Trigonometric ratios, graphs, radians and the sine and cosine rules Pure

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

Examiner's comment

In part (b), many drew a diagram to help and there were many correct answers. Those who did not draw a diagram found this more challenging. The vast majority achieved at least one mark for finding the length of OA or OB . Most understood the need to use the cosine rule, and this was often the rearranged form with cos(OAB) as the subject. A surprising number of candidates found the wrong angle of the triangle by substituting their vector lengths into the wrong places in the cosine rule formula, especially those without a diagram. Others made slips in the rearranging. Some candidates incorrectly assumed that the triangle was right-angled and attempted to find angle OAB using trigonometry in a right-angled triangle. For those with a good diagram, the 1 tan- approach worked well, and many successful strategies were seen. A few students used the scalar product approach, but again those without a diagram often used incorrect vectors and found the wrong angle.

Common mistakes

• OA = √73, AB = √29, OB = √40
• cosine rule with √40 opposite angle A gives 47.6°
• lengths in the wrong places find the wrong angle
• do not assume a right angle (sin⁻¹(√40/√73) = 47.75° is wrong)
• draw a diagram

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

Examiner's comment

This question was a good differentiator at the later end of the paper. Success with this question was somewhat mixed, with some learners scoring most marks in part (b), whilst part (a) proved to be more challenging as it tested not only their knowledge of trigonometric identities but also their algebraic manipulation skills. In part (a), most responses attempted the proof by using both compound angle expansions to reach an expression in sin x and cos x, usually gaining the method mark and in most cases also the first accuracy mark. Once a correct expression was achieved, some stronger responses reached the given result by simplifying the terms. The most common mistakes committed while implementing this approach, were sign errors, especially for the compound angle expansion of ( ) cos 30 x + ° , incorrect multiplication by 3 and bracketing errors, which were not condoned in this ‘show’ question. A smaller proportion of responses used the R-alpha method with ( ) cos 30 R x + °± alpha , which required learners to find both R and alpha correctly to reach ( ) 2cos 30 30 2cos x x + °- ° = as requested.

Common mistakes

• expand both compound angles (sign of cos(x + 30°) expansion), use sin 30° = ½ and cos 30° = √3/2 exactly
• bracket errors not condoned
• R-alpha method needs R = 2 and the angle

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

Examiner's comment

This question using differentiation to find a minimum was generally answered well. It was usually part (a) which provided most challenge, often resulting in work which was difficult to follow. In part (a), learners were required to use the given information about the area of the composite shape and show that the perimeter could be expressed in a specific form. The majority of responses formed a correct equation for the area of the shape, although there were some that omitted the rectangular area and others made an error with the sector area. Some responses used a longer approach of considering the sector as a proportion of a whole circle rather than using the more succinct sector area formula. The formula for the perimeter was also correctly stated by the majority, although the sides of the rectangle were sometimes omitted or internal sides were included. There were a variety of approaches which were successful in eliminating theta from the perimeter. The most common approach was to make theta the subject of the area and then substitute into the perimeter. A few unsuccessful attempts involved making r the subject of the area and substituting into the perimeter. There were some neat, logical solutions to this problem. However, there were also many very poorly presented solutions. It was common to see work jumping around the page and arrows being used to try to link the parts of the equations. There were responses that made errors in their algebra and when realising that they could not achieve the given form, worked back to track their error. These learners should be praised for checking their work and trying to identify the error, but this does lead to messy and unclear work, so the learners should be encouraged to rewrite the solution should they have time to make sure that they have a clear argument.

Common mistakes

• area ½r²θ + r²/10 = 240
• make rθ (or θ) the subject and substitute into P = rθ + 2r + r/5 (sector arc, two radii minus shared side plus rectangle sides)
• omitting the rectangle area or including internal sides
• work must be clearly set out

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

Examiner's comment

This proved to be a challenging question demanding good algebraic skills and attention to detail. For a final question on a paper, most made an attempt. In part (b), it was notable that those who did not reach a correct answer in part (a) often did not go on to attempt this, even though these marks were available to them. Even though they were told that the integral was a multiple of 2 cosec u some learners tried unsuccessfully to integrate their very complex answers to (a). A large number did not appear to know the integral of 2 cosec u or did not use the formula booklet correctly and some attempted very convoluted methods of integration. Those that did not score at all got muddled in the initial stage, choosing to use ln or some other given integration from the formula book that they had misread or misinterpreted. A significant number got confused between the integral of 2 cosec u andcosec u , which meant they ended up with expressions involving ln(tan ) u or ln(cosec cot ) u u + . However, several responses obtained the correct answer of 1 cot 4 u - , although a few omitted the negative sign, and others 23 stated the answer was 1 cot . 4 u There was a proportion who did not include the negative sign and consequently amended the resulting negative in their final answer. 24

Common mistakes

• ∫cosec²u du = −cot u (sign often lost
• confused with ∫cosec u)
• −¼cot(π/3) + ¼cot(π/6) = 1/(2√3) = √3/6
• can be attempted even if (a) was not completed

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

Examiner's comment

This was a fairly routine question involving the use of the cosine rule and area formulae in radians to find the area of a region. In part (a), most responses successfully used the cosine rule directly to find the relevant angle ABC. A small number attempted the cosine rule but used sine instead of cosine in the formula and so were not awarded both marks due to an incorrect formula. Some responses found other angles in triangle ABD first and used a combination of sine and cosine rules to find angle ABC; but those that attempted this were often less successful due to errors in their working. Responses occasionally did not gain the marks through lack of detail in the working required for a "show that" question. Some did not substitute values for a, b and c into the cosine rule, instead just quoting the formula before stating the given answer. Others who started correctly with ( )( ) 2 2 2 6.4 8 13 2 8 13 cos ABC = + - sometimes did not give an appropriate intermediate line such as cos ABC = 0.923 before stating the given answer.

Common mistakes

• cosine rule with values substituted and an intermediate line such as cos ABC = 0.923 before the given 0.394
• quoting the formula then the answer is not enough
• verification scores M1A0

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

Examiner's comment

This was a fairly routine question involving the use of the cosine rule and area formulae in radians to find the area of a region. Incorrect approaches involved finding the length of arc AC using the given angle 0.394 and then dividing the arc length by 8 to achieve the required angle. A very small number of responses worked on the incorrect assumption that triangle ABD was right angled. On rare occasions, learners worked in degrees and converted to radians, usually successfully, while a small number made no attempt at part (a) but used the given answer for part (b). The majority of learners successfully attempted part (b), most commonly by finding the area of triangle ABD, using the formula 1 sin 2 ab theta and subtracting the area of sector ABC using 2 1 2 r theta . A small proportion calculated angle DAB and used it to find the area of triangle ABD but most used the given angle ABC. Of those that did not gain full marks for this part, it was common that responses gained the first method mark for correctly finding either the area of the triangle or the area of the sector. However, many could not remember the other relevant formula needed and so could not go on to gain further credit, or mistakes were made in substituting values leading to the accuracy mark not being awarded. Other errors arose from using an incorrect formula for the area of the sector, including not squaring r, omitting the 1 2 or using the arc length formula instead.

Common mistakes

• area R = triangle ABD (½ × 8 × 13 sin 0.394) − sector ABC (½ × 8² × 0.394)
• sector formula errors (r not squared, ½ omitted, arc length used)
• triangle ABD is not right-angled
• awrt 7.34 to 7.37

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

Examiner's comment

This question, which required an understanding of transformation and periodicity of trigonometric functions, gave learners the opportunity to think carefully and demonstrate a strong understanding of the topic. Part (i) was generally well answered with many responses evidencing use of graphs to help 1 deduce that n was 3. The most common incorrect answer was , even though the n = 3 question specified that n was a natural number. Learners would be well advised to pay careful attention to the constraints placed on unknowns in all questions. In part (ii), a good number of responses demonstrated that there were 30 solutions to the equation, although fewer were able to communicate their reasoning well enough. A significant minority arrived incorrectly at 30 by mistakenly assuming that there were 15 solutions to sin(nx) = k and 15 solutions to sin(nx) = –k, or sometimes 18 and 12 solutions respectively (the correct number of solutions being 16 and 14 respectively). Such responses scored one of the two available marks. An uncommon but elegant approach was to rewrite the quadratic in sin3x as a linear equation in cos6x, making it easier to explain why there are 30 solutions in the given interval. Another was to identify 2 3 as the period of sin2 3x and then scale up accordingly, which demonstrated an excellent understanding of periodicity. A small number did not realise that n was still equal to 3 in part (ii), and gave an answer in terms of n. A common incorrect approach involved the misapprehension that squaring sinx would also square the number of solutions, leading to an incorrect answer of 36. Many did not take the negative solutions into account and simplify multiplied 6 by 2.5 to obtain the incorrect answer of 15. More learners would have benefited from a well-drawn and labelled graph to support their thinking and reasoning.

Common mistakes

• sin(nx) has n cycles in 0 to 2π, 2 solutions each, so n = 3
• n = 1/3 is not a natural number

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

Examiner's comment

This question, which required an understanding of transformation and periodicity of trigonometric functions, gave learners the opportunity to think carefully and demonstrate a strong understanding of the topic. Part (i) was generally well answered with many responses evidencing use of graphs to help 1 deduce that n was 3. The most common incorrect answer was , even though the n = 3 question specified that n was a natural number. Learners would be well advised to pay careful attention to the constraints placed on unknowns in all questions. In part (ii), a good number of responses demonstrated that there were 30 solutions to the equation, although fewer were able to communicate their reasoning well enough. A significant minority arrived incorrectly at 30 by mistakenly assuming that there were 15 solutions to sin(nx) = k and 15 solutions to sin(nx) = –k, or sometimes 18 and 12 solutions respectively (the correct number of solutions being 16 and 14 respectively). Such responses scored one of the two available marks. An uncommon but elegant approach was to rewrite the quadratic in sin3x as a linear equation in cos6x, making it easier to explain why there are 30 solutions in the given interval. Another was to identify 2 3 as the period of sin2 3x and then scale up accordingly, which demonstrated an excellent understanding of periodicity. A small number did not realise that n was still equal to 3 in part (ii), and gave an answer in terms of n. A common incorrect approach involved the misapprehension that squaring sinx would also square the number of solutions, leading to an incorrect answer of 36. Many did not take the negative solutions into account and simplify multiplied 6 by 2.5 to obtain the incorrect answer of 15. More learners would have benefited from a well-drawn and labelled graph to support their thinking and reasoning.

Common mistakes

• sin²(3x) = k² means sin 3x = ±k: 6 solutions per interval of length π, 5 intervals gives 30
• justification needed (16 + 14, not 15 + 15 or 18 + 12)
• 36 and 15 are common wrong answers
• a labelled sketch helps

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

Examiner's comment

The final question, which involved differentiation, integration by substitution and trigonometric identities, provided challenge in dealing with the algebra but, given its layout, proved accessible. The majority of learners attempted at least part of this question with some success due to the restart opportunities available. Part (c) proved to be more accessible for many learners, with many making an attempt after struggling with the earlier sections. Those who attempted this part often successful, regardless of their performance in parts (a) and (b). Common errors were with the coefficient, usually given as 1 2 - or 2, or when substituting the limits. Some tried to use double angle formulae to change the form of the integrand and therefore did not make progress with this part of the question, not realising that the given form in part (b) was designed to help them complete the integration. Despite the question asking the candidate to show all stages of their working, some responses did not show the substitution and so did not gain the last two marks.

Common mistakes

• ∫cos 2θ dθ = ½ sin 2θ (not −½ or 2)
• substitute the θ limits, not ±1
• area 1
• the substitution must be shown (calculator answer alone scores 0)

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

Examiner's comment

Part (a) was generally well done with some making good use of the diagram. After identifying the three forces, successful responses resolved vertically to form an equation with the vertical component of 5 N clearly found. This was a ‘Show that…’ question and so those who just wrote ‘3’ down, with no supporting evidence of how it had been obtained, received no reward. Very occasionally cosine was used instead of sine and a few learners used the value of g as 9.81, which meant that the accuracy mark was not awarded.

