The hardest Edexcel A-level Maths questions

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

Examiner's comment

This question, on two different kinds of proof, proved more demanding than expected. As a result, there were very few completely correct solutions to both parts of this question. In part (i) candidates were often unable to set up a proof by contradiction using appropriate language, failing to use words such as 'assume', 'there exists' or 'let' . Others simply set up a contradiction such as 'assume p and q are odd' without any reference to pq being even. Many candidates also failed to use different variables for their odd numbers p and q, implying that p and q were either the same odd number, or two consecutive odd numbers such as 2n+1 and 2n+3. Candidates who managed to write p and q using different variables were generally able to multiply out pq correctly and write it in the form 2(.....)+1. If this stage was reached, the conclusion was generally correct and an acceptable sentence was written down. A mark of 011 was possible from the mark scheme. In part (ii) most candidates expanded the left hand side and cancelled terms getting to 2xy < 8x2 or equivalent for the method mark. Many then lost the second mark by failing to identify the point at which the inequality reverses, simply writing down y < 4x followed by 'so y > 4x'. Other candidates did reverse the inequality at the right point but didn't give an explanation as to why, and so lost the final mark.

Common mistakes

• assume pq even AND p, q both odd ("assume/there exists")
• different variables p = 2m + 1, q = 2n + 1
• pq = 2(2mn + m + n) + 1 is odd, contradiction, then conclude

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

Examiner's comment

This question, on two different kinds of proof, proved more demanding than expected. As a result, there were very few completely correct solutions to both parts of this question. In part (i) candidates were often unable to set up a proof by contradiction using appropriate language, failing to use words such as 'assume', 'there exists' or 'let' . Others simply set up a contradiction such as 'assume p and q are odd' without any reference to pq being even. Many candidates also failed to use different variables for their odd numbers p and q, implying that p and q were either the same odd number, or two consecutive odd numbers such as 2n+1 and 2n+3. Candidates who managed to write p and q using different variables were generally able to multiply out pq correctly and write it in the form 2(.....)+1. If this stage was reached, the conclusion was generally correct and an acceptable sentence was written down. A mark of 011 was possible from the mark scheme. In part (ii) most candidates expanded the left hand side and cancelled terms getting to 2xy < 8x2 or equivalent for the method mark. Many then lost the second mark by failing to identify the point at which the inequality reverses, simply writing down y < 4x followed by 'so y > 4x'. Other candidates did reverse the inequality at the right point but didn't give an explanation as to why, and so lost the final mark.

Common mistakes

• expand and cancel to 2xy < 8x²
• dividing by x < 0 reverses the inequality, with the reason, to get y > 4x
• no incorrect lines

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. Part (a) required candidates to show that parallelogram PQRS was a rhombus. Although the requirement to 'show' is becoming more familiar to candidates, a significant number in this case didn't recognise the level of detail expected. So a lack of conclusion, or stating the sides were of length Ö29 without evidence of calculation, led to loss of credit. It was also necessary to conclude that the sides were equal, and not simply find the lengths and leave it to the reader to decide.

Common mistakes

• modulus of PQ = modulus of QR = √29 with calculations shown and conclusion "so rhombus"
• adjacent sides equal

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 on arithmetic sequences and series proved to be very discriminating. The proof of the sum of an arithmetic sequence is standard, but was not answered well by many candidates. Most candidates however were able to then made progress with the rest of the question, having been given the required formula. Good candidates often did well in this question however, with 17% of the cohort achieving full marks. In part(i) those who had learned the standard proof for the sum of the first n terms of an arithmetic series often gained full marks. It was however, unusual to see a well set out, fully correct proof. Candidates often failed to write a correct expression for Sn at the start of their proof, or sometimes having an incorrect first or last term or extra terms (most often a+nd) within their sum. Other errors included not displaying a sufficient number of terms, not writing S= or Sn =, and using commas instead of + sign between the terms. In part(ii) most realised that they could make headway even if they hadn’t attempted part (i). In (ii) (a) many of the candidates used a=20, d = -0.8 and Sn = 64 in the sum formula to score the M mark. Poor notation such as (n-1)-0.8 was frequently seen, but candidates usually recovered from this. Following a correct equation most were able to proceed correctly to the given quadratic for the accuracy mark. A common error was where candidates used +0.8 as the common difference rather than -0.8 Part (ii) (b) was generally answered without issues. Almost all candidates managed to get B1 for getting the two values of n, either by factorising or by using a calculator. Part (ii)(c) was not answered very well with most candidates identifying 10 as the valid solution but not adequately justifying this choice. The reasons provided were often too vague or not related to why 10 was better than 16. More tended to score the mark by explaining that at 16 weeks he would be saving a negative amount rather than using the fact that 10 was the smaller number or that he would have had enough by 10 weeks so would not continue saving until 16 weeks.

