| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| 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 covering routine techniques: sketching an absolute value function, algebraic manipulation to show equivalence of equations, sign-change method for locating roots, and applying an iterative formula. All parts follow predictable patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.06a 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 |
\includegraphics{figure_1}
Figure 1 shows a sketch of the curve with equation $y = e^{-x} - 1$.
\begin{enumerate}[label=(\alph*)]
\item Copy Fig. 1 and on the same axes sketch the graph of $y = \frac{1}{2}|x - 1|$. Show the coordinates of the points where the graph meets the axes. [2]
\end{enumerate}
The $x$-coordinate of the point of intersection of the graph is $\alpha$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $x = \alpha$ is a root of the equation $x + 2e^{-x} - 3 = 0$. [3]
\item Show that $-1 < \alpha < 0$. [2]
\end{enumerate}
The iterative formula $x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]$ is used to solve the equation $x + 2e^{-x} - 3 = 0$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Starting with $x_0 = -1$, find the values of $x_1$ and $x_2$. [2]
\item Show that, to 2 decimal places, $\alpha = -0.58$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q5 [11]}}