Common mistakes

• resolve vertically: 5 sin α + R = 2g
• the vertical component 5 × 3/5 = 3 must be shown, just writing 3 scores nothing
• sin/cos confusion
• g = 9.8 not 9.81

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

Examiner's comment

Finding the bearing of the direction of motion in part (c) proved more challenging for many learners. A good number were able to find an acute angle relating to their velocity vector. Use of basic trigonometry or the properties of a right-angled isosceles triangle were seen in roughly equal measure, but gaining the final mark for a correct bearing was often not achieved. Unrepresentative diagrams and a misunderstanding of bearings appeared to be the cause of this. A significant number did not need to use trigonometry, realising that the components indicated a north west direction and writing down 315o directly. A few learners first worked out the displacement vector and attempted to find the bearing of the position of the particle. This was not what the question had asked for and gained no credit.

Common mistakes

• v = −3i + 3j points north-west so bearing 315°
• unrepresentative diagrams and misread bearings lose the final mark
• using the displacement instead of the velocity scores nothing

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

Examiner's comment

In part (a) there were many excellent responses seen. Most responses gave an expression or an actual value for t by considering the horizontal motion and then substituted it into their equation for vertical motion. Many gained full marks, but some used a heavily rounded value for t and lost accuracy. Others were not clear whether they were taking upwards or downwards as positive in the vertical direction and this led to two common errors. The first was that both terms in their expression for vertical displacement had the same sign. The second was, if they were taking upwards as positive as most did, they replaced ‘s’ with H in their equation and erroneously reached the result H = 30 - and then just wrote H = 30 which resulted in not gained the final accuracy mark. Other errors included poor substitution into formulae, forgetting to resolve the velocity into its components and sin/cos confusion. Quoting and using the equation of the path was rarely seen. Some responses erroneously used half of their time from part (a), treating the motion as if it were symmetrical and gained no marks. The energy solution was rarely seen.

Common mistakes

• sin θ = 1/√5, cos θ = 2/√5
• t from horizontal motion 14cos θ × t = 40, then vertical motion
• keep a consistent positive direction (H = −30 must become 30 correctly)
• do not halve the time (not symmetrical)
• premature rounding of t

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

Examiner's comment

In part (b) the most popular and simplest approach was to take moments about A to derive the given expression for the normal reaction at B. The angle was given between the vertical and the rod, rather than the possibly more familiar case of between the horizontal plane and the rod. Although this was not a problem for the vast majority, a significant number used (90o - alpha ) for their angle which sometimes led to confusion, particularly for those who did not recognise the equivalence with sin alpha /cos alpha and instead tried to use the addition formula to write this in terms of alpha only. In addition to sin/cos confusion, errors seen included total omission or inconsistent inclusion of a, slips in copying from one line to the next and occasional omission of the weight of the rod. Since the requirement was to ‘show that tan S Mg ’ it was essential that working was complete and correct to achieve all the alpha = marks. A very small minority attempted to take moments about B but they often did not complete their solution by resolving forces in order to eliminate F and R. In the third part although many learners recognised that R (with no friction acting at B) was take moments about A, to follow from the work in part (b). Responses that did this usually scored some marks for their horizontal and/or vertical resolutions even if their moments equation was incorrect. Occasionally the reaction at the wall was considered to be perpendicular to the rod rather than to the wall which generally resulted in no marks being achieved. Errors in the moments equation included sin/cos confusion, inconsistent inclusion of a and incorrect perpendicular distances. Some omitted the weight of the rod leading to a missing term in both the moments equation and the vertical resolution. Fragmented respondes, simplifying terms before writing down a full moments equation, sometimes made slips in the process. Those who attempted to take moments about B often achieved the mark for their vertical resolution but often did not gain all the marks for their moments equation. Attempts to resolve parallel or perpendicular to the rod were rare and tended to have a term missing. Nevertheless, there were a fair number of entirely correct solutions seen with clear and concise working.

Common mistakes

• moments about A: S × 2a cos α = Mga sin α + 2Mg × 0.5a sin α
• angle is with the wall, not the ground (using 90° − α causes errors)
• sin/cos confusion
• inconsistent a
• at least one line of working before the given answer

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

Examiner's comment

Part (d) was the most challenging part of the question. It was often left blank or multiple half- finished attempts were seen with little clear structure. The most common approach was to take moments about A, to follow from the work in part (b). Responses that did this usually

Common mistakes

• include the 3M particle at B: N = 6Mg (not R from (a))
• moments about A give the wall reaction P = 2Mg = F
• limiting equilibrium so F = μN
• reaction at the wall is perpendicular to the wall, not the rod
• omitting the rod's weight
• incorrect perpendicular distances

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

This question involved a badge in the shape of a semicircle, less the area within a circular arc of the same radius as the semicircle, centred at one end of its diameter. Many candidates did not spot that triangle AOB was in fact an equilateral triangle which would have immediately provided the angle 3 . It was not uncommon for candidates to fail to find an appropriate angle, and these were unable to score any marks in this question. A significant proportion of the candidates found the angle 3 but it often came from considerable work on the triangle, included use of Pythagoras and trigonometry, instead of the intended, more straightforward, deduction from AOB being equilateral. Many candidates that found an appropriate angle only progressed as far as finding the area of a sector, the area of triangle AOB or the area of the semicircle, for which no credit was available unless further progress was made. Some candidates used Pythagoras to calculate the area of the right-angled triangle and occasionally missed dividing by 2. Very often candidates were able to find a correct expression for one of the segments in the circular arc for the second mark, which was also available for finding the total area of the unshaded region AOB, the latter scoring the mark much less often. To get this far, candidates needed to identify and find the area of the equilateral triangle within the cut- out section defined by the circular arc. The third mark out of the available 4 marks was much harder to obtain as this required an attempt at the required area by correctly combining their correct attempts at the required smaller areas that made up the whole shaded region. Several expressions were valid here, all of which had to arrive at the sum of a sector and the triangle in exact form for the final accuracy mark. The most common route involved subtracting their expression for the area of a segment, and since the area of the segment involved subtracting the area of the equilateral triangle, this calculation led to sign errors in final answers, losing the final accuracy mark. Candidates resorted to decimals very infrequently, although calculators could well have been used to obtain exact forms of the areas. Communication of method was, at times, very clear, but often this clarity was lacking and in combination with poor algebra/manipulation and layout of solutions this could have cost some candidates method marks. It was quite surprising how many candidates attempted this question by an attempt at integration. Of these attempts, the most common involved use of polar coordinates, presumably by further mathematics candidates. This was a demanding route, and while there were many attempts that found appropriate polar equations for the relevant circles, the limits chosen were often incorrect, while others faltered with the integration itself. As a result, many candidates attempting this method either failed to set up the problem correctly or lost their way in their processing, and very few gave a completely correct answer. Other attempts at integration, by writing e.g. 2 25 y x = - , required use of the substitution sin x = and, again, issues with selecting appropriate limits or proceeding with the following integration caused these candidates problems. Centres should encourage candidates to explore other routes to solving problems such as this but must discuss which routes are likely to be the most concise.

Common mistakes

• spot triangle AOB is equilateral so angle BAO = π/3
• area R = major sector − segment (or sector + triangle)
• radians
• exact (25/4)√3 + (25/6)π
• sector or triangle alone scores nothing

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

A very well answered question with many candidates using a clearly labelled force diagram and gaining full marks for (a) and (b). It is always good exam technique to annotate the diagram to support the answer. (a) Most candidates were able to resolve perpendicular to the plane to obtain R=mgcos and then replacecos to obtain the required answer. A few multiplied by sin instead of cos but this still earned the M mark. However, those who used tan or divided by sin or cos lost both marks. A few responses chose to use a decimal approximation for the angle which could score both marks in this part.

Common mistakes

• resolve perpendicular to the plane R = mg cos α = 12mg/13 with cos α = 12/13 from tan α = 5/12
• using tan or dividing by sin or cos scores nothing

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

Examiner's comment

A very well answered question with many candidates using a clearly labelled force diagram and gaining full marks for (a) and (b). It is always good exam technique to annotate the diagram to support the answer. (b) This was a “show that” question, so full and accurate working was required. The vast - = sin mg F ma majority wrote down an equation of motion parallel to the plane, and then used their answer from (a) to replace F, cancelled the m’s and factorised to obtain the given answer. Note that when there is a printed answer to show, candidates must give their answer in exactly the same form to guarantee earning all the marks. Common errors = mg sin F were setting , using an incorrect trigonometrical ratio, missing m’s or g’s, 5 using 𝜇= 12 .

Common mistakes

• mg sin α − F = ma with F = μ × (12mg/13)
• setting mg sin α = F (no acceleration)
• using μ = 5/12
• missing m or g
• both trig ratios substituted as fractions before the given answer

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

Examiner's comment

(a) This part of the question was worth three marks and required candidates to produce a rigorous proof to show that c = 3, either by using trigonometry or an isosceles triangle or both. It seemed that many candidates were telling themselves that the answer was obvious but were uncertain as to how to go about justifying the fact. Most candidates struggled to produce a rigorous proof, instead just stating, without justification, that for a bearing of 135o, the distance travelled in the x-direction would have the same magnitude as the distance travelled in the negative y-direction or equivalent. They were able to score two of the three marks available if they then went on the show that c = 3. Those candidates who drew a diagram and labelled it with all the information from the question were generally more successful in scoring all three marks than those who either did not draw a diagram or whose diagram did not include some of the information given. Some - 6 = o tan135 candidates correctly used or the equivalent, without a diagram. Common 2c errors included not being able to correctly sketch a bearing of 135o, or omissions of angles or sides in the diagram. The negative sign caused issues in several instances and was sometimes glossed over. There was also some poor algebra. A few candidates incorrectly wrote 2c + 6 = 0 to obtain c = 3.

Common mistakes

• at t = 4, r = 2ci − 6j
• bearing 135° means a 45° angle below east so 2c = 6, justified by tan 45° or an isosceles right-angled triangle on a correct diagram
• stating equal components without justification loses a mark
• 2c + 6 = 0 is wrong

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

Examiner's comment

(a)This part required the derivation of an equation for the flight path of a stone moving freely under gravity given the initial speed and angle of projection. There were many excellent responses seen where candidates used horizontal and vertical components of the = + 2 s ut at initial velocity in to find expressions for x and y in terms of t, with the 1 2 majority realising that the acceleration was zero in the horizontal direction. Occasionally the velocities were not resolved or the acceleration in both directions was taken as g, but such instances were rare. Most proceeded to eliminate t correctly and obtain the printed equation. Since it was a given answer any slips in working tended to be rectified including any previous cos/sin confusion. The final mark did require 9.8 for the value of g to be seen being used explicitly in the working. Although full marks were often achieved in this part of the question, there were candidates who failed to make any valid progress, often because they equated their vertical distance to zero, possibly proceeding to calculate t and then realising they had no strategy for finding a general equation. Those who quoted the general equation for a trajectory and substituted in the given angle achieved no credit, but this was only very rarely seen. Some omitted this part of the question entirely, but it did not preclude them from accessing all marks in the subsequent parts. An alternative but significantly less common approach was to quote the general equation of a parabola as y = ax2 + bx + c and use the coordinates of three specific points to find the constants. Most realised that c = 0 and achieved the first two marks. Generally, the points corresponding to the range and the greatest height were found using suvat equations, although other points were possible. Some attempted to use completing the square with the maximum point at (60, 22.5) but failed to deal properly with the coefficients. Occasionally the derivative at x = 0 was used successfully to find the value of b. There were a fair number of entirely correct solutions seen using this method but also some which lacked organisation and a clear strategy. made no progress in part (a) often achieved the marks here.

Common mistakes

• x = 35cos α × t (no horizontal acceleration) and y = 35sin α × t − ½gt², eliminate t
• g = 9.8 must be seen explicitly
• setting y = 0 gives no general equation
• quoting the trajectory formula scores nothing

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

Examiner's comment

(a) There were a good number of fully correct responses to this ‘Show that’ question. The majority of candidates found and used the trig ratios successfully, possibly because the answer was given in the question, which allowed them to identify whether they had calculated or used the wrong one. If a candidate gained one or two marks, but not all three, it was usually due to sin/cos confusion or making arithmetic errors when rearranging the equation. Some candidates lost all the marks because they omitted the a’s and so did not have a moments equation, but most included these initially and then cancelled appropriately. The final mark was occasionally lost because explicit use of 3 = was not seen or the candidate chose to give the force another name (such as RC) cos 5 and so did not arrive at the printed answer. A common error was using 2a instead of 1.5a as the distance. A few candidates gave the impression of having arrived at the required answer, but on closer inspection were not correct in their work. The most common 2 = = 0.5 sin S Mg Mg example of incorrect working was . Some tried to formulate a 5 proof based on the given answer that did not involve taking moments, but instead tried to resolve forces, and received no credit. Most candidates presented their answer in the given form, although S on the RHS was occasionally seen.

