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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Pre-U Certificate
MARK SCHEME for the May/June 2011 question paper
for the guidance of teachers
9795 FURTHER MATHEMATICS
9795/02
Paper 2 (Further Application of Mathematics),
maximum raw mark 120
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes must be read in conjunction with the question papers and the report on the
examination.
• Cambridge will not enter into discussions or correspondence in connection with these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE,
Pre-U, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
Page 2
1
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
( X 1 + X 2 + X 3 ) − (Y1 + Y2 + Y3 + Y4 ) ~ N( (3 × 30 − 4 × 20), (3 × 9 + 4 × 4) )
Paper
02
B1,B1
[2]
i.e. N(10,43)
(ii)
z=
0 − 10
= −1.52(5)
(Accept 1.52 to 1.53)
43
P( X 1 + X 2 + X 3 > Y1 + Y2 + Y3 + Y4 ) = 0.936
2
(i)
–12.25
P(X ≤ 5) = e

12.25 2 12.25 3 12.25 4 12.25 5 
1 + 12.25 +

+
+
+
2!
3!
4!
5! 

= 0.0174
(ii)
z=
5.5 − 12.25
= −1.929
3.5
[3]
M1A1
[3]
M1A1
(Allow 0.027(0))
(iii) Result is not reliable (error is approximately 50%).
The mean is not large enough.
(i) (a)
A1
A1
P(X ≤ 5) (from tables) = 1 – 0.9731 = 0.0269
3
M1A1
156.88 + 123.97
= 12.21 (AG)
15 + 10 − 2
(b) x A = 30.7
x B = 33.4
ν = 23 t 23 (0.975) = 2.069
s2 =
A1
[3]
B1
B1
[2]
M1A1
[2]
B1
B1B1
95% confidence limits are:
(33.4–30.7) ± 2.069 × 3.494
1 1
+
15 10
95% confidence interval is: (–0.252, 5.65) (Accept –0.251 for LB)
(ii) Since 0 ∈ CI there is not enough evidence to suggest that the claim is valid.
© University of Cambridge International Examinations 2011
M1A1√
A1
[6]
M1A1√
[2]
Page 3
4
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
( )= ∫
EX
2
∞
0
x2
e
θ
−
x
θ
∞
 2 −x 
= − x e θ  +

 0
∫
Paper
02
M1
dx
∞
0
∞
x

− 
θ
= − 2 xθe  +

 0
Syllabus
9795
∫
2 xe
∞
0
−
2θe
M1A1
x
θ dx
−
M1
x
θ dx
∞
x
− 

= − 2θ 2 e θ 

0
2
= 2θ
(ii) E(T) = θ 2
[(
A1
2
2
E(T ) = E k X 1 + X 2 + ....... + X n
[
2
2
2
)]
2
= k E ( X 1 ) + E( X 2 ) + ...... + E ( X n )
= knE ( X 2 )
⇒ θ 2 = 2knθ 2
1
⇒k=
2n
[5]
B1
M1
]
© University of Cambridge International Examinations 2011
A1
M1
A1
[5]
Page 4
5
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
P(X = r) = qr–1p ⇒ M (t ) =
∞
∑q
r −1
Syllabus
9795
pe tr
Paper
02
M1
1
(ii) (a)
(b)
=
p
(qe t ) r
∑
q 1
=
p
qe t
pe t
×
=
q 1 − qe t 1 − qe t
∞
A1
A1
(1 − qe t ) pe t + pe t .qe t
pe t
=
(1 − qe t ) 2
(1 − qe t ) 2
p
p
1
= 2 =
E( X ) = M ′(0) =
2
p
(1 − q)
p
M ′(t ) =
M ′′(t ) =
(1 − qe t ) 2 . pe t + 2 pe t (1 − qe t )qe t
(1 − qe t ) 4
p 3 + 2 p 2 q p + 2q 1 + q
=
= 2
(acf)
p4
p2
p
1+ q 1
q
1− p
Var ( X ) = 2 − 2 = 2
or
(OE)
p
p
p
p2
E ( X 2 ) = M ′′(0) =
(iii)
µ =6
(
and
σ 2 = 30
)
[3]
(AG)
(or σ = 30 )
P Y − 6 < 30 = P(Y ≤ 11)
11
5
= 1 −   = 0.865
6
© University of Cambridge International Examinations 2011
M1
M1A1
M1
A1
M1A1
[7]
B1
M1
A1
[3]
Page 5
6
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
F( x) =
∫
x
2x
 2x 
dx =   =
0π
π
 0 π
x2
Syllabus
9795
π

