ADVANCED GCE
4754A
MATHEMATICS (MEI)
Applications of Advanced Mathematics (C4) Paper A
Candidates answer on the Answer Booklet
OCR Supplied Materials:
•
8 page Answer Booklet
•
Graph paper
•
MEI Examination Formulae and Tables (MF2)
Other Materials Required:
None
Monday 1 June 2009
Morning
Duration: 1 hour 30 minutes
*4754A*
*
4
7
5
4
A
*
INSTRUCTIONS TO CANDIDATES
•
•
•
•
•
•
•
Write your name clearly in capital letters, your Centre Number and Candidate Number in the spaces provided
on the Answer Booklet.
Use black ink. Pencil may be used for graphs and diagrams only.
Read each question carefully and make sure that you know what you have to do before starting your answer.
Answer all the questions.
Do not write in the bar codes.
You are permitted to use a graphical calculator in this paper.
Final answers should be given to a degree of accuracy appropriate to the context.
INFORMATION FOR CANDIDATES
•
•
•
•
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised that an answer may receive no marks unless you show sufficient detail of the working to
indicate that a correct method is being used.
The total number of marks for this paper is 72.
This document consists of 4 pages. Any blank pages are indicated.
NOTE
•
This paper will be followed by Paper B: Comprehension.
© OCR 2009 [T/102/2653]
2R–9B09
OCR is an exempt Charity
Turn over
2
Section A (36 marks)
1
Express 4 cos θ − sin θ in the form R cos(θ + α ), where R > 0 and 0 < α < 12 π .
Hence solve the equation 4 cos θ − sin θ = 3, for 0 ≤ θ ≤ 2π .
[7]
2
Using partial fractions, find ä
[7]
3
A curve satisfies the differential equation
x
dx.
(x + 1)(2x + 1)
terms of x.
4
dy
= 3x2 y, and passes through the point (1, 1). Find y in
dx
[4]
The part of the curve y = 4 − x2 that is above the x-axis is rotated about the y-axis. This is shown in
Fig. 4.
Find the volume of revolution produced, giving your answer in terms of π .
[5]
y
x
Fig. 4
5
A curve has parametric equations
x = at3 ,
y=
where a is a constant.
Show that
a
,
1 + t2
−2
dy
=
.
dx 3t(1 + t2 )2
Hence find the gradient of the curve at the point (a, 12 a).
6
[7]
Given that cosec2 θ − cot θ = 3, show that cot2 θ − cot θ − 2 = 0.
Hence solve the equation cosec2 θ − cot θ = 3 for 0◦ ≤ θ ≤ 180◦ .
© OCR 2009
4754A Jun09
[6]
3
Section B (36 marks)
7
When a light ray passes from air to glass, it is deflected through an angle. The light ray ABC starts
at point A (1, 2, 2), and enters a glass object at point B (0, 0, 2). The surface of the glass object is a
plane with normal vector n. Fig. 7 shows a cross-section of the glass object in the plane of the light
ray and n.
n
B q
f
C
A
Fig. 7
−−→
(i) Find the vector AB and a vector equation of the line AB.
[2]
(ii) Write down the normal vector n, and hence calculate θ , giving your answer in degrees.
[5]
The surface of the glass object is a plane with equation x + ß = 2. AB makes an acute angle θ with the
normal to this plane.
The line BC has vector equation r =
normal to the plane.
(iii) Show that φ = 45◦ .
0
0! + µ
2
−2
−2 !. This line makes an acute angle φ with the
−1
[3]
(iv) Snell’s Law states that sin θ = k sin φ , where k is a constant called the refractive index. Find k.
[2]
The light ray leaves the glass object through a plane with equation x + ß = −1. Units are centimetres.
(v) Find the point of intersection of the line BC with the plane x + ß = −1. Hence find the distance
the light ray travels through the glass object.
[5]
[Question 8 is printed overleaf.]
Copyright Information
OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders
whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright
Acknowledgements Booklet. This is produced for each series of examinations, is given to all schools that receive assessment material and is freely available to download from our public
website (www.ocr.org.uk) after the live examination series.
If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1PB.
OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department
of the University of Cambridge.
© OCR 2009
4754A Jun09
4
8
Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain
approximations for π .
(i) Fig. 8.1 shows a regular 12-sided polygon inscribed in a circle of radius 1 unit, centre O. AB
is one of the sides of the polygon. C is the midpoint of AB. Archimedes used the fact that the
circumference of the circle is greater than the perimeter of this polygon.
A C B
1
1
O
Fig. 8.1
(A) Show that AB = 2 sin 15◦ .
