Problem source

In both parts of this question, $x$ is real and $0 < \theta < \pi$.
\begin{questionparts}
\item By completing the square, find in terms of $\theta$ the minimum value as $x$ varies of
$$9x^2 - 12x \cos \theta + 4.$$
Find also the maximum value as $x$ varies of $12x^2 \sin \theta - 9x^4$.
Hence determine the values of $x$ and $\theta$ that satisfy the equation
$$9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 = 0.$$
\item Sketch the curve
$$y = \frac{x^2}{x - \theta},$$
where $\theta$ is a constant. Deduce that either $\frac{x^2}{x - \theta} \leq 0$ or $\frac{x^2}{x - \theta} \geq 4\theta$.
By considering the numerator and denominator separately, or otherwise, show that
$$\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x} \leq 1.$$
Hence determine the values of $x$ and $\theta$ that satisfy the equation
$$\frac{x^2}{4\theta(x - \theta)} = \frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}.$$
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& y &= 9x^2 - 12x \cos \theta + 4 \\
&&&= (3x-2\cos \theta)^2+4-4\cos^2 \theta \\
&&&= (3x-2\cos \theta)^2 + 4 \sin^2 \theta
\end{align*}

Therefore the minimum is $4\sin^2 \theta$ when $x = \frac23 \cos \theta$.

\begin{align*}
&& y &= 12x^2 \sin \theta - 9x^4 \\
&&&=4\sin^2 \theta -(3x^2-2\sin\theta)^2
\end{align*}
Therefore the maximum is $4\sin^2 \theta$ when $x^2 =  \frac23\sin \theta$

Therefore

\begin{align*}
&& 0 &= 9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 \\
&& \underbrace{-9x^4+12x^2\sin \theta}_{\leq 4\sin^2 \theta } &= \underbrace{9x^2 - 12x \cos \theta + 4 }_{\geq 4 \sin^2 \theta}
\end{align*}

Therefore the equality cases must be achieved in both cases, ie $x = \frac23 \cos \theta$ and $x^2 = \frac23 \sin \theta$

\begin{align*}
&& x^2 &= \frac49\cos^2 \theta \\
&&&= \frac49(1-\sin^2 \theta) \\
&&&= \frac49(1-\frac94 x^2) \\
\Rightarrow && 2x^2 &= \frac49 \\
\Rightarrow && x &= \pm \frac{\sqrt{2}}3\\
\Rightarrow && \cos \theta &=\pm  \frac32 \frac{\sqrt{2}}3 \\
&&&= \pm \frac{1}{\sqrt{2}} \\
\Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4} \\
\Rightarrow && (x, \theta) &= \left (\frac{\sqrt{2}}{3}, \frac{\pi}{4} \right),  \left (-\frac{\sqrt{2}}{3}, \frac{3\pi}{4} \right)
\end{align*}

\item Sketching we obtain, noticing we can find the turning point by:

\begin{align*}
&& \frac{x^2}{x-\theta} &= \lambda \\
\Leftrightarrow && x^2 - \lambda x +\theta \lambda &= 0 \\
\Leftrightarrow && 0 &\leq \Delta = \lambda^2 -4\lambda \theta  \\
\Leftrightarrow && \lambda &\geq 4 \theta, \lambda \leq 0
\end{align*}

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-10};
    \def\xu{10};
    \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, smooth, domain=\xl:0.95, samples=100] 
            plot ({\x}, {\x*\x/(\x-1)});

        \draw[thick, blue, smooth, domain=1.05:\xu, samples=100] 
            plot ({\x}, {\x*\x/(\x-1)});

        \draw[red, dashed, thick] (1,\yl) -- (1, \yu);
    \end{scope}

    \filldraw (0,0) circle (1.5pt) node[above left] {$(0,0)$};
    \filldraw (2,4) circle (1.5pt) node[below] {$(2\theta,4\theta)$};
    
    % 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}

Notice that $\sin^2 \theta \cos^2 x \leq 1$ and $1 + cos^2 \theta \sin^2 x \geq 1$ and therefore we must have the inequality desired.

\begin{align*}
&& \underbrace{\frac{x^2}{4\theta(x - \theta)}}_{\geq 1 \text{ or } \leq 0} &= \underbrace{\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}}_{\in [0,1]} \\
\text{both}=0: && x = 0 &, \sin \theta = 0 \\ 
\text{both}=1: && x = 2\theta &, \sin^2 \theta = 1,\cos^2 x = 1 \\
&& 1 &= \cos^2 2 \theta  \\
&&&= (1-2 \sin^2 \theta)^2 \\
&&&= 1 \\
\Rightarrow && (x, \theta) &= \left(\frac{\pi}{2}, \pi\right)
\end{align*}


\end{questionparts}