| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Chord gradient with h (algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward differentiation from first principles question with standard chord gradient calculation and normal line application. Part (i) requires simple algebraic manipulation of the difference quotient, part (ii) tests conceptual understanding of limits, and part (iii) applies routine perpendicular gradient and line equation techniques. All steps are textbook-standard with no novel problem-solving required, making it easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y_1 = 50,\ y_2 = 2(5+h)^2\) | B1 | Finds \(y\) coordinates at 5 and \(5+h\) |
| \(\frac{(50 + 20h + 2h^2) - 50}{(5+h)-5}\) | M1 | Correct method to find gradient of a line segment; at least 3/4 values correct |
| \(20 + 2h\) | A1 | Fully correct working to give answer AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. "As \(h\) tends to zero, the gradient will be 20" | B1 | Indicates understanding of limit. e.g. refer to \(h\) tending to zero or substitute \(h=0\) into \(20+2h\) to obtain gradient at A |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient of normal \(= -\frac{1}{20}\) | B1 | |
| \(y - 50 = -\frac{1}{20}(x-5),\ x=0\) | M1 | Gradient of line must be numerical negative reciprocal of their gradient at A through their A. Any correct method e.g. labelled diagram. |
| \(50\frac{1}{4}\) | A1 | Correct coordinate in any form e.g. \(\frac{201}{4}, \frac{1005}{20}\) |
## Question 8:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y_1 = 50,\ y_2 = 2(5+h)^2$ | B1 | Finds $y$ coordinates at 5 and $5+h$ |
| $\frac{(50 + 20h + 2h^2) - 50}{(5+h)-5}$ | M1 | Correct method to find gradient of a line segment; at least 3/4 values correct |
| $20 + 2h$ | A1 | Fully correct working to give answer AG |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. "As $h$ tends to zero, the gradient will be 20" | B1 | Indicates understanding of limit. e.g. refer to $h$ tending to zero or substitute $h=0$ into $20+2h$ to obtain gradient at A |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of normal $= -\frac{1}{20}$ | B1 | |
| $y - 50 = -\frac{1}{20}(x-5),\ x=0$ | M1 | Gradient of line must be numerical negative reciprocal of their gradient at A through their A. Any correct method e.g. labelled diagram. |
| $50\frac{1}{4}$ | A1 | Correct coordinate in any form e.g. $\frac{201}{4}, \frac{1005}{20}$ |
---8 A curve has equation $y = 2 x ^ { 2 }$. The points $A$ and $B$ lie on the curve and have $x$-coordinates 5 and $5 + h$ respectively, where $h > 0$.\\
(i) Show that the gradient of the line $A B$ is $20 + 2 h$.\\
(ii) Explain how the answer to part (i) relates to the gradient of the curve at $A$.\\
(iii) The normal to the curve at $A$ meets the $y$-axis at the point $C$. Find the $y$-coordinate of $C$.
\hfill \mbox{\textit{OCR C1 2016 Q8 [7]}}