| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Sign Change with Function Evaluation |
| Difficulty | Standard +0.3 This is a straightforward sign change question requiring basic function evaluation, interval bisection, and recognition of a discontinuity. Part (a) is routine substitution, (b) is simple bisection, (c) requires iteration to 1 d.p., and (d) tests understanding that sign changes don't always indicate roots (discontinuity at x=2/3). All parts are standard textbook exercises with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases |
| A | B | |
| 1 | \(x\) | \(f ( x )\) |
| 2 | 1.5 | 3.1625 |
| 3 | 1.25 | 0.619977679 |
| 4 | 1.125 | - 0.250466087 |
| 5 |
| A | B | |
| 1 | x | f(x) |
| 2 | 0 | 0.5 |
| 3 | 1 | -1 |
| 4 | 0.5 | 1.5625 |
| 5 | 0.75 | -4.4336 |
| 6 | 0.6 | 4.5296 |
| 7 | 0.7 | -10.4599 |
| 8 | 0.65 | 19.5285 |
| 9 | 0.675 | -40.4674 |
| 10 | 0.6625 | 79.5301 |
| 11 | 0.66875 | -160.4687 |
| 10 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(1) = -1\) | B1 | Finding \(f(1)\) or \(f(2)\) |
| \(f(2) = 13.75\) or \(\frac{55}{4}\), so there is a change of sign | B1 | Completion to show change of sign with explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Value of \(x\) in the range \(1.125 < x < 1.25\) | B1 | Any value in this range. Candidates may give the range. |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(1.15) = -0.09\) so \(x \approx 1.2\) (cao) | B1 | Justified by their calculations (which may not necessarily use 1.15) |
| Answer | Marks | Guidance |
|---|---|---|
| There is a change of sign | E1 | Incorrect maths (e.g. it implies a y-intercept) B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Clear and correct explanation | E1 | E.g. the function is undefined for \(x = \frac{2}{3}\); accept 'discontinuous' or 'asymptote' for 'undefined'; Fig. 9.1 shows only one root; could refer to \(f(x)\) values diverging. isw after correct answer |
## Question 9:
### Part (a):
$f(1) = -1$ | B1 | Finding $f(1)$ or $f(2)$
$f(2) = 13.75$ or $\frac{55}{4}$, so there is a change of sign | B1 | Completion to show change of sign with explanation
**Total: [2]**
### Part (b):
Value of $x$ in the range $1.125 < x < 1.25$ | B1 | Any value in this range. Candidates may give the range.
**Total: [1]**
### Part (c):
$f(1.15) = -0.09$ so $x \approx 1.2$ (cao) | B1 | Justified by their calculations (which may not necessarily use 1.15)
**Total: [1]**
### Part (d)(i):
There is a change of sign | E1 | Incorrect maths (e.g. it implies a y-intercept) **B0**
**Total: [1]**
### Part (d)(ii):
Clear and correct explanation | E1 | E.g. the function is undefined for $x = \frac{2}{3}$; accept 'discontinuous' or 'asymptote' for 'undefined'; Fig. 9.1 shows only one root; could refer to $f(x)$ values diverging. isw after correct answer
**Total: [1]**
---9 This question is about the equation $\mathrm { f } ( x ) = 0$, where $\mathrm { f } ( x ) = x ^ { 4 } - x - \frac { 1 } { 3 x - 2 }$.\\
Fig. 9.1 shows the curve $y = f ( x )$.\\
Fig. 9.1\\
\includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-06_940_929_518_239}
\begin{enumerate}[label=(\alph*)]
\item Show, by calculation, that the equation $\mathrm { f } ( x ) = 0$ has a root between $x = 1$ and $x = 2$.
\item Fig. 9.2 shows part of a spreadsheet being used to find a root of the equation.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\begin{tabular}{ l | r | r | }
\hline
& \multicolumn{1}{|c|}{A} & \multicolumn{1}{c|}{B} \\
\hline
1 & \multicolumn{1}{c|}{$x$} & \multicolumn{1}{c|}{$f ( x )$} \\
\hline
2 & 1.5 & 3.1625 \\
\hline
3 & 1.25 & 0.619977679 \\
\hline
4 & 1.125 & - 0.250466087 \\
\hline
5 & & \\
\hline
\end{tabular}
\end{center}
\end{table}
Write down a suitable number to use as the next value of $x$ in the spreadsheet.
\item Determine a root of the equation $\mathrm { f } ( x ) = 0$. Give your answer correct to $\mathbf { 1 }$ decimal place.
\item Fig. 9.3 shows a similar spreadsheet being used to search for another root of $\mathrm { f } ( x ) = 0$.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 9.3}
\begin{tabular}{|l|l|l|}
\hline
& A & B \\
\hline
1 & x & f(x) \\
\hline
2 & 0 & 0.5 \\
\hline
3 & 1 & -1 \\
\hline
4 & 0.5 & 1.5625 \\
\hline
5 & 0.75 & -4.4336 \\
\hline
6 & 0.6 & 4.5296 \\
\hline
7 & 0.7 & -10.4599 \\
\hline
8 & 0.65 & 19.5285 \\
\hline
9 & 0.675 & -40.4674 \\
\hline
10 & 0.6625 & 79.5301 \\
\hline
11 & 0.66875 & -160.4687 \\
\hline
10 & & \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\roman*)]
\item Explain why it looks from rows 2 and 3 of the spreadsheet as if there is a root between 0 and 1.
\item Explain why this process will not find a root between 0 and 1 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q9 [6]}}