| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Pure Interval Bisection Only |
| Difficulty | Moderate -0.8 This is a straightforward application of the sign change method and interval bisection algorithm. Part (a) requires simple substitution to show f(1) and f(2) have opposite signs. Part (b) is mechanical repetition of bisection with no conceptual challenge. While it's a Further Maths topic, the execution is entirely routine and requires minimal problem-solving. |
| Spec | 1.09a Sign change methods: locate roots |
$$f(x) = 3^x - x - 6.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $f(x) = 0$ has a root $\alpha$ between $x = 1$ and $x = 2$. [2]
\item Starting with the interval $(1, 2)$, use interval bisection three times to find an interval of width 0.125 which contains $\alpha$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q12 [4]}}