| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Pure Interval Bisection Only |
| Difficulty | Easy -1.2 This is a straightforward application of the interval bisection method with minimal steps. Part (a) requires simple substitution to verify sign change, and part (b) involves just two iterations of bisection—a routine algorithmic procedure with no conceptual challenge or problem-solving required. |
| Spec | 1.09a Sign change methods: locate roots |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(1.6) = -1.304\), \(f(1.8) = 0.632\) | B1,B1 | Allow 1 dp throughout |
| Sign change, so root between | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(f(1.7)\) considered first | M1 | |
| \(f(1.7) = -0.387\), so root \(> 1.7\) | A1 | |
| \(f(1.75) = 0.109375\), so root \(\approx 1.7\) | m1A1 | ml for \(f(1.65)\) after error |
**(a)** $f(1.6) = -1.304$, $f(1.8) = 0.632$ | B1,B1 | Allow 1 dp throughout
Sign change, so root between | E1 |
**Total: 3 marks**
**(b)** $f(1.7)$ considered first | M1 |
$f(1.7) = -0.387$, so root $> 1.7$ | A1 |
$f(1.75) = 0.109375$, so root $\approx 1.7$ | m1A1 | ml for $f(1.65)$ after error
**Total: 4 marks**
---2
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$x ^ { 3 } + x - 7 = 0$$
has a root between 1.6 and 1.8.
\item Use interval bisection twice, starting with the interval in part (a), to give this root to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2007 Q2 [7]}}