AQA Paper 3 2021 June — Question 9 9 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2021
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeClassify nature of stationary points
DifficultyStandard +0.3 This is a straightforward calculus question requiring routine differentiation, second derivative test application, and basic transformations. While multi-part with 9 marks total, each step follows standard A-level procedures: finding f''(x) is mechanical, applying the second derivative test is textbook application, identifying increasing intervals from f'(x)>0 is routine, and recognizing the reflection transformation is standard. No novel problem-solving or geometric insight required—slightly easier than average due to its procedural nature.
Spec1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx
A function f is defined for all real values of \(x\) as $$f(x) = x^4 + 5x^3$$ The function has exactly two stationary points when \(x = 0\) and \(x = -\frac{15}{4}\)
    1. Find \(f''(x)\) [2 marks]
    2. Determine the nature of the stationary points. Fully justify your answer. [4 marks]
  2. State the range of values of \(x\) for which $$f(x) = x^4 + 5x^3$$ is an increasing function. [1 mark]
  3. A second function g is defined for all real values of \(x\) as $$g(x) = x^4 - 5x^3$$
    1. State the single transformation which maps f onto g. [1 mark]
    2. State the range of values of \(x\) for which g is an increasing function. [1 mark]
Question 9:
AnswerMarks
9(a)(i)Differentiates f(x) at least one
correct term
AnswerMarks Guidance
May be unsimplified1.1a M1
f′′(x)=12x2 +30x
f′′(x)=12x2 +30x
AnswerMarks Guidance
Obtains1.1b A1
Subtotal2
QMarking instructions AO
AnswerMarks
9(a)(ii)15
Substitutesx = − into their
4
f′′(x)
or
uses gradient test both sides of
15
x = −
AnswerMarks Guidance
41.1a M1
 15  15  15
f′′ − =12 − +30 −
   
 4   4   4 
225
= >0
4
Hence there is a minimum at
15
x = −
4
f′′(0)=0
f′′(1)=12+30>0and
f’’(-1) = 12 – 30 < 0
x=0
hence point of inflection at
Completes rigorous justification
15
for minimum at x = −
4
This must be correctly deduced
using shape of graph or
 15 225
f′′ − = >0
 
AnswerMarks Guidance
 4  42.1 R1
Substitutes two values either
side of x = 0 into their f′′(x)
or
uses gradient test both sides of
x =0
or
argues using the shape of a
quartic curve with two stationary
AnswerMarks Guidance
points1.1a M1
Completes rigorous justification
for point of inflection at x = 0
This must be correctly deduced
using the shape of the graph or
a completely correct test both
sides of the point
Other explanation
eg quartic with two stationary
points, one of the points must be
AnswerMarks Guidance
a point of inflection2.2a R1
Subtotal4
QMarking instructions AO
AnswerMarks
9(b)15
Deduces x>−
4
OE
AnswerMarks Guidance
Condone use of ’2.2a B1
x>−
4
′ ≥
AnswerMarks Guidance
Subtotal1
QMarking instructions AO
AnswerMarks
9(c)(i)Deduces the transformation is a
reflection in the y-axis
AnswerMarks Guidance
OE2.2a B1
Subtotal1
QMarking instructions AO
AnswerMarks
9(c)(ii)15
Deduces x >
4
Condone use of ’
FT their answer in part (b) only if
AnswerMarks Guidance
their value in (b) ′is≥ negative2.2a B1F
x >
4
AnswerMarks Guidance
Subtotal1
Question Total9
QMarking Instructions AO 
Question 9:
--- 9(a)(i) ---
9(a)(i) | Differentiates f(x) at least one
correct term
May be unsimplified | 1.1a | M1 | f′(x)=4x3 +15x2
f′′(x)=12x2 +30x
f′′(x)=12x2 +30x
Obtains | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(a)(ii) ---
9(a)(ii) | 15
Substitutesx = − into their
4
f′′(x)
or
uses gradient test both sides of
15
x = −
4 | 1.1a | M1 | 2
 15  15  15
f′′ − =12 − +30 −
   
 4   4   4 
225
= >0
4
Hence there is a minimum at
15
x = −
4
f′′(0)=0
f′′(1)=12+30>0and
f’’(-1) = 12 – 30 < 0
x=0
hence point of inflection at
Completes rigorous justification
15
for minimum at x = −
4
This must be correctly deduced
using shape of graph or
 15 225
f′′ − = >0
 
 4  4 | 2.1 | R1
Substitutes two values either
side of x = 0 into their f′′(x)
or
uses gradient test both sides of
x =0
or
argues using the shape of a
quartic curve with two stationary
points | 1.1a | M1
Completes rigorous justification
for point of inflection at x = 0
This must be correctly deduced
using the shape of the graph or
a completely correct test both
sides of the point
Other explanation
eg quartic with two stationary
points, one of the points must be
a point of inflection | 2.2a | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | 15
Deduces x>−
4
OE
Condone use of ’ | 2.2a | B1 | 15
x>−
4
′ ≥
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c)(i) ---
9(c)(i) | Deduces the transformation is a
reflection in the y-axis
OE | 2.2a | B1 | Reflection in the y-axis
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c)(ii) ---
9(c)(ii) | 15
Deduces x >
4
Condone use of ’
FT their answer in part (b) only if
their value in (b) ′is≥ negative | 2.2a | B1F | 15
x >
4
Subtotal | 1
Question Total | 9
Q | Marking Instructions | AO | Marks | Typical Solution
A function f is defined for all real values of $x$ as
$$f(x) = x^4 + 5x^3$$

The function has exactly two stationary points when $x = 0$ and $x = -\frac{15}{4}$

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Find $f''(x)$
[2 marks]

\item Determine the nature of the stationary points.

Fully justify your answer.
[4 marks]
\end{enumerate}

\item State the range of values of $x$ for which
$$f(x) = x^4 + 5x^3$$
is an increasing function.
[1 mark]

\item A second function g is defined for all real values of $x$ as
$$g(x) = x^4 - 5x^3$$
\begin{enumerate}[label=(\roman*)]
\item State the single transformation which maps f onto g.
[1 mark]

\item State the range of values of $x$ for which g is an increasing function.
[1 mark]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2021 Q9 [9]}}
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