| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find range using calculus |
| Difficulty | Standard +0.8 This question requires applying the product rule to a three-factor product involving exponentials, then finding stationary points of a cubic to determine the range. Part (a) is moderately challenging differentiation; part (b) requires analyzing the cubic's roots and evaluating g at critical points; part (c) is straightforward. The multi-step nature and need to connect differentiation to range analysis places this above average difficulty. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.07i Differentiate x^n: for rational n and sums1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Applies \(vu' + uv'\) to \((x^2 - x^3)e^{-2x}\) | M1 | Product rule: \(u=x^2-x^3\), \(v=e^{-2x}\). Accept quotient rule or implicit differentiation methods |
| \(g'(x) = (x^2-x^3)\times -2e^{-2x} + (2x-3x^2)\times e^{-2x}\) | M1A1 | |
| \(g'(x) = (2x^3 - 5x^2 + 2x)e^{-2x}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sets \((2x^3-5x^2+2x)e^{-2x} = 0 \Rightarrow 2x^3-5x^2+2x=0\) | M1 | |
| \(x(2x^2-5x+2)=0 \Rightarrow x = 0, \frac{1}{2}, 2\) | M1, A1 | |
| Sub \(x=\frac{1}{2}, 2\): \(g\!\left(\frac{1}{2}\right) = \frac{1}{8e}\), \(g(2) = -\frac{4}{e^4}\) | dM1, A1 | |
| Range: \(-\frac{4}{e^4} \leq g(x) \leq \frac{1}{8e}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(g(x)\) is NOT a one-to-one function / is a many-to-one function | B1 | Accept: \(g^{-1}(x)\) would be one-to-many |
## Question 7:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Applies $vu' + uv'$ to $(x^2 - x^3)e^{-2x}$ | M1 | Product rule: $u=x^2-x^3$, $v=e^{-2x}$. Accept quotient rule or implicit differentiation methods |
| $g'(x) = (x^2-x^3)\times -2e^{-2x} + (2x-3x^2)\times e^{-2x}$ | M1A1 | |
| $g'(x) = (2x^3 - 5x^2 + 2x)e^{-2x}$ | A1 | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sets $(2x^3-5x^2+2x)e^{-2x} = 0 \Rightarrow 2x^3-5x^2+2x=0$ | M1 | |
| $x(2x^2-5x+2)=0 \Rightarrow x = 0, \frac{1}{2}, 2$ | M1, A1 | |
| Sub $x=\frac{1}{2}, 2$: $g\!\left(\frac{1}{2}\right) = \frac{1}{8e}$, $g(2) = -\frac{4}{e^4}$ | dM1, A1 | |
| Range: $-\frac{4}{e^4} \leq g(x) \leq \frac{1}{8e}$ | A1 | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $g(x)$ is NOT a one-to-one function / is a many-to-one function | B1 | Accept: $g^{-1}(x)$ would be one-to-many |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{57ea7a94-6939-4c12-a6fd-01cd6af73310-12_632_873_294_532}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation
$$\mathrm { g } ( x ) = x ^ { 2 } ( 1 - x ) \mathrm { e } ^ { - 2 x } , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { g } ^ { \prime } ( x ) = \mathrm { f } ( x ) \mathrm { e } ^ { - 2 x }$, where $\mathrm { f } ( x )$ is a cubic function to be found.
\item Hence find the range of g .
\item State a reason why the function $\mathrm { g } ^ { - 1 } ( x )$ does not exist.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2015 Q7 [10]}}