| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector perimeter calculation |
| Difficulty | Standard +0.3 Part (a) requires setting up an arithmetic progression with 6 terms summing to 2π, using the constraint that largest = 4×smallest, then calculating arc length and perimeter - this is a multi-step problem but uses standard techniques. Part (b) involves routine GP algebra using the common ratio property and sum to infinity formula. Both parts are slightly above average due to the setup required, but follow well-practiced procedures without requiring novel insight. |
| 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 |
|---|---|---|
| (a) \(a + 5d = 4a\) or \(\frac{(a + 4a)}{2} \times 6\) | B1 | co |
| \(\frac{6}{2}(2a + 5d)\) or \(\frac{(a + 4a)}{2} \times 6 = 360\) | M1 A1 | Correct left-hand side. All correct. |
| Sim Eqns \(a = 24°\) or \(\frac{2\pi}{15}\) rads | A1 | Either answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Perimeter \(= 12.1\) | M1 A1 [6] | Correct use of arc length with \(\theta\) in rads. co |
| Answer | Marks | Guidance |
|---|---|---|
| (NB stating \(a, ar, ar^2\) as \(f(k)\) gets M1) | M1 A1 A1 [3] | Correct eqn for \(k\). Co condone inclusion of \(k = -3\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\to S_{\infty} = 27 + \frac{1}{3} = 81\) | M1 A1 [2] | Correct formula for \(S_{\infty}\) must have \(-1 \leq r \leq 1\). co |
**(a)** $a + 5d = 4a$ or $\frac{(a + 4a)}{2} \times 6$ | B1 | co
$\frac{6}{2}(2a + 5d)$ or $\frac{(a + 4a)}{2} \times 6 = 360$ | M1 A1 | Correct left-hand side. All correct.
Sim Eqns $a = 24°$ or $\frac{2\pi}{15}$ rads | A1 | Either answer.
Arc length $= 50$
Perimeter $= 12.1$ | M1 A1 [6] | Correct use of arc length with $\theta$ in rads. co
**(b)(i)** $\frac{k + 6}{2k + 3} = \frac{k}{k + 6}$
$\to k^2 - 9k - 36 = 0 \to k = 12$
(NB stating $a, ar, ar^2$ as $f(k)$ gets M1) | M1 A1 A1 [3] | Correct eqn for $k$. Co condone inclusion of $k = -3$.
**(ii)** $r = \frac{2}{3}, a = 27$
$\to S_{\infty} = 27 + \frac{1}{3} = 81$ | M1 A1 [2] | Correct formula for $S_{\infty}$ must have $-1 \leq r \leq 1$. co\begin{enumerate}[label=(\alph*)]
\item A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector. [6]
\item The first, second and third terms of a geometric progression are $2k + 3$, $k + 6$ and $k$, respectively. Given that all the terms of the geometric progression are positive, calculate
\begin{enumerate}[label=(\roman*)]
\item the value of the constant $k$, [3]
\item the sum to infinity of the progression. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q10 [11]}}