normal which resulted in an inappropriate equation. When finding the value of d d y x , some candidates substituted 4 for their value of t, confusing it with the x coordinate of point P. Some candidates did not follow the instructions stating that the question had to be answered ‘using parametric differentiation’ and, consequently, scored no marks even though they achieved the required result by obtaining the Cartesian equation of the curve. Some candidates tried to complete the question by unnecessarily converting their d d d d y t t x in terms of t to be in terms of x and/or y which often led to incorrect solutions due to algebraic and/or differentiation errors. There were many candidates who initially went wrong and restarted the question. A number of these ran out of space and incorrectly used the back pages of their answer booklet to continue this question rather than using supplementary paper.

Common mistakes

• find t = −1 from P(4, 2) (t = −5 is outside −2 ≤ t ≤ 1)
• dy/dx = −3t²/(2(t + 3)) = −3/4
• must use parametric differentiation (Cartesian scores only B1)
• 2(t + 3) ≠ 2t + 3
• tangent not normal
• show an intermediate step to 3x + 4y = 20

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

Most candidates attempted this question, with only a small number leaving it blank. Most candidates In part (a), candidates who set up a linear model correctly tended to go on to get the correct linear model required and performed well in the rest of the question. Those who were successful tended to use the ( ) - = - gradient and rather than attempting to use simultaneous equations. The most common y y m x x 1 1 error was to use the coordinates the wrong way round, which resulted in candidates’ incorrectly calculating 1 - the gradient as –8 instead of . However, a good proportion of candidates recovered by using the gradient 8 of –8 to achieve a formula in terms of V (i.e., d = 400 – 8V or equivalent). There were a few candidates who did not include the ‘+c’ part of a linear model and instead seem to think the relationship was proportional. These candidates were often limited to one mark in this question for correctly identifying the link of 1 litre = 8km. Where a linear equation was not recognised an exponential was the most common alternative. Candidates who achieved a formula in part (a) generally continued onto part (b) and used their formula to attempt the question correctly. A common error was to omit the units, which cost just the final mark in part

Common mistakes

• linear model V = ad + b (not proportional, not exponential)
• gradient −1/8 (using coordinates the wrong way round gives −8)
• V = 50 − d/8 or d = 400 − 8V
• answer must be in V and d

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

Examiner's comment

Most candidates attempted this question, with only a small number leaving it blank. Most candidates Candidates who achieved a formula in part (a) generally continued onto part (b) and used their formula to attempt the question correctly. A common error was to omit the units, which cost just the final mark in part (b). Of the candidates who had V and d the wrong way round in their linear model, none seemed to realise that their answers of 400L and 50km were unrealistic. In part b(ii) a small number of candidates used their answer to part (i) 50 as the new value for V, rather than set V = 0 and rearrange. Very few candidates achieved the mark for part (c). Most candidates recognised it was not acceptable but

Common mistakes

• (i) set d = 0: 50 litres
• (ii) set V = 0: 400 km
• units needed for the final mark
• using 50 as the new V in (ii)
• unrealistic answers such as 400 litres not noticed

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

Examiner's comment

Most candidates attempted this question, with only a small number leaving it blank. Most candidates achieved at least one mark overall, but few candidates achieved full marks due to part (c). Very few candidates achieved the mark for part (c). Most candidates recognised it was not acceptable but failed to comment on the significance of the difference. Centres must be clear with candidates that models do not have to be perfect to be appropriate, and so it is only if the model is significantly out that it becomes unviable to use. A small number of candidates calculated percentage difference, which was acceptable. Some candidates focussed incorrectly on practical reasons (e.g., different routes, fuel consumption, more accelerating and breaking, etc.). A small number of candidates worked out the remaining fuel (10L) but often did not correctly conclude the significance of this amount.

Common mistakes

• must say the model is poor because 320 km is significantly less than 400 km (or 10 litres would remain)
• just 320 ≠ 400 is not enough
• practical reasons (routes, braking) score nothing

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

This question on straight-line coordinate geometry was well attempted by the vast majority of candidates and there was a good proportion of candidates scoring full marks. This “show that” question in part (a) was generally very well done, with clear work to find the negative reciprocal gradient and then form a correct equation. The majority were able to manipulate this to the given form without errors, but the occasional slip in their processing resulted in the loss of the final accuracy mark. Those who took the alternative approach, which was essentially a verification process, candidates generally showed that the given line was perpendicular to the original line but very few verified that it went through the required point and hence only scored 1 mark out of 3.

