| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Stationary point via first principles |
| Difficulty | Moderate -0.8 This is a straightforward application of first principles differentiation with clear scaffolding. Part (a) requires basic algebra to find the gradient of chord AB, and part (b) simply asks students to explain that taking the limit as h→0 gives gradient 0, confirming a stationary point. The calculation is routine and the conceptual leap minimal, making this easier than average despite being from FP1. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y_B = (2+h)^3 - 12(2+h)\) | M1 | Correct substitution for \(y_B\) |
| \(= 8 + 12h + 6h^2 + h^3 - 24 - 12h\) | M1 | Expanding correctly |
| \(= -16 + 6h^2 + h^3\) | A1 | Simplified correctly |
| Gradient \(= \frac{y_B - y_A}{h} = \frac{6h^2 + h^3}{h} = 6h + h^2\) | A1 | Correct gradient shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(h \to 0\), gradient \(\to 0\) | M1 | Recognising limit as \(h \to 0\) |
| Therefore gradient at \(A\) is \(0\), so \(A\) is a stationary point | A1 | Correct conclusion stated |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y_B = (2+h)^3 - 12(2+h)$ | M1 | Correct substitution for $y_B$ |
| $= 8 + 12h + 6h^2 + h^3 - 24 - 12h$ | M1 | Expanding correctly |
| $= -16 + 6h^2 + h^3$ | A1 | Simplified correctly |
| Gradient $= \frac{y_B - y_A}{h} = \frac{6h^2 + h^3}{h} = 6h + h^2$ | A1 | Correct gradient shown |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $h \to 0$, gradient $\to 0$ | M1 | Recognising limit as $h \to 0$ |
| Therefore gradient at $A$ is $0$, so $A$ is a stationary point | A1 | Correct conclusion stated |5 A curve has equation $y = x ^ { 3 } - 12 x$.\\
The point $A$ on the curve has coordinates ( $2 , - 16$ ).\\
The point $B$ on the curve has $x$-coordinate $2 + h$.
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the line $A B$ is $6 h + h ^ { 2 }$.
\item Explain how the result of part (a) can be used to show that $A$ is a stationary point on the curve.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{763d89e4-861a-4754-a93c-d0902987673f-06_1894_1709_813_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2010 Q5 [6]}}