\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same set of axes, the graphs of
$$y = 2 - e^{-x} \text{ and } y = \sqrt{x}.$$ [3]
\end{enumerate}

[It is not necessary to find the coordinates of any points of intersection with the axes.]

Given that $f(x) = e^{-x} + \sqrt{x} - 2, x \geq 0$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item explain how your graphs show that the equation $f(x) = 0$ has only one solution, [1]
\item show that the solution of $f(x) = 0$ lies between $x = 3$ and $x = 4$. [2]
\end{enumerate}

The iterative formula $x_{n+1} = (2 - e^{-x_n})^2$ is used to solve the equation $f(x) = 0$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Taking $x_0 = 4$, write down the values of $x_1, x_2, x_3$ and $x_4$, and hence find an approximation to the solution of $f(x) = 0$, giving your answer to 3 decimal places. [4]
\end{enumerate}

\textbf{28a.} \begin{enumerate}[label=(\roman*)]
\item Given that $\cos(x + 30)° = 3 \cos(x - 30)°$, prove that $\tan x° = -\frac{\sqrt{3}}{2}$. [5]

\item \begin{enumerate}[label=(\alph*)]
\item Prove that $\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta$. [3]
\item Verify that $\theta = 180°$ is a solution of the equation $\sin 2\theta = 2 - 2 \cos 2\theta$. [1]
\item Using the result in part (a), or otherwise, find the other two solutions, $0 < \theta < 360°$, of the equation using $\sin 2\theta = 2 - 2 \cos 2\theta$. [4]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3  Q28 [23]}}