| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Stationary points then area/volume |
| Difficulty | Standard +0.3 Part (a) requires differentiating a product (straightforward using product rule), setting equal to zero, and solving a linear equation. Part (b) requires integration by parts twice (standard technique for xe^{-x/2} type functions) with clear limits. Both parts are routine applications of standard A-level techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct product or quotient rule | *M1 | \(\frac{dy}{dx} = \left(-\frac{1}{2}\right)(2-x)e^{-\frac{1}{2}x} - e^{-\frac{1}{2}x}\); M1 requires at least one of derivatives correct |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for \(x\) | DM1 | |
| Obtain \(x = 4\) | A1 | ISW |
| Obtain \(y = -2e^{-2}\), or exact equivalent | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Commence integration and reach \(a(2-x)e^{-\frac{1}{2}x} + b\int e^{-\frac{1}{2}x}\,dx\) | *M1 | Condone omission of \(dx\); \(-2(2-x)e^{-\frac{1}{2}x} + 4e^{-\frac{1}{2}x}\) or \(2xe^{-\frac{1}{2}x}\) |
| Obtain \(-2(2-x)e^{-\frac{1}{2}x} - 2\int e^{-\frac{1}{2}x}\,dx\) | A1 | OE |
| Complete integration and obtain \(2xe^{-\frac{1}{2}x}\) | A1 | OE |
| Use correct limits \(x=0\) and \(x=2\), correctly, having integrated twice | DM1 | Ignore omission of zeros and allow max of 1 error |
| Obtain answer \(4e^{-1}\), or exact equivalent | A1 | ISW |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d\left(2xe^{-\frac{1}{2}x}\right)}{dx} = 2e^{-\frac{1}{2}x} - xe^{-\frac{1}{2}x}\) | *M1 A1 | |
| \(\therefore 2xe^{-\frac{1}{2}x}\) | A1 | |
| Use correct limits \(x=0\) and \(x=2\), correctly, having integrated twice | DM1 | Ignore omission of zeros and allow max of 1 error |
| Obtain answer \(4e^{-1}\), or exact equivalent | A1 | ISW |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct product or quotient rule | *M1 | $\frac{dy}{dx} = \left(-\frac{1}{2}\right)(2-x)e^{-\frac{1}{2}x} - e^{-\frac{1}{2}x}$; M1 requires at least one of derivatives correct |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for $x$ | DM1 | |
| Obtain $x = 4$ | A1 | ISW |
| Obtain $y = -2e^{-2}$, or exact equivalent | A1 | |
| **Total** | **5** | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Commence integration and reach $a(2-x)e^{-\frac{1}{2}x} + b\int e^{-\frac{1}{2}x}\,dx$ | *M1 | Condone omission of $dx$; $-2(2-x)e^{-\frac{1}{2}x} + 4e^{-\frac{1}{2}x}$ or $2xe^{-\frac{1}{2}x}$ |
| Obtain $-2(2-x)e^{-\frac{1}{2}x} - 2\int e^{-\frac{1}{2}x}\,dx$ | A1 | OE |
| Complete integration and obtain $2xe^{-\frac{1}{2}x}$ | A1 | OE |
| Use correct limits $x=0$ and $x=2$, correctly, having integrated twice | DM1 | Ignore omission of zeros and allow max of 1 error |
| Obtain answer $4e^{-1}$, or exact equivalent | A1 | ISW |
| **Total** | **5** | |
**Alternative method for 10(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d\left(2xe^{-\frac{1}{2}x}\right)}{dx} = 2e^{-\frac{1}{2}x} - xe^{-\frac{1}{2}x}$ | *M1 A1 | |
| $\therefore 2xe^{-\frac{1}{2}x}$ | A1 | |
| Use correct limits $x=0$ and $x=2$, correctly, having integrated twice | DM1 | Ignore omission of zeros and allow max of 1 error |
| Obtain answer $4e^{-1}$, or exact equivalent | A1 | ISW |
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{19aff1b7-51b7-4d44-86e6-45dad32aa121-16_426_908_262_616}
The diagram shows the curve $y = ( 2 - x ) \mathrm { e } ^ { - \frac { 1 } { 2 } x }$, and its minimum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q10 [10]}}