| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find stationary points with unknown constants |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring standard techniques: substituting points to find constants, differentiating to find stationary points, polynomial factorization, and function transformation. While it has multiple parts (7 marks typical), each step follows routine AS-level procedures with clear scaffolding. The most challenging aspect is part (d) involving the transformation f(0.2x), but this is still a standard application of horizontal stretch. Slightly above average due to length and the transformation component, but no novel insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02w Graph transformations: simple transformations of f(x)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(\frac{dy}{dx} = -3\) at \(x = 2 \Rightarrow 12a + 60 - 39 = -3\) | M1 | Attempts to use \(\frac{dy}{dx} = -3\) at \(x = 2\) to form an equation in \(a\). Condone slips but expect to see two of the powers reduced correctly |
| \(12a + 60 - 39 = -3 \Rightarrow 12a = -24 \Rightarrow a = -2^*\) | A1* | Correct differentiation with one correct intermediate step before \(a = -2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses the fact that \((2, 10)\) lies on \(C\): \(10 = 8a + 60 - 78 + b\) | M1 | Attempts to use the fact that \((2,10)\) lies on \(C\) by setting up an equation in \(a\) and \(b\) with \(a = -2\) leading to \(b = \ldots\) |
| Subs \(a = -2\) into \(10 = 8a + 60 - 78 + b \Rightarrow b = 44\) | A1 | \(b = 44\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = -2x^3 + 15x^2 - 39x + 44 \Rightarrow f'(x) = -6x^2 + 30x - 39\) | B1 | \(f'(x) = -6x^2 + 30x - 39\) oe |
| Attempts to show that \(-6x^2 + 30x - 39\) has no roots. E.g. calculates \(b^2 - 4ac = 30^2 - 4 \times -6 \times -39 = -36\) | M1 | Correct attempt to show \(-6x^2 + 30x - 39\) has no roots. Could involve: finding \(b^2 - 4ac\); finding roots via quadratic formula; completing the square for \(-6x^2 + 30x - 39\) |
| States that as \(f'(x) \neq 0 \Rightarrow\) hence \(f(x)\) has no turning points \(*\) | A1* | Fully correct method with reason and conclusion. E.g. \(b^2 - 4ac = -36 < 0\), \(f'(x) \neq 0\) meaning no stationary points exist |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-2x^3 + 15x^2 - 39x + 44 \equiv (x-4)(-2x^2 + 7x - 11)\) | M1 | For an attempt at division (seen or implied). E.g. \(-2x^3 + 15x^2 - 39x + b \equiv (x-4)\left(-2x^2 \ldots \pm \frac{b}{4}\right)\) |
| \((x-4)(-2x^2 + 7x - 11)\) | A1 | Sight of the quadratic with no incorrect working seen can score both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces either intercept: \((0, 44)\) or \((20, 0)\) | B1ft | Follow through on their value for \(b\) |
| Deduces both intercepts \((0, 44)\) and \((20, 0)\) | B1ft | Follow through on their value for \(b\) |
# Question 16:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\frac{dy}{dx} = -3$ at $x = 2 \Rightarrow 12a + 60 - 39 = -3$ | M1 | Attempts to use $\frac{dy}{dx} = -3$ at $x = 2$ to form an equation in $a$. Condone slips but expect to see two of the powers reduced correctly |
| $12a + 60 - 39 = -3 \Rightarrow 12a = -24 \Rightarrow a = -2^*$ | A1* | Correct differentiation with one correct intermediate step before $a = -2$ |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses the fact that $(2, 10)$ lies on $C$: $10 = 8a + 60 - 78 + b$ | M1 | Attempts to use the fact that $(2,10)$ lies on $C$ by setting up an equation in $a$ and $b$ with $a = -2$ leading to $b = \ldots$ |
| Subs $a = -2$ into $10 = 8a + 60 - 78 + b \Rightarrow b = 44$ | A1 | $b = 44$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = -2x^3 + 15x^2 - 39x + 44 \Rightarrow f'(x) = -6x^2 + 30x - 39$ | B1 | $f'(x) = -6x^2 + 30x - 39$ oe |
| Attempts to show that $-6x^2 + 30x - 39$ has no roots. E.g. calculates $b^2 - 4ac = 30^2 - 4 \times -6 \times -39 = -36$ | M1 | Correct attempt to show $-6x^2 + 30x - 39$ has no roots. Could involve: finding $b^2 - 4ac$; finding roots via quadratic formula; completing the square for $-6x^2 + 30x - 39$ |
| States that as $f'(x) \neq 0 \Rightarrow$ hence $f(x)$ has no turning points $*$ | A1* | Fully correct method with reason and conclusion. E.g. $b^2 - 4ac = -36 < 0$, $f'(x) \neq 0$ meaning no stationary points exist |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-2x^3 + 15x^2 - 39x + 44 \equiv (x-4)(-2x^2 + 7x - 11)$ | M1 | For an attempt at division (seen or implied). E.g. $-2x^3 + 15x^2 - 39x + b \equiv (x-4)\left(-2x^2 \ldots \pm \frac{b}{4}\right)$ |
| $(x-4)(-2x^2 + 7x - 11)$ | A1 | Sight of the quadratic with no incorrect working seen can score both marks |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces either intercept: $(0, 44)$ or $(20, 0)$ | B1ft | Follow through on their value for $b$ |
| Deduces both intercepts $(0, 44)$ and $(20, 0)$ | B1ft | Follow through on their value for $b$ |\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$f ( x ) = a x ^ { 3 } + 15 x ^ { 2 } - 39 x + b$$
and $a$ and $b$ are constants.\\
Given
\begin{itemize}
\item the point $( 2,10 )$ lies on $C$
\item the gradient of the curve at $( 2,10 )$ is - 3\\
(a) (i) show that the value of $a$ is - 2\\
(ii) find the value of $b$.\\
(b) Hence show that $C$ has no stationary points.\\
(c) Write $\mathrm { f } ( x )$ in the form $( x - 4 ) \mathrm { Q } ( x )$ where $\mathrm { Q } ( x )$ is a quadratic expression to be found.\\
(d) Hence deduce the coordinates of the points of intersection of the curve with equation
\end{itemize}
$$y = \mathrm { f } ( 0.2 x )$$
and the coordinate axes.
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q16 [11]}}