| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: differentiation of exponential/logarithmic functions, finding stationary points, and applying a given iterative formula. Part (b) requires simple algebraic manipulation of f'(x)=0, parts (c) involves routine calculator work, and part (d) uses sign-change verification. All steps are procedural with no novel insight required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
4.
$$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$
\begin{enumerate}[label=(\alph*)]
\item Differentiate to find $\mathrm { f } ^ { \prime } ( x )$.
The curve with equation $y = \mathrm { f } ( x )$ has a turning point at $P$. The $x$-coordinate of $P$ is $\alpha$.
\item Show that $\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }$.
The iterative formula
$$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , \quad x _ { 0 } = 1$$
is used to find an approximate value for $\alpha$.
\item Calculate the values of $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 4 decimal places.
\item By considering the change of sign of $\mathrm { f } ^ { \prime } ( x )$ in a suitable interval, prove that $\alpha = 0.1443$ correct to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q4 [10]}}