Year: 2012
Paper: 3
Question Number: 5
Course: LFM Pure
Section: Trigonometry 2
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1554.6
Banger Comparisons: 6
\begin{questionparts}
\item The point with coordinates $(a, b)$, where $a$ and $b$ are rational numbers,is called an \textit{integer rational point} if both $a$ and $b$ are integers; a \textit{non-integer rational point} if neither $a$ nor $b$ is an integer.
\begin{itemize}
\item[\bf (a)] Write down an integer rational point and a non-integer rational point on the circle $x^2+y^2 =1$.
\item [\bf (b)] Write down an integer rational point on the circle
$x^2+y^2=2$.
Simplify
\[
(\cos\theta + \sqrt m \sin\theta)^2 + (\sin\theta - \sqrt m \cos\theta)^2 \,
\]
and hence obtain a non-integer rational point on the circle $x^2+y^2=2\,$.
\end{itemize}
\item The point with coordinates $(p+\sqrt 2 \, q\,,\, r+\sqrt 2 \, s)$, where $p$, $q$, $r$ and $s$ are rational numbers, is called: an \textit{integer $2$-rational point} if all of $p$, $q$, $r$ and $s$ are integers; a \textit{non-integer $2$-rational point} if none of $p$, $q$, $r$ and $s$ is an integer.
\begin{itemize}
\item[\bf (a)] Write down an integer $2$-rational point, and obtain a non-integer $2$-rational point, on the circle $x^2+y^2=3\,$.
\item [\bf(b)] Obtain a non-integer $2$-rational point on the circle $x^2+y^2=11\,$.
\item [\bf(c)]Obtain a non-integer $2$-rational point on the hyperbola $x^2-y^2 =7 $.
\end{itemize}
\end{questionparts}\begin{questionparts}
\begin{itemize}
\item[\bf (a)] $(1,0)$ is an integer rational point, $(\tfrac35, \tfrac45)$ is a non-integer rational point.
\item[\bf (b)] $(1,1)$ is an integer rational point.
\begin{align*}
&& (\cos \theta + \sqrt{m} \sin \theta)^2 + (\sin \theta - \sqrt{m} \cos \theta)^2 &= (1+m)(\cos^2 \theta + \sin^2 \theta) \\
&&&= 1+m
\end{align*}
Therefore $(\tfrac75, \tfrac{1}{5})$ is a non-integer rational point.
\end{itemize}
\item
\begin{itemize}
\item[\bf (a)] $(1, \sqrt2)$ is an integer $2$-rational point. $(\frac35 + \frac45\sqrt2, \frac45 - \frac{3}{5}\sqrt2)$ is a non-integer $2$-rational point.
\item [\bf(b)] First notice that $(\sqrt2)^2 +3^2 = 11$ so then consider $(1 + \tfrac32\sqrt2, 1-\tfrac32\sqrt2)$ will work as $\pi/4$ degree rotation.
\item [\bf(c)] First notice $3^2-(\sqrt2)^2 = 2$. Notice that $(k\sec \theta + \sqrt{m} \tan \theta)^2 - (k\tan \theta + \sqrt{m} \sec \theta)^2 = k^2-m$. Taking $k= 3$ we have $(3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)$
\end{itemize}
\end{questionparts}
Note: we can also find the additional point in the last part by considering lines through $(3, \sqrt2)$, for example $y = -\frac32x + \sqrt2 + \frac92$ would give the same point.
This was only very slightly more popular than question 4, though with the same level of success. A lot of candidates scored just the first 5 marks, getting as far as completing the simplification in part (i)(b), but then, being unable to apply it for the final result, and then making no progress with part (ii). The biggest problem was that candidates ignored the definitions given at the start of the question, most notably that "a and b are rational numbers". The other common problem was that candidates chose a simple value for θ such as π or 2π rather than for cos such as ½. In part (ii), quite frequently, candidates substituted specific values involving √2, and some then successfully found solutions. For part (ii)(c), a method using cosh and sinh was not unexpected, although the comparable one with sec and tan was quite commonly used too.