| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find stationary points of polynomial with exponential factor |
| Difficulty | Standard +0.3 This is a standard A-level curve sketching question involving product rule differentiation of a polynomial times exponential, finding stationary points by solving a quadratic, and applying transformations. Part (b) is a 'show that' which guides students through the differentiation. Part (d) requires understanding of transformations but is straightforward once turning points are found. Slightly easier than average due to the structured guidance and routine techniques. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(0) = (2(0) - 9(0) + 9)e^{-0} = 9\) | ||
| Therefore the coordinates are \((0,9)\) | A1 | (1) mark |
| Answer | Marks | Guidance |
|---|---|---|
| The product rule states: \(\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\) | ||
| And gives: \(f'(x) = (4x - 9)e^{-x} - (2x^2 - 9x + 9)e^{-x} = -(2x^2 - 13x + 18)e^{-x}\) | M1 | Attempts the product rule or quotient and uses \(e^{-x} \to ke^{-x}\), \(k \ne 0\) |
| A1 | A correct \(f'(x)\) which may be unsimplified: \(f'(x) = (4x - 9)e^{-x} - (2x^2 - 9x + 9)e^{-x}\) or \(f'(x) = \frac{e^x(4x - 9) + (2x^2 - 9x + 9)e^x}{e^{2x}}\) | |
| A1 | Proceeds correctly to given answer showing all necessary steps: \(f'(x) = -(2x^2 - 13x + 18)e^{-x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Turning points of \(C\) are given by \(f'(x) = 0 \Rightarrow -(2x^2 - 13x + 18)e^{-x} \Rightarrow 2x^2 - 13x + 18 = 0\) | ||
| \(x = \frac{9}{2}, x = 2\) | B1 | States the roots of \(f'(x) = 0\) as \(2x^2 - 13x + 18 = 0 \Rightarrow x = 2, \frac{9}{2}\) |
| Therefore \(y = -e^{-2}\) or \(y = 9e^{-\frac{9}{2}}\), and the stationary points are given by \((2, -e^{-2})\) and \(\left(\frac{9}{2}, 9e^{-\frac{9}{2}}\right)\) | M1 | Substitutes either \(x = 2\) or \(x = \frac{9}{2}\) into f(x) to find a y value |
| A1 | Obtains \((2, -e^{-2})\) and \(\left(\frac{9}{2}, 9e^{-\frac{9}{2}}\right)\) as the stationary points |
| Answer | Marks |
|---|---|
| The curve \(C\) is stretched vertically with scale factor \(a\), and vertically translated up \(b\) units. | |
| The y-intercept of \(C\) is \((9,0)\), which will be the maximum value for \(x \ge 0\). Therefore 9 will become \(9e^2 + 1 \Rightarrow a = e^2, b = 1\) | |
| B1 | Either \(a = e^2\) or \(b = 1\) |
| B1 | Both \(a = e^2\) and \(b = 1\) |
**Part a:** Find the coordinates of the point where $C$ crosses the y-axis
| $f(0) = (2(0) - 9(0) + 9)e^{-0} = 9$ | | |
| --- | --- | --- |
| Therefore the coordinates are $(0,9)$ | A1 | (1) mark |
**Part b:** Show that $f'(x) = -(2x^2 - 13x + 18)e^{-x}$
| The product rule states: $\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ | | |
| --- | --- | --- |
| And gives: $f'(x) = (4x - 9)e^{-x} - (2x^2 - 9x + 9)e^{-x} = -(2x^2 - 13x + 18)e^{-x}$ | M1 | Attempts the product rule or quotient and uses $e^{-x} \to ke^{-x}$, $k \ne 0$ |
| | A1 | A correct $f'(x)$ which may be unsimplified: $f'(x) = (4x - 9)e^{-x} - (2x^2 - 9x + 9)e^{-x}$ or $f'(x) = \frac{e^x(4x - 9) + (2x^2 - 9x + 9)e^x}{e^{2x}}$ |
| | A1 | Proceeds correctly to given answer showing all necessary steps: $f'(x) = -(2x^2 - 13x + 18)e^{-x}$ |
**Part c:** Hence find the exact coordinates of the turning points of $C$
| Turning points of $C$ are given by $f'(x) = 0 \Rightarrow -(2x^2 - 13x + 18)e^{-x} \Rightarrow 2x^2 - 13x + 18 = 0$ | | |
| --- | --- | --- |
| $x = \frac{9}{2}, x = 2$ | B1 | States the roots of $f'(x) = 0$ as $2x^2 - 13x + 18 = 0 \Rightarrow x = 2, \frac{9}{2}$ |
| Therefore $y = -e^{-2}$ or $y = 9e^{-\frac{9}{2}}$, and the stationary points are given by $(2, -e^{-2})$ and $\left(\frac{9}{2}, 9e^{-\frac{9}{2}}\right)$ | M1 | Substitutes either $x = 2$ or $x = \frac{9}{2}$ into f(x) to find a y value |
| | A1 | Obtains $(2, -e^{-2})$ and $\left(\frac{9}{2}, 9e^{-\frac{9}{2}}\right)$ as the stationary points |
**Part d:** Find the value of $a$ and the value of $b$
| The curve $C$ is stretched vertically with scale factor $a$, and vertically translated up $b$ units. | | |
| --- | --- | --- |
| The y-intercept of $C$ is $(9,0)$, which will be the maximum value for $x \ge 0$. Therefore 9 will become $9e^2 + 1 \Rightarrow a = e^2, b = 1$ | | |
| | B1 | Either $a = e^2$ or $b = 1$ |
| | B1 | Both $a = e^2$ and $b = 1$ |
**(1) mark + (3) marks + (3) marks + (2) marks = (9) marks total for Question 8**
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{802e56f7-5cff-491a-b90b-0759a9b35778-11_1112_1211_280_386}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with the equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \left( 2 x ^ { 2 } - 9 x + 9 \right) e ^ { - x } , \quad x \in R$$
The curve has a minimum turning point at $A$ and a maximum turning point at $B$ as shown in the figure above.\\
a. Find the coordinates of the point where $C$ crosses the $y$-axis.\\
b. Show that $\mathrm { f } ^ { \prime } ( x ) = - \left( 2 x ^ { 2 } - 13 x + 18 \right) e ^ { - x }$\\
c. Hence find the exact coordinates of the turning points of $C$.
The graph with equation $y = \mathrm { f } ( x )$ is transformed onto the graph with equation
$$y = a \mathrm { f } ( x ) + b , \quad x \geq 0$$
The range of the graph with equation $y = a \mathrm { f } ( x ) + b$ is $0 \leq y \leq 9 e ^ { 2 } + 1$\\
Given that $a$ and $b$ are constants.\\
d. find the value of $a$ and the value of $b$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q8 [9]}}