2020 Paper 3 Q1

Year: 2020
Paper: 3
Question Number: 1

Course: UFM Additional Further Pure
Section: Reduction Formulae

Difficulty: 1500.0 Banger: 1500.0

Problem

For non-negative integers \(a\) and \(b\), let \[ \mathrm{I}(a,b) = \int_0^{\frac{\pi}{2}} \cos^a x \cos bx \; \mathrm{d}x. \]
  1. Show that for positive integers \(a\) and \(b\), \[ \mathrm{I}(a,b) = \frac{a}{a+b} \, \mathrm{I}(a-1, b-1). \]
  2. Prove by induction on \(n\) that for non-negative integers \(n\) and \(m\), \[ \int_0^{\frac{\pi}{2}} \cos^n x \cos(n+2m+1)x \; \mathrm{d}x = (-1)^m \frac{2^n \, n! \, (2m)! \, (n+m)!}{m! \, (2n+2m+1)!}. \]

No solution available for this problem.

Examiner's report
— 2020 STEP 3, Question 1
Mean: 12.6 / 20 90% attempted Mean score about 63% = 12.6/20; second best attempted question by a small margin

This was the most popular question, being attempted by about 90% with a fair degree of success: the mean score of about 63% made it the second best attempted question by a small margin. In part (i), nearly all candidates understood that they would need to use integration by parts and one (or more than one, for some methods) compound angle formula. However, there were numerous manipulative errors in the integration or differentiation of the components, and even sign errors in using compound angle formulae. There were a number of different correct approaches which could be used, but they were essentially very similar to one or other of the methods in the mark scheme. Part (ii) was prescriptively worded, and it was a test of correctly expressed formalism. In spite of this, some candidates did not employ the principle of induction, some ignored that the induction was on n, and some overlooked 'non-negative' requiring the base case to be zero. Often the first component of proof by induction was omitted or incorrectly expressed. 'Assume (or suppose) the result is true for some particular k' would be an improvement on what quite a number wrote. The word 'assume (suppose)' was often not written by candidates and clearly the letter n could not be used for the assumption. Similarly, demonstrating that the base case works correctly needs to be thorough and with fully precise detail as patently it will work, and so a solution must be convincing.

In spite of the change to criteria for entering the paper, there was still a very healthy number of candidates, and the vast majority handled the protocols for the online testing very well. Just over half the candidates attempted exactly six questions, and whilst about 10% attempted a seventh, hardly any did more than seven. With 20% attempting five questions, and 10% attempting only four, overall, there were very few candidates not attempting the target number. There was a spread of popularity across the questions, with no question attracting more than 90% of candidates and only one less than 10%, but every question received a good number of attempts. Likewise, there was a spread of success on the questions, though every question attracted at least one perfect solution.

Source: Cambridge STEP 2020 Examiner's Report · 2020-p3.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

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Problem source
For non-negative integers $a$ and $b$, let
\[ \mathrm{I}(a,b) = \int_0^{\frac{\pi}{2}} \cos^a x \cos bx \; \mathrm{d}x. \]
\begin{questionparts}
\item Show that for positive integers $a$ and $b$,
\[ \mathrm{I}(a,b) = \frac{a}{a+b} \, \mathrm{I}(a-1, b-1). \]
\item Prove by induction on $n$ that for non-negative integers $n$ and $m$,
\[ \int_0^{\frac{\pi}{2}} \cos^n x \cos(n+2m+1)x \; \mathrm{d}x = (-1)^m \frac{2^n \, n! \, (2m)! \, (n+m)!}{m! \, (2n+2m+1)!}. \]
\end{questionparts}