Year: 2024
Paper: 3
Question Number: 2
Course: LFM Stats And Pure
Section: Quadratics & Inequalities
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% 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 Solve the inequalities
\begin{enumerate}
\item $\sqrt{4x^2 - 8x + 64} \leqslant |x+8|\,$,
\item $\sqrt{4x^2 - 8x + 64} \leqslant |3x-8|\,$.
\end{enumerate}
\item
\begin{enumerate}
\item Let $\mathrm{f}(x) = \sqrt{4x^2 - 8x + 64} - 2(x-1)$.
Show, by considering $\bigl(\sqrt{4x^2 - 8x + 64} + 2(x-1)\bigr)\mathrm{f}(x)$ or otherwise, that $\mathrm{f}(x) \to 0$ as $x \to \infty$.
\item Sketch $y = \sqrt{4x^2 - 8x + 64}$ and $y = 2(x-1)$ on the same axes.
\end{enumerate}
\item Find a value of $m$ and the corresponding value of $c$ such that the solution set of the inequality
\[\sqrt{4x^2 - 5x + 4} \leqslant |mx + c|\]
is $\{x : x \geqslant 3\}$.
\item Find values of $p$, $q$, $m$ and $c$ such that the solution set of the inequality
\[|x^2 + px + q| \leqslant mx + c\]
is $\{x : -5 \leqslant x \leqslant 1\} \cup \{x : 5 \leqslant x \leqslant 7\}$.
\end{questionparts}
Three quarters of the candidates attempted this question with a mean score of just under half marks. In part (i), candidates often omitted a justification that the LHS of the inequality was real and for noting that both sides are positive before squaring. Part (ii)(a) was generally done quite well, although some candidates ignored the suggested method and argued that because the lead terms cancel as x → ∞, f(x) → 0, not earning full marks. The sketch in (ii)(b) was not generally done very well. In general, sketches just need to have the same key features as the actual plot of the function. The asymptotes and symmetry about x = 1 were crucial here. Part (iii) was done fairly well by those that attempted it, most noticing that they should choose values of m to ensure that the x2 terms should cancel. There were not many significant attempts on part (iv). To start, it was relatively straightforward to state that as four critical values were required, the quadratic needed to cross the x-axis, but this was often missed. However, there were some very efficient and neat solutions to this part, and candidates who got on the right path initially executed it well. The most common error was failure to get the four roots attached to the correctly signed version of the quadratic. Candidates who used a diagram were generally much more successful with this.