Year: 2008
Paper: 1
Question Number: 6
Course: LFM Pure
Section: Integration
No solution available for this problem.
There were significantly more candidates attempting this paper this year (an increase of nearly 25%), but many found it to be very difficult and only achieved low scores. The mean score was significantly lower than last year, although a similar number of candidates achieved very high marks. This may be, in part, due to the phenomenon of students in the Lower Sixth (Year 12) being entered for the examination before attempting papers II and III in the Upper Sixth. This is a questionable practice, as while students have enough technical knowledge to answer the STEP I questions at this stage, they often still lack the mathematical maturity to be able to apply their knowledge to these challenging problems. Again, a key difficulty experienced by most candidates was a lack of the algebraic skill required by the questions. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many students were simply unable to progress on some questions because they did not know how to handle the algebra. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was also pleasing that one of the applied questions, question 13, attracted a very large number of attempts. However, the examiners were again left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers and other available resources to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
The function $\f$ is defined by
\[
\f(x) = \frac{\e^x-1}{\e-1},
\ \ \ \ \
x\ge0,
\]
and the function $\g$ is the inverse function to $\f$, so that
$\g(\f(x))=x$. Sketch $\f(x)$ and $\g(x)$ on the same axes.
Verify, by evaluating each integral, that
\[
\int_0^\frac12 \f(x) \,\d x + \int_0^k \g(x) \,\d x = \frac1 {2(\sqrt \e
+1)}\,,
\]
where $\displaystyle k= \frac 1{\sqrt\e+1}$, and explain this result by means of a diagram.
This was another popular question which gained a pleasing number of good marks. The sketch was generally done well. A significant number of candidates did not realise that f(0) = 0 and f(1) = 1, so either had non-intersecting graphs or graphs which were tangent to each other at the origin. A number of candidates sketched the graph of f(x) for all real x, in spite of the question stating x ≥ 0; they were not penalised for this. Most understood how the graphs of f(x) and g(x) related. The determination of g(x) algebraically was performed correctly by a majority of candidates. However, a disturbing number of candidates introduced absolute value signs, writing g(x) = ln |(e−1)x+1|. Whilst technically correct in this range (and therefore not penalised here), it is indicative of confusion about when absolute values are used with logs: when integrating 1/x they are required; when inverting exponentiation they are not. A smaller number made very significant errors in their handling of the logarithm function, writing such things as ln(eˣ − x + 1) = ln eˣ − ln x + ln 1. The majority of candidates correctly integrated f(x). A small minority bizarrely asserted that ∫₀^(1/2) f(x)dx = f(½) - f(0) which was somewhat disturbing. The integration of g(x) proved much more troublesome. Despite ∫ ln x dx being a standard integral and explicitly mentioned in both the STEP Specification and A2 Mathematics specifications, the introduction of the linear function of x flummoxed most candidates. Some differentiated instead of integrating, others just gave up. A small number either attempted to use parts or to substitute, and a good proportion of such attempts were successful. Some candidates confused differentiation with integration during this process and tried to use a mixture of parts and the product rule. Finally, of those who managed to reach this point, a decent number gave a very convincing explanation of why ∫f + ∫g = ½k.