Common mistakes

• n = 10 or 16

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

Examiner's comment

This question on arithmetic sequences and series proved to be very discriminating. The proof of the sum of an arithmetic sequence is standard, but was not answered well by many candidates. Most candidates however were able to then made progress with the rest of the question, having been given the required formula. Good candidates often did well in this question however, with 17% of the cohort achieving full marks. In part(i) those who had learned the standard proof for the sum of the first n terms of an arithmetic series often gained full marks. It was however, unusual to see a well set out, fully correct proof. Candidates often failed to write a correct expression for Sn at the start of their proof, or sometimes having an incorrect first or last term or extra terms (most often a+nd) within their sum. Other errors included not displaying a sufficient number of terms, not writing S= or Sn =, and using commas instead of + sign between the terms. In part(ii) most realised that they could make headway even if they hadn’t attempted part (i). In (ii) (a) many of the candidates used a=20, d = -0.8 and Sn = 64 in the sum formula to score the M mark. Poor notation such as (n-1)-0.8 was frequently seen, but candidates usually recovered from this. Following a correct equation most were able to proceed correctly to the given quadratic for the accuracy mark. A common error was where candidates used +0.8 as the common difference rather than -0.8 Part (ii) (b) was generally answered without issues. Almost all candidates managed to get B1 for getting the two values of n, either by factorising or by using a calculator. Part (ii)(c) was not answered very well with most candidates identifying 10 as the valid solution but not adequately justifying this choice. The reasons provided were often too vague or not related to why 10 was better than 16. More tended to score the mark by explaining that at 16 weeks he would be saving a negative amount rather than using the fact that 10 was the smaller number or that he would have had enough by 10 weeks so would not continue saving until 16 weeks.

Common mistakes

• 10 weeks: he has £64 by week 10 (and amounts go negative from week 14)
• reason must explain why not 16

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Almost a quarter of the candidates scored zero whilst 21% scored full marks on this

Common mistakes

• down the plane (a negative F from assuming up the plane means down)
• must be stated as the answer, not only shown on the diagram

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

Examiner's comment

This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% scoring full marks. Most students navigated parts (a) and (b) successfully but a few missed the information that “No students read all three magazines” and wasted a lot of time trying to find two equations in p and q. The conditional probability in part (c) proved to be a challenge for some; there were those who effectively

Common mistakes

• exactly one magazine = 0.08 + 0.09 + 0.36 = 0.53

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

Examiner's comment

students found the correct set of values with varying degrees of explanation on the way. Part (b) was often left blank though the first mark was available without values for a, b and c provided brackets were used correctly for the expression. Nearly 10% managed to score full marks for the question.

Common mistakes

• P(X₁ = X₂) = sum of squared probabilities (log₃₆a)² + (log₃₆b)² + (log₃₆c)² = 0.381
• brackets needed

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

Examiner's comment

It was very rare to see a response worthy of full marks and many candidates made no attempt to answer this question. It was clear that this new aspect of the specification was not well known by candidates. It was expected that candidates would consider odd and even numbers separately using a logical or an algebraic approach as detailed in the mark scheme. Almost all the candidates who gained any marks used either approach, or even a mixture of both. Marks were commonly lost by: using n (even) and n +1 (odd) which brought no success at all failing to appreciate the difference between “cannot be divided by 8” and “is not divisible by 8” errors in the expansion and simplification of (2k + 1)3 + 2 a lack of rigour in the explanation that 8k3 + 12k2 + 6k + 3 is not divisible by 8 failing to write a concluding statement (for the final mark) There were also a few attempts at proof by contradiction, some of which were quite impressive, considering the more difficult nature of this approach. Many embarked on a completely numerical approach (in some cases claiming to have demonstrated proof by exhaustion) and did not, of course, score any marks.

Common mistakes

• consider n even (2k) and n odd (2k + 1) and conclude for all n
• numerical examples score nothing
• n and n + 1 is not even/odd
• "8k³ + 2 cannot be divided by 8" is wrong (not exactly divisible)
• concluding statement needed

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

• a and b are parallel and in the same direction
• do not just restate the equation in words

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

Only the most able candidates on the paper made significant progress through this question Despite the question stating the test should be one-tailed in part (c) some candidates found a two-tail critical region or went for the left hand tail rather than the right hand one. Where the upper tail was found the critical region was usually correct although some gave a critical value rather than a region or wrote it as a probability i.e. P(X >= 9). Some candidates rewrote Julia’s hypothesis test correctly at this point, not appreciating the difference between this and a critical region approach. Part (d) was usually correctly done by those who were successful in part (c). On some occasions answers were given only to two significant figures. Just over 5% of candidates scored full marks on the final question on this paper.

