| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| 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 application of the chord gradient formula leading to differentiation from first principles. Part (a) requires substituting into y = x² - 6x + 5 and simplifying (y₂-y₁)/(x₂-x₁), which is algebraically routine. Part (b) asks for conceptual understanding of limits (as h→0) but doesn't require actual calculation since the answer is evident from the simplified form. Easier than average due to simple quadratic, clear structure, and minimal algebraic complexity. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks | Guidance |
|---|---|---|
| \(y_A = 4 - 12 + 5 = -3\) | B1 | |
| \(y_B = (2+h)^2 - 6(2+h) + 5 = 4 + 4h + h^2 - 12 - 6h + 5 = h^2 - 2h - 3\) | M1 | Expand correctly |
| Gradient \(= \frac{(h^2 - 2h - 3)-(-3)}{h} = \frac{h^2 - 2h}{h}\) | M1 | Correct difference over \(h\) |
| \(= h - 2\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| As \(h \to 0\), gradient of \(AB \to\) gradient of curve at \(A\) | M1 | Explanation of limiting process |
| Gradient at \(A = -2\) | A1 A1 |
# Question 2:
## Part (a)
| $y_A = 4 - 12 + 5 = -3$ | B1 | |
| $y_B = (2+h)^2 - 6(2+h) + 5 = 4 + 4h + h^2 - 12 - 6h + 5 = h^2 - 2h - 3$ | M1 | Expand correctly |
| Gradient $= \frac{(h^2 - 2h - 3)-(-3)}{h} = \frac{h^2 - 2h}{h}$ | M1 | Correct difference over $h$ |
| $= h - 2$ | A1 A1 | |
## Part (b)
| As $h \to 0$, gradient of $AB \to$ gradient of curve at $A$ | M1 | Explanation of limiting process |
| Gradient at $A = -2$ | A1 A1 | |
---2 A curve has equation
$$y = x ^ { 2 } - 6 x + 5$$
The points $A$ and $B$ on the curve have $x$-coordinates 2 and $2 + h$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $h$, the gradient of the line $A B$, giving your answer in its simplest form.
\item Explain how the result of part (a) can be used to find the gradient of the curve at $A$. State the value of this gradient.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2009 Q2 [8]}}