| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Trigonometric arithmetic progression |
| Difficulty | Standard +0.3 This question requires finding the common difference of an AP (straightforward subtraction), then using the trigonometric identity tan²θ = sec²θ - 1 to simplify, followed by a standard nth term calculation with substitution of θ = π/6. While it combines trigonometric identities with sequences, the steps are routine and well-signposted, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((d =) -\frac{\tan^2\theta}{\cos^2\theta} - \frac{1}{\cos^2\theta}\) | B1 | Allow sign error(s). Award only at form \((d =)\)... stage |
| \(-\frac{\sin^2\theta}{\cos^4\theta} - \frac{1}{\cos^2\theta}\) or \(\frac{-\sec^2\theta}{\cos^2\theta}\) | M1 | Allow sign error(s). Can imply B1 |
| \(\frac{-\sin^2\theta - \cos^2\theta}{\cos^4\theta}\) or \(\frac{-\frac{1}{\cos^2\theta}}{\cos^2\theta}\) | M1 | |
| \(-\frac{1}{\cos^4\theta}\) | A1 | AG, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{4}{3}\), \(d = -\frac{16}{9}\) | B1 | SOI, both required. Allow \(a = \frac{1}{3} \cdot \frac{1}{4}\), \(d = -\frac{1}{9} \cdot \frac{1}{16}\) |
| \(u_{13} = \frac{1}{\cos^2\theta} - \frac{12}{\cos^4\theta} = \frac{4}{3} + 12\left(\frac{-16}{9}\right)\) | M1 | Use of correct formula with *their* \(a\) and *their* \(d\). The first 2 steps could be reversed |
| \(-20\) | A1 | WWW |
## Question 7(a):
| $(d =) -\frac{\tan^2\theta}{\cos^2\theta} - \frac{1}{\cos^2\theta}$ | B1 | Allow sign error(s). Award only at form $(d =)$... stage |
|---|---|---|
| $-\frac{\sin^2\theta}{\cos^4\theta} - \frac{1}{\cos^2\theta}$ or $\frac{-\sec^2\theta}{\cos^2\theta}$ | M1 | Allow sign error(s). Can imply B1 |
| $\frac{-\sin^2\theta - \cos^2\theta}{\cos^4\theta}$ or $\frac{-\frac{1}{\cos^2\theta}}{\cos^2\theta}$ | M1 | |
| $-\frac{1}{\cos^4\theta}$ | A1 | AG, WWW |
## Question 7(b):
| $a = \frac{4}{3}$, $d = -\frac{16}{9}$ | B1 | SOI, both required. Allow $a = \frac{1}{3} \cdot \frac{1}{4}$, $d = -\frac{1}{9} \cdot \frac{1}{16}$ |
|---|---|---|
| $u_{13} = \frac{1}{\cos^2\theta} - \frac{12}{\cos^4\theta} = \frac{4}{3} + 12\left(\frac{-16}{9}\right)$ | M1 | Use of correct formula with *their* $a$ and *their* $d$. The first 2 steps could be reversed |
| $-20$ | A1 | WWW |
---7 The first and second terms of an arithmetic progression are $\frac { 1 } { \cos ^ { 2 } \theta }$ and $- \frac { \tan ^ { 2 } \theta } { \cos ^ { 2 } \theta }$, respectively, where $0 < \theta < \frac { 1 } { 2 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Show that the common difference is $- \frac { 1 } { \cos ^ { 4 } \theta }$.
\item Find the exact value of the 13th term when $\theta = \frac { 1 } { 6 } \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q7 [7]}}