| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring differentiation to find a tangent, finding x-intercepts by solving a quadratic, and calculating an area using integration. All steps are routine applications of basic techniques with clear guidance ('show that' statements reduce problem-solving demand). Slightly above average only due to the multi-part nature and final area calculation requiring subtraction of regions. |
| 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 equations1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(7 - 2x\) | M1 | Differentiation must be used |
| \(x = 2\), gradient \(= 3\) | A1 | |
| \(x = 2\), \(y = 4\) | B1 | |
| \(y - 4 = \text{their grad}(x - 2)\) | M1 | Or use of \(y = mx + c\) and subst \((2, \text{their } 4)\), dependent on diffn seen |
| Subst \(y = 0\) in their linear eqn | M1 | |
| Completion to \(x = \frac{2}{3}\) (ans given) | A1 | |
| (ii) \(f(1) = 0\) or factorising to \((x-1)(6-x)\) or \((x-1)(x-6)\) | 1 | Or using quadratic formula correctly to obtain \(x = 1\) |
| \(6\) www | 1 | |
| (iii) \(\frac{7}{2}x^2 - \frac{1}{3}x^3 - 6x\) | M1 | For two terms correct; ignore \(+c\) |
| Value at \(2\) − value at \(1\) | M1 | ft attempt at integration only |
| \(2\frac{1}{6}\) or \(2.16\) to \(2.17\) | A1 | |
| \(\frac{1}{2} \times \frac{4}{3} \times 4\) − their integral | M1 | |
| \(0.5\) o.e. | A1 |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $7 - 2x$ | M1 | Differentiation must be used |
| $x = 2$, gradient $= 3$ | A1 | |
| $x = 2$, $y = 4$ | B1 | |
| $y - 4 = \text{their grad}(x - 2)$ | M1 | Or use of $y = mx + c$ and subst $(2, \text{their } 4)$, dependent on diffn seen |
| Subst $y = 0$ in their linear eqn | M1 | |
| Completion to $x = \frac{2}{3}$ (ans given) | A1 | |
| **(ii)** $f(1) = 0$ or factorising to $(x-1)(6-x)$ or $(x-1)(x-6)$ | 1 | Or using quadratic formula correctly to obtain $x = 1$ |
| $6$ www | 1 | |
| **(iii)** $\frac{7}{2}x^2 - \frac{1}{3}x^3 - 6x$ | M1 | For two terms correct; ignore $+c$ |
| Value at $2$ − value at $1$ | M1 | ft attempt at integration only |
| $2\frac{1}{6}$ or $2.16$ to $2.17$ | A1 | |
| $\frac{1}{2} \times \frac{4}{3} \times 4$ − their integral | M1 | |
| $0.5$ o.e. | A1 | |5 Fig. 10 shows a sketch of the graph of $y = 7 x - x ^ { 2 } - 6$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-4_608_908_290_663}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and hence find the equation of the tangent to the curve at the point on the curve where $x = 2$.
Show that this tangent crosses the $x$-axis where $x = \frac { 2 } { 3 }$.\\
(ii) Show that the curve crosses the $x$-axis where $x = 1$ and find the $x$-coordinate of the other point of intersection of the curve with the $x$-axis.\\
(iii) Find $\int _ { 1 } ^ { 2 } \left( 7 x - x ^ { 2 } - 6 \right) \mathrm { d } x$.
Hence find the area of the region bounded by the curve, the tangent and the $x$-axis, shown shaded on Fig. 10.
\hfill \mbox{\textit{OCR MEI C2 Q5 [13]}}