Year: 2019
Paper: 1
Question Number: 1
Course: LFM Pure
Section: Differentiation
In 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.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A straight line passes through the fixed point $(1 , k)$ and has gradient $- \tan \theta$, where $k > 0$ and $0 < \theta < \frac{1}{2}\pi$. Find, in terms of $\theta$ and $k$, the coordinates of the points $X$ and $Y$ where the line meets the $x$-axis and the $y$-axis respectively.
\begin{questionparts}
\item Find an expression for the area $A$ of triangle $OXY$ in terms of $k$ and $\theta$. (The point $O$ is the origin.)
You are given that, as $\theta$ varies, $A$ has a minimum value. Find an expression in terms of $k$ for this minimum value.
\item Show that the length $L$ of the perimeter of triangle $OXY$ is given by
$$L = 1 + \tan \theta + \sec \theta + k(1 + \cot \theta + \cosec \theta).$$
You are given that, as $\theta$ varies, $L$ has a minimum value. Show that this minimum value occurs when $\theta = \alpha$ where
$$\frac{1 - \cos \alpha}{1 - \sin \alpha} = k.$$
Find and simplify an expression for the minimum value of $L$ in terms of $\alpha$.
\end{questionparts}$y = (-\tan \theta)(x-1)+k$ so when $x = 0$, $y = k + \tan \theta$, so $Y = (0, k+\tan \theta)$. When $y = 0$, $x = 1 + \frac{k}{\tan \theta}$
\begin{questionparts}
\item $A = \frac12 (k+\tan \theta)\left ( 1 + \frac{k}{\tan \theta} \right) = k + \frac12 \left (\tan \theta + \frac{k^2}{\tan \theta} \right)$ Notice that $x + \frac{k^2}{x} \geq 2 k$ by AM-GM, so the minimum is $k + \frac12 \cdot 2k = 2k$
\item $\,$
\begin{align*}
L &= k + \tan \theta + 1 + k \cot \theta + \sqrt{(k + \tan \theta)^2 + \left (1 + \frac{k}{\tan \theta} \right)^2} \\
&= k + \tan \theta + 1 + k \cot \theta + \sqrt{k^2 + 2 k \tan \theta +\tan^2 \theta + 1 + 2k \cot \theta + k^2\cot^2 \theta} \\
&= k + \tan \theta + 1 + k \cot \theta + \sqrt{\sec^2 \theta+ 2k \sec\theta\cosec \theta + k^2\cosec^2 \theta} \\
&= k + \tan \theta + 1 + k \cot \theta +\sec \theta + k\cosec \theta\\
&= 1 + \tan \theta + \sec \theta + k (1 + \cot \theta + \cosec \theta)
\end{align*}
\begin{align*}
&& \frac{\d L}{\d \theta} &= \sec^2 \theta + \tan \theta \sec \theta + k(-\cosec^2 \theta - \cot \theta \cosec \theta ) \\
\Rightarrow && 0 &=\sec^2 \alpha+ \tan \theta \sec \alpha+ k(-\cosec^2 \alpha- \cot \alpha\cosec \alpha) \\
\Rightarrow && k &= \frac{\sec^2 \alpha+ \tan \alpha\sec \alpha}{\cosec^2 \alpha+ \cot \alpha\cosec \alpha} \\
&&&= \frac{\sin^2 \alpha(1 + \sin \alpha)}{\cos^2 \alpha (1+ \cos \alpha)} \\
&&&= \frac{(1-\cos^2 \alpha)(1 + \sin \alpha)}{(1-\sin^2 \alpha )(1+ \cos \alpha)} \\
&&&= \frac{1-\cos \alpha}{1-\sin \alpha} \\
\Rightarrow && L &= 1 + \tan \alpha + \sec \alpha + \frac{1-\cos \alpha}{1-\sin \alpha} \left (1 + \cot \alpha + \cosec \alpha \right) \\
&&&= \frac{1+\tan \alpha + \sec \alpha -\sin \alpha-\sin \alpha \tan \alpha-\tan \alpha}{1-\sin \alpha} + \\
&&&\quad \quad \frac{1+\cot \alpha + \cosec \alpha-\cos \alpha-\cos \alpha \cot \alpha -\cot \alpha}{1-\sin \alpha} \\
&&&= \frac{2+\sec \alpha(1-\sin^2 \alpha)-\sin \alpha + \cosec \alpha(1-\cos^2 \alpha)-\cos \alpha}{1-\sin \alpha} \\
&&&= \frac{2+\cos\alpha-\sin \alpha + \sin\alpha-\cos \alpha}{1-\sin \alpha} \\
&&&= \frac{2}{1-\sin \alpha}
\end{align*}
\end{questionparts}
As intended, this was the most popular question on the paper and the one that elicited the highest average score. The set-up is a familiar one for A-level, though the "coefficients" involved are trigonometric throughout; it was this side of things that provided the only real degree of difficulty to the question. Given that there was no requirement for candidates to justify the natures of the two extrema involved in the question, the issue was how accurately candidates could manage to deploy the necessary trig. identities at the appropriate points. Interestingly, it was clear that those who made tough going of the working were the ones who handled the three reciprocal trig. functions – cosec, sec and cot – and their derivatives less confidently. For instance, there are those who immediately convert any trig. situation exclusively into sines and cosines … this generally works perfectly well for A-level questions. Here, it simply turned out to be a hindrance as the resulting terms would contain rational functions which required heavier-duty methods for dealing with them; and many such candidates got into a muddle somewhere along the way, especially with the square of the distance XY in part (ii), which should have been turned into a perfect square reasonably swiftly. In order to get the given result for k towards the end of (ii), a number of candidates resorted to verification, though this worked reasonably well provided they didn't somehow confuse θs with φs. Those who wrapped up the required final answer correctly were those who spotted the difference-of-two-squares factorisation embedded in there and spotted that (c + s)² = 1 + 2sc at the right moment (using the usual abbreviations for cos and sin).