Common mistakes

• perpendicular gradient −5/3 (negative reciprocal)
• use B(8, 0)
• at least one intermediate line before 5x + 3y = 40
• verification must show both perpendicular and passing through (8, 0)

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

Examiner's comment

This question on straight-line coordinate geometry was well attempted by the vast majority of candidates and there was a good proportion of candidates scoring full marks. Part (b) was also very well attempted, with the vast majority of candidates obtaining the correct coordinates for A and C. However, many failed to recognise that the base of the triangle was 18 and the height was simply the y coordinate of C and instead focused on the fact that AC and BC were perpendicular and used Pythagoras to find the lengths of AC and BC. Sometimes they lost the accuracy with this rather laborious method by using decimal answers, although some very accurate candidates were able to proceed using this method to the exact answer. Quite often candidates only obtained the x coordinate of C and were thus unable to proceed further. Very few attempts were seen that used discriminant approaches to find the required area.

Common mistakes

• A(−10, 0) from y = 0 on l1
• solve simultaneously for C, y = 135/17
• area = ½ × base 18 × height 135/17 = 1215/17
• using Pythagoras for AC and BC is laborious and decimals lose accuracy
• exact fraction required

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

Examiner's comment

Given that candidates often find vector problems quite challenging, this question proved to be quite accessible, and most candidates made good progress through parts (a) and (b) of this question, with part (c) more mixed. Many candidates did not successfully reach p = 5. Typical errors in (c)(i) were assuming AB = BC , setting the sum of the position vectors of A, B and C equal to zero or assuming that ( ) 18 12 2 k p - + = - + i j i j . A significant number of candidates omitted (c)(ii) after an unsuccessful attempt at (c)(i), and when responses were seen they were often lengthy and involved unnecessary attempts at calculating the areas of the triangles AOC and AOB. A common incorrect answer was 4:9, reached after assuming incorrectly that the two triangles were similar. Some candidates did, however, successfully reach a ratio of 2:3 either after area calculations, or by, for example, examining the relative lengths of AB and AC. Part (ii) did not rely on an answer to part (i) and so a reasonable number of candidates scored the final mark having been unable to find p, with some doing so very concisely. It was notable that many of the concise solutions to both parts came from a simple, well-constructed and clear diagram.

Common mistakes

• (i) AB parallel to BC (or gradient/line AB) gives p = 5
• wrong assumptions: AB = BC, position vectors summing to zero
• (ii) triangles share height from O so ratio = AC : AB = 2 : 3 (4 : 9 from assuming similarity is wrong)
• a clear diagram helps

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

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

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

This question was, on the whole, well-attempted and provided some differentiation between candidates. Most candidates made an attempt at part (b), and many were able to undertake some integration correctly. There were sometimes slips on the fractional powers and the coefficients. This question was effective at differentiating between those candidates who were able to clearly identify a strategy to determine the required area and many were successful in doing so. Most who were successful, employed the main mark scheme approach of determining the area under the curve between x = 0 and x = 4 followed by subtracting the area of the triangle however there were a number of different approaches seen, including finding the area under the curve from x = 0 to x = 2 together with the area between the curve and line between x = 2 and x = 4. Errors were perhaps a little more common in the second of these two approaches perhaps due sign errors when subtracting the two equations. Unfortunately, though, a number of candidates were not clear on the strategy required and so it was not uncommon to see attempts which stopped after calculating the area under the curve between x = 0 and x = 4 and thus a value of 76 9 was a fairly common incorrect final answer. Others using one of the alternative approaches simply found the integral of the difference between the two equations between x = 0 and x = 4 or between x = 0 and x = 2 and failed to go on the find the area under the curve separately. Most candidates had few issues identifying the intercept of the tangent with the x-axis but there was sometimes confusion in identifying the base and/or the height of the triangle when this was required to find the area. Some candidates lost the final A mark in this question by not giving the exact answer, giving it as a rounded decimal instead.