Common mistakes

• moments about A: S × 1.5a = Mg × a cos θ with cos θ = 3/5
• using 2a instead of 1.5a or a for the weight
• missing the a's (not a moment)
• resolving instead of moments scores nothing

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

Examiner's comment

(b) Many candidates scored all six marks in this part. Resolving vertically and horizontally was the most common and the most successful method. There were few errors apart from sin and cos confusion. Taking moments about C to give an equation in R and F and then resolving was the next most common method, but candidates were more likely to make errors using this approach, either by confusing sin with cos or by omitting a force entirely. Very few candidates resolved parallel and perpendicular to the rod and most of those who used this method failed to arrive at the correct answer due to the more challenging algebra. Occasionally candidates wrote dimensionally inconsistent equations by failing to include lengths in all of their moment terms or, more rarely, erroneously including lengths in their force terms. Errors included missing a g where there should have been one or not having the correct number of terms in a moment or a resolution equation. A surprisingly common error was for students to confuse the reaction force S 2 = F Mg with the reaction force at A, resulting in the incorrect equation Some 5 candidates lost method marks by resolving forces that did not need resolving (for and ). Whichever method Mg sin example, by writing an equation with both cos S was used, early substitution for F into their solutions was often seen, with almost all candidates realising that they needed to use F = μR. Some candidates clearly still incorrectly believe that μ cannot be greater than 1 and were ‘dismissive’ when their value was. It was not uncommon to see very messy responses, with multiple attempts and work crossed out. Many candidates made a good attempt at solving for mu , but this was often impeded by the lack of clarity over the previous equations. The mark scheme enabled candidates who used a correct method but who made sign errors, confused sin and cos or used incorrect distances to score half marks (generally, M1A0M1A0M1A0), while those with more fundamental misunderstandings were prevented from accessing the dependent method mark; this seemed to sort out the candidates well. When mu was found, most candidates gave the answer as a fraction or to an appropriate degree of accuracy.

Common mistakes

• resolve horizontally F = S sin θ and vertically R = Mg − S cos θ, then F = μR: μ = 8/19 ≈ 0.42
• confusing S with the reaction at A
• sin/cos confusion
• resolving forces that need no resolving
• μ may be greater than 1 in general

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

Examiner's comment

Part (a) of this question required the use of the cosine rule in combinations with bearings. The vast majority of candidates recognised the need to use the cosine rule, but were unsuccessful in labelling their diagram correctly, usually labelling the 72 or the 39 in the wrong place. This resulted in an incorrect attempt at applying the cosine rule and limited them to the method mark in part (a). A small proportion of candidates attempted to use the sine rule or Pythagoras, which were ultimately invalid approaches, but a further small proportion attempted this part using a vector approach with a good success rate.

Common mistakes

• angle ACB = 72° − 39° = 33°
• cosine rule AB² = 8.2² + 15.6² − 2(8.2)(15.6)cos 33°
• label the diagram correctly (72 or 39 put in the wrong place)
• sine rule or Pythagoras are invalid
• AB ≈ 9.8 km

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

Examiner's comment

Part (b) was generally well answered, with the most common, concise, explanation referring to it being unlikely that the engineer would be able to travel in a perfect straight line (either because of obstacles or because the road would not be straight). Other successful responses included challenging the assumption that the base of the masts was on the same horizontal plane. For those who did not score the mark, it was often because they were talking about the numbers being rounded or the bearings being inexact. Some candidates focused on speed and having to stop, i.e., referring to time rather than distance and appeared to misunderstand the requirement of the question.

Common mistakes

• the route will not be a straight line (roads, obstacles) or the bases may not be in the same horizontal plane
• comments on rounding or inexact bearings score nothing
• comments about time or speed miss the point

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

Examiner's comment

Generally most candidates found this question accessible, working confidently in radians throughout, although inadvisable conversions to degrees were not uncommon with many candidates unaccountably preferring the unwieldy 2 360 r to the elegant 2 1 2 r . In part (a) a rigorous solution was expected; a very large number of candidates lost the second mark by failing to explicitly show the process of dividing the arc length by the angle (perhaps the mark most commonly lost on this question). Candidates need to appreciate that every step of working needs to be clearly demonstrated in order to gain full marks for “show that” questions and that their final result needs to be what the question asked them to show in the first place. As the question was worth 2 marks, candidates should not expect to just state 2.3 27.6 r = and for this to be sufficient. Those who chose to convert 2.3 radians to degrees not only lost time but also invariably lost the accuracy mark. Others did show the division and did not label their answer so just an expression was seen.

Common mistakes

• s = rθ
• the division 27.6 ÷ 2.3 must be shown before OC = 12

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

Examiner's comment

Generally most candidates found this question accessible, working confidently in radians throughout, although inadvisable conversions to degrees were not uncommon with many candidates unaccountably preferring the unwieldy 2 360 r to the elegant 2 1 2 r . Part (b) was typically answered well, with many candidates recognising the straight line giving a total of radians and allowing them to solve the problem. Where candidates lost marks, this was often due to a lack of brackets, e.g. - 2.3 ÷ 2 = 0.421, or an incorrect joined statement such as 0.842... 2.3 0.421 2 - = = . A small minority worked in degrees which needed much more work to secure both marks.

Common mistakes

• (π − 2.3) ÷ 2 with brackets
• stay in radians

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

Examiner's comment

Generally most candidates found this question accessible, working confidently in radians throughout, although inadvisable conversions to degrees were not uncommon with many candidates unaccountably preferring the unwieldy 2 360 r to the elegant 2 1 2 r . There was a wide range of marks for part (c), although many candidates produced highly competent solutions and gained full marks. Most candidates seemed familiar with the area of a sector formula and the vast majority scored both marks for this. Candidates also seemed fully aware of the area of a triangle formula, although they did not always use a correct method to work out the length 15.7. Some candidates seemed not to realise that they needed to use the information provided, i.e. the length of the front of the stage, in order to obtain all the lengths needed for their area calculations. There were also some more complex and longer methods to find the area of a triangle which were not generally successful, but the most common error was using 12 as opposed to 15.7. Some candidates spent time working out the length of AB but then realised that this was not necessary.

Common mistakes

• area = sector + two triangles
• OB = (35 − 27.6)/2 + 12 = 15.7 (common error: using 12)
• ½ab sin C

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

Examiner's comment

This question involving the use of trigonometric identities and solving a trigonometric equation was generally very well answered with the exception of part (c). There were very few blank scripts. Part (c) was much less successful, with only a small number of candidates gaining any marks. Most candidates only identified one of the adaptations to the question: either the stretch or the increase in the range. The most common incorrect answer resulted from them multiplying their number of solutions by 5 to achieve a total of 10 solutions. Of those candidates who realised there were 30 solutions nearly all were able to give a satisfactory explanation.

Common mistakes

• 3x stretches: 3 periods per 360° and 1800° is 5 lots of 360°, so 15 × 2 = 30 solutions
• multiplying by 5 only (10) is the common error
• explanation required

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

Examiner's comment

A significant number of candidates found this question very challenging, with a fair proportion of blank answers, and another significant number working with various substitutions in (a) that were quickly abandoned. Those who were able to work with both the quadratic and trigonometric models confidently were able to score highly and demonstrated a good understanding of how mathematics can be used to model real life contextual problems. In part (c), a significant number of candidates had no real idea how to start this. Many attempts involved fruitlessly expanding cos(9 ) t + but most candidates who successfully found and did not do this. The candidates who realised that when 60 H = , cos(9 ) 1 t + = which meant 31 = had the most success. Candidates who did this typically then substituted 20 t = into cos(9 ) 1 t + = to find a value of . Candidates who found one value typically found the other. There were a number of candidates who found the two values that did not ‘find a complete equation’ as the question required. However, there were a pleasing number of fully correct solutions, and almost all candidates who found and continued to give the full model.

Common mistakes

• β = 31 (mid-height), α = 180 so that H = 2 at t = 0

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

Examiner's comment

For the most part candidates answered this question well. This was dependent on whether they correctly understood the iteration formula. In part (a) (i), although many scored this first mark, there was a common misunderstanding of how a recurrence relationship works with confusion as to the meaning of n. Rather than term position, it was often confused with 𝑢𝑛 so that the incorrect calculation ( ) 35 2 35 35 7cos 5 1 2 u = + - - was often seen, giving the correct answer but obviously not scoring the B mark. Also not uncommonly seen was n replaced by 1 n u + , with candidates stating ( ) 40 2 40 40 7cos 5 1 2 u = + - - . The errors stated above also affected the performance of some candidates in part (ii). Furthermore, a minority of candidates had difficulty with accuracy in the sign of the final term, struggling with the odd/even power of -1 . For example, when calculating 3 u , ( ) 2 3 2 40 7cos 5 1 40 7 5 38 2 u = + - - = - + = , was not uncommon. Occasionally, candidates were seen to be working with degrees rather than radians, also leading to loss of the accuracy mark. In part (ii) many candidates recognised how to construct the arithmetic calculation to find the required sum. The most common of these attempts were: ( ) 1 2 3 4 1 6 u u u u u + + + + and ( ) 2 3 4 1 6 7 u u u u + + + Credit was given to candidates who used such ideas with incorrect values from earlier in the question for 3 u and 4 u , although those who had a wrong answer in (a)(i) who used their incorrect value for 2 u (rather than the given 2 40 u = ), lost both marks. It is important that centres make candidates aware that they must use the ‘show that’ answer in subsequent calculations no matter what they achieve in their working. Occasionally, candidates used ( ) 25 1 2 3 4 1 6.25 r r u u u u u = = + + + but these usually did not add 1 to give a fully correct method. Some also incorrectly tried summing the series as though the terms formed an arithmetic progression. Some candidates incorrectly multiplied (u1 + u2 + u3 + u4) by 4 instead of by 6. A minority forgot to add 35 to their total. A very small minority of candidates generated all the terms of the whole sequence and added them together.

Common mistakes

• n is the term position: u₂ = 35 + 7cos(π/2) − 5(−1)¹ = 40 (using n = 35 scores 0 even though it gives 40)
• u₃ = 28, u₄ = 33
• work in radians
• sign of (−1)ⁿ

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

Examiner's comment

Candidates found part (b) more challenging, with many failing to link areas of mathematics, in this case trigonometry and series. Even where a candidate successfully found the sum, 9cos 36sin S x x = + , many did not relate it to part (a). Those who did recognise the form were usually able to get 4.5 " " R , usually 4.5 2 17 for part (i). Some seemed to misread the question and stated the minimum value, so lost the accuracy mark. A significant number of candidates found a correct expression in harmonic form but lost both the M1 and A1 as they kept a correct 9 17 embedded in their expression.

Common mistakes

• common difference sin x: S₉ = 9cos x + 36sin x = 4.5(2cos x + 8sin x), so max = 4.5R = 9√17 (not left embedded in an expression)
• occurs when x = α ≈ 1.33
• many did not link to (a)
• giving the minimum

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

A number of candidates lost all marks on this question by not attempting any solution. Most candidates correctly expanded the given expression and used sin2 x + cos2 x = 1 to simplify and score the first two marks. A small number incorrectly used (sin x - cos x)2 = sin2 x - cos2 x, scoring no marks. Others attempted to substitute values of 90 and 180 into the given expression instead of using algebraic manipulation. Whilst most candidates scored the first two marks, only the very best were able to score the final mark, with most candidates failing to offer a convincing reason for a contradiction. Candidates who drew a graph of sin x and cos x or sin 2x to support their understanding tended to score well. Successful candidates typically explained that 2x must be between 180 and 360, and hence sin 2x must be negative, which is a contradiction with their inequality. Those that worked with sin x cos x were generally more successful in achieving this mark than those who chose to use sin2x. However, it was clear that not all students understood what an obtuse angle was. Another typical correct conclusion involved candidates stating if x is obtuse, sin x must be positive and cos x must be negative, leading to a negative product and a contradiction. Of those who were able to offer a reason for the contradiction, some still lost the final A mark for failing to then reach a conclusion that the original statement was true or because of errors in the proof such as missing x’s or mixed variables.

