| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring standard differentiation of exponentials, finding tangent equations, and solving for a parameter. Part (c) requires geometric insight to connect the tangent condition to the number of solutions, which elevates it slightly above routine exercises, but the scaffolding through parts (a) and (b) makes it accessible for AS-level students. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Recalls and applies gradient rule | |
| for ekx | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| formula | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| FT their gradient. | 1.1b | A1F |
| Subtotal | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b) | Substitutes x = 0 and y = 0 into | |
| their line equation | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation | 1.1b | A1F |
| Subtotal | 2 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(c) | Deduces correct lower limit | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| on their answers to (a) and (b) | 2.2a | M1 |
| Obtains correct inequality | 1.1b | A1 |
| Subtotal | 3 | |
| Question Total | 8 | |
| Q | Marking Instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Recalls and applies gradient rule
for ekx | 1.2 | B1 | Gradient = 4e4a
Equation is
y – e4a = 4e4a(x – a)
Uses their gradient in line equation
formula | 1.1a | M1
Obtains correct equation, any form,
FT their gradient. | 1.1b | A1F
Subtotal | 3
--- 8(b) ---
8(b) | Substitutes x = 0 and y = 0 into
their line equation | 1.1a | M1 | x = 0 and y = 0 gives
0 = e4a(1 – 4a)
a =
1
Finds correct value of a for their
equation | 1.1b | A1F
Subtotal | 2 | 4
--- 8(c) ---
8(c) | Deduces correct lower limit | 2.2a | B1 | Any negative gradient cuts curve
0 ≤ m
With a = contact point is (
1 1
Gradient (0, 0) to ( is 4e
4 4 ,e)
1
4 ,e)
0 ≤ m < 4e
Deduces correct upper limit based
on their answers to (a) and (b) | 2.2a | M1
Obtains correct inequality | 1.1b | A1
Subtotal | 3
Question Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve $y = e^{4x}$ at the point $(a, e^{4a})$. [3 marks]
\item Find the value of $a$ for which this tangent passes through the origin. [2 marks]
\item Hence, find the set of values of $m$ for which the equation
$$e^{4x} = mx$$
has no real solutions. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2020 Q8 [8]}}