| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure with Technology (Further Pure with Technology) |
| Session | Specimen |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Arc length of polar curve |
| Difficulty | Challenging +1.8 This is a substantial multi-part Further Maths question combining polar curves, arc length integration, parametric differentiation, and surprisingly number theory. While individual components are standard (sketching polar curves, arc length formula, parametric chain rule), the extended nature, requirement to prove general results for a family of curves, and the programming/Wilson's theorem section make this significantly harder than typical A-level questions. The conceptual demand and multi-step reasoning across diverse topics justify placing this well above average. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve8.02l Fermat's little theorem: both forms8.02m Order of a modulo p: p-1 not necessarily least such n |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) | (A) |
| Answer | Marks |
|---|---|
| E | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks |
|---|---|
| I | N |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) | (B) |
| Curves pass through (0, 0) and (1,0) | B1 | |
| [1] | 1.1 | |
| 1 | (i) | (C) |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. Curves have cusps at the pole. | B1 | |
| [1] | 2.2b | B1 for one common feature |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (ii) | (A) |
| Answer | Marks |
|---|---|
| 0 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 1.1a | Correct formula seen |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (ii) | (B) |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (iii) | (A) |
| Answer | Marks |
|---|---|
| (cid:169)n(cid:185) (cid:169)n(cid:185) | B1 |
| [1] | I |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (iii) | (B) |
| Answer | Marks |
|---|---|
| AG | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks |
|---|---|
| I | Evidence of use of derivative |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (iv) | (cid:11) (cid:12) (cid:11) (cid:12) |
| Answer | Marks |
|---|---|
| parallel to the y-axis. | M1 |
| Answer | Marks |
|---|---|
| [6] | 3.1a |
| Answer | Marks |
|---|---|
| 2.4 | Set denominator of dy/dx=0 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (v) | E |
| Answer | Marks |
|---|---|
| no errors | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
Question 1:
1 | (i) | (A) | n = 2
n = 4
E | B1
B1
C
[2] | 1.1
1.1
I | N
E
M
1 | (i) | (B) | P
Curves pass through (0, 0) and (1,0) | B1
[1] | 1.1
1 | (i) | (C) | S
e.g. Curves are bounded
e.g. Curves are symmetrical about the line (cid:84)(cid:32)0
e.g. Curves have cusps at the pole. | B1
[1] | 2.2b | B1 for one common feature
Accept “Curves are symmetrical about the x-
axis”
1 | (ii) | (A) | 4(cid:83) (cid:167) dr (cid:183) 2
Length = (cid:179) (cid:168) (cid:184) (cid:14)r2d(cid:84)
(cid:169)d(cid:84)(cid:185)
0 | M1
M1
[2] | 1.2
1.1a | Correct formula seen
Correct limits seen
1 | (ii) | (B) | dr (cid:84) (cid:84)
(cid:32)(cid:16)sin cos3
d(cid:84) 4 4
16
(cid:32)
3 | A1
A1
[2] | 1.1
1.1 | N
Esoi
M
1 | (iii) | (A) | (cid:167) t (cid:183) (cid:167) t (cid:183)
x(cid:32)cosn cost(cid:15)y(cid:32)cosn sint
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)n(cid:185) (cid:169)n(cid:185) | B1
[1] | I
1.1
1 | (iii) | (B) | dy
dy dt
(cid:32)
dx dx
dt
(cid:167) t (cid:183)(cid:167) 1 (cid:167) t (cid:183)(cid:183) (cid:167) t (cid:183)
ncosn(cid:16)1 (cid:16) sin sint(cid:14)cosn cost
(cid:168) (cid:184)(cid:168) (cid:168) (cid:184)(cid:184) (cid:168) (cid:184)
(cid:169)n(cid:185)(cid:169) n (cid:169)n(cid:185)(cid:185) (cid:169)n(cid:185)
(cid:32)
(cid:167) t (cid:183)(cid:167) 1 (cid:167) t (cid:183)(cid:183) (cid:167) t (cid:183)
ncosn(cid:16)1 (cid:16) sin cost(cid:16)cosn sint
(cid:168) (cid:184)(cid:168) (cid:168) (cid:184)(cid:184) (cid:168) (cid:184)
(cid:169)n(cid:185)(cid:169) n (cid:169)n(cid:185)(cid:185) (cid:169)n(cid:185)
(cid:167) t (cid:183)(cid:11) (cid:11)t (cid:12) (cid:11)t (cid:12)(cid:12)
(cid:16)cosn(cid:16)1 sintsin (cid:16)costcos
(cid:168) (cid:184)
(cid:169)n(cid:185) n n
(cid:32)
(cid:167) t (cid:183)(cid:11) (cid:11)t (cid:12) (cid:11)t (cid:12)(cid:12)
(cid:16)cosn(cid:16)1 sintcos (cid:14)costsin
(cid:168) (cid:184)
(cid:169)n(cid:185) n n
(cid:11)t (cid:12) (cid:11)t (cid:12)
sintsin (cid:16)costcos
n n E
(cid:32)
(cid:11)t (cid:12) (cid:11)t (cid:12)
sintcos (cid:14)costsin
n n
AG | M1
A1
C
[2] | 1.1
2.1
I | Evidence of use of derivative
N
E
M
1 | (iv) | (cid:11) (cid:12) (cid:11) (cid:12)
t t
sintcos (cid:14)costsin (cid:32)0
n n
(cid:167)n(cid:14)1 (cid:183)
sin t (cid:32)0
(cid:168) (cid:184)
(cid:169) n (cid:185)
n(cid:14)1
t (cid:32)0,(cid:83),2(cid:83),3(cid:83)...