Common mistakes

• expand (sin x − cos x)² = sin²x − 2sinxcosx + cos²x (not sin²x − cos²x)
• use sin²x + cos²x = 1 to get −2sinxcosx < 0
• obtuse so sin x > 0, cos x < 0 (or 180° < 2x < 360°, sin2x < 0): contradiction, so sin x − cos x ≥ 1
• substituting 90° and 180° is not a proof
• many did not explain the contradiction

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

Examiner's comment

In part (a), the vast majority of candidates recognised that the horizontal component of velocity of the particle remained constant throughout the motion; this was then used, together with the horizontal distance, to form an equation in T and alpha and to derive the given result. The second part, however, provided a greater challenge and, although many occasional sign errors. They mostly then substituted the expression for T from part (a) to obtain an equation in alpha although slips were not uncommon at this stage. Those starting with different suvat equations tended to make little significant progress. To achieve the given result it was necessary to use the trig identity sec2 alpha = 1 + tan2 alpha (or equivalent work) to obtain an unsimplified quadratic equation in tan alpha before reaching the given equation. This was the main stumbling block for many and it was not unusual to see candidates make more than one attempt. Since the final equation was given and the demand of the question was “show that”, it was essential that all steps in the working were correct and the final answer was stated exactly as printed in order to achieve all the marks. In part (c) height as half the answer in part (a) or choosing sin alpha = 1 to give the maximum height in an otherwise correct suvat equation. Rounding errors were not uncommon leaving the final answer incorrect to 3 significant figures. In the final part, most candidates were able

Common mistakes

• horizontal velocity 28 cos α is constant: 28 cos α × T = 40

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

Examiner's comment

given result. The second part, however, provided a greater challenge and, although many realised that they needed to consider the vertical motion, a few did not know how to proceed, with some solving the given quadratic in tan alpha rather deriving it. However, many 1 = + 2 s ut at did make progress, using with a vertical distance of 20 although there were 2 occasional sign errors. They mostly then substituted the expression for T from part (a) to

Common mistakes

• vertical motion 20 = 28 sin α T − ½gT²
• eliminate T using (a)
• use sec²α = 1 + tan²α to reach a quadratic in tan α
• derive (not solve) the given equation and state it exactly

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

Examiner's comment

symbols in the same order, to achieve the final mark. Part (c) required a calculation of the coefficient of friction. There were many fully correct solutions involving vertical and horizontal resolution but many incorrect responses were also seen. The majority who tried other approaches, such as moments about B or resolving parallel and perpendicular to the rod, were usually unsuccessful as they would often miss out a term or there would be a sin/cos error. Some candidates attempted to resolve vertically and horizontally but included an extra force. It was not uncommon for the horizontal resolution F = S to be correct but an extra force would appear in the vertical resolution. Almost all recalled F = mu R and tried to use it in some way. Part (d) involved finding the magnitude of the many who, having made an error in part (c), were still able to apply Pythagoras’ theorem correctly with their values of F and S. However, some candidates did not understand what was required with R = mg, or an attempt to just combine R and F by adding, seen quite often. A small number failed to deal properly with the square root making an error with the simplification in terms of mg in an otherwise correct solution. In the final part, the

Common mistakes

• resolve vertically R = Mg and horizontally F = S
• F = μR in limiting equilibrium
• μ = 2/3
• extra forces in the vertical resolution
• moments about B or resolving along the rod often lose a term

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

Errors in part (b) were few and far between. The main problems that occurred included incorrect manipulation of the sine rule (usually scoring M1A0) and the use of the incorrect sides in the sine rule (scoring M0A0). There were some very long-winded routes seen, with the area formula attempted twice and the cosine rule used, both with limited success.

Common mistakes

• sides 5, 19, 21
• sine rule 5/sin θ = 19/sin 60° with each side opposite its angle (or cosine rule)
• θ = 13.2°

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

Examiner's comment

This vector question was set within the context of a rhombus. Surprisingly, many candidates were unable to use (or remember) the basic geometric of a rhombus and over 30% did not manage to make any progress. A vast array of approaches were in evidence in part (b) to find the area of the rhombus. The easiest and most direct approach was to find 1 2 PR QS x . A more common method used involved finding the length of one diagonal, followed by the cosine rule to find an internal angle and then use of the formula 'ab sin C '. Unfortunately most attempts using this method only scored 3 marks as it was usual for answers to be given as 22.7 rather than the one required, Ö517. Other methods seen involved use of Heron’s formula, the “shoelace” determinant method, scalar product and vector product….. with many successfully used. Students who added a “narrative” – so explaining what they attempted to do – were in a better position to gain credit if they suffered a processing error early in their solution. This has to be considered good practice. The most common incorrect solution, seen many times over was in stating area = 29 29 29 PQ PS x = x =

Common mistakes

• diagonals PR = PQ + QR = 7i + 3j − 6k, QS = −PQ + QR = 3i − 3j + 2k
• area = ½ × PR × QS = √517 (exact)
• the cosine-rule/½ab sin C route often gave 22.7 not √517
• over 30% made no progress

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

Examiner's comment

This question was another big discriminator. Good candidates found it familiar and accessible with some 25% scoring full marks. Strangely, for candidates who dropped marks, they seemed to be able to do either part (a) or part (b), but not both. For those candidates who attempted part (a) most recognised the need to use the compound angle expansions. Carelessness with the multiplication by 2, signs and use of the formulas was seen in a small minority of the scripts. Students who attempted to use sin (A+B) and cos (A+B) with a negative angle B, often introduced sign errors; it is clear that such a method is not to be recommended. Substituting exact numerical values was usually correct, however candidates occasionally fudged their working to get the given answer.

Common mistakes

• expand both compound angles
• exact values of sin/cos 30° and 60°
• collect terms and use tan x = sin x/cos x
• doubling slips and fudging to the given answer

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

Examiner's comment

This question, based on the minimum surface area of a children's toy proved to be a useful source of marks so late in the paper. In fact 25% of the cohort achieved full marks. Many candidates who could not access part (a) were able to use their skills in differentiation to find an appropriate value for the radius and prove that their value gave a minimum surface area. Unsurprisingly, part (a) was the most demanding part of this question. The key was to find an expression (using the volume of 240 cm3) for h in terms of r. Unfortunately some solutions featured the volume of a sphere, which proved a costly error, from which there was no return. Candidates produced their formula for surface area in a variety of ways – sometimes leaving it to the last line to substitute in the value for the angle, occasionally replacing 0.8 by 4/5. As with other “show” questions, sufficient detail has to be written to convince the reader that progress is genuine.

Common mistakes

• volume ½r² × 0.8 × h = 240 gives h = 600/r²
• S = two sectors 0.8r² + rectangles 2rh + curved face 0.8rh
• substitute to get the given result
• volume of a sphere used wrongly

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. It was good to see students who were unsuccessful at part (a) still attempting part (b), and in some cases scoring full marks. Most candidates were able to integrate the first two terms correctly, but a significant proportion were unable to integrate the third term to get an expression in 3 sin .t The majority of correct solutions came from the use of the reverse chain rule for the final term. Those who attempted to use integration by substitution or by parts were usually unsuccessful. Candidates who had successfully completed the integration often went on to score full marks.

Common mistakes

• 8t − 2 sin 4t + 16 sin³t by reverse chain rule on 48 sin²t cos t (by parts/substitution rarely worked)
• limits 0 and π/4: 2π + 4√2 exact

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

This was a challenging question for the majority of the candidates with very few scoring most 1 or 2 marks. These were the initial B1 mark for realising that 𝐴𝐴= 50, however all four of the marks in part (a). For those that attempted the question, most scored at many wrote that 𝐻𝐻= 50 and did not relate that 𝐻𝐻 meant 𝐴𝐴. The second B mark for finding a value for 𝑏𝑏 was very poorly answered, with many not realising a simple 720 was needed and attempted to use an ‘𝑅𝑅-𝛼𝛼’ method or failed to take 180 calculation of d𝐻𝐻 into account transformations of the sine function, showing a lack of understanding of d𝑡𝑡 with very limited success. A the properties of the sine wave. Some also tried to use small minority of students achieved b = 1/4. The best answers tended to use transformation arguments, recognising that the full cycle depicted must represent a scale factor of 4 though many students attempting this approach did not appreciate the effect of the modulus in determining the value of b, obtaining instead a value of b = ½. A common incorrect method came from consideration of the wheel taking 720 seconds to complete 1 revolution leading candidates to solve the equation 720b + 1.15 = 180 or from thinking that the maximum occurs at t = 360 leading to 360b + 1.15 = 90. Of those 𝜋𝜋 2𝜋𝜋 that managed to access the question, few mixed radians with degrees and managed to 720 or 720, which scored no marks. 1010 was the most common mark profile for this question, as a significant number of students went on to find 𝛼𝛼 correctly give an answer as (albeit occasionally in radians) as they were able to spot that at 𝑡𝑡= 0, 𝐻𝐻= 1. Sadly, a few that did achieve all three elements did not conclude with an equation of the model and so lost the (easy) last mark, with this final mark being scored very rarely, at less than 5% of the cohort (the second least commonly scored mark in the paper). In part (b), many candidates incorrectly referred to the fact that the passenger was 1 m above the ground and related the value of 𝑑𝑑 to this fact, for the most part because they had not appreciated that the original model already took this into account. Some responses referred to the fact that the height needed to be a positive value or could not There were many attempts at part (b) without any attempt at part (a). It was unfortunately quite common that responses to this part were very difficult to read and centres need to stress to candidates that work cannot be awarded marks if it is illegible.

Common mistakes

• A = 50
• the modulus means one revolution is half a sine period, so 180/b = 720 gives b = ¼ (not ½ or 1/720)
• α from 1 = 50 sin α, α = 1.15 (degrees)
• give the full equation H = modulus of 50 sin(¼t + 1.15)°

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

• a = −6√3, r = −1/√3 (modulus less than 1)
• S∞ = a/(1 − r) and rationalise the denominator to 9(1 − √3)
• using θ = π/6 gives 9(1 + √3), which does not match the given form
• steps must be shown, not calculator

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

Examiner's comment

Almost a quarter of the candidates scored zero whilst 21% scored full marks on this question. In part (a) the most common approach was to resolve perpendicular and parallel to the plane to first find a value for X (49) before finding a value for the friction. Very few used the most efficient method and resolved vertically. The resolving was generally done well with most candidates resolving all the forces that needed to be resolved and most without sin/cos confusion and sign errors. Many who reached the correct value for friction were able to give the correct direction of the force, including those that had assumed that it acted up the plane and had obtained a negative answer. Not all remembered to give the magnitude of the force and numerical calculation errors were seen in some working. mu R was seen in some responses which lost the last mark as friction is not limiting in this part of the question. Part (b) was generally done well, even by those that had struggled with the first part, with many candidates writing down a correct F = ma equation. The most common mistake was to take R as 68.6N, the value given in part (a), and not resolve to find the new value of R. Those that resolved perpendicular to the plane generally went on to use F = mu R correctly and then accurately calculate the acceleration of the block. A few candidates did not remove the force X. It was noticeable that candidates who produced well labelled diagrams and who presented their solutions in a logical, neat and concise manner often continued to produce a correct solution in both parts.

