| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a standard Further Maths calculus question combining differentiation (including chain rule for ln), algebraic manipulation to derive an equation, straightforward iteration with a calculator, and a change of sign method to verify a root. All techniques are routine and well-practiced, with no novel insight required. The 8 marks reflect length rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
The curve with equation $y = f(x)$ where
$$f(x) = x^2 + \ln(2x^2 - 4x + 5)$$
has a single turning point at $x = \alpha$
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ is a solution of the equation
$$2x^3 - 4x^2 + 7x - 2 = 0$$ [4]
\end{enumerate}
The iterative formula
$$x_{n+1} = \frac{1}{7}(2 + 4x_n^2 - 2x_n^3)$$
is used to find an approximate value for $\alpha$.
Starting with $x_1 = 0.3$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item calculate, giving each answer to 4 decimal places,
\begin{enumerate}[label=(\roman*)]
\item the value of $x_2$
\item the value of $x_4$
\end{enumerate} [2]
\end{enumerate}
Using a suitable interval and a suitable function that should be stated,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item show that $\alpha$ is 0.341 to 3 decimal places. [2]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2022 Q4 [8]}}