Common mistakes

• area under C from 0 to 4 (76/9) minus the triangle from x = 2 to 4 with height 13/3
• stopping at 76/9 is common
• confusing the y-coordinate 13/3 with the gradient 13/6
• exact answer 37/9

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

Examiner's comment

This question often enabled candidates to score marks in the first part, however, part (b) proved to be very good at discriminating between the most able candidates. Part (b) was challenging for many; non attempts or attempts which achieved no marks were common. Quite a few candidates scored the B mark for the gradient of the perpendicular and some went on to find the equation of the normal, but they did not find the point of intersection of the straight lines and so could not progress to the remaining marks. It was common to see one of the straight-line equations substituted into the equation of the circle. Those who substituted the normal equation and had also found the point of intersection of the straight lines, often correctly found the distance between the points correctly and gained full marks. Those who found the distance between the centre and the point of intersection of the straight lines, often forgot to subtract 3 for the radius, thus losing the last two marks. A large number of candidates found the y intercept of the line l, and then the distance between (0,3) and the centre. A very small number of candidates used a vector method.

Common mistakes

• the normal to l through the centre has gradient 2
• intersect with l at (6/5, −18/5)
• distance from the centre minus the radius 3 (often forgotten)
• using the y-intercept (0, −3) is wrong
• (19√5 − 15)/5 ≈ 5.50

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

Examiner's comment

This question proved to be a good differentiator between candidates. It provided a challenge for many candidates but equally there were a number of accessible marks here. Most candidates attempted part (a), many attempted part (b) and at least partially attempted part (c), usually attempting to find the lower limit for k. There were, however, very few completely correct answers to this question and a substantial minority made no attempt at any part, perhaps having run out of time by the time they reached the last question on the paper. Part (a) was mostly approached correctly with a good number scoring full marks. Candidates generally noticed the instruction to use parametric differentiation. Whilst d𝑥𝑥 d𝑡𝑡 was most often correct, d𝑦𝑦 d𝑡𝑡 proved a challenge for many who struggled to differentiate sec2𝑡𝑡. Many did not seem to know that these functions were standard derivatives presented in the formula booklet and instead there were some long-winded and overly complex attempts to convert to sine or cosine functions and such approaches often led to errors. However, even some of the most inefficient approaches managed to persevere through lengthy algebra to obtain a correct d𝑦𝑦 d𝑥𝑥 eventually. Almost all who attempted d𝑥𝑥 d𝑡𝑡 and d𝑦𝑦 d𝑡𝑡 knew and attempted d𝑦𝑦 d𝑥𝑥= d𝑦𝑦 d𝑡𝑡 ÷ d𝑥𝑥 d𝑡𝑡. Those who made errors when determining d𝑦𝑦 d𝑥𝑥 often exhibited an over-reliance on trying to match the printed result and thus ‘adjusted’ their d𝑦𝑦 d𝑥𝑥 in an attempt to obtain the required value for the normal gradient. Often this meant the loss of a method mark. Candidates would be better advised to utilise a correct method rather than incorrectly manipulating their answers to match the result they are attempting to achieve and caution should always be advised when working backwards from a given answer. Nonetheless, application of negative reciprocals and equations of lines was generally successfully attempted, and candidates were clearly confident in these areas. There were a few attempts at a Cartesian approach in part (a) which gained zero marks as the question had specified the use of parametric differentiation.

Common mistakes

• must use parametric differentiation
• dy/dt = 4sec²t tan t (sec²t is a standard result), dy/dx = 2tan t = 2 at t = π/4
• P(3, 7)
• normal gradient −½
• do not fudge dy/dx to fit the given answer

