Year: 2009
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Implicit equations and differentiation
There were significantly more candidates attempting this paper again this year (over 900 in total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20 marks, and the median mark was 48. The pure questions were the most popular as usual; about two-thirds of candidates attempted each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and question 5 (attempted by about one third). The mechanics questions were only marginally more popular than the probability and statistics questions this year; about one quarter of the candidates attempted each of the mechanics questions, while the statistics questions were attempted by about one fifth of the candidates. A significant number of candidates ignored the advice on the front cover and attempted more than six questions. In general, those candidates who submitted answers to eight or more questions did fairly poorly; very few people who tackled nine or more questions gained more than 60 marks overall (as only the best six questions are taken for the final mark). This suggests that a skill lacking in many students attempting STEP is the ability to pick questions effectively. This is not required for A-levels, so must become an important part of STEP preparation. Another "rubric"-type error was failing to follow the instructions in the question. In particular, when a question says "Hence", the candidate must make (significant) use of the preceding result(s) in their answer if they wish to gain any credit. In some questions (such as question 2), many candidates gained no marks for the final part (which was worth 10 marks) as they simply quoted an answer without using any of their earlier work. There were a number of common errors which appeared across the whole paper. These included a noticeable weakness in algebraic manipulations, sometimes indicating a serious lack of understanding of the mathematics involved. As examples, one candidate tried to use the misremembered identity cos β = sin √(1 − β²), while numerous candidates made deductions of the form "if a² + b² = c², then a + b = c" at some point in their work. Fraction manipulations are also notorious in the school classroom; the effects of this weakness were felt here, too. Another common problem was a lack of direction; writing a whole page of algebraic manipulations with no sense of purpose was unlikely to either reach the requested answer or gain the candidate any marks. It is a good idea when faced with a STEP question to ask oneself, "What is the point of this (part of the) question?" or "Why has this (part of the) question been asked?" Thinking about this can be a helpful guide. One aspect of this is evidenced by pages of formulæ and equations with no explanation. It is very good practice to explain why you are doing the calculation you are, and to write sentences in English to achieve this. It also forces one to focus on the purpose of the calculations, and may help avoid some dead ends. Finally, there is a tendency among some candidates when short of time to write what they would do at this point, rather than using the limited time to actually try doing it. Such comments gain no credit; marks are only awarded for making progress in a question. STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A curve has the equation
\[
y^3 = x^3 +a^3+b^3\,,
\]
where $a$ and $b$ are positive constants. Show that the tangent to the curve at the point $(-a,b)$ is
\[
b^2y-a^2x = a^3+b^3\,.
\]
In the case $a=1$ and $b=2$, show that the $x$-coordinates of the points where the tangent meets the curve satisfy
\[
7x^3 -3x^2 -27x-17 =0\,.
\]
Hence find positive integers $p$, $q$, $r$ and $s$ such that
\[
p^3 = q^3 +r^3 +s^3\,.
\]\begin{align*}
&& y^3 &= x^3 + a^3 + b^3 \\
\Rightarrow && 3y^2 \frac{\d y}{\d x} &= 3x^2 \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{x^2}{y^2}
\end{align*}
Therefore the tangent at the point $(-a,b)$ has gradient $\frac{a^2}{b^2}$, ie
\begin{align*}
&& \frac{y-b}{x+a} &= \frac{a^2}{b^2} \\
\Rightarrow && b^2y - b^3 &= a^2 x + a^3 \\
\Rightarrow && b^2 y-a^2 x &= a^3 + b^3
\end{align*}
Notice that tangent will be, $4y-x = 9$ so substituting this we obtain:
\begin{align*}
&& \left (\frac{9+x}{4} \right)^3 &= x^3 + 9 \\
\Rightarrow && 9^3 + 3 \cdot 9^2 x + 3 \cdot 9x^2 + x^3 &= 64x^3 + 64 \cdot 9 \\
\Rightarrow && 9 \cdot (9^2 - 8^2) + 9 \cdot (3 \cdot 9) + 9 \cdot 3x^2 -9 \cdot 7x^3 &= 0 \\
\Rightarrow && 7x^3-3x^2-27x-17 &= 0 \\
\Rightarrow && (x+1)^2(7x-17) &= 0 \tag{repeated root since tangent}
\end{align*}
So we have another point on the curve $y^3 = x^3 + 2^3 + 1^3$, namely $(\frac{17}7, \frac{17+9 \cdot 7}{28}) = (\frac{17}7, \frac{20}{7})$, so
\begin{align*}
20^3 &= 17^3 + 14^3 + 7^3
\end{align*}
This was the most popular question on the paper by far, being attempted by about 90% of the candidates. It was also found to be one of the most straightforward questions; almost 60% of the attempts successfully completed the first two parts of the question, gaining 10 of the available 20 marks in the process, and about 20% of the attempts gained 15 marks or above. In the first part, most candidates understood the procedures required to find the tangent to a curve given implicitly, although some went to the effort of taking the cube root of the original equation. Disturbingly, a few of them could not do this correctly, and wrote y = x + a + b. A much more common and equally concerning error was to differentiate the equation and correctly deduce that dy/dx = x²/y², and then use this as the gradient without substituting in the values of x and y at the point under consideration. The resulting purported equation for the tangent, y − b = (x²/y²)(x + a), is clearly not a linear equation. The candidates then rearranged it and figured out that some substitutions were necessary to reach the given answer, and fudged their way to it. It is always recommended that students go back through their answer to find the source of the error, rather than trying to fudge, as the latter is almost certainly going to get no credit. For the second part of the question, most candidates realised that they needed to solve the two equations simultaneously (although there were some substitution errors encountered at this point). The process of substituting one equation into another and rearranging to reach the given cubic was done well, although there was a reasonable amount of fudging present here, too. Also, some candidates attempted to perform the substitutions while keeping the as and bs present in the equations; it is much simpler to substitute at the start, so that one only needs to work with two variables (x and y). A sizeable number of candidates gave up at this point. This is unfortunate, as with the question saying "Hence", the obvious thing to do is to solve the cubic. Even without any idea of how to go beyond this, actually solving the cubic might have given a clue. Those who attempted this step were reasonably successful, primarily once they realised that x = −1 is a root of the cubic. Again, though, a number of candidates were let down by poor algebraic manipulation at this point. Finally, having reached the two solutions of the cubic, many candidates floundered, assuming that x = 17/7 could not possibly give them a solution. Unfortunately, x = −1 does not help, either. Those who did press ahead with x = 17/7 often struggled to determine the corresponding value of y, frequently leading to some very long pieces of arithmetic as they attempted to calculate x³ + a³ + b³ and then take the cube root, rather than using the tangent equation 4y − x = 9. Often, the long arithmetical approaches failed to reach the correct answer because of a simple slip, which was a shame. A significant number of candidates did, however, manage to correctly reach the final conclusion. As mentioned in the introduction, many candidates simply quoted a solution to this final equation (namely 6³ = 5³ + 4³ + 3³), and this was given no credit, as the question had specifically said "Hence".