| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: x⁴ and higher power terms |
| Difficulty | Moderate -0.8 This is a guided, multi-part question that walks students through the first principles definition of differentiation. Part (a) is routine binomial expansion, part (b) applies the difference quotient formula with scaffolding, and part (c) asks for explanation of taking the limit as h→0. While conceptually important, the heavy scaffolding and standard nature of each step makes this easier than average for A-level. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Expands (x + h)4, with correct | |
| powers of x and h. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains fully correct expansion. | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b) | Forms an expression for |
| Answer | Marks | Guidance |
|---|---|---|
| FT their expression from part (a) | 1.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(c) | Explains or shows that the |
| Answer | Marks | Guidance |
|---|---|---|
| working is shown. | 1.1b | E1F |
| Answer | Marks | Guidance |
|---|---|---|
| Uses the limit as h → 0 | 1.1b | E1 |
| Subtotal | 2 | |
| Question 7 Total | 5 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Expands (x + h)4, with correct
powers of x and h. | 1.1a | M1 | (x+ h)4
= x4 + 4x3h + 6x2h 2 + 4xh 3 + h 4
Obtains fully correct expansion. | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Forms an expression for
d i f f e r e n c e i n y
d i f f e r e n c e i n x
in terms of x and h
ACF
FT their expression from part (a) | 1.1b | B1F | x 4 + 4 x 3 h + 6 x 2 h 2 + 4 x h 3 + h 4 – x 4
h
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c) ---
7(c) | Explains or shows that the
answer to part (b) can be
(expanded and) simplified.
FT their parts (a) and (b) where
working is shown. | 1.1b | E1F | The expression from part (b) can
be simplified by cancelling h
We then apply the limit as h tends
to 0 to obtain the gradient.
Explains that the gradient is
found by letting h tend to 0
or
Uses the limit as h → 0 | 1.1b | E1
Subtotal | 2
Question 7 Total | 5
Q | Marking instructions | AO | Marks | Typical solutionPoints $P$ and $Q$ lie on the curve with equation $y = x^4$
The $x$-coordinate of $P$ is $x$
The $x$-coordinate of $Q$ is $x + h$
\begin{enumerate}[label=(\alph*)]
\item Expand $(x + h)^4$
[2 marks]
\item Hence, find an expression, in terms of $x$ and $h$, for the gradient of the line $PQ$
[1 mark]
\item Explain how to use the answer from part (b) to obtain the gradient function of $y = x^4$
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q7 [5]}}