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

Examiner's comment

This question proved to be a good differentiator between candidates. It provided a challenge for many candidates but equally there were a number of accessible marks here. Most candidates attempted part (a), many attempted part (b) and at least partially attempted part (c), usually attempting to find the lower limit for k. There were, however, very few completely correct answers to this question and a substantial minority made no attempt at any part, perhaps having run out of time by the time they reached the last question on the paper. Part (c) provided a particular challenge and usually only the lower limit was attempted. Most candidates appeared to completely miss the need to find an upper limit. Of those who did manage to find both limits, only a small minority of candidates got to the final solution with correct inequality interpretation. It is worth noting that candidates achieving full marks for this part very often had clear construction lines on their graph that enabled them to see the correct approach. Most often though, candidates gained the method mark for the lower limit, 43 8 , for identifying that two intersections were required and set up a correct inequality for the discriminant for the quadratic equation obtained by equating the equations of the line and the curve. Common errors in manipulation, particularly of the coefficient of ½ from the quadratic meant that a number of candidates did not earn the accuracy mark for 43/8. Many candidates stopped after obtaining a lower limit for k, but others explored the parameter limits at 𝑡𝑡= - 𝜋𝜋 4 and 𝑡𝑡 = 𝜋𝜋 3 to identify 𝑘𝑘 values of 13 2 and 23 2 - sqrt 3 for the line passing through the associated points. Very few, however, recognised 13 2 as an upper limit and a number of candidates assumed incorrectly that 13 2 < 𝑘𝑘 < 23 2 - sqrt 3. A minority of candidates incorrectly thought that setting d𝑦𝑦 d𝑥𝑥 = 0 or d𝑦𝑦 d𝑥𝑥 =- 1 2 would provide limiting criterion for k values. In summary this was an interesting question which gave rise to a wide variety of responses in all three parts. There were instances of both succinct and ingenious approaches seen, particularly in part (c) and it was pleasing to see tenacity in some response where candidates had inadvertently opted for less efficient approaches but had persevered nonetheless to obtain the required results.

Common mistakes

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

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

Examiner's comment

This 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

9 This question provided a good range of marks between the candidates, although nearly all candidates struggled to proceed to the final answer in part (c). There were also several blank responses on this question which may have been down to candidates either running out of time or just getting overwhelmed towards the end of a paper. In part (a), candidates were usually able to find the equation of the line. The odd error in rearranging their equation of the line or incorrectly sub substituting in the coordinates for x and y were the few situations where full marks were not scored.

Common mistakes

• radius NP is perpendicular to the tangent so gradient −3 through (7, 4): y = −3x + 25

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

Examiner's comment

9 This question provided a good range of marks between the candidates, although nearly all candidates struggled to proceed to the final answer in part (c). There were also several blank responses on this question which may have been down to candidates either running out of time or just getting overwhelmed towards the end of a paper. Part (b) was generally attempted well. Most knew and attempted to find the point where the two lines intersected each other and found the point P correctly. They were then able to find the radius and usually proceeded to find the equation of the circle. Sometimes the equation of the circle had the squared missing from the brackets or it was equated to the radius rather than 2r .

Common mistakes

• solve y = −3x + 25 with y = x/3 for P(15/2, 5/2)
• r² = (15/2 − 7)² + (5/2 − 4)² = 5/2
• equation (x − 7)² + (y − 4)² = 5/2 (not = r, not (√(5/2))²)
• missing squares on the brackets

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

In part (b), most students understood the need to show that the y-intercepts were the same for both the line and the curve, but often candidates struggled to write an adequate conclusion to verify they meet again, and hence lost this mark.

Common mistakes

• both C and l pass through (0, −23)
• a concluding statement is required

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

Examiner's comment

Part (b) was demanding for many candidates, with confusion as to how to correctly deal with the log terms. Most candidates who were successful found the value of by finding the gradient of the line. A less popular but usually successful approach was using simultaneous equations to eliminate . Solving the equation to find the value of caused some problems. The question required a complete equation for the model, so just stating the correct values of and without the equation written explicitly meant candidates scored 3 out of 4 marks, which was a very common error.

Common mistakes

• gradient b = 0.45/0.91 = 0.495
• intercept is log₁₀a so a = 10^0.346 = 2.22
• must write the complete equation T = 2.22l^0.495

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

This question focussed largely on the problem-solving assessment objective, AO3, and was those candidates who were familiar with linear models scored full marks. The algebra was generally well done, whilst the written component in part (b) really assessed who knew what the modelling aspect was all about. In part (a) the first part of the question was generally well done with most finding at least the gradient or the intercept. Some candidates worked using x and y, although they should make sure that their final answer is in terms of the given variables. One area of confusion was in the definition of the variable n. Some candidates did not read the detail of the question which stated that n was the number of years past 1997- not simply the date (number of years AD). Whilst they still found a linear equation, it was not the correct one linking A and n. There were some candidates who tried to form an arithmetic sequence, too.

