| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Pure Interval Bisection Only |
| Difficulty | Moderate -0.8 This is a straightforward application of interval bisection with clear sign changes and simple function evaluation. Part (a) requires only substituting two values and checking signs, while part (b) is mechanical repetition of the bisection algorithm twice. No conceptual difficulty or problem-solving insight needed—purely procedural execution of a standard numerical method. |
| Spec | 1.09a Sign change methods: locate roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(1) = \text{awrt } -0.7\) or \(f(1.4) = 1.9\) or awrt \(2.0\) | M1 | Either any one of \(f(1) = \text{awrt } -0.7\) or \(f(1.4) = 1.9\) or awrt \(2.0\) |
| Sign change (and \(f(x)\) is continuous) therefore a root \(\alpha\) exists between \(x = 1\) and \(x = 1.4\) | A1 | Both values correct, sign change and conclusion |
| Answer | Marks |
|---|---|
| \(f(1.2) = \text{awrt } 0.6\) | B1 |
| Attempt to find \(f(1.1)\) | M1 |
| \(f(1.1) = -0.06\) or awrt \(-0.07\) with \(1.1 \le \alpha \le 1.2\) or \(1.1 < \alpha < 1.2\) or \([1.1, 1.2]\) or \((1.1, 1.2)\) | A1 |
**(a)**
| $f(1) = \text{awrt } -0.7$ or $f(1.4) = 1.9$ or awrt $2.0$ | M1 | Either any one of $f(1) = \text{awrt } -0.7$ or $f(1.4) = 1.9$ or awrt $2.0$ |
| Sign change (and $f(x)$ is continuous) therefore a root $\alpha$ exists between $x = 1$ and $x = 1.4$ | A1 | Both values correct, sign change and conclusion |
**(b)**
| $f(1.2) = \text{awrt } 0.6$ | B1 | |
| Attempt to find $f(1.1)$ | M1 | |
| $f(1.1) = -0.06$ or awrt $-0.07$ with $1.1 \le \alpha \le 1.2$ or $1.1 < \alpha < 1.2$ or $[1.1, 1.2]$ or $(1.1, 1.2)$ | A1 | |
---\begin{enumerate}
\item $\mathrm { f } ( x ) = 2 x - 5 \cos x , \quad$ where $x$ is in radians.\\
(a) Show that the equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ in the interval $[ 1,1.4 ]$.\\[0pt]
(b) Starting with the interval [1,1.4], use interval bisection twice to find an interval of width 0.1 which contains $\alpha$.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q1 [5]}}