| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Vertical translation of cubic with factorisation |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic algebraic manipulation (expanding brackets), curve sketching from factored form, simple transformations (vertical translation), and routine factorization using the factor theorem. All parts are standard C1 techniques with no problem-solving insight required, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Translation | B1 | 0 for "shift" or "move" without stating translation; 0 if stretch also mentioned |
| \(\begin{pmatrix} 0 \\ -36 \end{pmatrix}\) | B1 | Or 36 down, or \(-36\) in \(y\) direction; if conflict between "\(-36\) in \(y\) direction" and wrong vector, award B0; 0 for "\(-36\) down" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-1 - 4 + 11 - 6 = 0\) | B1 | Or B1 for correct division by \((x+1)\) or for quadratic factor found by inspection, and conclusion that no remainder means \(g(-1)=0\); NB examiners must annotate |
| Attempt at division by \((x+1)\) as far as \(x^3 + x^2\) in working | M1 | Or inspection with at least two terms of three-term quadratic factor correct; or finding \(f(6)=0\); M0 for trials of factors unless correct answer found with clear working |
| Correctly obtaining \(x^2 - 5x - 6\) | A1 | Or \((x-6)\) found as factor |
| Factorising the correct quadratic \(x^2 - 5x - 6\) | M1 | For factors giving two terms correct or ft one error; M0 for formula without factors; for those using factor theorem to find \((x-6)\), M1 for working with cubic to find \((x+1)\) is repeated |
| \((x-6)(x+1)^2\) oe isw | A1 | Condone inclusion of "\(=0\)"; isw roots found even if stated as factors; just the answer \((x-6)(x+1)^2\) gets last 4 marks |
# Question 1:
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Translation | B1 | 0 for "shift" or "move" without stating translation; 0 if stretch also mentioned |
| $\begin{pmatrix} 0 \\ -36 \end{pmatrix}$ | B1 | Or 36 down, or $-36$ in $y$ direction; if conflict between "$-36$ in $y$ direction" and wrong vector, award B0; 0 for "$-36$ down" |
**[2 marks]**
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $-1 - 4 + 11 - 6 = 0$ | B1 | Or B1 for correct division by $(x+1)$ or for quadratic factor found by inspection, and conclusion that no remainder means $g(-1)=0$; NB examiners must annotate |
| Attempt at division by $(x+1)$ as far as $x^3 + x^2$ in working | M1 | Or inspection with at least two terms of three-term quadratic factor correct; or finding $f(6)=0$; M0 for trials of factors unless correct answer found with clear working |
| Correctly obtaining $x^2 - 5x - 6$ | A1 | Or $(x-6)$ found as factor |
| Factorising the correct quadratic $x^2 - 5x - 6$ | M1 | For factors giving two terms correct or ft one error; M0 for formula without factors; for those using factor theorem to find $(x-6)$, M1 for working with cubic to find $(x+1)$ is repeated |
| $(x-6)(x+1)^2$ oe isw | A1 | Condone inclusion of "$=0$"; isw roots found even if stated as factors; just the answer $(x-6)(x+1)^2$ gets last 4 marks |
**[5 marks]**
---1 You are given that $\mathrm { f } ( x ) = ( x + 3 ) ( x - 2 ) ( x - 5 )$.\\
(i) Sketch the curve $y = \mathrm { f } ( x )$.\\
(ii) Show that $\mathrm { f } ( x )$ may be written as $x ^ { 3 } - 4 x ^ { 2 } - 11 x + 30$.\\
(iii) Describe fully the transformation that maps the graph of $y = \mathrm { f } ( x )$ onto the graph of $y = \mathrm { g } ( x )$, where $\mathrm { g } ( x ) = x ^ { 3 } - 4 x ^ { 2 } - 11 x - 6$.\\
(iv) Show that $\mathrm { g } ( - 1 ) = 0$. Hence factorise $\mathrm { g } ( x )$ completely.
\hfill \mbox{\textit{OCR MEI C1 Q1 [12]}}