| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Moderate -0.3 This is a straightforward C1 differentiation question requiring basic derivative calculation, point-gradient form for tangent equation, and solving dy/dx = constant. All steps are routine applications of standard techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{1}{2} + x^{-2}\) | M1 A1 |
| \(\text{grad} = \frac{1}{2} + 2^{-2} = \frac{3}{4}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(x=2 \therefore y = \frac{7}{2}\) | B1 |
| \(y - \frac{7}{2} = \frac{3}{4}(x-2)\) | M1 |
| \(4y - 14 = 3x - 6\) | |
| \(3x - 4y + 8 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(B\), \(\text{grad} = \frac{3}{4}\) | ||
| \(\therefore \frac{1}{2} + x^{-2} = \frac{3}{4}\) | M1 | |
| \(x^2 = 4\) | ||
| \(x = 2\) (at \(A\)), \(-2\) | A1 | |
| \(\therefore B\left(-2, \frac{5}{2}\right)\) | A1 | (10) |
# Question 9:
## Part (i):
| $\frac{dy}{dx} = \frac{1}{2} + x^{-2}$ | M1 A1 | |
|---|---|---|
| $\text{grad} = \frac{1}{2} + 2^{-2} = \frac{3}{4}$ | M1 A1 | |
## Part (ii):
| $x=2 \therefore y = \frac{7}{2}$ | B1 | |
|---|---|---|
| $y - \frac{7}{2} = \frac{3}{4}(x-2)$ | M1 | |
| $4y - 14 = 3x - 6$ | | |
| $3x - 4y + 8 = 0$ | A1 | |
## Part (iii):
| At $B$, $\text{grad} = \frac{3}{4}$ | | |
|---|---|---|
| $\therefore \frac{1}{2} + x^{-2} = \frac{3}{4}$ | M1 | |
| $x^2 = 4$ | | |
| $x = 2$ (at $A$), $-2$ | A1 | |
| $\therefore B\left(-2, \frac{5}{2}\right)$ | A1 | **(10)** |
---9. A curve has the equation $y = \frac { x } { 2 } + 3 - \frac { 1 } { x } , x \neq 0$.
The point $A$ on the curve has $x$-coordinate 2 .\\
(i) Find the gradient of the curve at $A$.\\
(ii) Show that the tangent to the curve at $A$ has equation
$$3 x - 4 y + 8 = 0$$
The tangent to the curve at the point $B$ is parallel to the tangent at $A$.\\
(iii) Find the coordinates of $B$.\\
\hfill \mbox{\textit{OCR C1 Q9 [10]}}