| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Moderate -0.3 This is a straightforward multi-part coordinate geometry question combining line-curve intersection, differentiation for tangents, and solving simultaneous equations. All techniques are standard C2 material with clear scaffolding. Part (i) is shown rather than found independently, reducing difficulty. The question requires more steps than trivial exercises but involves no novel insights or tricky algebraic manipulation. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| 4 | (i) eqn of AB is y = 3x + 1 o.e. |
| Answer | Marks |
|---|---|
| (− 1/4) + 1[=1/4 as required] | M1 |
| Answer | Marks |
|---|---|
| A1 | 2 |
| Answer | Marks |
|---|---|
| or y = ¼ implies x = ¼ so at C x = − ¼ | SC3 for verifying that A, B and C are collinear and |
| Answer | Marks |
|---|---|
| 4 | (ii) y = 8x |
| Answer | Marks |
|---|---|
| lies on given line | M1 |
| Answer | Marks |
|---|---|
| A1 | ft their gradient |
| Answer | Marks |
|---|---|
| marks. | gradient must follow from evaluation of |
| Answer | Marks |
|---|---|
| 4 | (iii) their “8x – 4” = −2x − ¼ |
| y = −1 www | M1 |
| A1 | y4 y 1 |
| Answer | Marks |
|---|---|
| 8 2 | o.e. |
Question 4:
4 | (i) eqn of AB is y = 3x + 1 o.e.
their “3x + 1” = 4x2
(4x + 1) (x – 1) = 0 o.e. so x = −1/4
at C, x = −1/4, y = 4 (−1/4)2 or 3
(− 1/4) + 1[=1/4 as required] | M1
M1
M1
A1 | 2
y1
or equiv in y: y 4
3
or rearranging and deriving roots y = 4
or ¼
condone verification by showing lhs =
rhs o.e.
or y = ¼ implies x = ¼ so at C x = − ¼ | SC3 for verifying that A, B and C are collinear and
that C also lies on the curve
SC2 for verifying that A, B and C are collinear by
showing that gradient of AB = AC (for example) or
showing C lies on AB
solely verifying that C lies on the curve scores 0
4 | (ii) y = 8x
at A y = 8
eqn of tgt at A
y − 4 = their“8”(x − 1)
y = 8x - 4
at C y = 8 −1/4 [=−2]
y − ¼ = −2(x − (− ¼)) or other
unsimplified equivalent to obtain
given result.
allow correct verification that (-¼,¼)
lies on given line | M1
A1
M1
A1
M1
A1 | ft their gradient
NB if m = -2 obtained from given
answer or only showing that (− ¼, ¼)
lies on given line y = −2x − ¼ then 0
marks. | gradient must follow from evaluation of
condone unsimplified versions of y = 8x - 4
dependent on award of first M1
SC2 if equation of tangent and curve solved
simultaneously to correctly show repeated root
4 | (iii) their “8x – 4” = −2x − ¼
y = −1 www | M1
A1 | y4 y 1
or 4
8 2 | o.e.
[x = 3/8]\includegraphics{figure_3}
A is the point with coordinates (1, 4) on the curve $y = 4x^2$. B is the point with coordinates (0, 1), as shown in Fig. 10.
\begin{enumerate}[label=(\roman*)]
\item The line through A and B intersects the curve again at the point C. Show that the coordinates of C are $\left(-\frac{1}{4}, \frac{1}{4}\right)$. [4]
\item 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 $y = -2x - \frac{1}{4}$. [6]
\item The two tangents intersect at the point D. Find the $y$-coordinate of D. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q4 [12]}}