Common mistakes

• one-tailed upper tail of B(30, 0.15): P(X ≥ 9) = 0.0277 < 0.05, P(X ≥ 8) > 0.05
• CR X ≥ 9 (not a two-tailed or lower tail region, not P(X ≥ 9) as a probability, not just the critical value)

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

Examiner's comment

Only the most able candidates on the paper made significant progress through this question Part (d) was usually correctly done by those who were successful in part (c). On some occasions answers were given only to two significant figures. Just over 5% of candidates scored full marks on the final question on this paper.

Common mistakes

• actual significance level = P(X ≥ 9) = 0.0278 to 3 s.f.

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

Examiner's comment

The idea of proof is another new one in this updated specification. It was not well known and a disproportionate number of blank or barely started solutions indicated a lack of expertise in this topic. Candidates who scored full marks for part (i) most often did so via algebraic proof, letting n=2m (for even numbers) and then n=2m+1 (for odd numbers), although several candidates missed out on the final accuracy mark as they failed to insert a conclusion for all n. Weaker candidates tried to use m + 1 instead of 2m + 1 to represent odd numbers. Also many candidates only considered either odd or even numbers, but not both. Candidates who set out to use proof via contradiction tended to score few marks as they did not show an understanding of how to use that method of proof. Proofs via logic were rare and tended to appear as part of a mix of different methods where the candidates seemed to be unclear of the best way to proceed. Some candidates seemed to think that using random numbers to show that the expression is not divisible by 4 amounted to proof (receiving 0 marks). Students need more practice on this style of proof. In part (ii) more candidates approached the problem algebraically than graphically. They were generally able to set up the two equations / inequalities required, but errors often occurred in their use of algebra when attempting to solve them. Some candidates only set up one of the two required inequalities. Many students attempted the question by just substituting numbers in to the expression, with most of these deducing that the statement was always true. Where candidates drew a graph, they often did not describe why their graph indicated that the statement is sometimes true. Many graphs showed a line and a V shape but some were incorrect in the relative positioning of these shapes.

Common mistakes

• consider n = 2m and n = 2m + 1 (not m + 1), both cases
• conclude for all n
• trying numbers scores 0

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

Examiner's comment

The idea of proof is another new one in this updated specification. It was not well known and a disproportionate number of blank or barely started solutions indicated a lack of expertise in this topic. Candidates who scored full marks for part (i) most often did so via algebraic proof, letting n=2m (for even numbers) and then n=2m+1 (for odd numbers), although several candidates missed out on the final accuracy mark as they failed to insert a conclusion for all n. Weaker candidates tried to use m + 1 instead of 2m + 1 to represent odd numbers. Also many candidates only considered either odd or even numbers, but not both. Candidates who set out to use proof via contradiction tended to score few marks as they did not show an understanding of how to use that method of proof. Proofs via logic were rare and tended to appear as part of a mix of different methods where the candidates seemed to be unclear of the best way to proceed. Some candidates seemed to think that using random numbers to show that the expression is not divisible by 4 amounted to proof (receiving 0 marks). Students need more practice on this style of proof. In part (ii) more candidates approached the problem algebraically than graphically. They were generally able to set up the two equations / inequalities required, but errors often occurred in their use of algebra when attempting to solve them. Some candidates only set up one of the two required inequalities. Many students attempted the question by just substituting numbers in to the expression, with most of these deducing that the statement was always true. Where candidates drew a graph, they often did not describe why their graph indicated that the statement is sometimes true. Many graphs showed a line and a V shape but some were incorrect in the relative positioning of these shapes.

Common mistakes

• sometimes true: false for 9.25 < x < 9.5, true e.g. x = 12
• give one example of each
• substituting a few numbers led many to "always true"
• graph needs V shape and line crossing twice

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

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

Common mistakes

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

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

Examiner's comment

A number of candidates did not attempt this question. Whether this was due to lack of understanding or time constraints is unknown. Only a small percentage of candidates were able to secure full marks for part (c) and it was not uncommon to see the whole question left unanswered. This type of question was common on Core 3 and usually well answered.

Common mistakes

• cos 2y = √(1 − (x/4)²) from sin² + cos² = 1
• dy/dx = 1/(2√(16 − x²)), a = 2, b = 16