| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding maximum/minimum on curve |
| Difficulty | Standard +0.8 Part (a) requires differentiation using the product rule and chain rule, then solving e^(-x²/4) = 0 which needs insight that the exponential term never equals zero. Part (b) requires recognizing that the substitution x = √u transforms the integral into a standard form, involving careful manipulation of dx = du/(2√u) and bounds. This is above-average difficulty due to the non-standard substitution and multi-step reasoning required in both parts. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the correct product rule | \*M1 | Condone error in chain rule. |
| Obtain correct derivative in any form | A1 | e.g. \(\frac{dy}{dx} = -\frac{x^2}{2}e^{-\frac{x^2}{4}} + e^{-\frac{x^2}{4}}\) |
| Equate derivative to zero and solve for \(x\) | DM1 | |
| Obtain answer \(\left(\sqrt{2},\ \sqrt{2}e^{-\frac{1}{2}}\right)\) | A1 | Or exact equivalent. Can state the components separately. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(dx = \frac{1}{2}u^{-\frac{1}{2}}\,du\) | B1 | Or equivalent e.g. \(du = 2x\,dx\). Alternative substitution: \(u = -\frac{1}{4}x^2\) |
| Substitute for \(x\) and \(dx\) | M1 | |
| Obtain correct integral \(\frac{1}{2}\int e^{-\frac{1}{4}u}\,du\) | A1 | OE |
| Use correct limits in an integral of the form \(ae^{-\frac{1}{4}u}\) or \(ae^{-\frac{1}{4}x^2}\) | M1 | \(u=9\) and \(u=0\) or \(x=3\) and \(x=0\) |
| Obtain answer \(2 - 2e^{-\frac{9}{4}}\) | A1 | Or exact equivalent. |
| Alternative Method: | ||
| \(\int xe^{-\frac{1}{4}x^2}dx = ae^{-\frac{1}{4}x^2}\) | M1 | Recognition used. |
| \(a\) negative | A1 | |
| \(a = -2\) | A1 | |
| Use correct limits in an integral of the form \(ae^{-\frac{1}{4}x^2}\) | M1 | \(x=3\) and \(x=0\) |
| Obtain answer \(2 - 2e^{-\frac{9}{4}}\) | A1 | Or exact equivalent. |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the correct product rule | **\*M1** | Condone error in chain rule. |
| Obtain correct derivative in any form | **A1** | e.g. $\frac{dy}{dx} = -\frac{x^2}{2}e^{-\frac{x^2}{4}} + e^{-\frac{x^2}{4}}$ |
| Equate derivative to zero and solve for $x$ | **DM1** | |
| Obtain answer $\left(\sqrt{2},\ \sqrt{2}e^{-\frac{1}{2}}\right)$ | **A1** | Or exact equivalent. Can state the components separately. |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $dx = \frac{1}{2}u^{-\frac{1}{2}}\,du$ | **B1** | Or equivalent e.g. $du = 2x\,dx$. Alternative substitution: $u = -\frac{1}{4}x^2$ |
| Substitute for $x$ and $dx$ | **M1** | |
| Obtain correct integral $\frac{1}{2}\int e^{-\frac{1}{4}u}\,du$ | **A1** | OE |
| Use correct limits in an integral of the form $ae^{-\frac{1}{4}u}$ or $ae^{-\frac{1}{4}x^2}$ | **M1** | $u=9$ and $u=0$ or $x=3$ and $x=0$ |
| Obtain answer $2 - 2e^{-\frac{9}{4}}$ | **A1** | Or exact equivalent. |
| **Alternative Method:** | | |
| $\int xe^{-\frac{1}{4}x^2}dx = ae^{-\frac{1}{4}x^2}$ | **M1** | Recognition used. |
| $a$ negative | **A1** | |
| $a = -2$ | **A1** | |
| Use correct limits in an integral of the form $ae^{-\frac{1}{4}x^2}$ | **M1** | $x=3$ and $x=0$ |
| Obtain answer $2 - 2e^{-\frac{9}{4}}$ | **A1** | Or exact equivalent. |
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\includegraphics[max width=\textwidth, alt={}, center]{ce3c4a9c-bf83-4d28-96e2-ef31c3673dea-12_375_645_274_742}
The diagram shows the curve $y = x \mathrm { e } ^ { - \frac { 1 } { 4 } x ^ { 2 } }$, for $x \geqslant 0$, and its maximum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item Using the substitution $x = \sqrt { u }$, or otherwise, find by integration the exact area of the shaded region bounded by the curve, the $x$-axis and the line $x = 3$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q9 [9]}}