| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| 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 straightforward multi-part integration question requiring standard techniques: finding a derivative, equation of a tangent line, solving quadratics, and computing a definite integral. The 'hence' part requires recognizing that the shaded area equals the integral from 1 to 2 minus the triangle area, but this is a routine application of area decomposition. All steps are standard C2 procedures with no novel insight 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 equations1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(7 - 2x\) | 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 |
| subst \(y = 0\) in their linear eqn | M1 | seen |
| completion to \(x = \frac{2}{3}\) (ans given) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1) = 0\) 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{7}{2}x^2 - \frac{1}{3}x^3 - 6x\) | 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 10:
### Part i:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $7 - 2x$ | 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 |
| subst $y = 0$ in their linear eqn | M1 | seen |
| completion to $x = \frac{2}{3}$ (ans given) | A1 | | **[6]** |
### Part ii:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1) = 0$ 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 | | **[2]** |
### Part iii:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{7}{2}x^2 - \frac{1}{3}x^3 - 6x$ | 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]** |
---10 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]{838d6b15-69a9-4e67-bc36-5bf60254a767-4_609_908_1338_621}
\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 2009 Q10 [13]}}