| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.2 This is a polar curves question requiring a sketch and area calculation. While polar coordinates are a Further Maths topic (making it inherently harder on an absolute scale), this is a relatively standard application: sketching r = sin(θ/4) involves plotting key points and recognizing the petal pattern, and the area integral ½∫r²dθ is a direct formula application. The main challenge is careful handling of the period (8π, but restricted to [0,2π)) and correct setup of integration limits, but no novel insight is required—it's a textbook-style polar area problem. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks |
|---|---|
| (ii) | Closed curve containing the pole, O |
| Answer | Marks |
|---|---|
| 2 | B1 |
| Answer | Marks |
|---|---|
| (iii) | VA x = 3 |
| Answer | Marks |
|---|---|
| 2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 4 | Mark Scheme | Syllabus |
| Cambridge Pre-U – May/June 2015 | 9795 | 01 |
| Answer | Marks |
|---|---|
| (iii) | 1 |
| Answer | Marks |
|---|---|
| 3 3 | B1 |
Question 4:
--- 4 (i)
(ii) ---
4 (i)
(ii) | Closed curve containing the pole, O
Cusp at the pole
Essentially all correct, including at least (r, θ ) = (1, π )
( )2π
A = 1 ∫sin2 1θ dθ Condone missing 1 until final A1
2 2 2
0
2π( )
use of double-angle identity A = 1 ∫ 1 −1cosθ dθ
2 2 2
0
1[ ]
correct integration. = θ −sinθ
4
1π
=
2 | B1
B1
B1
[3]
M1
M1
A1
A1
[4]
5 (i)
(ii)
(iii) | VA x = 3
2x2 +5x−25 2x(x−3)+11(x−3) [+8]
y= =
x−3 x−3
y = 2x + 11 oblique asymptote
( )
dy (x−3)(4x+5)− 2x2 +5x−25
for = (good attempt at the Quotient Rule)
dx (x−3)2
d 8 8
correct unsimplified OR 2x+11+ =2−
dx x−3 (x−3)2
solving quadratic equation 4x2 – 7x – 15 = 2x2 + 5x – 25
i.e. 2x2 – 12x + 10 = 0 OR (x – 3) 2 = 4
(1, 9) and (5, 25) (Give one A1 for both x’s correct without either y)
General shape (with asymptotes
and turning points in approximately correct places)
2x2 + 5x – 25 = 0 solved to find x-intercepts
x=−5 or 21
2 | B1
M1
A1
[3]
M1
A1
M1
A1 A1
[5]
B1
M1
A1
[3]
Page 4 | Mark Scheme | Syllabus | Paper
Cambridge Pre-U – May/June 2015 | 9795 | 01
6 (i)
(ii)
(iii) | 1
zn = cosnθ +i sinnθ and z – n = cosnθ −i sinnθ ⇒ zn + ≡2cosnθ
zn
15 1 1 1
z+ ≡z5+ +5z3+ +10z+
z z5 z3 z
Repeated use of result ⇒ 32cos5θ ≡2cos5θ +10cos3θ +20cosθ
⇒ 16cos5θ ≡cos5θ +5cos3θ +10cosθ Answer Given
16cos5θ =±cosθ Condone sign error here to allow for first A1
cosθ = 0 ⇒ θ = 1π or 3 π (both)
2 2
cosθ = 1 ⇒ θ = 1π or 5 π (both)
2 3 3
cosθ = −1 ⇒ θ = 2π or 4π (both)
2 3 3
but allow one A1 for θ = 1π and θ = 2π if both 2nd answers missing
3 3 | B1
[1]
M1 A1
M1
A1
[4]
M1
A1
A1
A1
[4]A curve has polar equation $r = \sin \frac{1}{4}\theta$ for $0 \leqslant \theta < 2\pi$.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve. [3]
\item Determine the area of the region enclosed by the curve. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2015 Q4 [7]}}