| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Interval Bisection with Other Methods |
| Difficulty | Standard +0.3 This is a straightforward application of two standard numerical methods (interval bisection and linear interpolation) with clear instructions and minimal steps. The calculations are routine—evaluating a simple function at given points and applying formulaic procedures. While it's Further Maths content, the question requires only mechanical execution of algorithms rather than problem-solving or insight, making it slightly easier than an average A-level question. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(f(x)=3+x\sin\!\left(\frac{x}{4}\right)\); \(f(13)=1.593\), \(f(13.5)=-0.122\), so root in \([13, 13.5]\) | M1 A1 | M1: Evaluate \(f(13)\) and \(f(13.5)\) giving at least positive, negative OR evaluate \(f(13.5)\) and \(f(13.25)\) giving at least negative, positive. Do not award if using degrees |
| \(f(13.25)=0.746\), so root in \([13.25, 13.5]\) | A1 (3) | A1: \(f(13.5)=\) awrt \(-0.1\), \(f(13.25)=\) awrt \(0.7(5)\). A1: Correct interval \([13.25, 13.5]\) or equivalent form with or without boundaries |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\alpha-13}{14-\alpha}=\frac{1.593}{1.911}\) or \(\frac{\alpha-13}{1}=\frac{1.593}{1.593+1.911}\) | M1 A1 | M1: Attempt at linear interpolation on either side with correct signs. A1: Correct equivalent statement |
| \(\alpha(1.911+1.593)=1.593\times14+13\times1.911\) and \(\alpha=\frac{47.145}{3.504}=13.455\) | dM1 A1 (4)(7 marks) | dM1: Makes \(\alpha\) subject of formula. A1: cao. Award A0 for 13.456 and 13.454 |
| ALT: Gradient \(\frac{y_1-y_0}{x_1-x_0}=\frac{-1.911-1.593}{14-13}(=-3.504\ldots)\), use \(y-y_0=m(x-x_0)\) with \(x=13\) or \(14\), giving \(y=-3.504x+47.145\), substitute \(y=0\) | M1 A1 dM1 A1 | A1: Correct statement after substituting \(y=0\): \(0=-3.504x+47.145\). dM1: Makes \(x\) the subject. A1: cao. Award A0 for 13.456 and 13.454 |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $f(x)=3+x\sin\!\left(\frac{x}{4}\right)$; $f(13)=1.593$, $f(13.5)=-0.122$, so root in $[13, 13.5]$ | M1 A1 | M1: Evaluate $f(13)$ and $f(13.5)$ giving at least positive, negative OR evaluate $f(13.5)$ and $f(13.25)$ giving at least negative, positive. Do not award if using degrees |
| $f(13.25)=0.746$, so root in $[13.25, 13.5]$ | A1 (3) | A1: $f(13.5)=$ awrt $-0.1$, $f(13.25)=$ awrt $0.7(5)$. A1: Correct interval $[13.25, 13.5]$ or equivalent form with or without boundaries |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\alpha-13}{14-\alpha}=\frac{1.593}{1.911}$ or $\frac{\alpha-13}{1}=\frac{1.593}{1.593+1.911}$ | M1 A1 | M1: Attempt at linear interpolation on either side with correct signs. A1: Correct equivalent statement |
| $\alpha(1.911+1.593)=1.593\times14+13\times1.911$ and $\alpha=\frac{47.145}{3.504}=13.455$ | dM1 A1 (4)(7 marks) | dM1: Makes $\alpha$ subject of formula. A1: cao. Award A0 for 13.456 and 13.454 |
| **ALT:** Gradient $\frac{y_1-y_0}{x_1-x_0}=\frac{-1.911-1.593}{14-13}(=-3.504\ldots)$, use $y-y_0=m(x-x_0)$ with $x=13$ or $14$, giving $y=-3.504x+47.145$, substitute $y=0$ | M1 A1 dM1 A1 | A1: Correct statement after substituting $y=0$: $0=-3.504x+47.145$. dM1: Makes $x$ the subject. A1: cao. Award A0 for 13.456 and 13.454 |
---2. In the interval $13 < x < 14$, the equation
$$3 + x \sin \left( \frac { x } { 4 } \right) = 0 , \text { where } x \text { is measured in radians, }$$
has exactly one root, $\alpha$.\\[0pt]
\begin{enumerate}[label=(\alph*)]
\item Starting with the interval [13,14], use interval bisection twice to find an interval of width 0.25 which contains $\alpha$.\\[0pt]
\item Use linear interpolation once on the interval [13,14] to find an approximate value for $\alpha$. Give your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2015 Q2 [7]}}