| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a multi-part question requiring standard C2 techniques: finding intersection points by solving simultaneous equations, differentiation to find tangent equations, and finding intersection of two lines. While it has multiple steps (typical of a longer exam question), each individual step uses routine methods with no novel insight required. It's slightly easier than average because the algebra is straightforward and the question provides significant scaffolding (e.g., telling students what to show in part (i)). |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| equation of \(AB\) is \(y = 3x+1\) o.e. | M1 | |
| their "\(3x+1\)" \(= 4x^2\) | M1 | or equiv in \(y\): \(y = 4\left(\frac{y-1}{3}\right)^2\) or rearranging and deriving roots \(y=4\) |
| \((4x+1)(x-1)=0\) o.e. so \(x=-\frac{1}{4}\) | M1 | or \(\frac{1}{4}\); condone verification by showing lhs \(=\) rhs o.e. |
| at C, \(x=-\frac{1}{4}\), \(y = 4\times(-\frac{1}{4})^2\) or \(3\times(-\frac{1}{4})+1 [=\frac{1}{4}\) as required\(]\) | A1 | or \(y=\frac{1}{4}\) implies \(x=\pm\frac{1}{4}\) so at C \(x=-\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y' = 8x\) | M1 | |
| at A \(y'=8\) | A1 | |
| equation of tangent at A: \(y-4 =\) their \(8(x-1)\) | M1 | ft their gradient |
| \(y = 8x-4\) | A1 | |
| at C \(y' = 8 \times -\frac{1}{4} [=-2]\) | M1 | NB if \(m=-2\) obtained from given answer or only showing that \((-\frac{1}{4}, \frac{1}{4})\) lies on given line \(y=-2x-\frac{1}{4}\) then 0 marks |
| \(y - \frac{1}{4} = -2(x-(-\frac{1}{4}))\) or other unsimplified equivalent to obtain given result; allow correct verification that \((-\frac{1}{4}, \frac{1}{4})\) lies on given line | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| their "\(8x-4\)" \(= -2x - \frac{1}{4}\) | M1 | or \(\frac{y+4}{8} = \frac{y+\frac{1}{4}}{-2}\) |
| \(y = -1\) www | A1 |
**Question 10(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| equation of $AB$ is $y = 3x+1$ o.e. | **M1** | | **SC3** for verifying that A, B and C are collinear and that C also lies on the curve; **SC2** for verifying A, B and C are collinear by showing gradient of $AB=AC$ (for example) or showing C lies on AB; solely verifying that C lies on the curve scores 0 |
| their "$3x+1$" $= 4x^2$ | **M1** | or equiv in $y$: $y = 4\left(\frac{y-1}{3}\right)^2$ or rearranging and deriving roots $y=4$ | |
| $(4x+1)(x-1)=0$ o.e. so $x=-\frac{1}{4}$ | **M1** | or $\frac{1}{4}$; condone verification by showing lhs $=$ rhs o.e. | |
| at C, $x=-\frac{1}{4}$, $y = 4\times(-\frac{1}{4})^2$ or $3\times(-\frac{1}{4})+1 [=\frac{1}{4}$ as required$]$ | **A1** | or $y=\frac{1}{4}$ implies $x=\pm\frac{1}{4}$ so at C $x=-\frac{1}{4}$ | |
---
**Question 10(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y' = 8x$ | **M1** | | |
| at A $y'=8$ | **A1** | | |
| equation of tangent at A: $y-4 =$ their $8(x-1)$ | **M1** | ft their gradient | gradient must follow from evaluation of $\frac{dy}{dx}$; condone unsimplified versions of $y=8x-4$ |
| $y = 8x-4$ | **A1** | | |
| at C $y' = 8 \times -\frac{1}{4} [=-2]$ | **M1** | NB if $m=-2$ obtained from given answer or only showing that $(-\frac{1}{4}, \frac{1}{4})$ lies on given line $y=-2x-\frac{1}{4}$ then 0 marks | dependent on award of first **M1** |
| $y - \frac{1}{4} = -2(x-(-\frac{1}{4}))$ or other unsimplified equivalent to obtain given result; allow correct verification that $(-\frac{1}{4}, \frac{1}{4})$ lies on given line | **A1** | | **SC2** if equation of tangent and curve solved simultaneously to correctly show repeated root |
---
**Question 10(iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| their "$8x-4$" $= -2x - \frac{1}{4}$ | **M1** | or $\frac{y+4}{8} = \frac{y+\frac{1}{4}}{-2}$ | o.e. |
| $y = -1$ www | **A1** | | $[x = \frac{3}{8}]$ |
---10
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa8688bb-3608-45ab-85b7-bf84889dd189-3_645_793_1377_676}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
A is the point with coordinates $( 1,4 )$ on the curve $y = 4 x ^ { 2 }$. B is the point with coordinates $( 0,1 )$, as shown in Fig. 10.\\
(i) The line through A and B intersects the curve again at the point C . Show that the coordinates of C are $\left( - \frac { 1 } { 4 } , \frac { 1 } { 4 } \right)$.\\
(ii) Use calculus to find the equation of the tangent to the curve at A and verify that the equation of the tangent at C is $y = - 2 x - \frac { 1 } { 4 }$.\\
(iii) The two tangents intersect at the point D . Find the $y$-coordinate of D .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa8688bb-3608-45ab-85b7-bf84889dd189-4_773_1027_255_557}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows the curve $y = x ^ { 3 } - 3 x ^ { 2 } - x + 3$.\\
\hfill \mbox{\textit{OCR MEI C2 2011 Q10 [12]}}