| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Stationary point via first principles |
| Difficulty | Moderate -0.5 This is a straightforward application of differentiation from first principles with clear scaffolding. Part (a) requires computing the gradient of a chord using the difference quotient formula with given coordinates, involving routine algebra with cubics. Part (b) is a simple conceptual explanation about taking the limit as h→0. While it tests understanding of first principles, the algebraic manipulation is standard and the question structure guides students through each step. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AB = \frac{(-4+h)^3 - 48(-4+h) - \left((-4)^3 - 48(-4)\right)}{h}\) | M1 | Forms expression of correct form; condone sign error |
| \(= \frac{h^3 - 12h^2 + 48h - 64 - 48h + 192 - 128}{h}\) | B1 | Obtains correct expansion of \((-4+h)^3\) |
| \(= \frac{h^3 - 12h^2}{h}\) | A1 | Obtains correct expansion of numerator |
| \(= h^2 - 12h\) | R1 | Simplifies numerator and shows given result |
| Answer | Marks | Guidance |
|---|---|---|
| The gradient of the curve is given by \(\lim_{h \to 0} h^2 - 12h\) | E1 | Explains that as \(h \to 0\) gradient of line \(AB \to\) gradient of curve/tangent; must not use \(h = 0\) |
| As \(h \to 0\), \(h^2 - 12h \to 0\) therefore \(A\) must be a stationary point | E1 | Explains that \(\lim_{h \to 0} h^2 - 12h = 0\) therefore \(A\) must be a stationary point |
## Question 15(a):
Gradient of $AB = \frac{(-4+h)^3 - 48(-4+h) - \left((-4)^3 - 48(-4)\right)}{h}$ | M1 | Forms expression of correct form; condone sign error
$= \frac{h^3 - 12h^2 + 48h - 64 - 48h + 192 - 128}{h}$ | B1 | Obtains correct expansion of $(-4+h)^3$
$= \frac{h^3 - 12h^2}{h}$ | A1 | Obtains correct expansion of numerator
$= h^2 - 12h$ | R1 | Simplifies numerator and shows given result
---
## Question 15(b):
The gradient of the curve is given by $\lim_{h \to 0} h^2 - 12h$ | E1 | Explains that as $h \to 0$ gradient of line $AB \to$ gradient of curve/tangent; must not use $h = 0$
As $h \to 0$, $h^2 - 12h \to 0$ therefore $A$ must be a stationary point | E1 | Explains that $\lim_{h \to 0} h^2 - 12h = 0$ therefore $A$ must be a stationary point
**Total: 6 marks**
**TOTAL: 100 marks**15 A curve has equation $y = x ^ { 3 } - 48 x$
The point $A$ on the curve has $x$ coordinate - 4\\
The point $B$ on the curve has $x$ coordinate $- 4 + h$\\
15
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the line $A B$ is $h ^ { 2 } - 12 h$\\
15
\item Explain how the result of part (a) can be used to show that $A$ is a stationary point on the curve.\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-25_2488_1719_219_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-26_2488_1719_219_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-27_2488_1719_219_150}\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-28_2498_1721_213_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q15 [6]}}