| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Tangent with specified gradient |
| Difficulty | Moderate -0.3 Part (a) is a straightforward differentiation requiring rewriting x√x as x^(3/2) and applying the power rule—routine AS-level calculus. Part (b) requires finding where the tangent gradient equals 3, solving a simple equation, then substituting to find k. While multi-step, these are standard techniques with no novel insight required, making it slightly easier than average but not trivial. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| 5(a) | Expresses x√x in index form. |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 5(b) | Rearranges the equation of the |
| Answer | Marks | Guidance |
|---|---|---|
| PI by gradient = 3 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains gradient of line = 3 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| dx | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| value of the contact point | 1.1a | M1 |
| Deduces k = 6.5 | .2.2a | A1 |
| Subtotal | 5 | |
| Question 5 Total | 7 | |
| Q | Marking instructions | AO |
Question 5:
--- 5(a) ---
5(a) | Expresses x√x in index form.
PI by correct answer
ACF | 1.1a | M1 | 3
y = x 2
1
d y 3
= x 2
d x 2
Obtains the correct derivative.
ACF
ISW | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 5(b) ---
5(b) | Rearranges the equation of the
line to isolate the term in y or x
PI by gradient = 3 | 1.1a | M1 | 6x – 2y + 5 = 0
2y = 6x + 5
Gradient = 3
1
3
x 2 = 3
2
x = 4
From line 2y = 6 × 4 + 5
y = 14.5
14.5 = 4 × 2 + k
k = 6.5
Obtains gradient of line = 3 | 1.1b | A1
Equates their gradient of line to
dy
their expression for
dx | 3.1a | M1
Solves their equation correctly
d y
using their to obtain their x
d x
value of the contact point | 1.1a | M1
Deduces k = 6.5 | .2.2a | A1
Subtotal | 5
Question 5 Total | 7
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Given that $y = x\sqrt{x}$, find $\frac{dy}{dx}$
[2 marks]
\item The line, $L$, has equation $6x - 2y + 5 = 0$
$L$ is a tangent to the curve with equation $y = x\sqrt{x} + k$
Find the value of $k$
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q5 [7]}}