AQA C3 2011 June — Question 3 5 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
TopicFixed Point Iteration
TypeShow root in interval
DifficultyStandard +0.3 This is a straightforward iterative methods question requiring basic substitution to verify bounds, simple algebraic rearrangement using inverse function properties, and calculator-based iteration. All techniques are standard C3 content with no novel problem-solving required, making it slightly easier than average.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.06a Exponential function: a^x and e^x graphs and properties1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

The curve \(y = \cos^{-1}(2x - 1)\) intersects the curve \(y = e^x\) at a single point where \(x = \alpha\).
  1. Show that \(\alpha\) lies between 0.4 and 0.5. [2]
  2. Show that the equation \(\cos^{-1}(2x - 1) = e^x\) can be written as \(x = \frac{1}{2} + \frac{1}{2}\cos(e^x)\). [1]
  3. Use the iteration \(x_{n+1} = \frac{1}{2} + \frac{1}{2}\cos(e^{x_n})\) with \(x_1 = 0.4\) to find the values of \(x_2\) and \(x_3\), giving your answers to three decimal places. [2]

The curve $y = \cos^{-1}(2x - 1)$ intersects the curve $y = e^x$ at a single point where $x = \alpha$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 0.4 and 0.5. [2]

\item Show that the equation $\cos^{-1}(2x - 1) = e^x$ can be written as $x = \frac{1}{2} + \frac{1}{2}\cos(e^x)$. [1]

\item Use the iteration $x_{n+1} = \frac{1}{2} + \frac{1}{2}\cos(e^{x_n})$ with $x_1 = 0.4$ to find the values of $x_2$ and $x_3$, giving your answers to three decimal places. [2]
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2011 Q3 [5]}}