Year: 2008
Paper: 1
Question Number: 8
Course: LFM Pure
Section: 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: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item
The gradient $y'$ of a curve at a point $(x,y)$ satisfies
\[
(y')^2 -xy'+y=0\,.
\tag{$*$}
\]
By differentiating $(*)$ with respect to $x$, show that either
$y''=0$ or $2y'=x\,$.
Hence show that the curve is either a straight line of the form
$y=mx+c$, where $c=-m^2$, or the parabola $4y=x^2$.
\item
The gradient $y'$ of a curve at a point $(x,y)$ satisfies
\[
(x^2-1)(y')^2 -2xyy'+y^2-1=0\,.
\]
Show that the curve is either a straight line, the form of which you should specify, or a circle, the equation of which you should determine.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& 0 &= (y')^2 -xy'+y\\
\Rightarrow && 0 &= 2y' y'' -y' - xy'' + y' \\
&&&= 2y'y'' - xy'' \\
&&&= y'' (2y'-x)
\end{align*}
Therefore $y'' = 0 \Rightarrow y = mx + c$ or $y' = \frac12 x \Rightarrow x = \frac14x^2 + C$.
Plugging these into the original equation we have $m^2 - xm+mx+c = 0 \Rightarrow c = -m^2$
$\frac14 x^2 - \frac12 x^2 + \frac14x^2 + C = 0 \Rightarrow C = 0$. Therefore $4y = x^2$
\item \begin{align*}
&& 0 &= (x^2-1)(y')^2 -2xyy'+y^2-1 \\
\Rightarrow && 0 &= 2x(y')^2 +(x^2-1)2y'y'' - 2yy' - 2x(y')^2-2xyy''+2yy' \\
&&&= (x^2-1)2y'y'' -2xyy'' \\
&&&= 2y'' ((x^2-1)y'-xy)
\end{align*}
Therefore $y'' = 0$ so $y = mx + c$ or
\begin{align*}
&& \frac{\d y}{\d x} &= \frac{xy}{x^2-1} \\
\Rightarrow && \int \frac1y \d y &= \int \frac{x}{x^2-1} \d x \\
\Rightarrow && \ln |y| &= \frac12 \ln |x^2-1| + C \\
\Rightarrow && y^2 &= A(x^2-1)
\end{align*}
Suppose $y = mx+c$ then we must have $(x^2-1)m^2-2xm(mx+c)+(mx+c)^2 = -m^2+c^2 \Rightarrow c^2 = m^2$
If $y^2 = A(x^2-1)$ then $2yy' = 2xA$ and
\begin{align*}
&& 0 &= \frac{y^2}{A}\left ( \frac{xA}{y} \right)^2 - 2x^2A+A(x^2-1)-1 \\
&&&= x^2A-2x^2A+x^2A-A-1 \\
\Rightarrow && A &= -1
\end{align*}
Therefore $x^2 + y^2 = 1$
\end{questionparts}
This was another popular question, and many candidates achieved decent marks on this question. Many candidates were correctly able to differentiate (*), although a significant number ran into difficulties with the (y')² term, where things such as 2y' d/dx(y') = 2y' · y'' dy/dx were common errors. Although the rest of (*) was usually differentiated correctly by these candidates, since the rest of part (i) depended upon getting this first step correct, they floundered from then on. Many of these candidates nevertheless went on to gain additional marks by at least making a good start to part (ii). Also, candidates must remember to read the question and to follow its guidance; the question instructed them to differentiate (*), and those who tried rearranging it instead got nowhere. From y'' = 0, most candidates deduced that y' = m and substituted this back into (*) to determine y = mx − m². However, when working like this, it is vital to check that the purported y, call it ŷ say, satisfies dŷ/dx = y', since y and y' must be related by both the given differential equation and also by y' = dy/dx. There may be other arguments which would allow one not to differentiate the obtained y, but these would have to be given explicitly. The alternative method of determining that y = mx + c and then substituting this into (*) was noticeably less common, but avoided this subtlety. For the 2y'' = x case, similar comments again apply, although here it was concerning how many candidates integrated to get y = ¼x² without including a constant of integration. In part (ii), few candidates succeeded in correctly differentiating the differential equation. One of the most common errors was to claim that d/dx(y²) = 2y'. The few candidates who correctly differentiated the equation mostly applied the techniques from part (i) to solve the equation successfully. Several fudged the solution of the resulting equation (x² − 1)y' = xy by conveniently forgetting the absolute value signs when integrating (as shown in the sample solutions), but this error was not penalised on this occasion.