2011 Paper 2 Q1
Year: 2011
Paper: 2
Question Number: 1
Course: LFM Stats And Pure
Section: Curve Sketching
Problem
- Sketch the curve \(y=\sqrt{1-x} + \sqrt{3+x}\;\). Use your sketch to show that only one real value of \(x\) satisfies \[ \sqrt{1-x} + \sqrt{3+x} = x+1\,, \] and give this value.
- Determine graphically the number of real values of \(x\) that satisfy \[ 2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}\;. \] Solve this equation.
Solution
- Clearly the only solution is \(x = 1\)
- There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}
Examiner's report
— 2011 STEP 2, Question 1There were just under 1000 entries for paper II this year, almost exactly the same number as last year. After the relatively easy time candidates experienced on last year's paper, this year's questions had been toughened up significantly, with particular attention made to ensure that candidates had to be prepared to invest more thought at the start of each question – last year saw far too many attempts from the weaker brethren at little more than the first part of up to ten questions, when the idea is that they should devote 25-40 minutes on four to six complete questions in order to present work of a substantial nature. It was also the intention to toughen up the final "quarter" of questions, so that a complete, or nearly-complete, conclusion to any question represented a significant (and, hopefully, satisfying) mathematical achievement. Although such matters are always best assessed with the benefit of hindsight, our efforts in these areas seem to have proved entirely successful, with the vast majority of candidates concentrating their efforts on four to six questions, as planned. Moreover, marks really did have to be earned: only around 20 candidates managed to gain or exceed a score of 100, and only a third of the entry managed to hit the half-way mark of 60. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were attempted by almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the remaining questions were less popular, and some received almost no "hits". Overall, the highest scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts, but those who braved it scored very well). This at least is indicative that candidates are being careful in exercising some degree of thought when choosing (at least the first four) 'good' questions for themselves, although finding six successful questions then turned out to be a key discriminating factor of candidates' abilities from the examining team's perspective. Each of questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
\begin{questionparts}
\item
Sketch the curve $y=\sqrt{1-x} + \sqrt{3+x}\;$.
Use your sketch to show that only one real value of $x$ satisfies
\[
\sqrt{1-x} + \sqrt{3+x} = x+1\,,
\]
and give this value.
\item
Determine graphically the number of real values of $x$ that satisfy
\[
2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}\;.
\]
Solve this equation.
\end{questionparts}Solution source
\begin{questionparts}
\item
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){sqrt(1-(#1))+sqrt(3+(#1))};
\def\xl{-4};
\def\xu{2};
\def\yl{-1};
\def\yu{4};
% 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 styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=-3:1, samples=100]
plot (\x, {\functionf(\x)});
\filldraw (0,2) circle (1.5pt) node[right] {$2$};
\filldraw (-3,0) circle (1.5pt) node[below] {$-3$};
\filldraw (1,0) circle (1.5pt) node[below] {$1$};
\filldraw (-1,0) circle (1.5pt) node[below] {$-1$};
\filldraw (0,1) circle (1.5pt) node[right] {$1$};
\draw[thick, red] (\xl, {\xl+1}) -- (\xu, {\xu+1});
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
Clearly the only solution is $x = 1$
\item
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){sqrt(3-(#1))+sqrt(3+(#1))};
\def\functiong(#1){2*sqrt(1-(#1))};
\def\xl{-4};
\def\xu{4};
\def\yl{-1};
\def\yu{4};
% 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 styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=-3:3, samples=101]
plot (\x, {\functionf(\x)});
\draw[thick, red, smooth, domain=\xl:1, samples=101]
plot (\x, {\functiong(\x)});
\filldraw (0,2) circle (1.5pt) node[right] {$2$};
\filldraw (-3,0) circle (1.5pt) node[below] {$-3$};
\filldraw (3,0) circle (1.5pt) node[below] {$3$};
\filldraw (1,0) circle (1.5pt) node[below] {$1$};
\filldraw (-1,0) circle (1.5pt) node[below] {$-1$};
\filldraw (0,1) circle (1.5pt) node[right] {$1$};
\filldraw ({-(2+2*sqrt(11))/5},{\functiong(-(2+2*sqrt(11))/5)}) circle (1.5pt);
% \draw[thick, red] (\xl, {\xl+1}) -- (\xu, {\xu+1});
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
There is clearly only one solution, with $x \approx -2$
\begin{align*}
&& 4(1-x) &= 6+2\sqrt{9-x^2} \\
&& -2x-1 &=\sqrt{9-x^2} \\
\Rightarrow && 4x^2+4x+1 &= 9-x^2 \\
\Rightarrow && 0 &= 5x^2+4x-8 \\
&&x&= \frac{-2\pm 2\sqrt{11}}{5} \\
\Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right)
\end{align*}
\end{questionparts}
The first question is invariably set with the intention that everyone should be able to attempt it, giving all candidates something to get their teeth into and thereby easing them into the paper with some measure of success. As mentioned above, this was both a very popular question and a high-scoring one. Even so, there were some general weaknesses revealed in the curve-sketching department, as many candidates failed to consider (explicitly or not) things such as the gradient of the curve at its endpoints and, in particular, the shape of the curve at its peak (often more of a vertex than a maximum). It was also strange that surprisingly many who had the correct domain for the curve and had decided that the single point of intersection of line and curve was at x = 1 still managed to draw the line y = x + 1 not through the endpoint at x = 1. Most other features – domain, symmetry, coordinates of key points, etc. – were well done in (i). Unfortunately, those who simply resort to plotting points are really sending quite the wrong message about their capabilities to the examiners. Following on in (ii), the majority of candidates employed the expected methods and were also quite happy to plough into the algebra of squaring-up and rearranging; however, there were frequently many (unnecessarily) careless errors involved. The only other very common error was in the sketch of the half-parabola y² = 1 − x, due to a misunderstanding of the significance of the radix sign.