Year: 2011
Paper: 3
Question Number: 3
Course: UFM Pure
Section: Complex numbers 2
No solution available for this problem.
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Show that, provided $q^2\ne 4p^3$,
the polynomial
\[
\hphantom{(p\ne0, \ q\ne0)\hspace{2cm}}
x^3-3px +q
\hspace {2cm} (p\ne0, \ q\ne0)
\]
can be written in the
form
\[ a(x-\alpha)^3 + b(x-\beta)^3\,,
\]
where $\alpha$ and $\beta$ are the roots of the quadratic equation
$pt^2 -qt +p^2=0$, and $a$ and $b$ are constants which you should
express in terms of $\alpha$ and $\beta$.
Hence show that one solution of
the equation $x^3-24x+48=0\,$ is
\[
x= \frac{2 (2-2^{\frac13})}{1-2^{\frac13}}
\]
and obtain similar expressions for the other two solutions
in
terms of $\omega$, where $\omega = \mathrm{e}^{2\pi\mathrm{i}/3}\,$.
Find also the roots of
$x^3-3px +q=0$
when
$p=r^2$ and $q= 2r^3$ for some non-zero constant $r$.
The second most popular question, attempted by 80% of the cohort, with a similar level of success to question 2. The significance of the condition ≥ 4 was ignored by many candidates, and the fact that it does not apply in the last part of the question was often similarly overlooked. Whilst a and b were generally found correctly, the rest of the first part was often missing. Though there were frequent numerical errors, many candidates correctly found the given solution of the equation, though the other two eluded most, with a common error being to assume that the other two were xω and xω².