n
In the range 0(cid:100)t(cid:31)n(cid:83)roots are:
n 2n 3n n2
t (cid:32)0, (cid:83), (cid:83), (cid:83),..., (cid:83)
n(cid:14)1 n(cid:14)1 n(cid:14)1 n(cid:14)1
(cid:167)n(cid:14)1 (cid:183)
Numerator: (cid:16)cos (cid:168) t (cid:184) ≠0 for any of these values
(cid:169) n (cid:185)
therefore all n+1 are points where the tangent is
parallel to the y-axis. | M1
M1
M1
M1
A1
E1
C
[6] | 3.1a
1.1
1.1
3.1a
2.1
I
2.4 | Set denominator of dy/dx=0
N
Or equivalent expression.
E
M
dy (cid:16)1
Alternative based on (cid:32) is
dx tant
acceptable
1 | (v) | E
P
States looking for 5 and zooms in on origin
Shows 5 points clearly with appropriate sketches and
no errors | M1
A1
[2] | 1.1
2.41 A family of curves has polar equation $r = \cos n \left( \frac { \theta } { n } \right) , 0 \leq \theta < n \pi$, where $n$ is a positive even integer.
\begin{enumerate}[label=(\roman*)]
\item (A) Sketch the curve for the cases $n = 2$ and $n = 4$.\\
(B) State two points which lie on every curve in the family.\\
(C) State one other feature common to all the curves.
\item (A) Write down an integral for the length of the curve for the case $n = 4$.\\
(B) Evaluate the integral.
\item (A) Using $t = \theta$ as the parameter, find a parametric form of the equation of the family of curves.\\
(B) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sin t \sin \left( \frac { t } { n } \right) - \cos t \cos \left( \frac { t } { n } \right) } { \sin t \cos \left( \frac { t } { n } \right) + \cos t \sin \left( \frac { t } { n } \right) }$.
\item Hence show that there are $n + 1$ points where the tangent to the curve is parallel to the $y$-axis.
\item By referring to appropriate sketches, show that the result in part (iv) is true in the case $n = 4$.
\item (A) Create a program to find all the solutions to $x ^ { 2 } \equiv - 1 ( \bmod p )$ where $0 \leq x < p$. Write out your program in full in the Printed Answer Booklet.\\
(B) Use the program to find the solutions to $x ^ { 2 } \equiv - 1 ( \bmod p )$ for the primes
\begin{itemize}
\item $p = 809$,
\item $p = 811$ and
\item $p = 444001$.
\item State Wilson's Theorem.
\item The following argument shows that $( 4 k ) ! \equiv ( ( 2 k ) ! ) ^ { 2 } ( \bmod p )$ for the case $p = 4 k + 1$.
\end{itemize}
$$\begin{aligned}
( 4 k ) ! & \equiv 1 \times 2 \times 3 \times \ldots \times ( 2 k - 1 ) \times 2 k \times ( 2 k + 1 ) \times ( 2 k + 2 ) \times \ldots \times ( 4 k - 1 ) \times 4 k ( \bmod p ) \\
& \equiv 1 \times 2 \times 3 \times \ldots \times ( 2 k - 1 ) \times 2 k \times ( - 2 k ) \times ( - ( 2 k - 1 ) ) \times \ldots \times ( - 2 ) \times ( - 1 ) ( \bmod p ) \\
& \equiv ( ( 2 k ) ! ) ^ { 2 } ( \bmod p )
\end{aligned}$$
(A) Explain why ( $2 k + 2$ ) can be written as ( $- ( 2 k - 1 )$ ) in line ( 2 ).\\
(B) Explain how line (3) has been obtained.\\
(C) Explain why, if $p$ is a prime of the form $p = 4 k + 1$, then $x ^ { 2 } \equiv - 1 ( \bmod p )$ will have at least one solution.\\
(D) Hence find a solution of $x ^ { 2 } \equiv - 1 ( \bmod 29 )$.
\item (A) Create a program that will find all the positive integers $n$, where $n < 1000$, such that $( n - 1 ) ! \equiv - 1 \left( \bmod n ^ { 2 } \right)$. Write out your program in full.\\
(B) State the values of $n$ obtained.\\
(C) A Wilson prime is a prime $p$ such that $( p - 1 ) ! \equiv - 1 \left( \bmod p ^ { 2 } \right)$. Write down all the Wilson primes $p$ where $p < 1000$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure with Technology Q1 [19]}}