| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Factor condition (zero remainder) |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor and Remainder Theorem requiring only direct substitution into f(x) for various conditions. Part (i) involves routine evaluation at specific x-values (e.g., f(0)=k, f(2)=0, f(-1)=5), and part (ii) requires solving a cubic after finding one factor. All techniques are standard C1 material with no novel problem-solving required, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (i)(A) \(k=0\) | B1 |
| (i)(B) \(k=-2\) | B1 |
| (i)(C) \(f(2)=8-8-8+k=0 \Rightarrow k=8\) | M1 A1 |
| (i)(D) \(f(-1)=-1-2+4+k=5 \Rightarrow k=4\) | M1 A1 |
| (i)(E) \(k=7\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(k=8 \Rightarrow x^3-2x^2+2x+8=0\) | M1 | |
| \(\Rightarrow (x-2)(x^2-4)=0\) | A1 | |
| \(\Rightarrow x=2,\ 2,\ -2\) | A2 | \(-1\) each omission |
| Cubic graph cutting at \((-2,0)\) and touching at \((2,0)\) | B1 |
## Question 11:
**(i)(A)** $k=0$ | B1 |
**(i)(B)** $k=-2$ | B1 |
**(i)(C)** $f(2)=8-8-8+k=0 \Rightarrow k=8$ | M1 A1 |
**(i)(D)** $f(-1)=-1-2+4+k=5 \Rightarrow k=4$ | M1 A1 |
**(i)(E)** $k=7$ | B1 |
**(ii)**
$k=8 \Rightarrow x^3-2x^2+2x+8=0$ | M1 |
$\Rightarrow (x-2)(x^2-4)=0$ | A1 |
$\Rightarrow x=2,\ 2,\ -2$ | A2 | $-1$ each omission
Cubic graph cutting at $(-2,0)$ and touching at $(2,0)$ | B1 |
---11 In this question $\mathrm { f } ( x ) = x ^ { 3 } - 2 x ^ { 2 } - 4 x + k$.
\begin{enumerate}[label=(\roman*)]
\item You are asked to find the values of $k$ which satisfy the following conditions.\\
(A) The graph of $y = \mathrm { f } ( x )$ goes through the origin.\\
(B) The graph of $y = \mathrm { f } ( x )$ intersects with the $y$ axis at ( $0 , - 2$ ).\\
(C) ( $x - 2$ ) is a factor of $\mathrm { f } ( x )$.\\
(D) The remainder when $\mathrm { f } ( x )$ is divided by $( x + 1 )$ is 5 .\\
(E) The graph of $y = \mathrm { f } ( x )$ is as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{3b6291ef-bef9-49de-a20f-591e621bed65-3_373_788_2131_584}
\item Find the solution of the equation $\mathrm { f } ( x ) = 0$ when $k = 8$.
Sketch a graph of $y = \mathrm { f } ( x )$ in this case.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 Q11 [12]}}