June 2018: AS AS Paper 1: Pure Mathematics

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

Examiner's comment

A good starting question. Almost all students attempted this question with some success. Students who scored no marks often differentiated instead of integrating. A significant number of students scored full marks. For students who integrated, the majority scored at least one mark for increasing one of the powers by one. The most common error was to forget the '+c'. Other errors included: (1) Incorrect simplification of the coefficients with the first term having a coefficient of 8/3 or the second term having a coefficient of 9 (2) Incorrect index work especially on term two (3) Incorrect sign on term two.

Common mistakes

• forgetting + c is the most common error
• differentiating instead of integrating
• coefficient slips (8/3 on the first term, 9 on the second)
• index errors on 6√x (x^(3/2) ÷ 3/2 gives −4x^(3/2))
• answer must be simplified

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

• answer is sometimes true
• a numerical approach needs specific values calculated, e.g. x = −5 gives 4 < 25 (not true) and x = 5 gives 64 > 25 (true)
• just mentioning positive and negative numbers scores 0
• algebraically 6x + 9 > 0 so true only for x > −3/2

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

Examiner's comment

This question was very accessible to the vast majority of students with many scoring full marks. Errors seen in part (a) included finding the sum of the two vectors instead of the difference, or, finding the difference ‘the wrong way around’. Correct vector notation was seen in all but a few cases with incorrect notation such as (-9i,3j) being penalised.

Common mistakes

• AB = OB − OA = −9i + 3j
• common errors: adding the vectors or OA − OB
• incorrect notation such as (−9i, 3j) penalised

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

Examiner's comment

This question was very accessible to the vast majority of students with many scoring full marks. Part (b) was more successful with the vast majority of students knowing how to calculate the length of a vector and simplify the surd. A follow through on their vector from part (a) enabled most students access to both marks.

Common mistakes

• modulus √((−9)² + 3²) = √90 = 3√10 as a simplified surd
• ±3√10 loses the A mark

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

Students found it difficult to explain the errors in part (a) even though fully correct answers were seen

Common mistakes

• error 1: the subtraction law was used without first dealing with the coefficient 2 (2log₂x = log₂x²)
• error 2: log₂x = 3 means x = 2³ = 8, not 3²
• a full explanation of each error is needed

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

Examiner's comment

in (b). The error in line 2 caused the most problems with many students failing to mention the x log 3 should be 23 not 32 was more successfully dealt with. It is subtraction law. The second error 2 really important in questions such as these that a full explanation is given. Students were much more successful in part (b). Log work, in general, remains a difficult topic for less able students.

Common mistakes

• combine to log₂(x²/√x) = 3 or (3/2)log₂x = 3
• undo the log: x^(3/2) = 2³ so x = 4
• at least one intermediate step must be shown
• giving only the answer scores 1 mark

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

This question on the factor theorem proved to be a good source of marks for many students. Clearly they had been well prepared for this kind of question. In part (a), most students correctly substituted 2 x into the function and obtained the answer 0. A few students incorrectly used 2 x . The accuracy mark was often not awarded as explanations were insufficient or indeed absent. Many failed to use the correct terminology g (-2) = …, just substituting 2 x into the “expression” rather than the function. Additionally, a significant minority of students did not state any conclusion. A small number of students incorrectly used long division rather than the factor theorem but these students were allowed to score the marks in (b) following this approach.

Common mistakes

• evaluate g(−2) = 0 (not g(2))
• write g(−2) = ... rather than substituting into the expression
• a conclusion "so (x + 2) is a factor" is required
• long division is not the factor theorem

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

Examiner's comment

This question on the factor theorem proved to be a good source of marks for many students. Clearly they had been well prepared for this kind of question. The majority attempted part (b) using long division or inspection with most obtaining the correct quadratic expression 2 4 20 25 x x . Most then correctly factorised this quadratic. A small number did not state a final, fully factorised solution. Common errors of (x – 2.5)2 and (4x – 10)2 probably came from students using their calculators to find the solutions then attempting to work backwards.

Common mistakes

• divide by (x + 2) to get 4x² − 20x + 25, then factorise to (2x − 5)²: a = 2, b = −5
• (x − 2.5)² or (4x − 10)² are wrong

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

Examiner's comment

This question on the factor theorem proved to be a good source of marks for many students. Clearly they had been well prepared for this kind of question. Part (c) was usually attempted, but with limited success. In part (i) many achieved one mark, usually for 2 x (or x < - 2 which was condoned for the M mark). Errors from part (b) were often followed through correctly. Only more able (and careful) students gained full marks. In part (ii) many correctly use transformations to state x = -1 and x = 1.25. Some students multiplied the critical values by 2 (instead of dividing by 2) or in other cases they wrote down just one of the two solutions.

