2019 Paper 1 Q2
Year: 2019
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Implicit equations and differentiation
Problem
Solution
Examiner's report
— 2019 STEP 1, Question 2In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document; all of which are available from the STEP and Cambridge Examinations Board websites. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions for several months beforehand. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
The curve $C$ is given parametrically by the equations $x = 3t^2$, $y = 2t^3$. Show that the equation of the tangent to $C$ at the point $(3p^2 , 2p^3)$ is $y = px - p^3$.
Find the point of intersection of the tangents to $C$ at the distinct points $(3p^2 , 2p^3)$ and $(3q^2 , 2q^3)$. Hence show that, if these two tangents are perpendicular, their point of intersection is $(u^2 + 1 , -u)$, where $u = p + q$. The curve $\tilde{C}$ is given parametrically by the equations $x = u^2 + 1$, $y = -u$. Find the coordinates of the points that lie on both $C$ and $\tilde{C}$.
Sketch $C$ and $\tilde{C}$ on the same axes.Solution source
\begin{align*}
&& \frac{\d y}{\d x} &= \frac{\frac{\d y}{\d t}}{\frac{\d x}{\d t}} \\
&&&= \frac{6t^2}{6t} = t \\
\Rightarrow && \frac{y-2p^3}{x - 3p^2} &= p \\
\Rightarrow && y &= px-3p^3+2p^3 \\
&& y &= px - p^3
\end{align*}
The two lines will be
\begin{align*}
&& y &= px - p^3 \\
&& y &= qx - q^3 \\
\Rightarrow && p^3-q^3 &= (p-q)x \\
\Rightarrow && x &= p^2+pq+q^2 \\
&& y &= p(p^2+pq+q^2)-p^3 \\
&&&= pq(p+q) \\
&& (x,y) &= (p^2+pq+q^2,pq(p+q)) \\
\end{align*}
If the tangents are $\perp$ then $pq=-1$, so we have
\begin{align*}
&& (x,y) &= (p^2+2pq+q^2-pq, pq(p+q)) \\
&&&= ((p+q)^2-1, -(p+q)) \\
&&&= (u^2-1, -u)
\end{align*}
We have $x = y^2+1$ and $\left ( \frac{x}{3} \right)^3 = \left ( \frac{y}{2}\right)^2 \Rightarrow y^2 = \frac{4}{27}x^3$ so
\begin{align*}
&& 0 &= \frac{4}{27}x^3-x+1 \\
&&0&=4x^3-27x+27 \\
&&&= (x+3)(2x-3)^2
\end{align*}
So we have the points $(x,y) = \left (\frac32, \pm\frac{1}{\sqrt{2}}\right)$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
\def\xl{-1};
\def\xu{10};
\def\yl{-5};
\def\yu{5};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=-2:2, samples=100]
plot ({3*\x*\x}, {2*\x*\x*\x});
\draw[thick, red, smooth, domain=-5:5, samples=100]
plot ({\x*\x+1}, {-\x});
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\filldraw (1.5, {1/sqrt(2)}) circle (1.5pt) node[above] {$(\frac32, \frac1{\sqrt{2}})$};
\filldraw (1.5, {-1/sqrt(2)}) circle (1.5pt) node[below] {$(\frac32, -\frac1{\sqrt{2}})$};
\end{tikzpicture}
\end{center}
Given that there has been a question of a similar nature to this on several of the STEP I's of recent years, the demands of these sorts of coordinate geometry questions have become relatively routine. Attracting the interest of 60% of the candidature, this question drew the second highest mean score overall, just over half-marks. The first major stumbling block in this question was a lack of technical precision when taking square roots of equations. If done using parametric differentiation this was not needed, but too many students rearranged x = 3t² to form t = √(x/3). Many candidates also assumed that if 3p² = 3t² then p must equal t. Candidates should also be aware that not all letters are equivalent. In the first part of this question, t was a variable but p was a fixed value. Many candidates wrote showing a fundamental misunderstanding, although it led algebraically to the correct result in this instance. Lots of candidates did not read the question carefully. In the second paragraph it was required that the point of intersection of the tangents be found in general. Many students who clearly could have done this conflated it with the constraint that the tangents had to be perpendicular, which was a separate question. Finding the parameterised form of the locus of intersections of perpendicular tangents required some judicious algebra and use of pq = −1. This was generally done quite well, although many candidates were not aware of the difference of two cubes which made the algebra much nastier. It is often worth neatening up algebraic expressions before continuing to work with them. When this curve was intersected with the original curve it became clear that many candidates were not clear which x coordinate was related to which point. Correct simultaneous equations usually led to a disguised cubic. Candidates seemed quite good at spotting one solution and factorising to find the remaining ones. However, not all candidates explained why one solution was not possible. The final sketch was rarely done well. The semi-cubical parabola was meant to be unfamiliar, but most students could not piece together the information to realise that it must have a cusp. Overall, most candidates were able to engage and make progress with this question, albeit with several technical errors.