Year: 2013
Paper: 1
Question Number: 1
Course: LFM Stats And Pure
Section: Quadratics & Inequalities
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Use the substitution $\sqrt x = y$ (where $y\ge0$) to find the real root of the equation
\[
x + 3\, \sqrt x - \tfrac12 =0\,.
\]
\item Find all real roots of the following equations:
\begin{itemize}
\item $x+10\,\sqrt{x+2\, }\, -22 =0\,$;
\item $x^2 -4x + \sqrt{2x^2 -8x-3 \,}\, -9 =0\,$.
\end{itemize}
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& 0 &= x + 3\sqrt{x} - \frac12 \\
\sqrt{x} = y: && 0&= y^2 + 3y - \frac12 \\
\Rightarrow && y &= \frac{-3\pm\sqrt{3^2+2}}{2} \\
&&&= \frac{-3 \pm \sqrt{11}}{2} \\
y > 0: && x &= \left ( \frac{\sqrt{11}-3}{2} \right)^2
\end{align*}
\item \begin{itemize}
\item \begin{align*}
&& 0 &= x + 10\sqrt{x+2} - 22 \\
y = \sqrt{x+2}: && 0 &= y^2 - 2 + 10y - 22 \\
&&&= y^2 + 10y - 24 \\
&&&= (y-2)(y+12) \\
\Rightarrow && y &= 2, -12 \\
y > 0: && x &= 2
\end{align*}
\item Let $y = \sqrt{2x^2-8x-3}$, so \begin{align*}
&& 0 &= x^2 - 4x +\sqrt{2x^2-8x-3} - 9 \\
&& 0 &= \frac{y^2+3}{2} + y - 9 \\
&&&= \frac12 y^2 +y - \frac{15}{2} \\
&&&= \frac12 (y-3)(y+5) \\
\Rightarrow && y &= 3,-5 \\
y > 0: && 9 &= 2x^2-8x-3 \\
\Rightarrow && 0 &= 2x^2-8x-12 \\
&&&= 2(x^2-4x-6) \\
\Rightarrow && x &= 2 \pm \sqrt{10}
\end{align*}
\end{itemize}
\end{questionparts}
This question is all about using substitutions to simplify the working required to solve various increasingly complicated looking equations. It was the most popular question on the paper, essentially attempted by every candidate (as is the intention). The obvious pitfall of not realising that the square-root sign indicates the "non-negative square-root" of a quantity was clearly flagged at the outset. Thus, the only remaining hurdle to fully complete success lay in the need to check the validity of solutions once found. The mean score on this question was 14/20, and this question thus represented a successful entré to the paper for almost everyone. The use of the quadratic formula and the method of completing the square appeared in almost equal measure throughout the question, although a significant minority of candidates opted to rearrange and square in both (ii) and (iii). This was not a major obstacle to success in (ii) but led to a quartic equation in (iii) with which few candidates knew how to make successful progress. The final hurdle for most candidates lay in a final justification that any roots found (up to four of them, depending upon the method chosen) were genuinely valid. It is very easy to explain, without the use of direct verification, that the two roots found via the substitution method are good, but very few candidates made any attempt to justify their results.