Common mistakes

• n is years after 1997, not the year AD
• gradient −21/8 = −2.625
• A = 190 − 2.625n in the given variables

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

Examiner's comment

This question focussed largely on the problem-solving assessment objective, AO3, and was those candidates who were familiar with linear models scored full marks. The algebra was generally well done, whilst the written component in part (b) really assessed who knew what the modelling aspect was all about. In part (b) the common approach was to substitute 19 n = in to the model found in part (a), scoring 2 marks for a correct answer. Where candidates used the dates in part (a), they could still find the correct value for the year 2016 with their model. When commenting on the suitability of the model candidates need to make sure they comment on the model, as opposed to explaining what external factors may have been resulted in the figures being different. It was pleasing to see that less candidates made the comment that the model is not suitable because the value predicted and the actual value are not the same which does not demonstrate an understanding of a model. Equally, just comparing the values without making a judgement on suitability is also insufficient.

Common mistakes

• n = 19 gives A ≈ 140, much higher than 120 so the model is not suitable
• comment on the model, not external factors
• "the values are not the same" alone scores B0

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

A very significant majority of the candidates did not seem to know how to define a region using inequalities. It was common to see a response where the area of region R was attempted using integration. It was only a small minority of candidates who made a correct inequality statement at the end, even with the allowance of follow through on their quadratic curve and their line equations. It was often the case that this final mark was lost due to a candidate using the letter R rather than y in their inequality. Finding the equations of the boundaries of the region is essential when defining the inequalities, so candidates who went on to find an area were mostly still trying to do this and if successful would have only lost the final mark. A very small minority of candidates found the area below the line by using a trapezium area formula and so did not manage to gain the marks for the line equation. Finding the line equation correctly was achieved by a majority of candidates, although a slip with the line gradient did occur occasionally even if the diagram showed a convincingly positive gradient. Finding the equation of the quadratic curve was found to be more difficult, with a common error being f(x)=(x+2)2+13. Occasionally a candidate spotted that this was not consistent with the curve going through (0,25) and tried to adjust their answer, but not often in a correct way. Attempts at using f (x)=a(x+2)2+13 were often successful at finding a, whilst attempts at using f (x)=ax2+bx+c were much less often correct.

Common mistakes

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

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

Examiner's comment

Part (a) of this question was generally well answered, with most candidates being able locate P correctly. A few candidates launched into some lengthy algebra, in some cases resulting in values for x and y bearing no relation to the position of P on the diagram. Answers to part (b) from this cohort of candidates were very disappointing and displayed an overreliance on algebra and poor understand of modulus graphs. Very few candidates identified x = -10.6 as the solution to the equation. Instead most answers wrote down the solution to 3x+40 = 2(x+4) – 5 {x =-37} as well as x = -10.6, scoring M1A0. A few candidates used a calculator to directly write down the correct value of x to gain both marks – perhaps not a bad strategy for students who are weak at modulus graphs. A simple check on a calculator would also have established that only x = -10.6 was a solution to the printed equation. Part [c] of this question was one of the most challenging parts of this paper, with hardly any candidates gaining all 3 marks. Candidates seemed unable to ‘visualise’ the question and see directly that there would be an intersection if a was greater than 2, so the B mark was hardly ever gained (and consequently the A mark was immediately lost). The M mark for considering P and attempting a = 5/4 was occasionally gained, although as above this often entailed long winded algebraic methods e.g. solving simultaneous equations to find a.

Common mistakes

• a > 2 (steeper than the right branch)
• line through P gives a = 5/4, so a ≤ 1.25
• set notation {a : a ≤ 1.25} ∪ {a : a > 2}

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

Examiner's comment

This question was generally well answered. (a) Most candidates attempted to rearrange the equation of l1 into the form y = mx + c, with only a small minority stating that the gradient was 2. In a few cases a sign error crept in when the terms in the original equation were rearranged. Most candidates attempted the negative reciprocal, although a significant minority failed to change the sign. Relatively few candidates lost a mark by leaving the gradient unsimplified (as 4/2).

Common mistakes

• rearrange to y = −½x + ¾ first (the gradient is not 2)
• negative reciprocal must change sign
• simplify 4/2 to m = 2

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

Examiner's comment

This question was generally well answered. (b) Marks were lost owing to slips in the rearrangement of the equation of l1 by either a sign error or, more frequently, a failure to divide the constant by 4. Most candidates attempted to equate two equations rather than substituting their y = 2x + 7 into the original equation. Sign errors, particularly with the constant, were a frequent cause of lost marks as the candidates attempted to solve their equation for x. Many candidates continued, unnecessarily, to find the y-coordinate. A few candidates, who erroneously had parallel lines, still tried to find a point of intersection.

