| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Chord gradient with h (algebraic) |
| Difficulty | Standard +0.3 This is a straightforward application of the chord gradient formula leading to differentiation from first principles. Part (a) requires algebraic expansion of (−2+h)^4 using binomial theorem and simplification—routine but slightly tedious. Part (b) is conceptual recognition that taking h→0 gives the derivative, which is standard FP1 material. The algebra is mechanical rather than insightful, making this slightly easier than average. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | \(y = x^4 + x\) so \(\{y(-2+h) =\} (-2+h)^4 + (-2+h)\) | M1 |
| 2(a) | \(= h^4 - 8h^3 + 24h^2 - 32h + 16 - 2 + h\) | B1 |
| 2(a) | \(= h^4 - 8h^3 + 24h^2 - 31h + 14\) | A1F |
| 2(a) | Gradient = \(\frac{y_2 - y_1}{x_2 - x_1}\) | M1 |
| 2(a) | \(= \frac{h^4 - 8h^3 + 24h^2 - 31h + 14 - (14)}{-2+h-(-2)}\) | |
| 2(a) | \(= \frac{h^3 - 8h^3 + 24h^2 - 31h}{h} =\) | A1 |
| 2(a) | \(h^3 - 8h^2 + 24h - 31\) | |
| 2(b) | As \(h \to 0\), gradient of line in (a) \(\to\) gradient of curve at point \((-2, 14)\) | E1 |
| 2(b) | {Gradient of curve at point \((-2, 14)\) is} \(-31\) | E1 |
2(a) | $y = x^4 + x$ so $\{y(-2+h) =\} (-2+h)^4 + (-2+h)$ | M1 | $(-2+h)^4 + (-2+h)$ PI. Correct expansion of $(-2+h)^4$ as $h^4 - 8h^3 + 24h^2 - 32h + 16$ PI
2(a) | $= h^4 - 8h^3 + 24h^2 - 32h + 16 - 2 + h$ | B1 |
2(a) | $= h^4 - 8h^3 + 24h^2 - 31h + 14$ | A1F | Seen separately or as part of the gradient expression. Ft one incorrect term in expansion of $(-2+h)^4$
2(a) | Gradient = $\frac{y_2 - y_1}{x_2 - x_1}$ | M1 | Use of correct formula for gradient PI
2(a) | $= \frac{h^4 - 8h^3 + 24h^2 - 31h + 14 - (14)}{-2+h-(-2)}$ | |
2(a) | $= \frac{h^3 - 8h^3 + 24h^2 - 31h}{h} =$ | A1 | The four correct terms in any order. A0 if incorrect (constant/h) term ignored due printed form of answer
2(a) | $h^3 - 8h^2 + 24h - 31$ | |
2(b) | As $h \to 0$, gradient of line in (a) $\to$ gradient of curve at point $(-2, 14)$ | E1 | Lim $[c'3s(p+q+r+h^2+h^3)]$ OE. NB 'h=0' instead of 'h→ 0' gets E0. Dependent on previous E1 and printed form of answer in (a) obtained convincingly but then ft on c's p value
2(b) | {Gradient of curve at point $(-2, 14)$ is} $-31$ | E1 |2 A curve has equation $y = x ^ { 4 } + x$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the line passing through the point $( - 2,14 )$ and the point on the curve for which $x = - 2 + h$. Give your answer in the form
$$p + q h + r h ^ { 2 } + h ^ { 3 }$$
where $p , q$ and $r$ are integers.
\item Show how the answer to part (a) can be used to find the gradient of the curve at the point ( $- 2,14$ ). State the value of this gradient.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q2 [7]}}