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Hello and welcome to this pre-recorded event for Feedback on June 2015 AEA mathematics.
To help you get the most out of this feedback, you may want to have the following documents
handy as we will refer to them during the session.
The question paper so that you can view the questions.
The mark scheme - you will use this in activity 2.
The principal examiner's report - although the main points are quoted in this presentation.
The Activity 1 file - a set of tasks which should illuminate aspects of the paper.
The Activity 1 responses file which contains ideas which go with Activity 1.
The Activity 2 file which contains examples of students' work.
The Activity 2 responses file which contains the agreed marks.
The aims and objectives of this presentation are:
To receive feedback on national performance of candidates in the AEA exam of June 2015.
To consider the variation of candidates' performance on different questions and possible reasons
for this. To look at extracts from the principal examiner's
report. To share best teaching practice.
Here are some opening extracts from the principal examiner's report.
Many students were able to demonstrate competence in problem solving on this paper.
However, many found questions 3, 4c and 7b and 7c a severe challenge.
More on these later.
Question 1: Most candidates were able to score marks on both parts.
However, in some cases the asymptote was not labelled correctly.
The graph shown scores only 2 of the 3 available marks. Question 2: Most candidates were able to score 6 of the 9 marks available.
Typically this was done by squaring both sides, rearranging and squaring again to get a cubic equation. This could be easily solved by factorisation. However, only a few realised that squaring may have introduced extra solutions and suitable checks were only carried out by the better students. This is a key teaching point - this student has been taught to check solutions, so gaining an additional 3 marks. Question 3 proved to be very challenging: Many candidates were unable to score any marks. This was because some vision was required to see that the left hand side terms could be combined. Students then had to see that the sec x on the right hand side could be cancelled away once the sin 2x term had been expanded.
A second challenge was to solve the equation cos(2x + 78o) = sin x
Good students tended to approach it in the way shown here, although only the most able
managed to find all 4 solutions in the range 0 to 360o
At this stage you might want to mark the 5 examples of students work on question 3.
The marks actually awarded are given in the response file.
Question 4: Competent candidates were able to score marks on parts (a), (b) and (d).
For this level of candidate, part (a) was straightforward....
Here is a typical competent response.
... and part (b) was generally well done as it was a routine development of part (a). The work shown was typical. Some candidates factorised the given expression first, used part (a) and then multiplied out the two separate expansions. We shall look at part (c) in a moment. Part (d) was a development of part (b), requiring
term by term integration. Good candidates saw that the even powered
terms were not going to contribute to the final answer and saved themselves some work.
This is especially important in a non-calculator examination such as AEA.
As stated earlier, it was part (c), which caused the most difficulty. For some candidates
the problem was either not knowing the conditions for convergence of a binomial expansion or
failing to see how these could be used to establish the required result.
Successful candidates knew that condition and also how to approach showing the required
result. This slide shows one possible approach and
the next slide a second.
This approach uses the graph of y = mod 5x + x2 and looks at its value in the interval
[-0.5, 0.5]. The blue line is y = 4, so it is clear that
the value of mod 5x + x2 is less than 4. Tucked away at the bottom of the screen is
the comment that another approach involves the factorisation of 4 + 5x + x2 as
(4 + x)(1 + x) and then looking at the convergence of the two terms in the expansion of