Common mistakes

• resolving vertically is quickest (most found X = 49 first)
• sin α = 3/5, cos α = 4/5
• friction is not limiting here so μR is wrong
• F = 9.8 N must be positive
• sin/cos interchanged gives 9.8 in the wrong direction

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

Examiner's comment

Almost a quarter of the candidates scored zero whilst 21% scored full marks on this is not limiting in this part of the question. Part (b) was generally done well, even by those that had struggled with the first part, with many candidates writing down a correct F =

Common mistakes

• X removed so R must be re-resolved: R = 5g cos α, not 68.6
• do not reuse F = 9.8 from (a)
• F = 0.5R
• 5g sin α − F = 5a gives a = 1.96 m s⁻²

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

Examiner's comment

In the second part, it was required to derive a given expression for T (tension). Although some candidates struggled to make a valid start and basically just wrote down the answer and stated ’by resolving ’, it was well done by a fair number who set up a correct equation for moments about A and hence deduced the given result. Alternative approaches by either taking moments about other points or resolving forces were very rare and almost never successful, mostly because of missing terms. A few candidates failed to use perpendicular distances for the moments of the weight terms or failed to include the weight of the rod leading to a missing term. Part (c) was generally well answered by those candidates who

Common mistakes

• take moments about A to avoid F and R: T × 2a = Mga cos θ + 2Mg × 1.5a cos θ
• use perpendicular distances
• include the rod's own weight
• just quoting the result "by resolving" scores nothing

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

Examiner's comment

leading to a missing term. Part (c) was generally well answered by those candidates who resolved vertically, substituted for T and re-arranged to obtain the given answer thereby gaining full marks in a few lines of working. Candidates who used alternative equations as a starting point were generally unsuccessful, usually a result of a missing term in one or both equations used, this often involving friction. In part (d) those candidates who

Common mistakes

• resolve vertically R + T cos θ = 3Mg with T = 2Mg cos θ and cos θ = 3/5
• other equations need F eliminated and often miss a term

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

Examiner's comment

or both equations used, this often involving friction. In part (d) those candidates who resolved horizontally to give F = T sinƟ usually managed to score all four marks for deriving the exact value of the coefficient of friction. There were a variety of equations, resolving or taking moments, which would have led directly to a value for F since the tension and normal reaction were known at that stage; however, these involved more working and therefore greater opportunity for error. Nevertheless, successful solutions by these methods were seen on occasion. Most of those candidates who found an expression for F were able to achieve the independent method mark for the use of F = μR even if their method for finding F was incorrect. Since all the answers were given in this question, candidates needed to ensure that they showed sufficient detail in their working to warrant being awarded all the available marks. In particular, candidates should be encouraged to state their resolved forces and moments as equations, rather than considering them in isolation, and to ensure the substitution stages are shown.

Common mistakes

• resolve horizontally F = T sin θ (sin θ = 4/5)
• F = μR with R = 57Mg/25 must be used, not just quoted
• show substitution as the answer is given

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

Examiner's comment

In part (a), the majority of candidates found an equation for the horizontal motion in terms of t. Most then used this to substitute t =120/Ucos alpha into 0 = Usin alpha t – ½ gt2, the vertical motion equation for the whole trajectory. This was generally rearranged to give the required answer although cancelling the common factor of t from the quadratic before substitution led to a neater solution. A minority chose to consider just half the motion with v = u + at and v = 0 at the greatest height which also led to a neat solution. Some candidates failed to identify that the vertical displacement would be zero over the whole motion, and some were inconsistent in considering the time for the whole or half trajectory. A small but surprising number of answers were seen where 120 m was used as the vertical distance. It should again be emphasised that candidates, when required to derive a given result, should write the final answer in exactly the same form as that stated in the question. In the second part, many candidates correctly arrived at an equation in from part (a) and, of those who did, some struggled with the required algebra. Many who correctly identified the value of alpha as tan-1 (1/3) used the decimal value for the angle in degrees rather than calculating an exact value for sin alpha with which to substitute, and lost the final mark. A few candidates set up an equation by using s = ut + ½at2 with s = 10 and substituting in the time to travel 60 m horizontally, but again had trouble finishing off for the last 2 marks having more difficulty with the slightly more complicated algebra required. Part (c) required a comparison of the initial velocity in the model without air

Common mistakes

• horizontal U cos α × t = 120 and vertical U sin α × t − ½gt² = 0 (or half-way with 0 = U sin α − gt and 60 m)
• eliminate t
• inconsistent whole/half times
• 120 m used as a vertical distance
• final line must match the given answer

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

Examiner's comment

in the question. In the second part, many candidates correctly arrived at an equation in Usin alpha using v2 = u2 + 2as although a failure to resolve the vertical velocity giving U2 = 196 was not uncommon. Most were unable to progress further and so only gained 2 out of a possible 4 marks. For example, they just left it there or found the “correct” answer without any justification for the value of alpha . Some did not realise they could use the answer from part (a) and, of those who did, some struggled with the required algebra. Many who

Common mistakes

• 0 = (U sin α)² − 2g × 10 (not U² = 196 unresolved)
• divide by the result of (a) to get tan α = 1/3 then exact sin²α = 1/10
• using α = 18.43° loses the final mark
• verification from U² = 1960 not accepted

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

Examiner's comment

This was another question where the modal mark was full marks, this time achieved by over 40% of the candidates Part (b) The vast majority of candidates made some progress in this part, realising that the lengths of the vectors would be required to solve this problem. A small minority then simply stopped, having done no more than finding the lengths of the sides of the triangle ABC, with some assuming that the triangle was right angled. However most then went on to use the cosine rule to try to show the required result. In almost all cases a correct cosine rule was stated and used, so that following on from a previous accuracy error, the method marks could be awarded. Fully correct calculations were necessary in order to gain the final accuracy mark and very many candidates managed to do this and so gained full marks for this question. A few candidates, presumably those who had studied further maths, used the scalar correct way around. Many using this method showed that cos ABC = - 9 product. This method would be perfectly acceptable, as long as the vectors are used the 10 but failed to explain why the answer should be positive.

Common mistakes

• find magnitudes √50, √18, √14 and use the cosine rule with the sides in the right positions
• scalar product (BA.BC, not AB.BC) gives −9/10 if the direction is wrong

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

Examiner's comment

Question 7 was testing a candidate’s ability to apply trigonometry to solve a problem relating to a parallelogram. This was one of the most challenging questions on the paper and was largely as a result of the candidates having to draw their own diagram. In part (a) there were many incorrect diagrams to begin with, which included incorrect labelling of the vertices. Some tried to find the required angle by splitting the parallelogram into a rectangle and two triangles. However, they made little progress with this and often just stopped. Other candidates who did correctly set up an equation using the sine rule, then forgot to find the obtuse angle. Some subtracted the acute angle from 360°.

Common mistakes

• draw your own diagram
• area of parallelogram = 14 × 7 × sin SPQ = 50 (not ½ab sin C)
• obtuse angle is 180° − arcsin(50/98) = 149.32°
• not 360° minus the acute angle

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

Examiner's comment

Question 7 was testing a candidate’s ability to apply trigonometry to solve a problem relating to a parallelogram. This was one of the most challenging questions on the paper and was largely as a result of the candidates having to draw their own diagram. In part (b) candidates often tried to apply Pythagoras to the two given sides when there was not a right angle and although some did proceed with the intention of applying the cosine 6 rule, the sine function or omission of the square root sometimes prevented a correct answer being achieved.

Common mistakes

• cosine rule SQ² = 14² + 7² − 2 × 14 × 7 cos 149.32°, then square root: 20.3 cm
• Pythagoras is wrong as there is no right angle

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

Examiner's comment

This was found to be another discriminating question. Those who were familiar with the topic made good progress yet there were many other attempts that scored no marks at all. Part (b) A significant number of candidates used the result of (a) correctly and started their solution with tan2x=3sin2x. However, it was also the case that tanx =3 sin2x was seen several times as well. Of those candidates who started correctly with tan2x = 3sin2x, the significant majority simply cancelled out the resulting factor of sin2x and in so doing neglected to consider the possibility that sin2x=0 and in so doing the solution x = 90° was omitted from the fully correct list of three answers. Many candidates however, did state that cos2x =⅓ and went on to get 35.3° and 144.7° and so only one mark was lost.

Common mistakes

• replace θ by 2x: tan 2x = 3 sin 2x (not tan x)
• factorise sin 2x(1 − 3cos 2x) = 0, do not cancel sin 2x (loses x = 90°)
• x = 35.3°, 90°, 144.7°

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

Examiner's comment

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

Common mistakes

• 5α + 40° = 720° − 53.1° for the fourth positive solution, giving α = 125°
• list the solutions in order and rearrange for α
• make the final answer clear

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

Examiner's comment

Performance on this question was variable and approximately 25% of candidates failed to score any marks. (a) The majority of candidates were successful with this part. Common errors were stating that angle AOB was or , sometimes following through to gain marks in (b) and (c).

Common mistakes

• angle AOB = (π − θ)/2 in radians
• (180 − θ)/2 scores B0

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

Examiner's comment

Performance on this question was variable and approximately 25% of candidates failed to score any marks. marks in (b) and (c). (b) Most candidates followed the correct strategy for finding the area of the logo by finding the sum of the areas of sectors OCD, OAB and OEF. Common errors were omitting the brackets round the 2r when finding the area of the sector OCD and algebraic errors when simplifying their expression for the sum of the two smaller sectors OAB and OEF. A small number of candidates showed insufficient working when simplifying their area of the sum of the two smaller sectors to arrive at the given result.

Common mistakes

• area = two small sectors + ½(2r)²θ
• keep the brackets round 2r
• first term must be seen expanded as ½r²π − ½r²θ

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

Examiner's comment

Performance on this question was variable and approximately 25% of candidates failed to score any marks. marks in (b) and (c). (c) Again, most candidates attempted this part well. A common errors was not to include 4r in the expression for the perimeter of the logo and there were some algebraic errors when simplifying the expression for the sum of the perimeters of sectors OAB and OEF.

Common mistakes

• perimeter must include the four straight edges 4r (OA, OF, BC, ED)
• r(4 + π + θ)

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

Examiner's comment

This question was fairly well answered by a number of candidates, who understood the need to use the sum to infinity formula, but a sizeable majority made little or no progress, not realising for example, that the sum started from n = 2 and not n =1 – hence they often did not get the first mark for a correct first term and/or common ratio. 9/16 was seen fairly often, but not always attributed to the first term, with some candidates thinking this was the value of the sum. One candidate misread the infinity symbol as an 8 and applied the sum to 8 terms correctly, but was obviously not able to score any marks. A number of students found the sum to infinity, using n = 1 to infinity and then subtracting the first term. This was done well by candidates who went down this route. Unfortunately, many students achieved no marks on this part due to a wide variety of errors at the beginning. Calculating cos (360) on the calculator in radians instead of degrees was seen. Some tried to split the summation incorrectly into ∑ ቀ 3 4ቁ 𝑛𝑛 ∞ 2 x ∑ cos 180𝑛𝑛 ∞ 1 and find the sum to infinity of the trigonometric terms separately. A few wrote ( 3 4)2cos (360) but did not evaluate the cos(360). There were some very good solutions but in many responses there was often a lack of clear structure to their proof.

Common mistakes

• cos(180n)° = (−1)ⁿ so r = −3/4
• first term is (3/4)² = 9/16 because the sum starts at n = 2
• calculator in radians gives wrong values
• do not split into two separate sums

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). Part (b) proved more challenging for many candidates, with some unable to make a start as they were unable to process 8 = cosec3 𝜃𝜃 to arrive at either a value for sin 𝜃𝜃 or 𝜃𝜃. Of those that did arrive at a value for sin 𝜃𝜃 or 𝜃𝜃, many were able to work in exact form using their value to find an exact value for their 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑, others were unable to proceed or attempted to give a decimal answer. 24 sqrt 3 and 8 sqrt 3 were common incorrect final answers. There were, however, some excellent and concise responses to this question.

Common mistakes

• cosec³θ = 8 gives sinθ = ½, θ = π/6
• work in exact values
• −24√3 (not 24√3 or 8√3)

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

Examiner's comment

This question discriminated well between candidates that got this far, but it did appear that many candidates had simply run out of time, perhaps indicative of a generally lower understanding than in a June series because of it being a resit paper following the pandemic. Part (b)(i) was usually well attempted by those who had answered part (a), however, some candidates did not recognise the need to double 𝑅𝑅 and arrived at 3 + 𝑅𝑅 rather than 3 + 2𝑅𝑅 instead. Others simply assumed that the maximum was 3, 5 or 7, from a poor understanding of combining two trigonometric functions and the purpose of using the compound angle formulae to combine these two functions. Part (b)(ii) proved to cause problems for many candidates, with the majority incorrectly assuming that 0.5𝑡𝑡+ 𝛼𝛼= 0, and proceeded to ignore that this equation results in a negative value of 𝑡𝑡 (usually -0.9) by simply making this value of 𝑡𝑡 positive. They had failed to recognise that the first maximum would occur when 0.5𝑡𝑡+ 𝛼𝛼= 2𝜋𝜋. Perhaps the use of a sketch or graphical calculators would have helped these candidates. An alternative approach to part (b) required candidates to differentiate 𝐻𝐻, either in its original form or having used part (a) and solve the equation d𝐻𝐻 d𝑡𝑡= 0. This approach was more complicated and resulted in needing to solve tan(0.5𝑡𝑡) = - 1 2 and find the third solution, in order to avoid finding the maximum when 𝑡𝑡< 0 or the first minimum. Generally, candidates taking this approach were unsuccessful in scoring the accuracy mark, although the method mark was scored reasonably frequently.

