| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Derivative then integrate by parts |
| Difficulty | Standard +0.3 Part (i) requires straightforward product rule differentiation and algebraic verification. Part (ii) is a standard integration by parts problem (∫x cos(x/2) dx) requiring two applications, which is routine for P3 level. The 10 marks reflect length rather than conceptual difficulty—this is a textbook exercise testing standard techniques without requiring problem-solving insight. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks |
|---|---|
| (i) Use product rule | M1 |
| Obtain derivative in any correct form | A1 |
| Differentiate first derivative using the product rule | M1 |
| Obtain second derivative in any correct form, e.g. \(-\frac{1}{4}\sin\frac{1}{4}x - \frac{1}{4}x\cos\frac{1}{4}x - \frac{1}{2}\sin\frac{1}{2}x\) | A1 |
| Verify the given statement | A1 |
| Total: 5 marks | |
| (ii) Integrate and reach \(kx\sin\frac{1}{2}x + I[\sin\frac{1}{2}x \, dx]\) | M1* |
| Obtain \(2x\sin\frac{1}{2}x - 2[\sin\frac{1}{2}x \, dx]\), or equivalent | A1 |
| Obtain indefinite integral \(2x\sin\frac{1}{2}x + 4\cos\frac{1}{2}x\) | A1 |
| Use correct limits \(x = 0, x = \pi\) correctly | M1(dep*) |
| Obtain answer \(2\pi - 4\), or exact equivalent | A1 |
| Total: 5 marks |
**(i)** Use product rule | M1 |
Obtain derivative in any correct form | A1 |
Differentiate first derivative using the product rule | M1 |
Obtain second derivative in any correct form, e.g. $-\frac{1}{4}\sin\frac{1}{4}x - \frac{1}{4}x\cos\frac{1}{4}x - \frac{1}{2}\sin\frac{1}{2}x$ | A1 |
Verify the given statement | A1 |
| Total: 5 marks |
**(ii)** Integrate and reach $kx\sin\frac{1}{2}x + I[\sin\frac{1}{2}x \, dx]$ | M1* |
Obtain $2x\sin\frac{1}{2}x - 2[\sin\frac{1}{2}x \, dx]$, or equivalent | A1 |
Obtain indefinite integral $2x\sin\frac{1}{2}x + 4\cos\frac{1}{2}x$ | A1 |
Use correct limits $x = 0, x = \pi$ correctly | M1(dep*) |
Obtain answer $2\pi - 4$, or exact equivalent | A1 |
| Total: 5 marks |
---\includegraphics{figure_8}
The diagram shows the curve $y = x\cos\frac{1}{2}x$ for $0 \leqslant x \leqslant \pi$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ and show that $4\frac{d^2y}{dx^2} + y + 4\sin\frac{1}{2}x = 0$. [5]
\item Find the exact value of the area of the region enclosed by this part of the curve and the $x$-axis. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2014 Q8 [10]}}