[2]
(B) Use a double angle formula topexpress cos 30◦ in terms of sin 15◦ . Using the exact value of
√
[4]
cos 30◦ , show that sin 15◦ = 12 2 − 3.
(C ) Use this result to find an exact expression for the perimeter of the polygon.
p
√
Hence show that π > 6 2 − 3.
[2]
(ii) In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each
side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of
the circle is less than the perimeter of this polygon.
D F E
1
O
Fig. 8.2
(A) Show that DE = 2 tan 15◦ .
[2]
(B) Let t = tan 15◦ . Use a double angle formula to express tan 30◦ in terms of t.
√
Hence show that t2 + 2 3 t − 1 = 0.
√
(C ) Solve this equation, and hence show that π < 12 2 − 3.
[3]
[4]
(iii) Use the results in parts (i)(C ) and (ii)(C) to establish upper and lower bounds for the value of π ,
giving your answers in decimal form.
[2]
© OCR 2009
4754A Jun09
4754
Mark Scheme
June 2009
4754 (C4) Applications of Advanced Mathematics
Section A
⇒
⇒
4 cos θ − sin θ = R cos(θ + α )
= R cos θ cos α − R sin θ sin α
⇒ R cos α = 4, R sin α = 1
⇒ R2 = 12 + 42 = 17, R = √17 = 4.123
tan α = ¼
⇒ α = 0.245
√17 cos(θ + 0.245) = 3
cos(θ + 0.245) = 3/√17
θ + 0.245 = 0.756, 5.527
⇒
θ = 0.511, 5.282
1
x
A
B
=
+
( x + 1)(2 x + 1) x + 1 (2 x + 1)
⇒
x = A(2x + 1) + B(x + 1)
x = –1 ⇒ –1 = –A ⇒ A = 1
x = – ½ ⇒ – ½ = ½ B ⇒ B = –1
⇒
x
1
1
=
−
( x + 1)(2 x + 1) x + 1 (2 x + 1)
x
1
1
∫ ( x + 1)(2 x + 1) dx = ∫ x + 1 − (2 x + 1) dx
= ln(x + 1) – ½ ln(2x + 1) + c
3
dy
= 3x 2 y
dx
⇒
∫
⇒
⇒
ln y = x3 + c
When x = 1, y = 1, ⇒ ln 1 = 1 + c ⇒ c = –1
ln y = x3 – 1
⇒
y = ex
4
When x = 0, y = 4
⇒
dy
= 3x 2 d x
y ∫
3
−1
4
V = π ∫ x dy
2
correct pairs
B1
M1
A1
R = √17 = 4.123
tan α = ¼ o.e.
α = 0.245
M1
θ + 0.245 = arcos 3/√17
A1A1
[7]
2
⇒
M1
ft their R, α for method
(penalise extra solutions in the range (-1))
M1
M1
A1
A1
correct partial fractions
substituting, equating coeffts or cover-up
A=1
B = –1
B1
B1
A1
[7]
ln(x + 1) ft their A
-½ ln(2x + 1) ft their B
cao – must have c
M1
separating variables
A1
B1
condone absence of c
c = –1 oe
A1
[4]
o.e.
