| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Linear Interpolation Only |
| Difficulty | Moderate -0.8 This is a straightforward application of two basic techniques: sign change to confirm a root exists (simple substitution), followed by a single linear interpolation calculation using a standard formula. Both are routine FP1 procedures requiring minimal problem-solving, making this easier than average A-level questions. |
| Spec | 1.09a Sign change methods: locate roots |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(0.5) = -0.875\), \(f(1) = 1\) Change of sign, so root between | B1, E1 | Root exists between 0.5 and 1 |
| (b) Complete line interpolation method. Estimated root = \(\frac{11}{15} = 0.73\) | M2, 1; A1 | M1 for partially correct method; Allow \(\frac{11}{15}\) as answer |
**(a)** $f(0.5) = -0.875$, $f(1) = 1$ Change of sign, so root between | B1, E1 | Root exists between 0.5 and 1
**(b)** Complete line interpolation method. Estimated root = $\frac{11}{15} = 0.73$ | M2, 1; A1 | M1 for partially correct method; Allow $\frac{11}{15}$ as answer
**Total for Q1: 5 marks**
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\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$x ^ { 3 } + 2 x - 2 = 0$$
has a root between 0.5 and 1 .
\item Use linear interpolation once to find an estimate of this root. Give your answer to two decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2006 Q1 [5]}}