Common mistakes

• H = 3 + 2√5 cos(0.5t + 0.464), so maximum 3 + 2R = 7.47 m (not 3 + R)
• first maximum when 0.5t + α = 2π (0 gives negative t)
• t = 11.6

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

Examiner's comment

Candidates need to make sure that they read and then answer questions carefully. In part (a) an equation of motion for A was required. Most realized that this required the use of Newton’s second law on the 3m mass. However, some ignored this requirement and went straight to a whole system approach to the problem. Errors in part (a) were mainly due to the omission of a relevant force or a misunderstanding of direction. It was essential that the frictional force on A and the tension were in the same direction i.e. had the same sign in the equation. In part (b) either an equation of motion for B was required or an equation

Common mistakes

• equation for A only (not the whole system): 3mg sin α − F − T = 3ma
• friction and tension both act up the slope (same sign)
• missing a force or wrong direction

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

Examiner's comment

Candidates need to make sure that they read and then answer questions carefully. In part in the equation. In part (b) either an equation of motion for B was required or an equation of motion for the whole system. This was generally well answered provided that the 1 reaction force from the plane on A had been found and used with F to give F R = R mu = 6 . The question was a “Show that” question so full marks were only awarded if there was a fully correct solution, with all the steps included. Errors involved sin/cos confusion, the omission of g, an incorrect direction for the reaction force, omission of a force in an equation of motion or an incorrect reaction such as R = 3mg. Some candidates did make algebraic errors in solving their equations. A few candidates assumed that the tension in the string was mg. Those candidates who attempted a whole system approach were mostly successful. In part (c), most of the graphs given were correct but there are still candidates which asked for “ how this would affect the working in part (b)”. Those who referred to how it would affect the answer received no credit.

Common mistakes

• R = 3mg cos α (not 3mg)
• F = R/6
• sin α = 3/5, cos α = 4/5 (sin/cos confusion)
• T − mg = ma for B or whole-system equation
• tension is not mg
• full working needed for the given answer

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

Examiner's comment

Part (a) required candidates “ To show that…”. Consequently, full marks were only awarded if there was a fully correct solution, with all the steps included. Thus successful candidates needed one moments equation together with a vertical resolution and/or a horizontal resolution, depending on which point they had used for their moments equation. These equations, combined with F , could then lead to full marks being R mu <= earned. Many candidates assumed that the beam was in limiting equilibrium i.e. that , produced an equation and then changed it to the required inequality. These F R mu = candidates lost the final mark. Other errors included omission of g, sin/cos confusion in moments equations and assuming that the reaction at A and the reaction at B were the same. Candidates tended to be more successful in part (b), where a similar strategy was

Common mistakes

• one moments equation (e.g. about A: mga cos θ = 2Sa sin θ) plus resolving R = mg, F = S
• use F ≤ μR to get the inequality (assuming limiting F = μR and then switching to ≥ loses the final mark)
• omitting g
• sin/cos confusion
• reactions at A and B are not equal

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

Examiner's comment

same. Candidates tended to be more successful in part (b), where a similar strategy was required, together with F , since the beam was now in limiting equilibrium. A R = mu common error was to have the friction force acting towards the wall. This lead to a k value of – 0.1 and the negative sign should have flagged up an error to candidates.

Common mistakes

• limiting so F = μR = ½mg
• friction now acts away from the wall (friction towards the wall gives k = −0.1, a negative k should flag the error)
• moments equation plus horizontal resolution N = kmg − F
• k = 0.9

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

Examiner's comment

Candidates were instructed to use g = 10 in this question but were only penalised once if another value was used. In part (a), the majority of successful candidates employed the 1 most efficient method using a single application of to find the time to the 2 s ut at = + 2 ground. Other successful methods found the time to the top of the flight and then the time to go from the top to the ground and added. However, sign errors were common resulting candidates who used 𝑣𝑣2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎 to find the vertical component of velocity of the in 2/4 marks being scored for this part. A few candidates did not resolve the speed of 65 m s-. This lost the M mark and consequently they scored zero for part (a). In part (b), 1 particle at A were generally more successful than those who used the time found in part (a). Many solutions finished here and did not consider the need to look at the horizontal component as well. Those who had both components usually went on to apply Pythagoras theorem and calculate the actual speed of 75 m s-. For part (c), there were several 1

Common mistakes

• resolve the 65: vertical component 65 sin α = 25 (tan α = 5/12)
• s = ut + ½at² vertically with s = −70 and g = 10 (sign errors common)
• up-and-down methods possible
• t = 7 s

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

Examiner's comment

Candidates were instructed to use g = 10 in this question but were only penalised once if candidates who used 𝑣𝑣2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎 to find the vertical component of velocity of the in 2/4 marks being scored for this part. A few candidates did not resolve the speed of 65 m s-. This lost the M mark and consequently they scored zero for part (a). In part (b), 1 particle at A were generally more successful than those who used the time found in part (a). Many solutions finished here and did not consider the need to look at the horizontal component as well. Those who had both components usually went on to apply Pythagoras theorem and calculate the actual speed of 75 m s-. For part (c), there were several 1

Common mistakes

• need both components: horizontal 65 cos α = 60 (often omitted) and vertical 45 (v² = u² + 2as is more reliable than using t)
• Pythagoras gives 75 m/s

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

Examiner's comment

Candidates found the first part of this question challenging, but they were more successful with the second part, with many fully correct responses. There were a number of common mistakes in candidates’ approaches to part (a). Firstly, many candidates tried to find the resultant vector by adding the two given vectors, rather than subtracting them. Secondly, candidates seemed unclear on exactly what angle the question was looking for, with lots of time being spent on finding various different (but unfortunately not relevant) angles in their diagrams. Finally, some candidates took a rather long-winded 4 approach to find the correct angles (through use of the cosine rule to find one part of the required answer, then additional work using tan) as they had not spotted the more straightforward use of 7 arctan 3 . Even those who found a relevant angle were often unable to proceed to find a bearing.

Common mistakes

• displacement is (−3i − 5j) − (4i − 2j) = −7i − 3j (subtract, not add)
• find the right angle (arctan 7/3) and convert to a bearing 180° + 66.8° = 246.8°

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

Examiner's comment

With the exception of the first 2 marks, this was quite a polarising question that either scored very well or little at all. Those who went through to part (b) with an incorrect diagram and understanding of the question found it very difficult to make sense of the sides. It was only after abandoning this line of thought and starting afresh did some candidates manage to salvage the question. In part (a), the first couple of marks were scored by the vast majority of candidates using the sine rule to find 51.1 degrees. From there, an overwhelming number neglected to solve for the obtuse angle; those who did left both answers. Many even labelled their obtuse angle on the diagram as 51.1 degrees. Candidates should always be checking solutions to equations to ensure they make sense in context and are well defined. As the 128.9 degrees was appearing as a complementary angle in part (b), it was important to state it unambiguously here as credit could not be gained for it seen later in the question. In part (b), depending on the answer to part (a), there were a couple of possible configurations that would lead to different answers. After labelling the triangle, candidates had the choice of the sine rule or cosine rule to find the missing side AD. Those who 6 mislabelled in part (a) tended to over work this, believing they had only found AC and still had to find another side, so it was condoned that they may have started again, allowing full marks to be achieved if they reached the answer of 42 m. Candidates should keep sight of the bigger picture by looking at the sum of angles in the right most triangle. Hardly any noticed that they had two base angles of the isosceles triangle of 128.9 degrees. There should always be an internal consistency, even when some individual value might have been calculated incorrectly.

Common mistakes

• sine rule gives 51.1°
• ambiguous case: angle ACB is obtuse so 128.9° (state it here, not just later)

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

Examiner's comment

With the exception of the first 2 marks, this was quite a polarising question that either scored very well or little at all. Those who went through to part (b) with an incorrect diagram and understanding of the question found it very difficult to make sense of the sides. It was only after abandoning this line of thought and starting afresh did some candidates manage to salvage the question. In part (b), depending on the answer to part (a), there were a couple of possible configurations that would lead to different answers. After labelling the triangle, candidates had the choice of the sine rule or cosine rule to find the missing side AD. Those who 6 mislabelled in part (a) tended to over work this, believing they had only found AC and still had to find another side, so it was condoned that they may have started again, allowing full marks to be achieved if they reached the answer of 42 m. Candidates should keep sight of the bigger picture by looking at the sum of angles in the right most triangle. Hardly any noticed that they had two base angles of the isosceles triangle of 128.9 degrees. There should always be an internal consistency, even when some individual value might have been calculated incorrectly.

Common mistakes

• triangle BCD is isosceles (base angles 51.1°)
• AD ≈ 15.09 by cosine or sine rule
• total 41.09 m rounds UP to 42 m of steel to buy

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

Examiner's comment

This was the second modelling question on the paper. Part (b) was also well done with most of those candidates giving the exact answer for the temperature. Occasionally candidates obtained an answer of 8° C from 5 + 1 + 2, which they obtained from taking the maximum values of sin and cos.

Common mistakes

• θ = 5 + √5 sin(πt/12 + 1.107 − 3), so maximum 5 + √5 = 7.24 °C
• adding the maxima of sin and cos separately (5 + 1 + 2 = 8) is wrong

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

Examiner's comment

This question was attempted by most candidates with many being able to secure nearly all the marks in part (c). It was rare, however, for full marks to be scored overall, with most struggling with the transformations in part (b). In part (a), candidates occasionally got their signs mixed up, or did not appreciate the vertical stretch of scale factor 3 and therefore 1 d = -. Some even had their d being positive, despite the diagram showing that P was below the horizontal axis.

Common mistakes

• (−180, −3): stretch factor 3 gives d = −3
• c is negative

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

Examiner's comment

This question was attempted by most candidates with many being able to secure nearly all the marks in part (c). It was rare, however, for full marks to be scored overall, with most struggling with the transformations in part (b). In part (b), candidates could still score marks for acceptable follow throughs as long as d was a negative value. This enabled some candidates to score usually one of the marks. Some mixed up the horizontal stretch of scale factor 4 and scale factor 1 4 in part (i) and mixed up a translation of 36 units in the positive x-direction with 36 units in the negative x-direction. There were some candidates who only stated the x coordinate whereas the question asked for the coordinates, so they were unable to score in this part.

Common mistakes

• (i) x/4 is a horizontal stretch factor 4: (−720, −3) (not ¼)
• (ii) translation +36 in x: (−144, −3)
• give both coordinates

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

Examiner's comment

This proved to be a challenging question with very few candidates achieving full marks (a) Candidates used a variety of approaches. The most common started by using algebra to find the coordinates of the points where the circles meet. This was achieved correctly by a majority of those who did the question this way. Then the required angle was found either by trigonometry in a right angled triangle and doubling an angle, or by the cosine rule in triangle AOB. Errors in this approach included processing x2+y2=100 into x+y=10. Although a minority of candidates’ answers stopped after finding coordinates for A or B, there were a significant number who went on to score the remaining 2 marks. However, some did not seem to realise that length OA=10 as it is a radius of C1, and had to use the distance formula to find this length. Another approach that was seen, but not so frequently was to use knowledge of circle equations in finding the radii and centres, to establish the lengths of the sides of triangle OAX, with X being the centre of circle C2. Using the cosine rule then led to half of the required angle. When used, this method often gained all four marks. (See relevant diagram below)

Common mistakes

• solve the circle equations simultaneously, x = 9.5 (x² + y² = 100 is not x + y = 10)
• OA = 10 is the radius
• cos α = 9.5/10, AOB = 2α = 0.635

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

Examiner's comment

This proved to be a challenging question with very few candidates achieving full marks overall, and with many more difficulties encountered in (b) rather than (a). (b) It was very common to see just the first mark scored in this part. This was due to many candidates wrongly assuming that the given angle AOB=0.635 was the same size as the angle AXB at the centre of circle C2. For the minority of candidates who tried a correct method to find the size of angle AXB, or half of it, most of these achieved the correct values of either 1.03 or 0.516. Even for the minority of candidates who found one of these angles correctly, getting the final answer correct proved a challenge and only a very small minority managed this. Note that it was possible to do part (b) even if (a) was not attempted. However, it seemed that failure to find a way to do (a), led in nearly every case to no valid attempt at (b).

