| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Standard +0.2 This is a standard C3 question combining sketching exponential/root functions, sign-change verification, and iterative formula application. All parts are routine: (a) requires basic curve sketching, (b) is a simple observation about intersections, (c) is straightforward substitution to show sign change, and (d) involves calculator iteration with no conceptual challenge. Slightly easier than average due to its mechanical nature and clear structure. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.06a Exponential function: a^x and e^x graphs and properties1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
\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]
[It is not necessary to find the coordinates of any points of intersection with the axes.]
Given that f(x) = $e^{-x} + \sqrt{x} - 2$, $x \geq 0$,
\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}
\hfill \mbox{\textit{Edexcel C3 Q4 [10]}}