AQA AS Paper 2 2024 June — Question 8 5 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind stationary points
DifficultyStandard +0.8 This requires finding dy/dx for a function with negative power, setting it to zero, and proving no real solutions exist (likely by showing discriminant is negative or using inequality arguments). It's a multi-step proof question requiring algebraic manipulation and reasoning beyond routine differentiation, but the techniques are standard AS-level calculus and algebra.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx

Prove that the graph of the curve with equation $$y = x^3 + 15x - \frac{18}{x}$$ has no stationary points. [5 marks]

Question 8:
AnswerMarks
81 8
Rewrites as 18x–1
x
AnswerMarks Guidance
PI by correct differentiation.1.2 B1
d y 1 8
= 3x2 + 15 +
d x x 2
1 8
For stationary point 3x2 + 15 + = 0
x 2
3x4 + 15x2 + 18 = 0
x4 + 5x2 + 6 = 0
(x2 + 2)(x2 + 3) = 0
x2 = –2 or x2 = –3
No real solutions for x
No stationary points
Differentiates, at least one term
correct
AnswerMarks Guidance
ACF3.1a M1
d y 1 8
Obtains = 3x2 + 15 +
AnswerMarks Guidance
d x x 21.1b A1
Equates their derivative to 0 and
solves for x2 or x
Or
AnswerMarks Guidance
Uses x2 > 01.1a B1F
Deduces from correct working
that there are no stationary
points with a reason such as:
Cannot square root negative
values (of x) OE
Or solutions for x are
imaginary/not real
Or shows that quartic > 0
dy
Or shows that > 0
dx
Do not accept cannot be
solved/cannot factorise
AnswerMarks Guidance
Can score B1 M1 A1 B0 E12.2a E1
Question 8 Total5
QMarking instructions AO
Question 8:
8 | 1 8
Rewrites as 18x–1
x
PI by correct differentiation. | 1.2 | B1 | y = x3 + 15x – 18x–1
d y 1 8
= 3x2 + 15 +
d x x 2
1 8
For stationary point 3x2 + 15 + = 0
x 2
3x4 + 15x2 + 18 = 0
x4 + 5x2 + 6 = 0
(x2 + 2)(x2 + 3) = 0
x2 = –2 or x2 = –3
No real solutions for x
No stationary points
Differentiates, at least one term
correct
ACF | 3.1a | M1
d y 1 8
Obtains = 3x2 + 15 +
d x x 2 | 1.1b | A1
Equates their derivative to 0 and
solves for x2 or x
Or
Uses x2 > 0 | 1.1a | B1F
Deduces from correct working
that there are no stationary
points with a reason such as:
Cannot square root negative
values (of x) OE
Or solutions for x are
imaginary/not real
Or shows that quartic > 0
dy
Or shows that > 0
dx
Do not accept cannot be
solved/cannot factorise
Can score B1 M1 A1 B0 E1 | 2.2a | E1
Question 8 Total | 5
Q | Marking instructions | AO | Marks | Typical solution
Prove that the graph of the curve with equation
$$y = x^3 + 15x - \frac{18}{x}$$
has no stationary points.
[5 marks]

\hfill \mbox{\textit{AQA AS Paper 2 2024 Q8 [5]}}