Common mistakes

• major arc of C₁ is 10(2π − 0.635)
• angle AXB at the centre of C₂ is 1.03, not 0.635 (very common error)
• add √40(2π − 1.03)
• perimeter 89.7

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

Examiner's comment

Part (b), however was very often left completely blank. Many who attempted it failed to see the connection with (a) and got lost in complicated trig expansions which led nowhere. Of those who did use part (a) to answer part (b), the most common solution was simply 𝑥𝑥=25 after cos 𝑥𝑥 had been cancelled, thus losing the solution given by cos 𝑥𝑥 =0. Only a small minority of the most able found all 3 solutions.

Common mistakes

• use (a) to get cos x cot x = cos x cot(3x − 50°)
• do not cancel cos x without cos x = 0 (x = 90°)
• cot x = cot(3x − 50°) gives x = 25° and, as cot has period 180°, x = 115°
• three solutions

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)(ii), many knew that the integral was a multiple of cos3 t but had trouble getting to the printed answer of 20 with correct steps. As the answer was given, it was a requirement to see the limits calculated separately as 0 – ( -20 ) = 20. Integration by parts was a common approach for unsuccessful candidates, while others attempted to use a substitution, often not using the necessary u = cosx. Those candidates with an understanding of the integration of f’(x)[f(x)]n were generally very successful here. Some lost the penultimate accuracy mark usually due to a sign error.

Common mistakes

• reverse chain rule (or u = cos t) gives −20cos³t
• limits must be shown as 0 − (−20) = 20
• by parts is the wrong approach

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

Examiner's comment

The first part was well done by the majority of candidates who resolved perpendicular to the plane to give a correct expression for the normal reaction. In part (b), most resolved

Common mistakes

• resolve perpendicular to the plane: R = mg cos α = 4mg/5 with cos α = 4/5 from tan α = 3/4
• mg cos 36.86° (decimal angle) is A0
• sin/cos confusion

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

Examiner's comment

the plane to give a correct expression for the normal reaction. In part (b), most resolved parallel to the plane and then used F= mu R to derive the given answer for mu . There was occasional sin/cos confusion seen in the relevant equations but this was fairly rare. Horizontal resolution was an alternative valid approach. The third part was found to be

Common mistakes

• resolve parallel to the plane F = mg sin α (or horizontally R sin α = F cos α)
• F = μR must be used, not just stated, to reach the given 3/4
• limiting friction acts up the plane (on the point of sliding down)

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

Examiner's comment

A significant number of candidates made little or no progress with this question; a number of responses showed a complete lack of understanding of the nature and directions of the forces acting on the ladder. In part (a) a fair number attempted to take moments about A to derive the given reaction at C although sometimes perpendicular distances were not used and this invalidated the method. The most straightforward approach for part (b) was

Common mistakes

• moments about A with perpendicular distances: N × 4a/sin α = Mg × 3a cos α
• AC = 5a, not 6a
• reaction at the smooth rail is perpendicular to the ladder

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

Examiner's comment

A significant number of candidates made little or no progress with this question; a number of responses showed a complete lack of understanding of the nature and directions of the used and this invalidated the method. The most straightforward approach for part (b) was to resolve horizontally and vertically. Some candidates chose to take moments about various points or to resolve perpendicular and/or parallel to the ladder. These methods often led to errors including missing terms and incorrect distances. Some thought the normal reaction at C was horizontal. Those who attempted to use a reaction perpendicular to the ladder generally used this in F= mu R and were unaware that this is only valid for a normal reaction. Confusion between sin/cos in resolving the reaction at C horizontally and vertically was surprisingly common. There were, nevertheless, some well-expressed fully correct solutions seen.

Common mistakes

• resolve horizontally F = (9Mg/25) sin α and vertically R + (9Mg/25) cos α = Mg
• the reaction at C is perpendicular to the ladder, not horizontal
• F = μR only with the normal reaction at A (not 9Mg/25)
• sin/cos confusion
• μ = 18/49 ≈ 0.37

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

Examiner's comment

In part (a), many candidates made some valid progress in creating horizontal and vertical distance equations in terms of t and U. Most tried to eliminate or find t to solve for U but sometimes lacked systematic working and gave up before reaching an answer or simply stated the given answer following incorrect or incomplete work. Those who did not complete a solution for U could go on to gain full marks for part (b) which was generally

Common mistakes

• horizontal U cos 45° t = 100 and vertical U sin 45° t − ½gt² = −25 (sign of the 25)
• eliminate t systematically to reach U = 28
• no approximations or g = 9.81

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

Examiner's comment

complete a solution for U could go on to gain full marks for part (b) which was generally well attempted. Nearly all recognised that the vertical velocity was zero at the highest point. A few failed to use the vertical component of U and so did not achieve the method mark. Others assumed the highest point occurred when the horizontal distance was 50 m, half the range, which was not the case here since the start and end points were not on the same level. The most common error was in giving the answer as a height above the level of projection rather than above ground level. For part (c) there were certain statements

Common mistakes

• vertical velocity zero at the top: 0 = (28 sin 45°)² − 2gh gives 20 m
• add the 25 m cliff, answer 45 m above the ground
• use the vertical component of U
• the top is not at half the range (50 m)

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

Examiner's comment

This question proved to be more challenging for a number of candidates although many were still able to score full marks. In part (a) it was important for candidates to pay attention to the scale of the graph. Most candidates who did draw a straight-line graph were able to make the link between one point of intersection and one real root to the equation, however, a significant minority of these were unable to construct the correct straight-line graph with enough accuracy to gain both marks for this part of the question.

Common mistakes

• rearrange to cos x = 2x + ½ and draw that line (intercept ½, gradient 2: watch the scale)
• one intersection so one root, a reason is needed

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

Examiner's comment

This question was accessible to the vast majority of candidates, most of whom were able to, at the least, make a start in both part a) and part b). In part a), most candidates could recall the correct formula for the area of a triangle. After a correct initial statement for the area of the triangle, a number of candidates lost a factor of 𝑥, simplifying 1 .2 .3 x x to 3x thus obtaining a linear equation in 𝑥 rather than a quadratic. 2 The rigour demanded in a ‘show that’ question tended to cost some candidates marks when they failed to demonstrate the individual steps in their calculations. In particular, it was quite common for 3 sin60 candidates to fail to demonstrate that they had used within their calculation. Candidates 2 should be made aware that for questions such as these where they are asked to ‘show that’, it is imperative that they demonstrate each step of their method on paper. Part b) tended to be less problematic and candidates were often more successful here than in part a). The vast majority of candidates were able to correctly quote and use the cosine rule. Occasionally, candidates became confused with the side lengths and errors arose in the substitution of 2 became 3(2 sqrt 3) 2 x into the cosine formula. Other errors arose in the squaring, sometimes (3 x 2 sqrt 3) for example. Finally, and worryingly, a very common mistake was not knowing about ‘order of frequently seen operations’ with 156 144cos60 12cos60

Common mistakes

• ½ × 2x × 3x × sin 60° = 18√3 (keep both x's: 3x² not 3x)
• sin 60° = √3/2 must be seen
• x² = 12 so x = √12 = 2√3

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

Examiner's comment

This question was accessible to the vast majority of candidates, most of whom were able to, at the least, make a start in both part a) and part b). In part a), most candidates could recall the correct formula for the area of a triangle. After a correct initial statement for the area of the triangle, a number of candidates lost a factor of 𝑥, simplifying 1 .2 .3 x x to 3x thus obtaining a linear equation in 𝑥 rather than a quadratic. 2 The rigour demanded in a ‘show that’ question tended to cost some candidates marks when they failed to demonstrate the individual steps in their calculations. In particular, it was quite common for 3 sin60 candidates to fail to demonstrate that they had used within their calculation. Candidates 2 should be made aware that for questions such as these where they are asked to ‘show that’, it is imperative that they demonstrate each step of their method on paper. Part b) tended to be less problematic and candidates were often more successful here than in part a). The vast majority of candidates were able to correctly quote and use the cosine rule. Occasionally, candidates became confused with the side lengths and errors arose in the substitution of 2 became 3(2 sqrt 3) 2 x into the cosine formula. Other errors arose in the squaring, sometimes (3 x 2 sqrt 3) for example. Finally, and worryingly, a very common mistake was not knowing about ‘order of frequently seen operations’ with 156 144cos60 12cos60

Common mistakes

• cosine rule with sides 6√3 and 4√3
• (3 × 2√3)² squared correctly
• order of operations: 156 − 72 = 84, not (156 − 144)cos 60°
• BC = 2√21

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

Examiner's comment

This was a poorly answered question at all levels. In part (a) it was expected that candidates would recognise the fact that the vectors would need to lie in the same direction but many attempted to write out the given expression in words. There was slightly more success in part (b) with most attempting a diagram of sorts to get a feel for the problem. Unfortunately a lack of understanding of vectors 6 m-n 3 n 3 m meant that and was usually followed by . There were however, some excellent and well thought out solutions showing a full understanding of the question,

Common mistakes

• draw the triangle with 6 opposite the 30° and 3 adjacent
• sine rule sin 30°/6 = sin θ/3 gives θ = 14.5°
• the angle between m and m − n is 180 − 30 − 14.5 = 135.5°
• no negative lengths

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

Examiner's comment

Q03 proved to be the most accessible question on this paper with many candidates obtaining full marks. In Q03(a), many candidates explained that the angle 40 should have been converted 1 2 r Some 2 . to its equivalent in radians in order for it to be applied to the formula candidates’ explanations lacked sufficient detail, e.g. they stated ‘the angle is in degrees’ but failed to mention that the angle used in the formula should have been expressed in radians. A few candidates did not give a reason in Q03(a), but correctly calculated the area of sector AOB in Q3(b). Q03(b) was also well-answered. Most candidates converted 40 to radians and to the given formula applied a correct 2 1 2 r Some candidates applied a correct 2 . 9 or Errors included converting 40 to either 40 2 r formula with 40 . 360 360 1 2 9 1 or 2 2 (5 ) ; ; sin 2 r applying and applying incorrect formulae such as 2 9 2 1 Most candidates gave the correct answer in exact form as 25 9 and some 2 . r , 2 gave an answer 8.73 which was rounded to 3 significant figures, while others gave both 2 cm in their answer to exact and rounded answers. Most candidates gave the units Q03(b). On this occasion, a lack of units in their final answer was condoned.

Common mistakes

• ½r²θ needs θ in radians
• "the angle is in degrees" alone is not enough, must say it should be in radians

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

Examiner's comment

Q03 proved to be the most accessible question on this paper with many candidates obtaining full marks. In Q03(a), many candidates explained that the angle 40 should have been converted 1 2 r Some 2 . to its equivalent in radians in order for it to be applied to the formula candidates’ explanations lacked sufficient detail, e.g. they stated ‘the angle is in degrees’ but failed to mention that the angle used in the formula should have been expressed in radians. A few candidates did not give a reason in Q03(a), but correctly calculated the area of sector AOB in Q3(b). Q03(b) was also well-answered. Most candidates converted 40 to radians and to the given formula applied a correct 2 1 2 r Some candidates applied a correct 2 . 9 or Errors included converting 40 to either 40 2 r formula with 40 . 360 360 1 2 9 1 or 2 2 (5 ) ; ; sin 2 r applying and applying incorrect formulae such as 2 9 2 1 Most candidates gave the correct answer in exact form as 25 9 and some 2 . r , 2 gave an answer 8.73 which was rounded to 3 significant figures, while others gave both 2 cm in their answer to exact and rounded answers. Most candidates gave the units Q03(b). On this occasion, a lack of units in their final answer was condoned.

