| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Trigonometric arithmetic progression |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on geometric and arithmetic progressions with trigonometric terms. Part (a)(i) requires simple algebraic manipulation using the sum to infinity formula (S = a/(1-r)) to find the common ratio, then calculating the second term. Part (a)(ii) and (b) involve direct application of standard formulas with given values. The trigonometric context adds minimal complexity since specific angle values are provided. This is slightly easier than average as it's mostly formula application with clear scaffolding. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (a)(i) \(\frac{\cos\theta}{1-r} = \frac{1}{\cos\theta}\) | B1 | |
| (a)(i) \(1 - r = \cos^2\theta\) leading to \(r = 1 - \cos^2\theta\) | M1 | Eliminate fractions |
| (a)(i) \(r = \sin^2\theta\) leading to 2nd term \(= \cos\theta\sin^2\theta\) | A1 | AG |
| (a)(ii) \(S_{12} = \frac{\cos\!\left(\frac{\pi}{3}\right)\!\left[1 - \left(\sin^2\!\left(\frac{\pi}{3}\right)\right)^{12}\right]}{1 - \sin^2\!\left(\frac{\pi}{3}\right)} = \frac{0.5\left[1-(0.75)^{12}\right]}{1-0.75}\) | M1 | Evidence of correct substitution, use of \(S_n\) formula and attempt to evaluate |
| (a)(ii) \(1.937\) | A1 | |
| (b) \([d =]\ \cos\theta\sin^2\theta - \cos\theta\) | M1 | Use of \(d = u_2 - u_1\) |
| (b) \(-\frac{1}{8}\) | A1 | |
| (b) \(\text{[85th term ]} = \frac{1}{2} + 84 \times -\frac{1}{8}\) | M1 | Use of \(a + 84d\) with a calculated value of \(d\) |
| (b) \(-10\) | A1 |
## Question 9:
| Answer | Marks | Guidance |
|--------|-------|----------|
| (a)(i) $\frac{\cos\theta}{1-r} = \frac{1}{\cos\theta}$ | B1 | |
| (a)(i) $1 - r = \cos^2\theta$ leading to $r = 1 - \cos^2\theta$ | M1 | Eliminate fractions |
| (a)(i) $r = \sin^2\theta$ leading to 2nd term $= \cos\theta\sin^2\theta$ | A1 | AG |
| (a)(ii) $S_{12} = \frac{\cos\!\left(\frac{\pi}{3}\right)\!\left[1 - \left(\sin^2\!\left(\frac{\pi}{3}\right)\right)^{12}\right]}{1 - \sin^2\!\left(\frac{\pi}{3}\right)} = \frac{0.5\left[1-(0.75)^{12}\right]}{1-0.75}$ | M1 | Evidence of correct substitution, use of $S_n$ formula and attempt to evaluate |
| (a)(ii) $1.937$ | A1 | |
| (b) $[d =]\ \cos\theta\sin^2\theta - \cos\theta$ | M1 | Use of $d = u_2 - u_1$ |
| (b) $-\frac{1}{8}$ | A1 | |
| (b) $\text{[85th term ]} = \frac{1}{2} + 84 \times -\frac{1}{8}$ | M1 | Use of $a + 84d$ with a calculated value of $d$ |
| (b) $-10$ | A1 | |9 The first term of a progression is $\cos \theta$, where $0 < \theta < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item For the case where the progression is geometric, the sum to infinity is $\frac { 1 } { \cos \theta }$.
\begin{enumerate}[label=(\roman*)]
\item Show that the second term is $\cos \theta \sin ^ { 2 } \theta$.
\item Find the sum of the first 12 terms when $\theta = \frac { 1 } { 3 } \pi$, giving your answer correct to 4 significant figures.
\end{enumerate}\item For the case where the progression is arithmetic, the first two terms are again $\cos \theta$ and $\cos \theta \sin ^ { 2 } \theta$ respectively.
Find the 85 th term when $\theta = \frac { 1 } { 3 } \pi$.\\
\includegraphics[max width=\textwidth, alt={}, center]{54f3f051-e124-470d-87b5-8e25c35248a9-16_547_421_264_863}
The diagram shows a sector $A B C$ which is part of a circle of radius $a$. The points $D$ and $E$ lie on $A B$ and $A C$ respectively and are such that $A D = A E = k a$, where $k < 1$. The line $D E$ divides the sector into two regions which are equal in area.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q9 [9]}}