2014 Paper 2 Q7

Year: 2014
Paper: 2
Question Number: 7

Course: LFM Pure
Section: Modulus function

Difficulty: 1600.0 Banger: 1486.9
modulus function curve sketching graphical solution piecewise linear number of solutions inequality quadrilateral absolute value equations

Problem

  1. The function \(\f\) is defined by \(\f(x)= |x-a| + |x-b| \), where \(a < b\). Sketch the graph of \(\f(x)\), giving the gradient in each of the regions \(x < a\), \(a < x < b\) and \(x > b\). Sketch on the same diagram the graph of \(\g(x)\), where \(\g(x)= |2x-a-b|\). What shape is the quadrilateral with vertices \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\)?
  2. Show graphically that the equation \[ |x-a| + |x-b| = |x-c|\,, \] where \(a < b\), has \(0\), \(1\) or \(2\) solutions, stating the relationship of \(c\) to \(a\) and \(b\) in each case.
  3. For the equation \[ |x-a| + |x-b| = |x-c|+|x-d|\,, \] where \(a < b\), \(c < d\) and \(d-c < b-a\), determine the number of solutions in the various cases that arise, stating the relationship between \(a\), \(b\), \(c\) and \(d\) in each case.

Solution

  1. \(\,\)
    TikZ diagram
    \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
  2. There are no solutions if \(a < c < b\):
    TikZ diagram
    There is one solution if \(a=c\) or \(a = b\)
    TikZ diagram
    And there are two solution if \(c \not \in [a,b]\)
    TikZ diagram
    There is exactly one solution unless....
    TikZ diagram
    ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
    TikZ diagram
Examiner's report
— 2014 STEP 2, Question 7
Below Average

This was another of the less popular pure maths questions. The nature of this question meant that many solutions involved a series of sketches of graphs with very little written explanation. Most candidates were able to identify that the sloping edges of g(x) would have the same gradient as the sloping edges of f(x), but many did not have both sloping edges overlapping for the two graphs. In some cases only one sloping edge of g(x) was drawn. A large number of candidates who correctly sketched the graphs identified the quadrilateral as a rectangle, rather than a square. In the second part of the question, sketches of the case with one solution often did not have the graph of y = |f(x)| meeting the y-axis at one corner of the square identified in part (i), although many candidates were able to identify the different cases that could occur. Unfortunately in the final part of the question very few candidates used the result from the first part of the question and so considered a number of possibilities that do not exist for any values of a, b, c and d.

There were good solutions presented to all of the questions, although there was generally less success in those questions that required explanations of results or the use of diagrams and graphs to reach the solution. Algebraic manipulation was generally well done by many of the candidates although a range of common errors such as confusing differentiation and integration and simple arithmetic slips were evident. Candidates should also be advised to use the methods that are asked for in questions unless it is clear that other methods will be accepted (such as by the use of the phrase "or otherwise").

Source: Cambridge STEP 2014 Examiner's Report · 2014-full.pdf
Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1486.9

Banger Comparisons: 3

Show LaTeX source
Problem source
\begin{questionparts}
  \item The function $\f$ is defined by $\f(x)= |x-a| + |x-b| $, where $a < b$.  Sketch the graph of $\f(x)$, giving the gradient in each of the regions $x < a$, $a < x < b$ and $x > b$.  Sketch on the same diagram the graph of $\g(x)$, where $\g(x)= |2x-a-b|$.
    What shape is the quadrilateral with vertices $(a,0)$, $(b,0)$, $(b,\f(b))$ and $(a, \f(a))$? 
  \item Show graphically that the equation
    \[
    |x-a| + |x-b| = |x-c|\,,
    \]
    where $a < b$, has $0$, $1$ or $2$ solutions, stating the relationship of $c$ to $a$ and $b$ in each case.
  \item For the equation
    \[
    |x-a| + |x-b| = |x-c|+|x-d|\,,
    \]
    where $a < b$, $c < d$ and $d-c < b-a$, determine the number of solutions in the various cases that arise, stating the relationship between $a$, $b$, $c$ and $d$ in each case.
  \end{questionparts}
Solution source
\begin{questionparts}

\item $\,$ 
\begin{center}
    \begin{tikzpicture}
    \def\a{-1};
    \def\b{2};
    \def\functionf(#1){abs(2*(#1)-\a-\b};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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] (\xl, {\a+\b-2*\xl}) -- (\a ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= -2$};
        \draw[thick, blue] (\a ,\b-\a) -- (\b ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= 0$};
        \draw[thick, blue] (\b, {\b-\a}) -- (\xu ,{2*\xu-\a-\b})
            node[pos=0.5, sloped, below] {\tiny gradient $= 2$};

        \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
            plot (\x, {\functionf(\x)});

        \filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$}; 
        \filldraw ({\b}, 0) circle (1.5pt) node[below] {$b$}; 
        \filldraw ({(\a+\b)/2}, 0) circle (1.5pt) node[below] {$\frac{a+b}{2}$}; 
        \filldraw ({0}, {\b-\a}) circle (1.5pt) node[above right] {$b-a$}; 
    \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}

$(a,0)$, $(b,0)$, $(b,\f(b))$ and $(a, \f(a))$ forms a rectangle.

