Year: 2009
Paper: 1
Question Number: 8
Course: LFM Pure
Section: Introduction to trig
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: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item
The equation of the circle $C$ is
\[
(x-2t)^2 +(y-t)^2 =t^2,
\]
where $t$ is a positive number. Show that $C$ touches the line $y=0\,$. Let $\alpha$ be the acute angle between the $x$-axis and the line joining the origin to the centre of $C$. Show that $\tan2\alpha=\frac43$ and deduce that $C$ touches the line $3y=4x\,$.
\item Find the equation of the incircle of the triangle formed by the lines $y=0$, $3y=4x$ and $4y+3x=15\,$.
\textbf{Note:} The \textit{incircle} of a triangle is the circle, lying totally inside the triangle, that touches all three sides.
\end{questionparts}\begin{questionparts}
\item This is a circle centre $(2t,t)$ with radius $t$. Therefore it is exactly $t$ away from the line $y = 0$ so just touches that line.
Not that $\tan \alpha = \frac{t}{2t} = \frac12$ so $\tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2\alpha} = \frac{1}{1-\frac14} = \frac43$. Therefore the line $y = \frac43x$ or $3y = 4x$ is tangent to $C$.
\item Note that $3y=4x$ and $4y+3x=15$ are perpendicular, so this is a right-angled triangle with incenter $(2t,t)$ for some $t$ and hypotenuse $15$
We can find the third coordinate when $3y-4x = 0$ and $4y+3x = 15$ meet, ie $(\frac{9}{5}, \frac{12}5)$
The incentre lies on the bisector of the right angle at this point, which is the line through $(\frac{9}{5}, \frac{12}5)$ and $(\frac{15}{2}, 0)$, so
\begin{align*}
&& \frac{2t-\frac{12}{5}}{t - \frac{9}{5}} &= \frac{-\frac{12}{5}}{\frac{15}2-\frac95} \\
\Rightarrow && \frac{10t-12}{5t-9} &= \frac{-24}{57} = -\frac{8}{19} \\
\Rightarrow && 190t - 12 \cdot 19 &= -40t + 72 \\
\Rightarrow && t &= 2
\end{align*}
Therefore the center is $(4, 2)$ and the equation is $(x-4)^2+(y-2)^2=2^2$
\end{questionparts}
The first part of this question was generally done very well, but the second half stumped most candidates. Most students confidently used the double angle formula for tan 2α, and most realised that the line y = ¾x is tangent to the circle because it is at an angle of 2α to the x-axis and passes through the origin. There were some candidates who misremembered or misapplied the double angle formula, or were unable to explain why the circle touches the line. In questions such as these, a good sketch is very helpful. The second part of the question was done relatively poorly in comparison; fewer than a quarter of the candidates gained more than one mark for it. Even after drawing a good sketch, many candidates showed little idea of how to proceed. They wrote pages of calculations with no direction, and so gained no credit. It is not worth performing calculations without a strategy; how will knowing the coordinates of the vertices, for example, help find the value of t? (This was a very common calculation, perhaps because it is something the candidates are very comfortable with.) About half of the attempts noted that the two given lines were perpendicular, but then did not do anything useful with this fact. Many students attempted to substitute the equation of the line 4y+3x = 15 into the equation of the circle to eliminate y, but then went on to eliminate t by asserting that x = 2t. This cannot succeed, as the point of tangency with 4y +3x = 15 clearly does not lie on the vertical line through the centre of the circle. Some candidates, though, successfully used the ideas of part (i) to find the equation of the angle bisector of the x-axis and the side AB, and then found the value of t which had (2t, t) lying on this bisector. Another very common successful approach was to find the distance of the point (2t, t) from the line 4y + 3x = 15 using the formula given in the formula book for the distance of a point from a line. It is encouraging to see a significant number of candidates offering creative (and correct) solutions to a problem such as this.