Common mistakes

• θ = 40π/180 = 2π/9, area = 25π/9 or 8.73 cm²
• or πr² × 40/360
• not 40/360 × 2π in ½r²θ

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

Examiner's comment

In part (a), many candidates earned the first mark for correctly finding R, but there were a number who resolved incorrectly, using sin instead of cos . Almost all candidates obtained the second mark, 2 F R very often from simply stating somewhere on their paper. Most candidates who found the 3 equations of motion correctly did proceed to find T. The most successful methods were either adding the two initial equations to eliminate T, finding a and substituting back into one of the equations or setting both equations to 6ma and finding T directly. Candidates who rearranged their equations to give a and then equated the quotients more frequently made errors when solving for T. Common errors were inconsistent use or loss of m, including an extra g in the acceleration term or forgetting to include g in their weight and, on the RHS of the equation, putting 3a and 2a rather than 3ma and 2ma. A significant number of candidates set their equation of motion to zero, finding the tension which would keep A at rest on the slope and received little credit. It was surprising how many candidates simply wrote 12 mg T at the end of their incorrect working, assuming it would not be checked. In the second part, who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.

Common mistakes

• R = 2mg cos α (not sin α)
• F = ⅔R
• separate equations for A and B (or whole system) with 2ma and 3ma, not 2a, 3a or = 0
• exact trig ratios from tan α = 5/12 (no decimal angle)
• do not just write the given answer

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

Examiner's comment

12 mg T at the end of their incorrect working, assuming it would not be checked. In the second part, 5 many candidates tried to describe what would happen without any supporting calculations. Many talked about the continued motion of A, ignoring the actual question. Others simply said that it would continue moving as B had more mass than A, ignoring the effects of the plane. Many who showed calculations did successfully compare the component of the weight acting down the slope with the maximum friction, although others continued to include tension or actually compared friction with tension, ignoring the weight altogether. A few candidates showed a deeper understanding of the problem and simply Some candidates did not notice that the terms required were available from the compared and tan . equation of motion for A. Some candidates did not read the question carefully and considered what happened immediately after B reached the ground, rather than when block A came to rest. Some used an equation of motion rather than considering forces separately. A minority showed their lack of understanding by using tension in their explanation. The final part was generally well done, although who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.

Common mistakes

• compare Fmax = 16mg/13 with weight component down the slope 10mg/13 (or μ with tan α) numerically
• no tension once B has landed
• consider A at rest, not just after B lands

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

Examiner's comment

second part involved finding the components of the resultant force on the ramp at the point of contact with the ground and hence its magnitude. This proved challenging for many and some made little valid progress. A clearly labelled diagram would have helped to identify forces and distances. Most used a normal reaction and a horizontal frictional force at A; however, the latter was sometimes omitted which significantly reduced the number of marks available for subsequent equations. Some employed components parallel and perpendicular to the ramp; such attempts were rare but generally successful. A variety of resolution and moments equations were attempted but sometimes components were omitted, forces not resolved or perpendicular distances not used. Working was often difficult to decipher. Those who attempted vertical and horizontal resolutions and a ‘moments about A’ equation tended to be the more successful although a significant number stopped when they had found the normal reaction only. Some introduced a coefficient of friction which was not necessary to answer the question (even giving it the value 2 from the previous question). Occasionally those who completed a correct solution by 3 squaring and adding components to find the magnitude of the resultant failed to achieve the final accuracy mark for not rounding to 2 or 3 significant figures following the use of g = 9.8 m s-2. Some who had achieved very few marks for the rest of the question managed to secure the mark in part (c)

Common mistakes

• friction at A is horizontal and must be included
• at least one moments equation (e.g. M(A): 20g × 4cos θ = 5N) with perpendicular distances
• three equations then magnitude √(F² + R²) ≈ 67 N (2 or 3 s.f. after g = 9.8)
• μ is not needed
• stopping at R

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

Examiner's comment

In part (a), the majority of candidates obtained the first mark for the horizontal component and most candidates also made a good attempt to find the vertical component, although sometimes it was seen as part of their working for part (b). Of those who found the horizontal and vertical components of the

Common mistakes

• horizontal 20cos 30° and vertical 20sin 30° − 2g = −9.6
• need both speed (19.8 m/s) and direction (29° below horizontal)
• do not give the answer in i and j (not defined)

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

Examiner's comment

part of their working for part (b). Of those who found the horizontal and vertical components of the velocity, only a proportion proceeded to calculate the speed and of those, very few found the direction. This was despite the clear lead from the question which had given each velocity as a speed with a direction. A sizeable minority broke the universal rule that answers must be given in terms of quantities which are given in the question, giving their velocity in terms of i and j, which had not been defined in this question. The second part proved to be quite discriminating although a significant number of candidates did successfully find the correct values of both and u. The most common error was for candidates to simply equate both the horizontal and vertical displacements at the time of collision and ignore the 50 m horizontal distance altogether. Others assumed that the particles met halfway between A and B. However, only a minority of candidates made errors when resolving their velocities. In the final part, a substantial number of candidates correctly identified a limitation of the model with the size

Common mistakes

• horizontal distances sum to 50 m at t = 2 (not equal, not meeting halfway)
• vertical displacements equal gives u sin θ = 20 sin 30°
• θ = 52°, u = 12.6

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

Examiner's comment

This question was well answered by the vast majority of students. The formulae for arc length and area of a sector were well known and applied, and although the 4 was occasionally seen on the incorrect side of the equation, this was rare. Most students worked in radians and used simultaneous equations to find the value of before proceeding to find an exact (positive) value for r. Algebraic and numerical slips, however, when manipulating equations were not an uncommon reason for losing marks. If method errors were made in this question, it was usually to miss the half from the sector area or to confuse the area of a sector with that of a triangle.

Common mistakes

• ½r²θ = 11 and 2r + rθ = 4rθ give θ = 2/3 and r = √33 (exact, positive)
• missing the ½ in the sector area or using a triangle area
• 4 on the wrong side of the perimeter equation

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

Examiner's comment

This proved to be a very demanding question. The idea of there being two values for or cos was lost on many students. Most students ignored the need for two values in (a) and hence in part (b), which involved choosing the obtuse angle, no marks could be awarded. In (a) almost all students were able to apply the area formula to achieve an acceptable answer for sin theta or theta . However, few then had a strategy for obtaining cos theta . Even those who produced the two correct angles often failed to find the values of cos theta . The use of cos2 theta + sin2 theta = 1 was quite rare. Another acceptable method was to sketch a right angled triangle and use trigonometric ratios. Perhaps because of the request for values for cos theta some students incorrectly quoted 1 cos 2ab C for the area.

Common mistakes

• ½ × 10 × 5 × sin θ = 15 gives sin θ = 3/5
• then cos²θ = 1 − sin²θ gives cos θ = ±4/5 (two values: acute and obtuse angle)
• many give only one value
• ½ab cos C is wrong

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

Examiner's comment

This proved to be a very demanding question. The idea of there being two values for or cos was lost on many students. Most students ignored the need for two values in (a) and hence in part (b), which involved choosing the obtuse angle, no marks could be awarded. In (b), few responses demonstrated the need to select the obtuse angle (or negative cosine) to ensure BC was the longest side. Some substituted both angles (acute and obtuse) but needed to compare their lengths to gain credit. The most common response seen was simply to use theta = 36.9 in the cosine rule which scored zero for this part. A surprising number of otherwise correct solutions lost the final mark by giving a decimal answer rather than sqrt 205.

Common mistakes

• BC is longest, so it is opposite the obtuse angle: use cos θ = −4/5 in the cosine rule
• BC = √205 exact (a decimal loses the A mark)
• using θ = 36.9° scores 0

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

Examiner's comment

This was another modelling question, this time modelling the height of a roller coaster above Part (b) was found by almost all students, even those who could not attempt part (a). In this case the exact answer 5e 0.1 was accepted as well as 5.53 metres. Fewer students scored both t marks in (c), but many were aware that it would occur when sin(0.25 ) The most 1.

Common mistakes

• maximum when sin(0.25t) = 1: 5e^0.1 = 5.53 m (exact 5e^0.1 accepted)

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

Examiner's comment

This was another modelling question, this time modelling the height of a roller coaster above t marks in (c), but many were aware that it would occur when sin(0.25 ) The most 1. or else use 450 rather than 5 thus , , common errors were to find the first value using 2 2 arriving at the incorrect answer of 1800 seconds.

Common mistakes

• second maximum when 0.25t = 5π/2 (radians), T = 10π ≈ 31.4 s
• using π/2 (first maximum) or 450 in degrees (giving 1800) is wrong

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

Examiner's comment

part (b) scoring one out of the three marks. Those who managed to sketch the correct part of the parabola rarely were able to explain why the domain (and range) were restricted. Stating that ''because 0 2 t '' was not enough to explain why for instance 1 5 x . The best then 3 2sint students were able to explain that as 1 sin 1 t would have a maximum and a minimum value of 3 2 value of 3 2 5 1.

Common mistakes

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

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

Examiner's comment

In part (c), some students correctly deduced that the given equation had 3 real solutions for 7 10 , with a few students incorrectly deducing that there were 6 real solutions. Most students misunderstood the requirement of the question by finding 23.1, 26.2, 29.4, with those students who did not go on to state that there were 3 real solutions losing the mark for this part.

Common mistakes

• equation is −f(tan θ) = 0 so tan θ = 2 only
• one solution per period π gives 3 solutions in 7π ≤ θ < 10π
• listing 23.1, 26.2, 29.4 without stating 3 scores B0
• 6 is a common wrong answer

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

Examiner's comment

Part (a) was well done by the majority of students. Virtually all attempted to resolve vertically and then use ‘F = ma’ horizontally to find the acceleration. A few omitted the vertical component of the tension when finding the reaction which simplified the problem and resulted in the loss of 3 of the available 6 marks. ‘F = 0.14R’ was almost invariably used to deduce the value of the acceleration. Full marks were often achieved for this part of the question. In the second part, many clear and concise explanations were seen for why the acceleration would be

Common mistakes

• resolve vertically R + 40 sin α = 20g (vertical component of tension often omitted, losing 3 marks)
• 40 cos α − F = 20a with F = 0.14R
• a = 0.40 m s⁻²

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

Examiner's comment

In part (a), most students used a ‘moments about A’ equation to deduce the given expression for the tension in the rope. Since the answer was given it was important that the preceding working was sufficient and accurate but generally this was well done. In part (b), most used a horizontal resolution

Common mistakes

• moments about A: T × 2a sin α = Mga + 3Mgx with sin α = 3/5 seen
• enough working for the given answer

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

Examiner's comment

sufficient and accurate but generally this was well done. In part (b), most used a horizontal resolution and the given expression for T to calculate a value of x. Errors in the straightforward algebraic manipulation were surprisingly common, sometimes leading to a dimensionally incorrect value. The most popular approach to part (c) was to resolve vertically to find the vertical component of the

Common mistakes

• resolve horizontally T cos α = 2Mg using the given T (or moments about C)
• x = 2a/3
• algebra slips often gave a dimensionally wrong answer

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

Examiner's comment

most popular approach to part (c) was to resolve vertically to find the vertical component of the reaction. Occasionally a weight was omitted or the tension was not resolved, but there were a fair number of valid attempts. Again, slips in simplifying the expression were fairly prevalent and sometimes an incorrect value of x was being carried forward. Those who reached an answer which was a multiple of Mg generally gained the final method mark for dividing the components of reaction to find tan beta . Although there were some entirely correct solutions seen, there was evidence of confusion and lack of a systematic approach in other attempts.

Common mistakes

• resolve vertically (or moments about B) to find Y = 5Mg/2
• tan β = Y/X with X = 2Mg gives 5/4
• errors: omitting a weight, not resolving T

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

Examiner's comment

In the first part, the majority of candidates used ‘v2 = u2 + 2as’ with a correct distance to derive the given expression for U2. A few attempted equations involving t but not always successfully. In part

Common mistakes

• vertical motion to the top: 0 = (U sin α)² − 2g(3 − 2)
• distance 1 m, not 3