| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Quadratic inequalities |
| Difficulty | Easy -1.2 This is a straightforward algebraic manipulation followed by a simple quadratic equation. Part (i) requires expanding brackets and collecting terms (routine algebra), while part (ii) is a basic two-term quadratic requiring only rearrangement and square root. Both parts are below-average difficulty with no problem-solving insight needed. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-1)(x-2)(x-3) - (x^3 - x^2 + 11x - 12)\) | M1 | |
| \(= (x^3 - 6x^2 + 11x - 6) - (x^3 - x^2 + 11x - 12)\) | A1 | |
| \(= -5x^2 - 6 + 12 = 6 - 5x^2\) | E1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(6 - 5x^2 = 0 \Rightarrow x = \pm\sqrt{\frac{6}{5}}\) | M1 A1 | Total: 2 |
# Question 9:
## Part (i):
$(x-1)(x-2)(x-3) - (x^3 - x^2 + 11x - 12)$ | M1 |
$= (x^3 - 6x^2 + 11x - 6) - (x^3 - x^2 + 11x - 12)$ | A1 |
$= -5x^2 - 6 + 12 = 6 - 5x^2$ | E1 | **Total: 3**
## Part (ii):
$6 - 5x^2 = 0 \Rightarrow x = \pm\sqrt{\frac{6}{5}}$ | M1 A1 | **Total: 2**
---9 (i) Show that $( x - 1 ) ( x - 2 ) ( x - 3 ) - \left( x ^ { 3 } - x ^ { 2 } + 11 x - 12 \right) = 6 - 5 x ^ { 2 }$.\\
(ii) Solve the equation $6 - 5 x ^ { 2 } = 0$.
\hfill \mbox{\textit{OCR MEI C1 Q9 [5]}}