| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find x-coordinates with given gradient |
| Difficulty | Moderate -0.3 This is a straightforward C1 differentiation question with standard parts: finding a gradient (routine), finding coordinates using the normal gradient (one extra step with perpendicular gradients), and numerical approximation to the derivative using chords. Part (iii) requires recognizing that chord gradients approach the tangent gradient, and part (iv) uses dy/dx = 2kx at x=1. All techniques are standard for C1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | Marks | Guidance |
| \(\frac{dy}{dx} = 4x\) | B1 | \(4x\) |
| At \(x=3\), \(\frac{dy}{dx} = 12\) | B1 2 | \(12\) |
| (ii) | Marks | Guidance |
| Gradient of tangent \(= -8\) | M1 | \(\frac{dy}{dx} = -8\) |
| \(4x = -8\) | A1 | \(x = -2\) |
| \(x = -2\) | ||
| \(y = 8\) | A1 3 | \(y = 8\) |
| (iii) | Marks | Guidance |
| Gradient \(= 6\) | B1 1 | Gradient = or approaches 6 |
| (iv) | Marks | Guidance |
| \(\frac{dy}{dx} = 2kx\) | M1 | \(\frac{dy}{dx} = 2kx\) |
| \(x = 1\) | M1 | \(\frac{dy}{dx} = 2k\) |
| \(\frac{dy}{dx} = 2k\) | ||
| \(k = 3\) | A1 \(\sqrt{3}\) | \(k = 3\) |
| 9 |
**(i)** | **Marks** | **Guidance**
---|---|---
$\frac{dy}{dx} = 4x$ | B1 | $4x$
At $x=3$, $\frac{dy}{dx} = 12$ | B1 2 | $12$
**(ii)** | **Marks** | **Guidance**
---|---|---
Gradient of tangent $= -8$ | M1 | $\frac{dy}{dx} = -8$
$4x = -8$ | A1 | $x = -2$
$x = -2$ | |
$y = 8$ | A1 3 | $y = 8$
**(iii)** | **Marks** | **Guidance**
---|---|---
Gradient $= 6$ | B1 1 | Gradient = or approaches 6
**(iv)** | **Marks** | **Guidance**
---|---|---
$\frac{dy}{dx} = 2kx$ | M1 | $\frac{dy}{dx} = 2kx$
$x = 1$ | M1 | $\frac{dy}{dx} = 2k$
$\frac{dy}{dx} = 2k$ | |
$k = 3$ | A1 $\sqrt{3}$ | $k = 3$ | **CWO**
| **9** |9 (i) Find the gradient of the curve $y = 2 x ^ { 2 }$ at the point where $x = 3$.\\
(ii) At a point $A$ on the curve $y = 2 x ^ { 2 }$, the gradient of the normal is $\frac { 1 } { 8 }$. Find the coordinates of $A$.
Points $P _ { 1 } \left( 1 , y _ { 1 } \right) , P _ { 2 } \left( 1.01 , y _ { 2 } \right)$ and $P _ { 3 } \left( 1.1 , y _ { 3 } \right)$ lie on the curve $y = k x ^ { 2 }$. The gradient of the chord $P _ { 1 } P _ { 3 }$ is 6.3 and the gradient of the chord $P _ { 1 } P _ { 2 }$ is 6.03.\\
(iii) What do these results suggest about the gradient of the tangent to the curve $y = k x ^ { 2 }$ at $P _ { 1 }$ ?\\
(iv) Deduce the value of $k$.
\hfill \mbox{\textit{OCR C1 2005 Q9 [9]}}