2012 Paper 3 Q3
Year: 2012
Paper: 3
Question Number: 3
Course: LFM Pure
Section: Simultaneous equations
Difficulty: 1700.0
Banger: 1468.7
simultaneous equations
curve sketching
parabola
tangency
quartic
repeated root
polynomial factorisation
intersections with axes
Problem
It is given that the two curves
\[
y=4-x^2
\text{ and }
m x = k-y^2\,,
\]
where \(m > 0\), touch exactly once.
- In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes:
- \(k < 0\, \);
- \(0 < k < 16\), \(k/m < 2\,\);
- \(k > 16\), \(k/m > 2\,\);
- \(k > 16\), \(k/m < 2\,\).
- Now set \(m=12\). Show that the \(x\)-coordinate of any point at which the two curves meet satisfies \[ x^4-8x^2 +12x +16-k=0\,. \] Let \(a\) be the value of \(x\) at the point where the curves touch. Show that \(a\) satisfies \[ a^3 -4a +3 =0 \] and hence find the three possible values of \(a\). Derive also the equation \[ k= -4a^2 +9a +16\,. \] Which of the four sketches in part (i) arise?
Solution
- \(\,\)
- \(\,\)
- \(\,\)
- \(\,\)
- \(\,\)
- Suppose \(m = 12\) \begin{align*} && y &= 4-x^2 \\ && 12x &= k-y^2 \\ \Rightarrow && 12 x&=k-(4-x^2)^2 \\ &&&= k-16+8x^2-x^4 \\ \Rightarrow && 0 &= x^4- 8x^2+12x+16-k \end{align*} When the curves touch, we will have repeated root, ie \(a\) is a root of \(4x^3-16x+12 \Rightarrow a^3-4a+3 =0\). \begin{align*} &&0 &= a^3-4a+3 \\ &&&= (a-1)(a^2+a-3) \\ \Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2} \end{align*} \begin{align*} && 0 &= a^4-8a^2+12a+16-k \\ \Rightarrow && k &= a(a^3-8a+12)+16 \\ &&&= a(4a-3-8a+12)+16 \\ &&&= -4a^2+9a+16 \\ \\ \Rightarrow && a = 1& \quad k = 21 \\ && k &= -4(3-a)+9a+16 = 13a+4\\ && a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\ && a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\ \end{align*} So we have type (a), and (d).
Examiner's report
— 2012 STEP 3, Question 3
Mean: ~9.5 / 20 (inferred)
~67% attempted (inferred)
Inferred ~9.5/20 from 'just less than half marks'; inferred 67% from 'two thirds'
From the paper introduction
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Source: Cambridge STEP 2012 Examiner's Report
· 2012-full.pdf
Rating Information
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1468.7
Banger Comparisons: 2
Show LaTeX source
Problem source
It is given that the two curves
\[
y=4-x^2
\text{ and }
m x = k-y^2\,,
\]
where $m > 0$, touch exactly once.
\begin{questionparts}
\item In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes:
\begin{enumerate}
\item $k < 0\, $;
\item $0 < k < 16$, $k/m < 2\,$;
\item $k > 16$, $k/m > 2\,$;
\item $k > 16$, $k/m < 2\,$.
\end{enumerate}
\item
Now set $m=12$. Show that the $x$-coordinate of any point at which the two curves meet satisfies
\[
x^4-8x^2 +12x +16-k=0\,.
\]
Let $a$ be the value of $x$ at the point where the curves touch. Show that $a$ satisfies
\[
a^3 -4a +3 =0
\]
and hence find the three possible values of $a$. Derive also the equation
\[
k= -4a^2 +9a +16\,.
\]
Which of the four sketches in part (i) arise?
