| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Classify nature of stationary points |
| Difficulty | Moderate -0.3 Part (a) is a standard AS-level task requiring differentiation, solving f'(x)=0, and using the second derivative test—routine but multi-step. Part (b) tests understanding of transformations, which is straightforward coordinate manipulation once part (a) is complete. Overall slightly easier than average due to being a standard textbook-style question with no novel insight required. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (x-2)(x^2 - 6x + 9) = x^3 - 8x^2 + 21x - 18\) | B1 | Multiplies out \(f(x)\) correctly |
| Differentiates with at least one term of \(3x^2 - 16x + 21\) correct | M1 | |
| \(f'(x) = 0\) for a turning point | E1 | |
| Sets differential equal to zero and solves for two \(x\) values: \(3x^2 - 16x + 21 = 0\) | M1 | PI |
| \(x = \frac{7}{3}\) and \(3\) | A1 | |
| \(y = \frac{4}{27}\) and \(0\) | M1 | Substitutes \(x\) values into \(f(x)\) |
| \(f''(x) = 6x - 16\); \(f''\left(\frac{7}{3}\right) = -2 < 0\); \(f''(3) = 2 > 0\) | M1 | Second derivative, or gradient test each side, or shape of cubic argument |
| Maximum at \(\left(\frac{7}{3}, \frac{4}{27}\right)\), Minimum at \((3, 0)\) | R1 | Correct nature at correct coordinates, clearly identifying which is max/min. Not necessary to obtain E1 first |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{4}{3}, -\frac{104}{27}\right)\) | B1F | Deduces at least one fully correct coordinate. FT their coordinates |
| \((2, -4)\) | B1 | Both coordinates correct. CSO |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (x-2)(x^2 - 6x + 9) = x^3 - 8x^2 + 21x - 18$ | B1 | Multiplies out $f(x)$ correctly |
| Differentiates with at least one term of $3x^2 - 16x + 21$ correct | M1 | |
| $f'(x) = 0$ for a turning point | E1 | |
| Sets differential equal to zero and solves for two $x$ values: $3x^2 - 16x + 21 = 0$ | M1 | PI |
| $x = \frac{7}{3}$ and $3$ | A1 | |
| $y = \frac{4}{27}$ and $0$ | M1 | Substitutes $x$ values into $f(x)$ |
| $f''(x) = 6x - 16$; $f''\left(\frac{7}{3}\right) = -2 < 0$; $f''(3) = 2 > 0$ | M1 | Second derivative, or gradient test each side, or shape of cubic argument |
| Maximum at $\left(\frac{7}{3}, \frac{4}{27}\right)$, Minimum at $(3, 0)$ | R1 | Correct nature at correct coordinates, clearly identifying which is max/min. Not necessary to obtain E1 first |
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{4}{3}, -\frac{104}{27}\right)$ | B1F | Deduces at least one fully correct coordinate. FT their coordinates |
| $(2, -4)$ | B1 | Both coordinates correct. CSO |
---9
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of the turning points of $C$.\\
Determine the nature of each turning point.\\
Fully justify your answer.\\
9
\item State the coordinates of the turning points of the curve
$$y = \mathrm { f } ( x + 1 ) - 4$$
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2019 Q9 [10]}}