| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Classify nature of stationary points |
| Difficulty | Standard +0.3 This is a straightforward calculus optimization problem requiring differentiation of powers (including fractional powers), finding stationary points, and using the second derivative test. While it has multiple steps and requires careful algebraic manipulation of the √x term (x^{3/2}), all techniques are standard AS-level procedures with no novel problem-solving required. The 9 marks for part (a) reflect the need to show all working rather than conceptual difficulty. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| 10(a) | Verifies that x = 1 gives y =3 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Expresses x√𝑥 as 𝑥2 | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| one term correct) | AO1.1a | M1 |
| Correctly differentiates | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Must be explicitly seen | AO2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| or correct verification of x = 1 | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b)) | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| differential) | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| out award of this mark | AO2.1 | R1 |
| (b) | States coordinates | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| States minimum point | AO1.1b | B1 |
| Total | 11 | |
| Q | Marking Instructions | AO |
Question 10:
--- 10(a) ---
10(a) | Verifies that x = 1 gives y =3 | AO1.1b | B1 | 3
𝑦 = 2×12−8×12+8×1+1
=3
3
𝑦 = 2𝑥2−8𝑥2+8𝑥+1
𝑑𝑦
= 4𝑥−12√𝑥+8
𝑑𝑥
For stationary point
4(√𝑥)2−12√𝑥+8=0
√𝑥 = 1 𝑜𝑟 √𝑥 = 2
𝑥 = 1 𝑜𝑟 𝑥 = 4
𝑑2𝑦 6
= 4−
𝑑𝑥2 √𝑥
𝑑2𝑦
x = 1 gives =−2
𝑑𝑥2
Negative so maximum when x = 1
Maximum at (1, 3)
3
Expresses x√𝑥 as 𝑥2 | AO1.1b | B1
Attempts to differentiate (at least
one term correct) | AO1.1a | M1
Correctly differentiates | AO1.1b | A1
𝑑𝑦
Explains = 0 for stationary or
𝑑𝑥
maximum point
Must be explicitly seen | AO2.4 | E1
𝑑𝑦
Shows solution of = 0 to give
𝑑𝑥
𝑥 = 1 (and 𝑥 = 4)
(May be awarded for work seen in
(b))
or correct verification of x = 1 | AO1.1b | B1
Differentiates again
(May be awarded for work seen in
(b)) | AO1.1a | M1
Shows that x = 1 gives a negative
value (in a correct second
differential) | AO1.1b | A1
Concludes that maximum point is
at (1, 3).
Constructs rigorous mathematical
argument to show the required
result.
Failure to score E1 does not rule
out award of this mark | AO2.1 | R1
(b) | States coordinates | AO1.1b | B1 | (4, 1)
Minimum
States minimum point | AO1.1b | B1
Total | 11
Q | Marking Instructions | AO | Marks | Typical SolutionA curve has equation $y = 2x^2 - 8x\sqrt{x} + 8x + 1$ for $x \geq 0$
\begin{enumerate}[label=(\alph*)]
\item Prove that the curve has a maximum point at $(1, 3)$
Fully justify your answer.
[9 marks]
\item Find the coordinates of the other stationary point of the curve and state its nature.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2018 Q10 [11]}}