| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a structured multi-part calculus question requiring differentiation, solving a quartic (which factors as a quadratic in x²), second derivative test, and recognizing horizontal tangents. While it has 11 marks total and multiple parts, each step follows standard AS-level techniques with clear scaffolding. The quartic factorization is straightforward, and part (d) is a direct observation from the working. Slightly above average due to length and the need to justify the nature of stationary points, but no novel insights required. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Differentiates, at least one term | |
| correct. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct derivative. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation. | 2.1 | R1 |
| Subtotal | 3 | 𝑥𝑥 −2𝑥𝑥 −3 = 0 |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 8(b) | Factorises or solves using |
| Answer | Marks | Guidance |
|---|---|---|
| PI | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 2.2a | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 8(c) | Differentiates their again, at |
| Answer | Marks | Guidance |
|---|---|---|
| ± √3 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(d) | Deduces y = 0 | 2.2a |
| Subtotal | 1 | |
| Question 8 Total | 11 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Differentiates, at least one term
correct. | 1.1a | M1 | 𝑑𝑑𝑦𝑦 2 9
= 3𝑥𝑥 −6− 2
𝑑𝑑𝑥𝑥 𝑥𝑥
For stationary point
𝑑𝑑𝑑𝑑
𝑑𝑑𝑥𝑥 = 0
2 9
3𝑥𝑥 −6− 2 = 0
4 2 𝑥𝑥
3𝑥𝑥 −6𝑥𝑥 −9 = 0
4 2
Obtains correct derivative. | 1.1b | A1
Sets correct derivative = 0 and
rearranges to obtain given
equation. | 2.1 | R1
Subtotal | 3 | 𝑥𝑥 −2𝑥𝑥 −3 = 0
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Factorises or solves using
calculator.
PI | 1.1a | M1 | (
2 2
( 𝑥𝑥 − g3i)v(e𝑥𝑥s s+ta1ti)on=a0ry points
2 at ±
𝑥𝑥 −3)= 0
has n√o3 real solutions
so2 there are only two stationary
(𝑥𝑥 +1) = 0 points
Obtains two correct factors or
obtains two correct solutions.
ACF | 1.1b | A1
Concludes that as there are only
2 solutions, there are only 2
stationary points.
OE | 2.2a | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 8(c) ---
8(c) | Differentiates their again, at
𝑑𝑑𝑑𝑑
least one of the two non-zero
𝑑𝑑𝑥𝑥
terms correct, or uses values to
test the sign of close to their
𝑑𝑑𝑑𝑑
𝑑𝑑𝑥𝑥
OE
± √3 | 1.1a | M1 | 2
𝑑𝑑 𝑦𝑦 18
2 = 6𝑥𝑥+ 3
𝑑𝑑𝑥𝑥 𝑥𝑥
At ( , 0) is positive therefore
2
𝑑𝑑 𝑑𝑑
this is a minimum point
2
√3 𝑑𝑑𝑥𝑥
At (– , 0) is negative
2
𝑑𝑑 𝑑𝑑
therefore this is a maximum point
2
√3 𝑑𝑑𝑥𝑥
Makes consistent deduction
about the nature of one of their
stationary points from sign of
their or the sign of close
2
𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑
to their 2
𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 | 1.1a | M1
±√3
States correct coordinates for
one stationary point.
ACF | 1.1b | B1
Obtains the correct exact
coordinates of both stationary
points, along with their correct
natures (from correct
2
𝑑𝑑 𝑑𝑑
2 | 1.1b | A1
𝑑𝑑𝑥𝑥 )
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 8(d) ---
8(d) | Deduces y = 0 | 2.2a | B1 | y = 0
Subtotal | 1
Question 8 Total | 11
Q | Marking instructions | AO | Marks | Typical solutionA curve has equation
$$y = x^3 - 6x + \frac{9}{x}$$
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinates of the stationary points of the curve satisfy the equation
$$x^4 - 2x^2 - 3 = 0$$
[3 marks]
\item Deduce that the curve has exactly two stationary points.
[3 marks]
\item Find the coordinates and nature of the two stationary points.
Fully justify your answer.
[4 marks]
\item Write down the equation of a line which is a tangent to the curve in two places.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2022 Q8 [11]}}