Common mistakes

• substitute y = 2x + 7 into 2x + 4y − 3 = 0
• slips dividing the constant by 4 or with signs
• y coordinate not needed
• parallel lines (wrong m) have no intersection

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

Examiner's comment

This was a relatively straightforward question, but the success of candidates tended to be centre dependent, due to it being one of the newer aspects of the specification. In (a), candidates used a variety of successful methods to gain all three marks: the most commonly seen method involved using the given values to, first find the gradient, and then one of the straight-line graph forms to arrive at H = mt + c. Other methods included formation of simultaneous equations in m and c which could be solved in a variety of ways. The words ‘Using a linear model’, however, led some candidates to make the erroneous assumption of 𝐻= 𝑎𝑡𝑏 or 𝐻= 𝑎(𝑏𝑡). In part (b), most candidates achieving a linear model in (a) were able to recognise that the tree’s initial height, (consistent with their equation), was the value of ‘c’ in their linear model. The second mark was for a convincing statement suggesting that the values of 1.42m and 1.4m were similar and therefore supported the model. This mark was often lost due to lack of the correct deduction, or by thinking that these two values needed to be exactly the same and so rejecting the model.

Common mistakes

• gradient (3.28 − 2.35)/(6 − 3) = 0.31 then c = 1.42
• H = 0.31t + 1.42
• a linear model, not H = atᵇ or abᵗ

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

Examiner's comment

This was a relatively straightforward question, but the success of candidates tended to be centre In part (b), most candidates achieving a linear model in (a) were able to recognise that the tree’s initial height, (consistent with their equation), was the value of ‘c’ in their linear model. The second mark was for a convincing statement suggesting that the values of 1.42m and 1.4m were similar and therefore supported the model. This mark was often lost due to lack of the correct deduction, or by thinking that these two values needed to be exactly the same and so rejecting the model.

Common mistakes

• initial height is the intercept 1.42 m (t = 0)
• compare with 140 cm (units) and conclude the values are close so the fact supports the model
• they need not be equal

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

Examiner's comment

Q07 discriminated well between the medium and higher ability candidates. Lower ability candidates struggled to make much progress with this modelling question and most of them scored no more than one mark. In Q07(a), most candidates wrote down a correct equation for the model in the form , y kx c with constants k and c. A minority struggled to give a suitable form of the linear equation, with some omitting the fixed cost constant, c. Some candidates gave ; y x incorrect answers, such as equations involving exponential models; or differential equations. Q07(b) proved challenging for those candidates who struggled to grasp the concept ‘profit sales cost ’, with many of them incorrectly assuming that the model from Q07(a) was for the profit made. Many of these candidates used (800, 500) and either (300, 80) or (300, 80) to form two linear equations, with some erroneously finding a gradient of 0.84. Those candidates who understood the link between profit and cost usually found the total cost to the factory of £1100 and £680 for 800 and 300 bars of soap respectively. Most used (800, 1100) and (300, 680) to form two linear equations and proceeded to find the gradient, 0.84, and y-intercept, 428. A large proportion concluded by stating 0.84 428. y x the given equation A few candidates, however, did not complete their analysis with a suitable conclusion, and so lost the final mark. Some candidates used (800, 1100) and (300, 680)to verify the model 0.84 428. y x Many of these were successful but often failed to give a suitable conclusion stating that the given model was true. Q07(c) gave a significant number of candidates their only creditable access to Q07. Some candidates gave a correct interpretation such as 0.84 represented ‘the cost of making an extra bar of soap’. Incorrect explanations included that 0.84 represented ‘the selling price of a bar of soap’ or ‘the profit per bar of soap made’. There was a mixed response to Q07(d). The most popular approach was to form and or 2 0.84 428, n n solve a correct 2 which mostly led to the correct 0.84 428 n n n n n 369. 369, answer A final answer left as an inequality such as or 368 without reference to the statement that 369 was the number of bars required, was not n n sufficient for the final mark. Some candidates used trials of both and 368 369 n 369. in a complete trial and error method leading to a correct conclusion of

