| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve quadratic inequality |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question on quadratic functions covering standard C1 content: factorising to find roots, sketching parabolas, completing the square for the vertex, solving a simple quadratic inequality, and applying basic transformations. All parts are routine textbook exercises requiring no problem-solving insight, making it easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.02m Graphs of functions: difference between plotting and sketching1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch of parabola] | B1 | Sketch |
| \((2, 0)\) | B1 | \(x\) axis |
| \((4, 0)\) | B1 | \(y\) axis |
| \((0, 6)\) | B1 | \(x\) coordinate |
| Lowest point \(= (3, -1)\) | B1 | \(y\) coordinate. Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 6x + 8 < 0\) for \(2 < x < 4\) | M1 A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch: same shape, translated by \(-3\)] | B1 B1 | Same shape; Translation by \(-3\). Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch] | B1 B1 | |
| The transformation is \(y = f(x+3)\) moved down by 2 | B1 | Same shape, moved down 2 |
| So from \(y = f(x)\) it is "back 3 and down 2" or \(\begin{pmatrix}-3\\2\end{pmatrix}\) | Total: 3 |
# Question 10:
## Part (i):
[Sketch of parabola] | B1 | Sketch
$(2, 0)$ | B1 | $x$ axis
$(4, 0)$ | B1 | $y$ axis
$(0, 6)$ | B1 | $x$ coordinate
Lowest point $= (3, -1)$ | B1 | $y$ coordinate. **Total: 5**
## Part (ii):
$x^2 - 6x + 8 < 0$ for $2 < x < 4$ | M1 A1 | **Total: 2**
## Part (iii):
[Sketch: same shape, translated by $-3$] | B1 B1 | Same shape; Translation by $-3$. **Total: 2**
## Part (iv):
[Sketch] | B1 B1 |
The transformation is $y = f(x+3)$ moved down by 2 | B1 | Same shape, moved down 2
So from $y = f(x)$ it is "back 3 and down 2" or $\begin{pmatrix}-3\\2\end{pmatrix}$ | | **Total: 3**
---10 (i) A quadratic function is given by $\mathrm { f } ( x ) = x ^ { 2 } - 6 x + 8$.\\
Sketch the graph of $y = \mathrm { f } ( x )$, giving the coordinates of the points where it crosses the axes. Mark the lowest point on the curve, and give its coordinates.\\
(ii) Solve the inequality $x ^ { 2 } - 6 x + 8 < 0$.\\
(iii) On the same graph, sketch $y = \mathrm { f } ( x + 3 )$.\\
(iv) The graph of $y = \mathrm { f } ( x + 3 ) - 2$ is obtained from the graph of $y = \mathrm { f } ( x )$ by a transformation. Describe the transformation and sketch the curve on the same axes as in (i) and (iii) above. Label all these curves clearly.
\hfill \mbox{\textit{OCR MEI C1 Q10 [12]}}