Problem source

Prove that, if $\vert \alpha\vert < 2\sqrt{2},$ then there is no
value of $x$ for which
\begin{equation}
x^2 -{\alpha}\vert x \vert + 2 < 0\;.
\tag{$*$}
\end{equation}
Find the solution set of $(*)$ for ${\alpha}=3\,$.
For ${\alpha} > 2\sqrt{2}\,$,  the sum of the lengths of the intervals in which  $x$ satisfies  $(*)$ is denoted by $S\,$. Find $S$ in terms of ${\alpha}$  and deduce that $S < 2{\alpha}\,$.
Sketch the graph of $S\,$ against $\alpha \,$.

Solution source

There are two cases to consider by they are equivalent to $x^2 \pm \alpha x + 2 < 0$, which has no solution solutions if $\Delta < 0$, ie if $\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}$.

If $\alpha = 3$, we have

\begin{align*}
&& 0 & > x^2-3x+2 \\
&&&= (x-2)(x-1) \\
\Rightarrow && x & \in (1,2) \\
\\
&& 0 &> x^2+3x+2 \\
&&& = (x+2)(x+1) \\
\Rightarrow && x &\in (-2,-1)
\end{align*}

Both cases work here, so $x \in (-2, -1) \cup (1,2)$.


\begin{align*}
&& 0 &> x^2 \pm \alpha x + 2 \\
&&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4}
\end{align*}

The potential intervals therefore are $(\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})$ and $(\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})$. Neither of these intervals overlap with $0$, since $\alpha^2 > \alpha^2-8$, and their lengths are both $\sqrt{\alpha^2-8}$, therefore $S = 2\sqrt{\alpha^2-8} < 2\alpha$


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){abs((#1)-1)};
    \def\xl{-1};
    \def\xu{10};
    \def\yl{-1};
    \def\yu{2*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, smooth, domain={sqrt(8)+0.00001}:\xu, samples=101] 
            plot ({\x}, {2*sqrt(\x*\x-8)});

        \draw[thick, red, dashed] (0, 0) -- (\xu, {2*\xu});
        
        % \draw[thick, blue, smooth, domain=\yl:-1, samples=100] 
        %     plot ({sqrt(\x*\x-1)-0.25}, {\x});
        % \draw[thick, blue, smooth, domain=\yl:-1, samples=100] 
        %     plot ({-(sqrt(\x*\x-1)-0.25)}, {\x});
        
    \end{scope}


    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$\alpha$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$S$};
    
    \end{tikzpicture}
\end{center}