Common mistakes

• y = kx + c with constants
• omitting the fixed cost c
• exponential or differential equation models wrong

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

Examiner's comment

Q07 discriminated well between the medium and higher ability candidates. Lower ability candidates struggled to make much progress with this modelling question and most of them scored no more than one mark. In Q07(a), most candidates wrote down a correct equation for the model in the form , y kx c with constants k and c. A minority struggled to give a suitable form of the linear equation, with some omitting the fixed cost constant, c. Some candidates gave ; y x incorrect answers, such as equations involving exponential models; or differential equations. Q07(b) proved challenging for those candidates who struggled to grasp the concept ‘profit sales cost ’, with many of them incorrectly assuming that the model from Q07(a) was for the profit made. Many of these candidates used (800, 500) and either (300, 80) or (300, 80) to form two linear equations, with some erroneously finding a gradient of 0.84. Those candidates who understood the link between profit and cost usually found the total cost to the factory of £1100 and £680 for 800 and 300 bars of soap respectively. Most used (800, 1100) and (300, 680) to form two linear equations and proceeded to find the gradient, 0.84, and y-intercept, 428. A large proportion concluded by stating 0.84 428. y x the given equation A few candidates, however, did not complete their analysis with a suitable conclusion, and so lost the final mark. Some candidates used (800, 1100) and (300, 680)to verify the model 0.84 428. y x Many of these were successful but often failed to give a suitable conclusion stating that the given model was true. Q07(c) gave a significant number of candidates their only creditable access to Q07. Some candidates gave a correct interpretation such as 0.84 represented ‘the cost of making an extra bar of soap’. Incorrect explanations included that 0.84 represented ‘the selling price of a bar of soap’ or ‘the profit per bar of soap made’. There was a mixed response to Q07(d). The most popular approach was to form and or 2 0.84 428, n n solve a correct 2 which mostly led to the correct 0.84 428 n n n n n 369. 369, answer A final answer left as an inequality such as or 368 without reference to the statement that 369 was the number of bars required, was not n n sufficient for the final mark. Some candidates used trials of both and 368 369 n 369. in a complete trial and error method leading to a correct conclusion of

Common mistakes

• cost = revenue − profit: (800, 1100) and (300, 680), not (800, 500) and (300, −80)
• solve for k = 0.84, c = 428 and state the equation as a conclusion

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

Examiner's comment

Q07 discriminated well between the medium and higher ability candidates. Lower ability candidates struggled to make much progress with this modelling question and most of them scored no more than one mark. In Q07(a), most candidates wrote down a correct equation for the model in the form , y kx c with constants k and c. A minority struggled to give a suitable form of the linear equation, with some omitting the fixed cost constant, c. Some candidates gave ; y x incorrect answers, such as equations involving exponential models; or differential equations. Q07(b) proved challenging for those candidates who struggled to grasp the concept ‘profit sales cost ’, with many of them incorrectly assuming that the model from Q07(a) was for the profit made. Many of these candidates used (800, 500) and either (300, 80) or (300, 80) to form two linear equations, with some erroneously finding a gradient of 0.84. Those candidates who understood the link between profit and cost usually found the total cost to the factory of £1100 and £680 for 800 and 300 bars of soap respectively. Most used (800, 1100) and (300, 680) to form two linear equations and proceeded to find the gradient, 0.84, and y-intercept, 428. A large proportion concluded by stating 0.84 428. y x the given equation A few candidates, however, did not complete their analysis with a suitable conclusion, and so lost the final mark. Some candidates used (800, 1100) and (300, 680)to verify the model 0.84 428. y x Many of these were successful but often failed to give a suitable conclusion stating that the given model was true. Q07(c) gave a significant number of candidates their only creditable access to Q07. Some candidates gave a correct interpretation such as 0.84 represented ‘the cost of making an extra bar of soap’. Incorrect explanations included that 0.84 represented ‘the selling price of a bar of soap’ or ‘the profit per bar of soap made’. There was a mixed response to Q07(d). The most popular approach was to form and or 2 0.84 428, n n solve a correct 2 which mostly led to the correct 0.84 428 n n n n n 369. 369, answer A final answer left as an inequality such as or 368 without reference to the statement that 369 was the number of bars required, was not n n sufficient for the final mark. Some candidates used trials of both and 368 369 n 369. in a complete trial and error method leading to a correct conclusion of

