| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with trigonometric functions |
| Difficulty | Standard +0.8 This question requires finding a maximum using differentiation (involving chain rule and quotient/sec² derivative), then integrating a trigonometric function that needs the identity sin²(x/2) = (1-cos x)/2. The integration produces logarithmic terms requiring careful substitution limits. Multi-step with non-routine trigonometric manipulation pushes it above average difficulty. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate to obtain \(4\cos\frac{1}{2}x - \frac{1}{2}\sec^2\frac{1}{2}x\) | B1 | |
| Equate to zero and find value of \(\cos\frac{1}{2}x\) | M1 | |
| Obtain \(\cos\frac{1}{2}x = \frac{1}{2}\) and confirm \(\alpha = \frac{2}{3}\pi\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrate to obtain \(-16\cos\frac{1}{2}x \ldots\) | B1 | |
| \(\ldots + 2\ln\cos\frac{1}{2}x\) or equivalent | B1 | |
| Using limits \(0\) and \(\frac{2}{3}\pi\) in \(a\cos\frac{1}{2}x + b\ln\cos\frac{1}{2}x\) | M1 | |
| Obtain \(8 + 2\ln\frac{1}{2}\) or exact equivalent | A1 | [4] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate to obtain $4\cos\frac{1}{2}x - \frac{1}{2}\sec^2\frac{1}{2}x$ | B1 | |
| Equate to zero and find value of $\cos\frac{1}{2}x$ | M1 | |
| Obtain $\cos\frac{1}{2}x = \frac{1}{2}$ and confirm $\alpha = \frac{2}{3}\pi$ | A1 | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate to obtain $-16\cos\frac{1}{2}x \ldots$ | B1 | |
| $\ldots + 2\ln\cos\frac{1}{2}x$ or equivalent | B1 | |
| Using limits $0$ and $\frac{2}{3}\pi$ in $a\cos\frac{1}{2}x + b\ln\cos\frac{1}{2}x$ | M1 | |
| Obtain $8 + 2\ln\frac{1}{2}$ or exact equivalent | A1 | [4] |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{4c71f68a-efb9-4408-bf03-874e0d4426d5-2_458_807_1786_667}
The diagram shows the curve
$$y = 8 \sin \frac { 1 } { 2 } x - \tan \frac { 1 } { 2 } x$$
for $0 \leqslant x < \pi$. The $x$-coordinate of the maximum point is $\alpha$ and the shaded region is enclosed by the curve and the lines $x = \alpha$ and $y = 0$.\\
(i) Show that $\alpha = \frac { 2 } { 3 } \pi$.\\
(ii) Find the exact value of the area of the shaded region.
\hfill \mbox{\textit{CAIE P3 2012 Q5 [7]}}