2010 Paper 1 Q2
Year: 2010
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Differentiation
Problem
- Show that, in the case \(a=0\) and \(b= \frac12\), there is one stationary point on either side of the curve's vertical asymptote, and sketch the curve.
- Sketch the curve in the case \( a=\tfrac{9}{2}\) and \(b=0\,\).
Solution
- If \(a = 0, b = \frac12\), we have \(x^2-\frac12x -\frac12 = 0 \Rightarrow (2x+1)(x-1) = 0 \Rightarrow x = -\frac12, x=1\). The asymptote is at \(x = \frac12\) so they are on either side.
- \(\,\)
Examiner's report
— 2010 STEP 1, Question 2There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Show LaTeX source
Problem source
The
curve
$\displaystyle y=\Bigl(\frac{x-a}{x-b}\Bigr)\e^{x}$,
where $a$ and $b$ are constants,
has two stationary points.
Show that
\[
a-b<0 \ \ \ \text{or} \ \ \ a-b>4 \,.
\]
\begin{questionparts}
\item[(i)] Show that, in the case $a=0$ and $b= \frac12$,
there is one stationary point on either side of the
curve's vertical asymptote, and sketch the curve.
\item[(ii)] Sketch the curve in the case
$ a=\tfrac{9}{2}$ and $b=0\,$.
\end{questionparts}Solution source
\begin{align*}
&& y &= \left ( \frac{x-a}{x-b} \right )e^x \\
&&y'& = \left ( \frac{x-a}{x-b} \right )e^x + \left ( \frac{(x-b)-(x-a)}{(x-b)^2}\right )e^x \\
&&&= \left ( \frac{(x-b)(x-a) +a-b}{(x-b)^2} \right)e^x \\
&&&= \left ( \frac{x^2-(a+b)x+a-b+ab}{(x-b)^2} \right)e^x \\
&& 0 &< \Delta = (a+b)^2 - 4 \cdot 1 \cdot (a-b+ab) \\
&&&= a^2+2ab+b^2-4a+4b-4ab \\
&&&= a^2-2ab+b^2-4a+4b\\
&&&= (a-b)^2-4(a-b) \\
&&&= (a-b)(a-b-4) \\
\end{align*}
Considered as a quadratic in $a-b$ we can see $a-b < 0$ or $a-b > 4$
\begin{questionparts}
\item If $a = 0, b = \frac12$, we have $x^2-\frac12x -\frac12 = 0 \Rightarrow (2x+1)(x-1) = 0 \Rightarrow x = -\frac12, x=1$. The asymptote is at $x = \frac12$ so they are on either side.
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){((#1))/((#1)-0.5)*exp(#1)};
\def\xl{-3};
\def\xu{3};
\def\yl{-5};
\def\yu{20};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the reusable styles to keep code clean
\tikzset{
x=\xscale cm, y=\yscale cm,
axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
grid/.style={thin, dashed, gray!30},
curveA/.style={very thick, color=cyan!70!black, smooth},
curveB/.style={very thick, color=orange!90!black, smooth},
dot/.style={circle, fill=black, inner sep=1.2pt},
labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
}
% Draw background grid
\draw[grid] (\xl,\yl) grid[xstep=1,ystep=5] (\xu,\yu);
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
% Define the bounding region with clip
\begin{scope}
\clip (\xl,\yl) rectangle (\xu,\yu);
\draw[curveA, domain=\xl:0.45, samples=150]
plot ({\x},{\functionf(\x)});
\draw[curveA, domain=0.55:\xu, samples=150]
plot ({\x},{\functionf(\x)});
\draw[curveB, dashed] (0.5, \yl) -- (0.5, \yu);
\end{scope}
% Annotate Function Names
\node[curveA, labelbox] at ({-1}, {2}) {$y = \frac{x}{x-\frac12}e^x$};
\end{tikzpicture}
\end{center}
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){((#1)-4.5)/((#1))*exp(#1)};
\def\xl{-3};
\def\xu{6};
\def\yl{-20};
\def\yu{15};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the reusable styles to keep code clean
\tikzset{
x=\xscale cm, y=\yscale cm,
axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
grid/.style={thin, dashed, gray!30},
curveA/.style={very thick, color=cyan!70!black, smooth},
curveB/.style={very thick, color=orange!90!black, smooth},
dot/.style={circle, fill=black, inner sep=1.2pt},
labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
}
% Draw background grid
\draw[grid] (\xl,\yl) grid[xstep=1,ystep=5] (\xu,\yu);
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
% Define the bounding region with clip
\begin{scope}
\clip (\xl,\yl) rectangle (\xu,\yu);
\draw[curveA, domain=\xl:{-0.05}, samples=150]
plot ({\x},{\functionf(\x)});
\draw[curveA, domain={0.05}:\xu, samples=150]
plot ({\x},{\functionf(\x)});
\draw[curveB, dashed] (0, \yl) -- (0, \yu);
\end{scope}
% Annotate Function Names
\node[curveA, labelbox] at ({-1}, {2}) {$y = \frac{x-\frac92}{x}e^x$};
\end{tikzpicture}
\end{center}
\end{questionparts}
This was a fairly popular question, attempted by around two-thirds of candidates. It was pleasing to see how many of them were correctly able to differentiate the expression given at the start of the question. Again, the earlier comments on algebraic accuracy bear repeating at this point: a number of candidates became unstuck here through algebraic or sign errors, and this was a repeating theme throughout this question. The inequalities proved challenging: setting the derivative equal to zero was an obvious step, and then the resulting quadratic is crying out for consideration of the discriminant. Even many of those who got this far failed to adequately explain their solution of the quadratic inequality, jumping straight to the given answer. Moving on to the graph sketches, it became obvious that different candidates have different understandings of what is expected. At the very least, a graph sketch should indicate (where possible) the coordinates of stationary points and axis crossings, as well as asymptotic behaviour. Many candidates, pleased with their success at finding the x-coordinates of the stationary points, then stopped and did not calculate the y-coordinates. The latter would have made the decision about the nature of the stationary points essentially trivial and would have helped them draw more accurate sketches, especially in part (ii). On a positive note, most of the candidates who reached this part of the question determined the turning points and vertical asymptotes correctly, though some thought that part (ii) had no turning points (despite having shown that it does earlier on). The mean score on this question was noticeably lower than on several of the other pure mathematics questions, suggesting that graph-sketching is an area which requires more attention from candidates during their preparation.