Common mistakes

• (i) curve on or below the axis: x ≤ −2 or x = 2.5 (the repeated root where the curve touches the axis is often missed)
• (ii) g(2x) = 0 halves the roots: x = −1, 1.25 (not doubled)

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

Part (a) was standard ‘book work’ on the binomial expansion. This was usually well done but applying this expansion within a new expression was met with limited success. Many students did not attempt parts (b) and (c). The majority of students attempted part (a) of this question. Most were able to apply the binomial expansion and gain the correct unsimplified quadratic form. Common errors included missing out the negative in the expansion, so 512+144x+18x2 was a common incorrect answer. Some students failed to simplify correctly leaving their coefficients as fractions.

Common mistakes

• 512 − 144x + 18x²
• the negative sign of −x/16 is often lost (512 + 144x + 18x²)
• coefficients must be simplified

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

Examiner's comment

A common error in parts (b) and (c) was to set up a pair of (incorrect) simultaneous equations in a and b. Solutions of this type, that is via attempting to solve 512(a + bx) = 128 does show a lack of understanding between constants and variables. Students who did successfully find a = 0.25 rarely could handle the more complex process required to find b. Students scoring full marks in this question were rare.

Common mistakes

• constant term: 512a = 128 so a = ¼
• do not treat (a + bx) as a variable and form wrong simultaneous equations such as 512(a + bx) = 128

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

Examiner's comment

A common error in parts (b) and (c) was to set up a pair of (incorrect) simultaneous equations in a and b. Solutions of this type, that is via attempting to solve 512(a + bx) = 128 does show a lack of understanding between constants and variables. Students who did successfully find a = 0.25 rarely could handle the more complex process required to find b. Students scoring full marks in this question were rare.

Common mistakes

• x term comes from two products: 512b − 144a = 36 so b = 9/64
• rarely done correctly

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

Examiner's comment

Overall this question was well done, with many students taking advantage of being able to attempt (b) without completing (a). The few who restarted unnecessarily in (b) lost valuable time in the examination. In (a) almost all students substituted for tan theta correctly, with just a few using the identity upside down or attempting some version of sin theta = tan theta cos theta . It was common for the first accuracy mark to be lost for 4cos2 theta – 1 instead of 4cos2 theta – cos theta after multiplication by cos theta . Whilst many correctly used sin2𝜃+cos2𝜃=1 it was disappointing to see sin𝜃+cos𝜃 =1 being quoted. The final mark was sometimes lost through bracketing errors or poor notation e.g. sin theta 2 for sin2 theta .

Common mistakes

• tan θ = sin θ/cos θ
• multiply through by cos θ (4cos²θ − cos θ, not 4cos²θ − 1)
• use sin²θ = 1 − cos²θ (not sin θ + cos θ = 1)
• notation sin²θ not sinθ²

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

Although this was another new question, it was less well attempted than others with more students leaving this question blank, or starting the question but not gaining any marks. (a) where students correctly identified the link with p, q and the formula, they were able to correctly work out the values of p and q. The most common method was to convert to powers of 10. A common error was incorrectly rounding p = 63095. to 6310, instead of 63100. Although this was not penalised in part (a), it did often lead to a wrong answer in part (c). A small number of students converted the question into logs and a few also substituted values in for t and used logs to solve. Part (b) was attempted by the majority of students, even for those who failed to attempt part (a). Part (i) was well answered with the majority of students identifying p as the initial value of the painting. A few students did not put their statement in the context of the model. In part (ii) common mistakes were to state it was the increase in value, rather than to link it to the rate or proportional increase, or to state it was the gradient of the line. Very few students explained that the value of the painting would rise by 12.2% a year or that 1.122 was the decimal multiplier representing the year on year increase in value. part (a) often led to the loss of the accuracy mark here.

Common mistakes

• log V = log p + t log q, so p = 10^4.8 = 63 100 and q = 10^0.05 = 1.122 (4 s.f.)
• rounding p to 6310 is a common slip that spoils (c)

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

Examiner's comment

Although this was another new question, it was less well attempted than others with more students leaving this question blank, or starting the question but not gaining any marks. Part (b) was attempted by the majority of students, even for those who failed to attempt part (a). Part (i) was well answered with the majority of students identifying p as the initial value of the painting. A few students did not put their statement in the context of the model. In part (ii) common mistakes were to state it was the increase in value, rather than to link it to the rate or proportional increase, or to state it was the gradient of the line. Very few students explained that the value of the painting would rise by 12.2% a year or that 1.122 was the decimal multiplier representing the year on year increase in value.

Common mistakes

• p is the value on 1 January 1980 (initial value)
• q is the proportional increase per year (rises 12.2% a year), not "the amount" it rises by or the gradient of the line

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

Examiner's comment

Although this was another new question, it was less well attempted than others with more students leaving this question blank, or starting the question but not gaining any marks. penalised in part (a), it did often lead to a wrong answer in part (c). A small number of students converted the question into logs and a few also substituted values in for t and used logs to solve. Part (c) could be answered without values for p and q and some students scored just marks in this part of the question. When attempted it was generally done well. The follow through error mentioned in

Common mistakes

• t = 30 (not 2010)
• V ≈ £2 000 000 to the nearest hundred thousand
• can use log V = 0.05 × 30 + 4.8 without p and q

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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