\item

There are no solutions if $a < c < b$:


\begin{center}
    \begin{tikzpicture}
    \def\a{-1};
    \def\b{2};
    \def\c{0.1};
    \def\functionf(#1){abs((#1)-\c};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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] (\xl, {\a+\b-2*\xl}) -- (\a ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= -2$};
        \draw[thick, blue] (\a ,\b-\a) -- (\b ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= 0$};
        \draw[thick, blue] (\b, {\b-\a}) -- (\xu ,{2*\xu-\a-\b})
            node[pos=0.5, sloped, below] {\tiny gradient $= 2$};

        \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
            plot (\x, {\functionf(\x)});

        \filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$}; 
        \filldraw ({\b}, 0) circle (1.5pt) node[below] {$b$}; 
        \filldraw ({\c}, 0) circle (1.5pt) node[below] {$c$}; 
        \filldraw ({0}, {\b-\a}) circle (1.5pt) node[above right] {$b-a$}; 
    \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 one solution if $a=c$ or $a = b$


\begin{center}
    \begin{tikzpicture}
    \def\a{-1};
    \def\b{2};
    \def\c{2};
    \def\functionf(#1){abs((#1)-\c};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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] (\xl, {\a+\b-2*\xl}) -- (\a ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= -2$};
        \draw[thick, blue] (\a ,\b-\a) -- (\b ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= 0$};
        \draw[thick, blue] (\b, {\b-\a}) -- (\xu ,{2*\xu-\a-\b})
            node[pos=0.5, sloped, below] {\tiny gradient $= 2$};

        \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
            plot (\x, {\functionf(\x)});

        \filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$}; 
        \filldraw ({\b}, 0) circle (1.5pt) node[below] {$b$}; 
        % \filldraw ({\c}, 0) circle (1.5pt) node[below] {$c$}; 
        \filldraw ({0}, {\b-\a}) circle (1.5pt) node[above right] {$b-a$}; 
    \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}

And there are two solution if $c \not \in [a,b]$


\begin{center}
    \begin{tikzpicture}
    \def\a{-1};
    \def\b{2};
    \def\c{2.5};
    \def\functionf(#1){abs((#1)-\c};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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] (\xl, {\a+\b-2*\xl}) -- (\a ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= -2$};
        \draw[thick, blue] (\a ,\b-\a) -- (\b ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= 0$};
        \draw[thick, blue] (\b, {\b-\a}) -- (\xu ,{2*\xu-\a-\b})
            node[pos=0.5, sloped, below] {\tiny gradient $= 2$};

        \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
            plot (\x, {\functionf(\x)});

        \filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$}; 
        \filldraw ({\b}, 0) circle (1.5pt) node[below] {$b$}; 
        \filldraw ({\c}, 0) circle (1.5pt) node[below] {$c$}; 
        \filldraw ({0}, {\b-\a}) circle (1.5pt) node[above right] {$b-a$}; 
    \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 exactly one solution unless....

\begin{center}
    \begin{tikzpicture}
    \def\a{-1};
    \def\b{2};
    \def\c{1.5};
    \def\d{3.5};
    \def\functionf(#1){abs((#1)-\c};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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] (\xl, {\a+\b-2*\xl}) -- (\a ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= -2$};
        \draw[thick, blue] (\a ,\b-\a) -- (\b ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= 0$};
        \draw[thick, blue] (\b, {\b-\a}) -- (\xu ,{2*\xu-\a-\b})
            node[pos=0.5, sloped, below] {\tiny gradient $= 2$};

        \draw[thick, red, dashed] (\xl, {\c+\d-2*\xl}) -- (\c ,\d-\c);
        \draw[thick, red, dashed] (\c ,\d-\c) -- (\d ,\d-\c);
        \draw[thick, red, dashed] (\d, {\d-\c}) -- (\xu ,{2*\xu-\d-\c});

        % \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
            % plot (\x, {\functionf(\x)});

        \filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$}; 
        \filldraw ({\b}, 0) circle (1.5pt) node[below] {$b$}; 
        \filldraw ({\c}, 0) circle (1.5pt) node[below] {$c$}; 
        \filldraw ({\d}, 0) circle (1.5pt) node[below] {$d$}; 
        \filldraw ({0}, {\b-\a}) circle (1.5pt) node[above right] {$b-a$}; 
        \filldraw ({0}, {\d-\c}) circle (1.5pt) node[above right] {$d-c$}; 
    \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 are infinitely many solutions when the gradients line up perfectly, ie when $a+b=c+d$


\begin{center}
    \begin{tikzpicture}
    \def\a{-1};
    \def\b{2};
    \def\c{-0.5};
    \def\d{1.5};
    \def\functionf(#1){abs((#1)-\c};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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] (\xl, {\a+\b-2*\xl}) -- (\a ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= -2$};
        \draw[thick, blue] (\a ,\b-\a) -- (\b ,\b-\a)
            node[pos=0.5, sloped, below] {\tiny gradient $= 0$};
        \draw[thick, blue] (\b, {\b-\a}) -- (\xu ,{2*\xu-\a-\b})
            node[pos=0.5, sloped, below] {\tiny gradient $= 2$};

        \draw[thick, red, dashed] (\xl, {\c+\d-2*\xl}) -- (\c ,\d-\c);
        \draw[thick, red, dashed] (\c ,\d-\c) -- (\d ,\d-\c);
        \draw[thick, red, dashed] (\d, {\d-\c}) -- (\xu ,{2*\xu-\d-\c});

        % \draw[thick, red, dashed, domain=\xl:\xu, samples =100]
            % plot (\x, {\functionf(\x)});

        \filldraw ({\a}, 0) circle (1.5pt) node[below] {$a$}; 
        \filldraw ({\b}, 0) circle (1.5pt) node[below] {$b$}; 
        \filldraw ({\c}, 0) circle (1.5pt) node[below] {$c$}; 
        \filldraw ({\d}, 0) circle (1.5pt) node[below] {$d$}; 
        \filldraw ({0}, {\b-\a}) circle (1.5pt) node[above right] {$b-a$}; 
        \filldraw ({0}, {\d-\c}) circle (1.5pt) node[above right] {$d-c$}; 
    \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{questionparts}
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