B1
M1
must have integral, π, x² and dy soi
0
4
= π ∫ (4 − y )dy
M1
0
4
1 ⎤
⎡
= π ⎢4 y − y 2 ⎥
2 ⎦0
⎣
= π(16 – 8) = 8π
B1
A1
[5]
13
must have π ,their (4-y), their numerical y
limits
1 2⎤
⎡
⎢⎣ 4 y − 2 y ⎥⎦
4754
5
⇒
⇒
6
⇒
⇒
⇒
⇒
Mark Scheme
dy
= −a(1 + t 2 )−2 .2t
dt
dx
= 3at 2
dt
dy dy / dt
−2at
=
=
dx dx / dt 3at 2 (1 + t 2 )2
−2
*
=
3t (1 + t 2 ) 2
At (a, ½ a) , t = 1
gradient = −2 = –1/6
3 × 22
cosec 2 θ = 1 + cot 2 θ
1 + cot2θ – cot θ = 3 *
cot2 θ – cot θ – 2 = 0
(cot θ – 2)(cot θ + 1) = 0
cot θ = 2, tan θ = ½ , θ = 26.57°
cot θ = –1, tan θ = –1, θ = 135°
June 2009
M1
A1
B1
(1+t²)
M1
ft
−2
× kt for method
E1
M1
A1
[7]
finding t
E1
clear use of 1+cot²θ =cosec²θ
M1
A1
M1
A1
A1
[6]
factorising or formula
roots 2, –1
cot = 1/tan used
θ = 26.57°
θ = 135°
(penalise extra solutions in the range (-1))
14
4754
Mark Scheme
June 2009
Section B
7(i)
(ii)
⎛ −1 ⎞
→
⎜ ⎟
AB = ⎜ −2 ⎟
⎜0 ⎟
⎝ ⎠
⎛0⎞
⎛1 ⎞
⎜ ⎟
⎜ ⎟
r = ⎜0⎟ + λ ⎜ 2⎟
⎜2⎟
⎜0⎟
⎝ ⎠
⎝ ⎠
⎛1 ⎞
⎜ ⎟
n = ⎜0⎟
⎜1 ⎟
⎝ ⎠
⎛1 ⎞ ⎛1 ⎞
⎜ ⎟⎜ ⎟
⎜ 0 ⎟ .⎜ 2 ⎟
⎜1 ⎟ ⎜ 0 ⎟
1
cos θ = ⎝ ⎠ ⎝ ⎠ =
2 5
10
⇒
θ = 71.57°
⎛ −1⎞ ⎛ −2 ⎞
⎜ ⎟⎜ ⎟
⎜ 0 ⎟ . ⎜ −2 ⎟
⎜ ⎟⎜ ⎟
(iii) cos φ = ⎝ −1⎠ ⎝ −1 ⎠ = 2 + 1 = 1
2 9
3 2
2
⇒
φ = 45° *
(iv) sin 71.57° = k sin 45°
⇒
k = sin 71.57° / sin 45° = 1.34
(v)
⇒
⇒
⇒
B1
B1
[2]
or equivalent alternative
B1
B1
M1
M1
A1
[5]
M1
A1
correct vectors (any multiples)
scalar product used
finding invcos of scalar product divided by
two modulae
72° or better
ft their n for method
±1/√2 oe exact
E1
[3]
M1
A1
[2]
ft on their 71.57°
oe
M1
soi
M1
A1
A1
subst in x+z = -1
B1
[5]
www dep on μ=1
⎛0⎞
⎛ −2 ⎞
⎜ ⎟
⎜ ⎟
r = ⎜ 0 ⎟ + μ ⎜ −2 ⎟
⎜ 2⎟
⎜ −1 ⎟
⎝ ⎠
⎝ ⎠
x= -2μ , z=2-μ
x + z = –1
–2μ + 2 – μ = –1
3 μ = 3, μ = 1
point of intersection is (–2, –2, 1)
distance travelled through glass
= distance between (0, 0, 2) and (–2, –2, 1)
= √(22 + 22 + 12) = 3 cm
15
4754
8(i)
Mark Scheme
(A)
⇒
⇒
⇒
⇒
⇒
cos 30° = 1 – 2 sin2 15°
cos 30° = √3/2
√3/2 = 1 – 2 sin2 15°
2 sin2 15° = 1 – √3/2 = (2 – √3)/2
sin215° = (2 – √3)/4
⇒
sin 15° =
(B)
(ii)
360°÷24=15°
CB/OB = sin 15°
CB = 1 sin 15°
AB = 2CB = 2 sin 15°*
M1
E1
[2]
B1
M1
M1
(A) tan 15° = FE/OF
⇒
FE = tan 15°
⇒
DE = 2FE = 2tan 15°
M1
E1
[2]
E1
[2]
(B) tan 30 = 2 tan15 = 2t
1 − tan 2 15 1 − t 2
tan 30 = 1/√3
2t
1
⇒
=
⇒ 2 3t = 1 − t 2
1− t2
3
⇒
t2 + 2√3 t – 1 = 0 *
B1
M1
E1
[3]
−2 3 ± 12 + 4
= 2− 3
2
⇒
⇒
circumference < perimeter
2π < 24(2 – √3)
π < 12(2 – √3) *
simplifying
E1
[4]
(C) Perimeter = 12 × AB = 24 × ½ √(2 – √3)
= 12√(2 – √3)
circumference of circle > perimeter of polygon
⇒
2π > 12√(2 – √3)
⇒
π > 6√(2 – √3)
t=
AB=2AC or 2CB
∠AOC= 15°
oe
B1
2− 3 1
=
2− 3 *
4
2
(C)
June 2009
M1 A1
(iii) 6√(2 – √3) < π < 12(2 – √3)
⇒
3.106 < π < 3.215
16
M1
using positive root
E1
[4]
from exact working
B1 B1
[2]
3.106, 3.215