| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a standard C3 product rule question with routine steps: finding intercepts by substitution, differentiating using product rule, and solving a quadratic to find stationary points. While it requires multiple techniques, each step follows textbook procedures with no novel insight needed, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| Either \(y = 2\) or \((0, 2)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 2\), \(y = (8 - 10 + 2)e^{-2} = 0e^{-2} = 0\) | B1 | |
| \((2x^2 - 5x + 2) = 0 \Rightarrow (x-2)(2x-1) = 0\) | M1 | |
| Either \(x = 2\) (for possibly B1 above) or \(x = \frac{1}{2}\) | A1 | If candidate believes \(e^{-x} = 0\) solves to \(x = 0\) or gives extra solution \(x = 0\), withhold final accuracy mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = (4x-5)e^{-x} - (2x^2 - 5x + 2)e^{-x}\) | M1A1A1 | M1: (their \(u'\))\(e^{-x}\) + \((2x^2-5x+2)\)(their \(v'\)); A1: any one term correct; A1: both terms correct |
| Answer | Marks | Guidance |
|---|---|---|
| \((4x-5)e^{-x} - (2x^2 - 5x + 2)e^{-x} = 0\) | M1 | Set \(\frac{dy}{dx}\) from (c) equal to 0 |
| \(2x^2 - 9x + 7 = 0 \Rightarrow (2x-7)(x-1) = 0\) | M1 | Factorise or eliminate \(e^{-x}\) correctly and attempt to factorise 3-term quadratic |
| \(x = \frac{7}{2}, 1\) | A1 | |
| When \(x = \frac{7}{2}\), \(y = 9e^{-\frac{7}{2}}\); when \(x = 1\), \(y = -e^{-1}\) | ddM1A1 | ddM1: attempt to use at least one \(x\)-coordinate on \(y = (2x^2-5x+2)e^{-x}\); A1: both exact values required, cao |
## Question 5:
### Part (a)
| Either $y = 2$ or $(0, 2)$ | B1 | |
### Part (b)
| When $x = 2$, $y = (8 - 10 + 2)e^{-2} = 0e^{-2} = 0$ | B1 | |
| $(2x^2 - 5x + 2) = 0 \Rightarrow (x-2)(2x-1) = 0$ | M1 | |
| Either $x = 2$ (for possibly B1 above) or $x = \frac{1}{2}$ | A1 | If candidate believes $e^{-x} = 0$ solves to $x = 0$ or gives extra solution $x = 0$, withhold final accuracy mark |
### Part (c)
| $\frac{dy}{dx} = (4x-5)e^{-x} - (2x^2 - 5x + 2)e^{-x}$ | M1A1A1 | M1: (their $u'$)$e^{-x}$ + $(2x^2-5x+2)$(their $v'$); A1: any one term correct; A1: both terms correct |
### Part (d)
| $(4x-5)e^{-x} - (2x^2 - 5x + 2)e^{-x} = 0$ | M1 | Set $\frac{dy}{dx}$ from (c) equal to 0 |
| $2x^2 - 9x + 7 = 0 \Rightarrow (2x-7)(x-1) = 0$ | M1 | Factorise or eliminate $e^{-x}$ correctly and attempt to factorise 3-term quadratic |
| $x = \frac{7}{2}, 1$ | A1 | |
| When $x = \frac{7}{2}$, $y = 9e^{-\frac{7}{2}}$; when $x = 1$, $y = -e^{-1}$ | ddM1A1 | ddM1: attempt to use at least one $x$-coordinate on $y = (2x^2-5x+2)e^{-x}$; A1: both exact values required, cao |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{52f73407-14c5-46e6-b911-aa096b9b5893-08_701_1125_246_443}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with the equation $y = \left( 2 x ^ { 2 } - 5 x + 2 \right) \mathrm { e } ^ { - x }$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the point where $C$ crosses the $y$-axis.
\item Show that $C$ crosses the $x$-axis at $x = 2$ and find the $x$-coordinate of the other point where $C$ crosses the $x$-axis.
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence find the exact coordinates of the turning points of $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2010 Q5 [12]}}