| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2005 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Show that derivative equals expression |
| Difficulty | Moderate -0.8 Part (i) is a straightforward application of the chain rule to differentiate sin²x, which is a standard textbook exercise. Part (ii) requires solving sin 2x = 0.5, which is routine trigonometry. Part (iii) uses the standard double angle identity cos 2x = 1 - 2sin²x and then integrates, which is a common technique taught explicitly in P2. All three parts are standard procedures with no problem-solving insight required, making this easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Differentiate using the chain or product rule | M1 | |
| Obtain given answer correctly | A1 | 2 |
| (ii) Use correct method for solving \(\sin 2x = 0.5\) | M1 | |
| Obtain answer \(x = \frac{\pi}{12}\) (or 0.262 radians) | A1 | |
| Obtain answer \(x = \frac{5\pi}{12}\) (or 1.31 radians) and no others in range | A1 | 3 |
| (iii) Replace integrand by \(\frac{1}{2} - \frac{1}{2}\cos 2x\), or equivalent | B1 | |
| Integrate and obtain \(\frac{1}{2}x - \frac{1}{4}\sin 2x\), or equivalent | B1/* B1/* | |
| Use limits \(x = 0\) and \(x = \frac{\pi}{2}\) correctly | M1 | |
| Obtain final answer 1.57 (or \(\frac{1}{4}\pi\)) | A1 | 5 |
(i) Differentiate using the chain or product rule | M1 |
Obtain given answer correctly | A1 | 2
(ii) Use correct method for solving $\sin 2x = 0.5$ | M1 |
Obtain answer $x = \frac{\pi}{12}$ (or 0.262 radians) | A1 |
Obtain answer $x = \frac{5\pi}{12}$ (or 1.31 radians) and no others in range | A1 | 3
(iii) Replace integrand by $\frac{1}{2} - \frac{1}{2}\cos 2x$, or equivalent | B1 |
Integrate and obtain $\frac{1}{2}x - \frac{1}{4}\sin 2x$, or equivalent | B1/* B1/*
Use limits $x = 0$ and $x = \frac{\pi}{2}$ correctly | M1 |
Obtain final answer 1.57 (or $\frac{1}{4}\pi$) | A1 | 57\\
\includegraphics[max width=\textwidth, alt={}, center]{d527d21f-0ab5-40fa-8cfd-ebfb4aba0a87-3_493_863_264_641}
The diagram shows the part of the curve $y = \sin ^ { 2 } x$ for $0 \leqslant x \leqslant \pi$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin 2 x$.\\
(ii) Hence find the $x$-coordinates of the points on the curve at which the gradient of the curve is 0.5 . [3]\\
(iii) By expressing $\sin ^ { 2 } x$ in terms of $\cos 2 x$, find the area of the region bounded by the curve and the $x$-axis between 0 and $\pi$.
\hfill \mbox{\textit{CAIE P2 2005 Q7 [10]}}