Year: 2022
Paper: 3
Question Number: 3
Course: LFM Pure
Section: Implicit equations and differentiation
No solution available for this problem.
One question was attempted by well over 90% of the candidates two others by about 90%, and a fourth by over 80%. Two questions were attempted by about half the candidates and a further three questions by about a third of the candidates. Even the other three received attempts from a sixth of the candidates or more, meaning that even the least popular questions were markedly more popular than their counterparts in previous years. Nearly 90% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The curve $C_1$ has equation
\[ ax^2 + bxy + cy^2 = 1 \]
where $abc \neq 0$ and $a > 0$.
Show that, if the curve has two stationary points, then $b^2 < 4ac$.
\item The curve $C_2$ has equation
\[ ay^3 + bx^2y + cx = 1 \]
where $abc \neq 0$ and $b > 0$.
Show that the $x$-coordinates of stationary points on this curve satisfy
\[ 4cb^3 x^4 - 8b^3 x^3 - ac^3 = 0\,. \]
Show that, if the curve has two stationary points, then $4ac^6 + 27b^3 > 0$.
\item Consider the simultaneous equations
\begin{align*}
ay^3 + bx^2 y + cx &= 1 \\
2bxy + c &= 0 \\
3ay^2 + bx^2 &= 0
\end{align*}
where $abc \neq 0$ and $b > 0$.
Show that, if these simultaneous equations have a solution, then $4ac^6 + 27b^3 = 0$.
\end{questionparts}
This was very popular, being attempted by over 90%, but not very successfully, with a mean score of about 5.5/20. In part (i), candidates generally obtained a correct equation for x or y, but then failed to properly justify the manipulation of the inequality. Whilst the quartic was frequently correctly obtained in part (ii), there were a number of different incorrect assumptions or assertions made regarding the two stationary points being repeated roots or the value of the quartic having different signs at the two stationary points. It was also common that the case when c is negative was not considered. Whilst it was not uncommon for candidates to argue incorrectly for part (iii) that the three equations were equivalent to the curve C2 in part (ii) having one stationary point, (often using dy/dx = 0/0), in contrast, a pleasing number of candidates who made little progress in (ii) past obtaining the quartic, approached part (iii) by simply attempting to solve the equations by elimination, earning full or close to full marks.