\end{questionparts}Solution source
\begin{questionparts}
\item \begin{enumerate}
\item $\,$
\begin{center}
\begin{tikzpicture}[scale=1]
\def\functionf(#1){4-(#1)^2)};
\def\xl{-8};
\def\xu{8};
\def\yl{-8};
\def\yu{8};
% 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);
\filldraw (0, 4) circle (1.5pt) node[right]{$4$};
\filldraw (2, 0) circle (1.5pt) node[below]{$2$};
\filldraw (-2, 0) circle (1.5pt) node[below]{$-2$};
\filldraw (-3, 0) circle (1.5pt) node[below]{$\frac{k}{m}$};
\draw[thick, blue, smooth, domain=\yl:\yu, samples=100]
plot ({(-1.5-\x*\x)*2}, {\x});
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\node[blue, above, rotate=70] at (-1, {\functionf(-1)}) {\tiny $y =4-x^2$};
\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}
\item $\,$
\begin{center}
\begin{tikzpicture}[scale=1]
\def\functionf(#1){4-(#1)^2)};
\def\xl{-8};
\def\xu{8};
\def\yl{-8};
\def\yu{8};
% 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);
\filldraw (0, 4) circle (1.5pt) node[right]{$4$};
\filldraw (2, 0) circle (1.5pt) node[below]{$2$};
\filldraw (-2, 0) circle (1.5pt) node[below]{$-2$};
\filldraw (1, 0) circle (1.5pt) node[below]{$\frac{k}{m}$};
\filldraw (0, {sqrt(8)}) circle (1.5pt) node[right]{$\sqrt{k}$};
\filldraw (0, {-sqrt(8)}) circle (1.5pt) node[right]{$-\sqrt{k}$};
\draw[thick, blue, smooth, domain=\yl:\yu, samples=100]
plot ({(8-\x*\x)/8}, {\x});
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\node[blue, above, rotate=70] at (-1, {\functionf(-1)}) {\tiny $y =4-x^2$};
\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}
\item $\,$
\begin{center}
\begin{tikzpicture}[scale=1]
\def\functionf(#1){4-(#1)^2)};
\def\xl{-8};
\def\xu{8};
\def\yl{-8};
\def\yu{8};
% 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);
\filldraw (0, 4) circle (1.5pt) node[right]{$4$};
\filldraw (2, 0) circle (1.5pt) node[below]{$2$};
\filldraw (-2, 0) circle (1.5pt) node[below]{$-2$};
\filldraw ({20/7}, 0) circle (1.5pt) node[below]{$\frac{k}{m}$};
\filldraw (0, {sqrt(20)}) circle (1.5pt) node[right]{$\sqrt{k}$};
\filldraw (0, {-sqrt(20)}) circle (1.5pt) node[right]{$-\sqrt{k}$};
\draw[thick, blue, smooth, domain=\yl:\yu, samples=100]
plot ({(20-\x*\x)/7}, {\x});
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\node[blue, above, rotate=70] at (-1, {\functionf(-1)}) {\tiny $y =4-x^2$};
\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}
\item $\,$
\begin{center}
\begin{tikzpicture}[scale=1]
\def\functionf(#1){4-(#1)^2)};
\def\xl{-8};
\def\xu{8};
\def\yl{-8};
\def\yu{8};
% 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);
\filldraw (0, 4) circle (1.5pt) node[right]{$4$};
\filldraw (2, 0) circle (1.5pt) node[below]{$2$};
\filldraw (-2, 0) circle (1.5pt) node[below]{$-2$};
\filldraw ({20/12}, 0) circle (1.5pt) node[below]{$\frac{k}{m}$};
\filldraw (0, {sqrt(20)}) circle (1.5pt) node[right]{$\sqrt{k}$};
\filldraw (0, {-sqrt(20)}) circle (1.5pt) node[right]{$-\sqrt{k}$};
\draw[thick, blue, smooth, domain=\yl:\yu, samples=100]
plot ({(20-\x*\x)/12}, {\x});
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\node[blue, above, rotate=70] at (-1, {\functionf(-1)}) {\tiny $y =4-x^2$};
\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}
\end{enumerate}
\item Suppose $m = 12$
\begin{align*}
&& y &= 4-x^2 \\
&& 12x &= k-y^2 \\
\Rightarrow && 12 x&=k-(4-x^2)^2 \\
&&&= k-16+8x^2-x^4 \\
\Rightarrow && 0 &= x^4- 8x^2+12x+16-k
\end{align*}
When the curves touch, we will have repeated root, ie $a$ is a root of $4x^3-16x+12 \Rightarrow a^3-4a+3 =0$.
\begin{align*}
&&0 &= a^3-4a+3 \\
&&&= (a-1)(a^2+a-3) \\
\Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2}
\end{align*}
\begin{align*}
&& 0 &= a^4-8a^2+12a+16-k \\
\Rightarrow && k &= a(a^3-8a+12)+16 \\
&&&= a(4a-3-8a+12)+16 \\
&&&= -4a^2+9a+16 \\
\\
\Rightarrow && a = 1& \quad k = 21 \\
&& k &= -4(3-a)+9a+16 = 13a+4\\
&& a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\
&& a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\
\end{align*}
So we have type (a), and (d).
\end{questionparts}
Two thirds of candidates attempted this question, but generally, with only moderate success earning just less than half marks. The vast majority of candidates (more than 85%) did not observe that, regardless of the case, the two parabolas "touch exactly once", dropping 4 or 5 marks immediately. However, most managed to obtain the three results in part (ii), though a few seemed to forget to derive that for k. Unaccountably, many threw away the final marks, only considering the case k ≥ 1.