1989 Paper 3 Q4

Year: 1989
Paper: 3
Question Number: 4

Course: LFM Stats And Pure
Section: Curve Sketching

Difficulty: 1700.0 Banger: 1500.0

Problem

Sketch the curve whose cartesian equation is \[ y=\frac{2x(x^{2}-5)}{x^{2}-4}, \] and give the equations of the asymptotes and of the tangent to the curve at the origin. Hence, or otherwise, determine (giving reasons) the number of real roots of the following equations:
  1. \(4x^{2}(x^{2}-5)=(5x-2)(x^{2}-4)\);
  2. \(4x^{2}(x^{2}-5)^{2}=(x^{2}-4)^{2}(x^{2}+1)\);
  3. \(4z^{2}(z-5)^{2}=(z-4)^{2}(z+1)\).

Solution

TikZ diagram
The gradient at the origin is \(\frac{5}{2}\) which we can observe by looking at the taylor series
  1. \begin{align*} && 4x^{2}(x^{2}-5)&=(5x-2)(x^{2}-4) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{5x-2}{2x} = \frac52 -\frac1x \end{align*} Therefore it will have two roots as the hyperbola will intersect our graph in two places. (In the upper left and lower right quadrants).
  2. \begin{align*} && 4x^{2}(x^{2}-5)^{2}&=(x^{2}-4)^{2}(x^{2}+1) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{(x^2-4)(x^2+1)}{2x} \end{align*}
    TikZ diagram
    No solutions \begin{align*} && 4z^{2}(z-5)^{2}&=(z-4)^{2}(z+1) \\ && \frac{2z(z^2-5)}{z^2-4} &= \frac{(z+5)(z-4)(z+1)}{(z-5)(z+4)} \end{align*}
    TikZ diagram
    5 solutions
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
Sketch the curve whose cartesian equation is 
\[
y=\frac{2x(x^{2}-5)}{x^{2}-4},
\]
and give the equations of the asymptotes and of the tangent to the
curve at the origin. 
Hence, or otherwise, determine (giving reasons) the number of real
roots of the following equations: 
\begin{questionparts}
\item $4x^{2}(x^{2}-5)=(5x-2)(x^{2}-4)$; 
\item $4x^{2}(x^{2}-5)^{2}=(x^{2}-4)^{2}(x^{2}+1)$; 
\item $4z^{2}(z-5)^{2}=(z-4)^{2}(z+1)$. 
\end{questionparts}
Solution source
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \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}{1}
    \pgfmathsetmacro{\yscale}{\xrange/\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=-5:-2.1, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.9:1.9, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.1:5, samples=100] 
            plot (\x, {\functionf(\x)});  
        \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}

The gradient at the origin is $\frac{5}{2}$ which we can observe by looking at the taylor series

\begin{questionparts}
\item \begin{align*}
&& 4x^{2}(x^{2}-5)&=(5x-2)(x^{2}-4) \\
&& \frac{2x(x^2-5)}{x^2-4} &= \frac{5x-2}{2x} = \frac52 -\frac1x
\end{align*}

Therefore it will have two roots as the hyperbola will intersect our graph in two places. (In the upper left and lower right quadrants).

\item \begin{align*}
&& 4x^{2}(x^{2}-5)^{2}&=(x^{2}-4)^{2}(x^{2}+1) \\
&& \frac{2x(x^2-5)}{x^2-4} &= \frac{(x^2-4)(x^2+1)}{2x}
\end{align*}

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\functiong(#1){((#1)^2-4)*((#1)^2+1)/(2*(#1))};
    \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}{1}
    \pgfmathsetmacro{\yscale}{\xrange/\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=-5:-2.1, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.9:1.9, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.1:5, samples=100] 
            plot (\x, {\functionf(\x)});  

        \draw[thick, green, smooth, domain=-5:-0.1, samples=100] 
            plot (\x, {\functiong(\x)});  
        \draw[thick, green, smooth, domain=0.1:5, samples=100] 
            plot (\x, {\functiong(\x)});  
        \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}

No solutions

 \begin{align*}
&& 4z^{2}(z-5)^{2}&=(z-4)^{2}(z+1) \\
&& \frac{2z(z^2-5)}{z^2-4} &= \frac{(z+5)(z-4)(z+1)}{(z-5)(z+4)}
\end{align*}

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\functiong(#1){((#1)+5)*((#1)-4)*((#1)+1)/((#1)-5)/((#1)+4)};
    \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}{1}
    \pgfmathsetmacro{\yscale}{\xrange/\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=-5:-2.1, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.9:1.9, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.1:5, samples=100] 
            plot (\x, {\functionf(\x)});  

        \draw[thick, green, smooth, domain=-5:-4.1, samples=100] 
            plot (\x, {\functiong(\x)});  
        \draw[thick, green, smooth, domain=-3.9:4.8, samples=100] 
            plot (\x, {\functiong(\x)});  
        \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}

5 solutions

\end{questionparts}