| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Shaded region area with quadratic |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: factorizing a quadratic, finding a normal using differentiation, calculating a triangle area, and integrating to find area under a curve. All steps are routine C2 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((2x-3)(x-4)\) | M1 | or \((11 \pm\sqrt{121-96})/4\) |
| \(x = 4\) or \(1.5\) | A1A1 | if M0, then B1 for showing \(y=0\) when \(x=4\) and B2 for \(x=1.5\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y' = 4x - 11\) | M1 | condone one error |
| \(= 5\) when \(x = 4\) c.a.o. | A1 | |
| grad of normal \(= -1/\text{their } y'\) | M1f.t. | |
| \(y[ - 0] = \text{their} -0.2(x-4)\) | M1 | or \(0 = \text{their } (-0.2)x4 + c\) dep on normal attempt |
| y-intercept for their normal | B1f.t. | s.o.i. normal must be linear or |
| area \(= \frac{1}{2} \times 4 \times 0.8\) c.a.o. | A1 | integrating their \(f(x)\) from 0 to 4 M1 |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{2}{3}x^3 - \frac{11}{2}x^2 + 12x\) | M1 | condone one error, ignore \(+ c\) |
| attempt difference between value at 4 and value at 1.5 | M1 | ft their (i), dep on integration attempt. c.a.o. |
| \([-]5\frac{5}{24}\) o.e. or \([-]5.2(083..)\) | A1 | |
| [3] |
## Question 1:
**Part i:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(2x-3)(x-4)$ | M1 | or $(11 \pm\sqrt{121-96})/4$ |
| $x = 4$ or $1.5$ | A1A1 | if M0, then B1 for showing $y=0$ when $x=4$ and B2 for $x=1.5$ |
| | | **[3]** |
**Part ii:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y' = 4x - 11$ | M1 | condone one error |
| $= 5$ when $x = 4$ c.a.o. | A1 | |
| grad of normal $= -1/\text{their } y'$ | M1f.t. | |
| $y[ - 0] = \text{their} -0.2(x-4)$ | M1 | or $0 = \text{their } (-0.2)x4 + c$ dep on normal attempt |
| y-intercept for their normal | B1f.t. | s.o.i. normal must be linear or |
| area $= \frac{1}{2} \times 4 \times 0.8$ c.a.o. | A1 | integrating their $f(x)$ from 0 to 4 M1 |
| | | **[6]** |
**Part iii:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{2}{3}x^3 - \frac{11}{2}x^2 + 12x$ | M1 | condone one error, ignore $+ c$ |
| attempt difference between value at 4 and value at 1.5 | M1 | ft their (i), dep on integration attempt. c.a.o. |
| $[-]5\frac{5}{24}$ o.e. or $[-]5.2(083..)$ | A1 | |
| | | **[3]** |
---1 Fig. 12 is a sketch of the curve $y = 2 x ^ { 2 } - 11 x + 12$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44b860fb-040f-4d3f-94d8-42eae41c0e2d-1_468_940_285_830}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
(i) Show that the curve intersects the $x$-axis at $( 4,0 )$ and find the coordinates of the other point of intersection of the curve and the $x$-axis.\\
(ii) Find the equation of the normal to the curve at the point $( 4,0 )$.
Show also that the area of the triangle bounded by this normal and the axes is 1.6 units ${ } ^ { 2 }$.\\
(iii) Find the area of the region bounded by the curve and the $x$-axis.
\hfill \mbox{\textit{OCR MEI C2 Q1 [12]}}