| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Linear Interpolation Only |
| Difficulty | Moderate -0.5 This is a straightforward application of sign change and linear interpolation. Part (a) requires simple substitution to verify sign change, and part (b) is a direct application of the linear interpolation formula with no complications. While it's FP1, the technique itself is mechanical and requires no problem-solving insight, making it easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2.5) = 1.499...\) | M1 | Either \(f(2.5)\) = awrt 1.5 or \(f(3)\) = awrt \(-0.91\) |
| \(f(3) = -0.9111...\) | ||
| Sign change (positive, negative) and \(f(x)\) continuous, therefore root | A1 | Both \(f(2.5)\) = awrt 1.5 and \(f(3)\) = awrt \(-0.91\), sign change and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{3-\alpha}{0.91113026188} = \dfrac{\alpha - 2.5}{1.4994494182}\) | M1 A1ft | Correct linear interpolation method; accept equivalent equation - ensure signs are correct |
| \(\alpha = \dfrac{3 \times 1.499... + 2.5 \times 0.9111...}{1.499...+0.9111...}\) | ||
| \(\alpha = 2.81\) (2 d.p.) | A1 | cao |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2.5) = 1.499...$ | M1 | Either $f(2.5)$ = awrt 1.5 or $f(3)$ = awrt $-0.91$ |
| $f(3) = -0.9111...$ | | |
| Sign change (positive, negative) and $f(x)$ continuous, therefore root | A1 | Both $f(2.5)$ = awrt 1.5 and $f(3)$ = awrt $-0.91$, sign change and conclusion |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{3-\alpha}{0.91113026188} = \dfrac{\alpha - 2.5}{1.4994494182}$ | M1 A1ft | Correct linear interpolation method; accept equivalent equation - ensure signs are correct |
| $\alpha = \dfrac{3 \times 1.499... + 2.5 \times 0.9111...}{1.499...+0.9111...}$ | | |
| $\alpha = 2.81$ (2 d.p.) | A1 | cao |
---2.
$$\mathrm { f } ( x ) = \cos \left( x ^ { 2 } \right) - x + 3 , \quad 0 < x < \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ in the interval $[ 2.5,3 ]$.\\[0pt]
\item Use linear interpolation once on the interval [2.5,3] to find an approximation for $\alpha$, giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q2 [5]}}