Year: 2008
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Implicit equations and differentiation
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: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The variables $t$ and $x$ are related by $t=x+ \sqrt{x^2+2bx+c\;} \,$, where $b$ and $c$ are constants
and $b^2 < c$. Show that
\[
\frac{\d x}{\d t} = \frac{t-x}{t+b}\;,
\]
and hence integrate $\displaystyle \frac1 {\sqrt{x^2+2bx+c}}\,$.
Verify by direct integration that your result holds also in the case $b^2=c$ if $x+b > 0$ but that your result does not hold in the case $b^2=c$ if $x+b < 0\,$.\begin{align*}
&& t &= x+ \sqrt{x^2+2bx+c} \\
&& \frac{\d t}{\d x} &= 1 + \frac{x+b}{\sqrt{x^2+2bx+c}} \\
&&&= \frac{x + \sqrt{x^2+2bx+c} + b}{\sqrt{x^2+2bx+c}} \\
&&&= \frac{t+b}{t-x} \\
\Rightarrow && \frac{\d x}{\d t} &= \frac{t-x}{t+b} \\
\\
&& \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{t-x} \frac{t-x}{t+b} \d t \\
&&&= \int \frac{1}{t+b} \d t \\
&&&= \ln (t + b) +C \\
&&&= \ln \left (x + \sqrt{x^2+2bx+c} + b \right) + C
\end{align*}
If $b^2 = c$ then we have $x^2+2bx+b^2 = (x+b)^2$ so $\sqrt{x^2+2bx+c^2} = x+b$ (if $x+b>0$), so
\begin{align*}
&& \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{x+b} \d x\\
&&&= \ln (x + b) + C \\
&&&= \ln(x+b) + \ln 2 + C' \\
&&&= \ln (2(x+b)) + C' \\
&&&= \ln \left(x + b + \sqrt{(x+b)^2} \right)+C'\\
&&&= \ln \left(x + b + \sqrt{x^2+2bx+c} \right)+C'\\
\end{align*}
If $x+b < 0$ then the antiderivative is $\ln 0$.
\begin{align*}
&& \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= -\int \frac{1}{x+b} \d x\\
&&&= -\ln |x + b| + C \\
\end{align*}
which are clearly different.
This was by far the most popular question on the paper, with about six out of every seven candidates attempting it. The very first part involving implicit differentiation was generally done very well with most candidates scoring full marks for this part. A majority of candidates then went on to successfully see how to apply this result to the required integral, although a sizeable minority failed to understand that they were being asked to perform a substitution. Some candidates resorted to the formula book and quoted the standard integral ∫ 1/√(x² + a²) dx; however, this gained no credit as the question explicitly said 'hence'. Having reached ∫ 1/(t + b) dt, the vast majority of candidates became unstuck. Firstly, after integrating, some did not substitute back t = x + ··· to get an expression in terms of x. The fundamental problem, though, was that the candidates were mostly unaware of the need to use absolute values when integrating 1/x: almost everyone gave the intermediate answer as ln(t + b) + c rather than ln |t + b| + c. It turns out that in this case, t + b is always positive so the absolute values may be replaced by parentheses, but this requires explicit justification (which no-one gave). This lack of appreciation of absolute values prevented all but the strongest candidates from making a decent attempt at the last part of the question, the consideration of the case c = b². Some candidates successfully substituted this in to the earlier result as instructed, but many claimed that √(2x² + 2x + b²) = x + b. However, the correct expression is |x + b|, which is x + b when this is positive, but −(x + b) when x + b < 0. Only the tiny handful of candidates who appreciated this subtlety managed to correctly explain the distinction between these two cases.