A student was attempting to prove that $x = \frac{1}{2}$ is the only real root of
$$x^3 + \frac{1}{4}x - \frac{1}{2} = 0.$$

The attempted solution was as follows.

$$x^3 + \frac{1}{4}x = \frac{1}{2}$$

$$\therefore \quad x(x^2 + \frac{1}{4}) = \frac{1}{2}$$

$$\therefore \quad x = \frac{1}{2}$$

or

$$x^2 + \frac{1}{4} = \frac{1}{2}$$

i.e.

$$x^2 = -\frac{1}{4} \quad \text{no solution}$$

$$\therefore \quad \text{only real root is } x = \frac{1}{2}$$

\begin{enumerate}[label=(\alph*)]
\item Explain clearly the error in the above attempt.
[2]

\item Give a correct proof that $x = \frac{1}{2}$ is the only real root of $x^3 + \frac{1}{4}x - \frac{1}{2} = 0$.
[3]
\end{enumerate}

The equation
$$x^3 + \beta x - \alpha = 0 \quad \text{(I)}$$
where $\alpha$, $\beta$ are real, $\alpha \neq 0$, has a real root at $x = \alpha$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find and simplify an expression for $\beta$ in terms of $\alpha$ and prove that $\alpha$ is the only real root provided $|\alpha| < 2$.
[6]
\end{enumerate}

An examiner chooses a positive number $\alpha$ so that $\alpha$ is the only real root of equation (I) but the incorrect method used by the student produces 3 distinct real "roots".

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the range of possible values for $\alpha$.
[7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2002 Q7 [18]}}