| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Geometric area leads to quadratic |
| Difficulty | Moderate -0.8 This is a straightforward application question requiring students to set up the area formula for a triangle (½ × base × height = 9), expand and simplify to reach the given quadratic, then factorize and solve. The factorization of 2x² - x - 21 = (2x - 7)(x + 3) is routine, and rejecting the negative solution requires minimal reasoning. Below average difficulty as it's a standard textbook-style problem with clear scaffolding. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \frac{1}{2}(x+1)(2x-3) = 9\) | M1 | Area formula used |
| \((x+1)(2x-3) = 18\) | ||
| \(2x^2 - x - 3 = 18\) | A1 | Expand correctly |
| \(2x^2 - x - 21 = 0\) | A1 | Shown correctly (given result) |
| Answer | Marks | Guidance |
|---|---|---|
| \((2x-7)(x+3) = 0\) | M1 | Correct factorisation |
| \(x = \frac{7}{2}\) or \(x = -3\) | A1 | Both solutions |
| \(x = \frac{7}{2}\) (reject \(x=-3\) as lengths must be positive) | A1 | Reject negative, height \(= \frac{9}{2}\) cm, base \(= 4\) cm |
## Question 10:
**(i)**
Area $= \frac{1}{2}(x+1)(2x-3) = 9$ | M1 | Area formula used
$(x+1)(2x-3) = 18$ | |
$2x^2 - x - 3 = 18$ | A1 | Expand correctly
$2x^2 - x - 21 = 0$ | A1 | Shown correctly (given result)
**(ii)**
$(2x-7)(x+3) = 0$ | M1 | Correct factorisation
$x = \frac{7}{2}$ or $x = -3$ | A1 | Both solutions
$x = \frac{7}{2}$ (reject $x=-3$ as lengths must be positive) | A1 | Reject negative, height $= \frac{9}{2}$ cm, base $= 4$ cm
---10 The triangle shown in Fig. 10 has height $( x + 1 ) \mathrm { cm }$ and base $( 2 x - 3 ) \mathrm { cm }$. Its area is $9 \mathrm {~cm} ^ { 2 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4d8caf0f-7594-42cb-bd40-e6c11e2b6832-3_444_1088_351_715}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
(i) Show that $2 x ^ { 2 } - x - 21 = 0$.\\
(ii) By factorising, solve the equation $2 x ^ { 2 } - x - 21 = 0$. Hence find the height and base of the triangle.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI C1 2007 Q10 [5]}}