| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find range using calculus |
| Difficulty | Standard +0.8 This P3 question requires finding range from a graph, differentiating a product of three functions (exponential and polynomial), and then using calculus to find exact values where a horizontal line intersects the curve at exactly two points. The differentiation is algebraically demanding, and part (c) requires identifying turning points and their nature to determine the precise range—going beyond routine application of the product rule. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0) = (0-3)^2 = 9\) | M1 | Substitute \(x=0\) and find a value for \(y\) |
| \(0 \leqslant f(x) \leqslant 9\) | A1 | Accept with \(y\) or just f; accept interval \([0,9]\) or set notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = -2xe^{-x^2}(2x^2-3)^2 + e^{-x^2} \times 8x(2x^2-3)\) | M1A1 | M1: Attempts product rule and chain rule achieving \(\pm Pxe^{-x^2}(2x^2\pm3)^2 + e^{-x^2}\times Qx(2x^2\pm3)\) with \(P,Q>0\) |
| \(= 2x(2x^2-3)e^{-x^2}\left(-(2x^2-3)+4\right) = 2xe^{-x^2}(2x^2-3)(7-2x^2)\) | dM1A1 | dM1: Takes out factor of \(2xe^{-x^2}(2x^2-3)\) to obtain factor \((\pm C \pm Dx^2)\); A1: achieves fully simplified form with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 = \frac{3}{2}, \frac{7}{2} \Rightarrow f\!\left(\sqrt{\frac{7}{2}}\right) = e^{-\frac{7}{2}}\!\left(2\times\frac{7}{2}-3\right)^2 = 16e^{-\frac{7}{2}}\) | M1A1 | M1: Attempts to find a \(y\) value at a valid non-zero root of \(f'(x)=0\) |
| \(16e^{-\frac{7}{2}} < k < 9\) | dM1A1 | dM1: Attempts inside region using \(\leqslant\), \(y\)-intercept from (a) and positive \(y\) value from maxima; A1: accept awrt 0.483 for lower bound; allow \(k \in \left(\frac{16}{e^{7/2}}, 9\right)\) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0) = (0-3)^2 = 9$ | M1 | Substitute $x=0$ and find a value for $y$ |
| $0 \leqslant f(x) \leqslant 9$ | A1 | Accept with $y$ or just f; accept interval $[0,9]$ or set notation |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = -2xe^{-x^2}(2x^2-3)^2 + e^{-x^2} \times 8x(2x^2-3)$ | M1A1 | M1: Attempts product rule and chain rule achieving $\pm Pxe^{-x^2}(2x^2\pm3)^2 + e^{-x^2}\times Qx(2x^2\pm3)$ with $P,Q>0$ |
| $= 2x(2x^2-3)e^{-x^2}\left(-(2x^2-3)+4\right) = 2xe^{-x^2}(2x^2-3)(7-2x^2)$ | dM1A1 | dM1: Takes out factor of $2xe^{-x^2}(2x^2-3)$ to obtain factor $(\pm C \pm Dx^2)$; A1: achieves fully simplified form with no errors |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 = \frac{3}{2}, \frac{7}{2} \Rightarrow f\!\left(\sqrt{\frac{7}{2}}\right) = e^{-\frac{7}{2}}\!\left(2\times\frac{7}{2}-3\right)^2 = 16e^{-\frac{7}{2}}$ | M1A1 | M1: Attempts to find a $y$ value at a valid non-zero root of $f'(x)=0$ |
| $16e^{-\frac{7}{2}} < k < 9$ | dM1A1 | dM1: Attempts inside region using $\leqslant$, $y$-intercept from (a) and positive $y$ value from maxima; A1: accept awrt 0.483 for lower bound; allow $k \in \left(\frac{16}{e^{7/2}}, 9\right)$ |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08291ac1-bdd4-4241-8959-7c89318fa5eb-18_554_1129_248_468}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = e ^ { - x ^ { 2 } } \left( 2 x ^ { 2 } - 3 \right) ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the range of f
\item Show that
$$\mathrm { f } ^ { \prime } ( x ) = 2 x \left( 2 x ^ { 2 } - 3 \right) \mathrm { e } ^ { - x ^ { 2 } } \left( A - B x ^ { 2 } \right)$$
where $A$ and $B$ are constants to be found.
Given that the line $y = k$, where $k$ is a constant, $k > 0$, intersects the curve at exactly two distinct points,
\item find the exact range of values of $k$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q7 [10]}}