| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Parameter values from curve properties |
| Difficulty | Challenging +1.2 Part (a) is trivial recall. Part (b)(i) requires finding f'(x) and showing f'(0)=0, which is straightforward differentiation. Part (b)(ii) is the challenging component: students must find both turning points, evaluate f at these points, and establish inequalities for three real roots (requiring one turning point above x-axis, one below). This demands coordinated reasoning across multiple steps and careful inequality manipulation, elevating it above standard textbook exercises but not requiring truly novel insight. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks |
|---|---|
| 7(a) | Sketches any cubic graph, |
| Answer | Marks | Guidance |
|---|---|---|
| places | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| positive coefficient of x3 | 1.2 | B1 |
| Answer | Marks |
|---|---|
| 7(b)(i) | f′( ) |
| Answer | Marks | Guidance |
|---|---|---|
| correct - either 3x2or 6px | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| f′( x )=3x2 +6pxand obtains 0 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| x = 0 | 2.4 | R1 |
| Answer | Marks |
|---|---|
| 7(b)(ii) | Deduces that turning point at |
| Answer | Marks | Guidance |
|---|---|---|
| points | 2.2a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| f x | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| f (−2p )=4p3 +q | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone ≤ | 2.2a | R1 |
| Deduces −4p3 <q<0 CAO | 2.2a | R1 |
| Total | 10 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Sketches any cubic graph,
crossing the x-axis in three
places | 1.2 | B1
Sketches any cubic graph with a
positive coefficient of x3 | 1.2 | B1
--- 7(b)(i) ---
7(b)(i) | f′( )
Differentiates to obtain x
Two terms with at least one
correct - either 3x2or 6px | 1.1a | M1 | f′( )
For a turning point x = 0
f ( x )= x3 +3px2 +q
f′( x )=3x2 +6px
3x2 +6px =0
3x ( x+2p )=0
x =0
x = −2p
Since one of the roots is x = 0 there
must be a turning point on the y axis
Solves 3x2 +6px =0to obtain
x = 0 or x = −2p
or
Substitutes x = 0 in
f′( x )=3x2 +6pxand obtains 0 | 1.1b | A1
Obtains the correct two roots
x = 0 and x = −2p OE
and states why there must be a
turning point referring to root
x = 0 | 2.4 | R1
--- 7(b)(ii) ---
7(b)(ii) | Deduces that turning point at
x = -2p is a maximum or
deduces that turning point x = 0
is a minimum
May have been seen in part
(b)(i)
Accept a sketch showing correct
relative positions of turning
points | 2.2a | B1 | Since p > 0
x=−2p
is the maximum
x = 0 is the minimum
( )=q
f 0
f (−2p )=(−2p )3 +3p (−2p )2 +q
=4p3 +q
−4p3 <q<0
Substitutes their x = -2p into
( )
f x | 1.1a | M1
( )=qand
Obtains correct f 0
f (−2p )=4p3 +q | 1.1b | A1
Deduces
either q<0or −4p3 <q
Condone ≤ | 2.2a | R1
Deduces −4p3 <q<0 CAO | 2.2a | R1
Total | 10
Q | Marking instructions | AO | Mark | Typical solution\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of any cubic function that has both three distinct real roots and a positive coefficient of $x^3$ [2 marks]
\item The function f(x) is defined by
$$f(x) = x^3 + 3px^2 + q$$
where $p$ and $q$ are constants and $p > 0$
\begin{enumerate}[label=(\roman*)]
\item Show that there is a turning point where the curve crosses the $y$-axis. [3 marks]
\item The equation $f(x) = 0$ has three distinct real roots.
By considering the positions of the turning points find, in terms of $p$, the range of possible values of $q$. [5 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2019 Q7 [10]}}