| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a standard A-level question on differentiation using the product rule, finding stationary points by solving a quadratic, and determining ranges from a sketch. Part (a) is routine verification, part (b) requires solving f'(x)=0 (straightforward quadratic), and part (c) involves reading from the graph with simple transformations. Slightly easier than average due to the 'show that' scaffolding and straightforward algebraic steps. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = 4(x^2-2)e^{-2x}\); differentiates to \(e^{-2x} \times 8x + 4(x^2-2) \times -2e^{-2x}\) | M1, A1 | Attempts product rule with \(e^{-2x} \to ke^{-2x}\), \(k \neq 0\) |
| \(f'(x) = 8e^{-2x}\{x-(x^2-2)\} = 8(2+x-x^2)e^{-2x}\) | A1* | Show that question; all bracketing must be correct; no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States roots of \(f'(x) = 0\): \(x = -1, 2\) | B1 | |
| Substitutes one \(x\) value to find a \(y\) value | M1 | Allow decimals: \(-4e^2 \approx -29.6\), \(8e^{-4} \approx 0.147\) |
| Stationary points are \((-1, -4e^2)\) and \((2, 8e^{-4})\) | A1 | Must be scored in (b); extra solutions e.g. \(x=0\) penalised |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Range \([-8e^2, \infty)\) i.e. \(g(x) \geqslant -8e^2\) | B1ft | Follow through on \(2\times\) minimum \(y\) value from (b) if negative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(2f(0)-3 = 2\times-8-3 = (-19)\) as lower bound, or \(2\times\)"\(8e^{-4}\)"\(-3\) as upper bound | M1 | |
| Range \([-19, 16e^{-4}-3]\) | A1 | Must be exact; i.e. \(-19 \leqslant y \leqslant 16e^{-4}-3\) |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = 4(x^2-2)e^{-2x}$; differentiates to $e^{-2x} \times 8x + 4(x^2-2) \times -2e^{-2x}$ | M1, A1 | Attempts product rule with $e^{-2x} \to ke^{-2x}$, $k \neq 0$ |
| $f'(x) = 8e^{-2x}\{x-(x^2-2)\} = 8(2+x-x^2)e^{-2x}$ | A1* | Show that question; all bracketing must be correct; no errors |
---
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States roots of $f'(x) = 0$: $x = -1, 2$ | B1 | |
| Substitutes one $x$ value to find a $y$ value | M1 | Allow decimals: $-4e^2 \approx -29.6$, $8e^{-4} \approx 0.147$ |
| Stationary points are $(-1, -4e^2)$ and $(2, 8e^{-4})$ | A1 | Must be scored in (b); extra solutions e.g. $x=0$ penalised |
---
## Question 9(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range $[-8e^2, \infty)$ i.e. $g(x) \geqslant -8e^2$ | B1ft | Follow through on $2\times$ minimum $y$ value from (b) if negative |
## Question 9(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $2f(0)-3 = 2\times-8-3 = (-19)$ as lower bound, or $2\times$"$8e^{-4}$"$-3$ as upper bound | M1 | |
| Range $[-19, 16e^{-4}-3]$ | A1 | Must be exact; i.e. $-19 \leqslant y \leqslant 16e^{-4}-3$ |
---9.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{dc0ac5df-24a7-41b5-8410-f0e9b332ba64-22_602_752_246_657}
\end{center}
\section*{Figure 2}
Figure 2 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = 4 \left( x ^ { 2 } - 2 \right) \mathrm { e } ^ { - 2 x } \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ^ { \prime } ( x ) = 8 \left( 2 + x - x ^ { 2 } \right) \mathrm { e } ^ { - 2 x }$
\item Hence find, in simplest form, the exact coordinates of the stationary points of $C$.
The function g and the function h are defined by
$$\begin{array} { l l }
\mathrm { g } ( x ) = 2 \mathrm { f } ( x ) & x \in \mathbb { R } \\
\mathrm {~h} ( x ) = 2 \mathrm { f } ( x ) - 3 & x \geqslant 0
\end{array}$$
\item Find (i) the range of $g$\\
(ii) the range of h
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q9 [9]}}