Common mistakes

• 0.84 is the cost (£) of making each extra bar
• not the selling price or profit per bar

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

Examiner's comment

Q07 discriminated well between the medium and higher ability candidates. Lower ability candidates struggled to make much progress with this modelling question and most of them scored no more than one mark. In Q07(a), most candidates wrote down a correct equation for the model in the form , y kx c with constants k and c. A minority struggled to give a suitable form of the linear equation, with some omitting the fixed cost constant, c. Some candidates gave ; y x incorrect answers, such as equations involving exponential models; or differential equations. Q07(b) proved challenging for those candidates who struggled to grasp the concept ‘profit sales cost ’, with many of them incorrectly assuming that the model from Q07(a) was for the profit made. Many of these candidates used (800, 500) and either (300, 80) or (300, 80) to form two linear equations, with some erroneously finding a gradient of 0.84. Those candidates who understood the link between profit and cost usually found the total cost to the factory of £1100 and £680 for 800 and 300 bars of soap respectively. Most used (800, 1100) and (300, 680) to form two linear equations and proceeded to find the gradient, 0.84, and y-intercept, 428. A large proportion concluded by stating 0.84 428. y x the given equation A few candidates, however, did not complete their analysis with a suitable conclusion, and so lost the final mark. Some candidates used (800, 1100) and (300, 680)to verify the model 0.84 428. y x Many of these were successful but often failed to give a suitable conclusion stating that the given model was true. Q07(c) gave a significant number of candidates their only creditable access to Q07. Some candidates gave a correct interpretation such as 0.84 represented ‘the cost of making an extra bar of soap’. Incorrect explanations included that 0.84 represented ‘the selling price of a bar of soap’ or ‘the profit per bar of soap made’. There was a mixed response to Q07(d). The most popular approach was to form and or 2 0.84 428, n n solve a correct 2 which mostly led to the correct 0.84 428 n n n n n 369. 369, answer A final answer left as an inequality such as or 368 without reference to the statement that 369 was the number of bars required, was not n n sufficient for the final mark. Some candidates used trials of both and 368 369 n 369. in a complete trial and error method leading to a correct conclusion of

Common mistakes

• solve 2n > 0.84n + 428
• answer 369 bars (whole number, not n > 368.9 or a decimal)

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

Examiner's comment

Most students attempted this question and many gained all marks. The majority of students rearranged the given equation to make y the subject and hence found the gradient. Those who did not, often stated + or – 3 as the gradient. Most correctly found the gradient of the second line using the two coordinates. There were occasional sign slips. Many students unnecessarily continued to find the equation of the line for these coordinates after working out their gradient. The comment on parallel, perpendicular or neither was attempted by most students with varying success. Explanation on why they were not perpendicular was often unclear and errors included stating that they should be reciprocals but omitting the word negative and that the sum of the gradients should be -1. For parallel, a small number of students stated that the gradients should be multiples of each other. A few students stated neither but no reasons were given.

Common mistakes

• rearrange to y = ¾x + 2.5 for gradient 3/4 (not ±3)
• gradient of l2 = −3/2
• neither: gradients are not equal and their product is not −1 (not "reciprocals" without "negative", not "sum is −1")
• finding the equation of l2 is unnecessary

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

Part (a) was a ''show that'' question in which students used the idea of perpendicular gradients to find the equation of line PA. It was very straightforward and well done with almost all students scoring the 3 marks. Part (b) required the students to find the equation of the circle C

Common mistakes

• radius perpendicular to tangent so gradient of PA is −½
• line through A(7, 5)
• complete to 2y + x = 17

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

Examiner's comment

students scoring the 3 marks. Part (b) required the students to find the equation of the circle C using the information given. Again most students attempted this with ease, finding firstly the coordinates of point P, and then the radius PA, before writing down the equation of the circle. Errors witnessed tended to be arithmetic, although some students complicated the question by attempting to solve 2 2 simultaneously in an attempt to 2 and 7 5 x y r 2 1 y x find r. There were numerous ways to find the value of k in part (c). One of the best involved

Common mistakes

• solve y = 2x + 1 with 2y + x = 17 to get P(3, 7)
• r² = PA² = 20
• equation (x − 7)² + (y − 5)² = 20, not (√20)²

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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