Edexcel C3 — Question 18 11 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks11
PaperDownload PDF ↗
TopicFixed Point Iteration
TypeApply iteration to find root (pure fixed point)
DifficultyStandard +0.3 This is a structured, multi-part question on fixed point iteration with clear guidance at each step. Parts (a)-(c) involve routine sketching, algebraic manipulation to show equivalence of equations, and sign-change verification. Parts (d)-(e) require straightforward iteration with a calculator. While it covers several techniques, each step is standard and well-scaffolded, making it slightly easier than the average A-level question.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function1.06a Exponential function: a^x and e^x graphs and properties1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method
\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).
  1. 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]
The \(x\)-coordinate of the point of intersection of the graphs is \(\alpha\).
  1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
  2. Show that \(-1 < \alpha < 0\). [2]
The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).
  1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
  2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]
\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 graphs 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  Q18 [11]}}
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