| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| 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 the interval bisection method with minimal computational complexity. Part (a) requires simple substitution to verify sign change (f(1) = 3+3-7 = -1, f(2) = 9+6-7 = 8), and part (b) involves only two iterations of bisection with simple arithmetic. The method is purely mechanical with no conceptual challenges or problem-solving required beyond following the standard algorithm. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(1) = -1\), \(f(2) = 8\) | M1 | Either any one of \(f(1) = -1\) or \(f(2) = 8\) |
| Sign change (positive, negative) and \(f(x)\) is continuous, therefore a root \(\alpha\) is between \(x=1\) and \(x=2\) | A1 | Both values correct, sign change and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(1.5) = 2.696152423...\) \(\Rightarrow 1 < \alpha < 1.5\) | B1 | \(f(1.5) =\) awrt 2.7 (or truncated to 2.6) |
| Attempt to find \(f(1.25)\) | M1 | |
| \(f(1.25) = 0.698222038...\) \(\Rightarrow 1 < \alpha < 1.25\) | A1 | \(f(1.25) =\) awrt 0.7 with \(1 < \alpha < 1.25\) or \([1, 1.25]\) or \((1, 1.25)\) or equivalent in words |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = -1$, $f(2) = 8$ | M1 | Either any one of $f(1) = -1$ or $f(2) = 8$ |
| Sign change (positive, negative) and $f(x)$ is continuous, therefore a root $\alpha$ is between $x=1$ and $x=2$ | A1 | Both values correct, sign change and conclusion |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1.5) = 2.696152423...$ $\Rightarrow 1 < \alpha < 1.5$ | B1 | $f(1.5) =$ awrt 2.7 (or truncated to 2.6) |
| Attempt to find $f(1.25)$ | M1 | |
| $f(1.25) = 0.698222038...$ $\Rightarrow 1 < \alpha < 1.25$ | A1 | $f(1.25) =$ awrt 0.7 with $1 < \alpha < 1.25$ or $[1, 1.25]$ or $(1, 1.25)$ or equivalent in words |
> **Note:** In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks.
---1.
$$f ( x ) = 3 ^ { x } + 3 x - 7$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ between $x = 1$ and $x = 2$.
\item Starting with the interval $[ 1,2 ]$, use interval bisection twice to find an interval of width 0.25 which contains $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q1 [5]}}