0 < x < 
2

2 sin −1 y
π
2

dG 
⇒ g( y) =
= π 1− y2
dy 
0
⇒ G( y) =
B1B1
B1B1
0 < y <1
0 < y <1
Paper
02
M1A1
[6]
(AG)
Otherwise
OR
 π
For X ~ U  0,  and y = sin x
 2
∫
π
2
0
B1
B1
2
dx = 1
π
x = 0 ⇒ y = 0 and x =
π
⇒ y =1
2
dy
dx
1
1
= cos x ⇒
=
=
dx
dy cos x
1− y2
∫
(b)
∫
A1
1
2
1
.
dy = 1
0π
1− y2
2


f( y ) = π 1 − y 2
0

(ii) (a)
M1A1
M
0
2
π 1− y
2
dy =
0 < y <1
Otherwise
=−
∫
M1
M1A1
(CAO)
y
2
.
dy
0π
1− y2
M1A1
[5]
M1
1
1
2
2
1− y2  =
 0 π
π 
[6]
(AG)
1
2
2
1
arcsin M =
π
2
1
π 
M = sin   =
2
4
E(Y ) =
A1
(CAO)
© University of Cambridge International Examinations 2011
M1A1
[3]
Page 6
7
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
(i) Uses conservation of energy:
0.3 × 10 × 1.5(cos 40 0 − cos 75 0 ) =
⇒ v = 3.90 (AG)
Syllabus
9795
1
× 0.3v 2 where speed is v ms–1.
2
v 2 15.216...
=
= 10.1(4) ms–2
r
1.5
Magnitude of acceleration =
8
(=2.425)
(b) Uses impulse = change in momentum.
0
Im = 0.3(7 sin 60 + 0.4 × 7 sin 60) = 2.55 Ns
(ii)
Magnitude of velocity =
(7 cos 60 0 ) 2 + (0.4 × 7 sin 60 0 ) 2 = 4.26 ms–1.
 0.4 × 7 sin 60 0
0
 7 cos 60
Direction = tan −1 
9
(i)
(ii)
[3]
B1
B1
(6.428 2 + 10.14 2 ) = 12.0 ms–2.
(i) (a) Component of velocity perpendicular to wall = 0.4 × 7 sin 60 0
M1A1
A1
(g = 9.8 gets M1A1A0).
(ii) Transverse component = − g sin 40 0 = −6.428 = −6.43 ms–2 (Ignore sign).
Radial component =
Paper
02

 = 34.7 0

v A = (10 sin 42 o )i + (10 cos 42 o ) j and v B = 15i
o
o
(Accept –8.31i + 7.43j).
A v B = (10 sin 42 − 15)i + (10 cos 42 ) j
rA = (10 sin 42 o )ti + (10 cos 42 o )tj and rB = 15ti + 13 j
o
o
A rB = (10 sin 42 − 15)ti + (10 cos 42 t − 13) j
B1√
[3]
M1A1
[2]
M1
A1
[2]
M1A1
M1A1
[4]
M1A1
[2]
M1A1
[2]
(Accept –8.31ti + [7.43t – 13]j).
(iii) By considering relative motion, one object is taken to be at rest and the other B1
moving relative to it. Hence, when closest together, the relative position vector
is perpendicular to the relative velocity vector; (i.e. dot product = 0).
(iv)
r . v B = (10 sin 42 o − 15) 2 t + 10 cos 42 o (10 cos 42 o t − 13) = 0
⇒ (100 sin 2 42 o − 300 sin 42 o + 225)t + 100 cos 2 42 o t − 130 cos 42 o = 0
M1A1
130 cos 42 o
⇒t =
= 0.777 or 47 minutes (i.e. 1247)
(325 − 300 sin 42 o )
M1A1
A B A
© University of Cambridge International Examinations 2011
[1]
[4]
Page 7
10
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
20 x
, where T1 is the tension in AP (or BP)
2
10(3 − x)
T2 =
, where T2 is the tension in the other string.
1
10 x = 30 − 10 x ⇒ x = 1.5
T1 =
So AP is 3.5 metres.
Paper
02
M1A1
A1
A1
[4]
(ii) Apply Newton II:
10(1.5 − y ) 20(1.5 + y )
−
= 0.5 &y&
1
2
⇒ &y& = −40 y
( ⇒ ϖ = 40 )
M1A1
(AG)
(iii) Use y = a cosϖt
[3]
M1A1
− 0.5 = cos( 40t ) (a= 1)
π
2π
= 40t ⇒ t =
(= 0.331)
3
3 10
11
A1
M1A1
[4]
(i) Take x and y axes along and perpendicular to plane.
y = V sin θ t − 0.5 g cos α t 2
2V sin θ
y = 0 on landing ⇒ t =
g cos α
M1
M1A1
 2V sin θ  1
 2V sin θ 
 − g sin α 

x = v cos θ 
 g cos α  2
 g cos α 
2V 2 sin θ
⇒x=
(cos θ cos α − sin θ sin α )
g cos 2 α
2
M1A1
(up plane).
A1
[6]
B1
[1]
(AG)
(ii)
2V 2 sin θ
(cos θ cos α + sin θ sin α )
g cos 2 α
(iii)
4(cos θ cos α − sin θ sin α ) = cos θ cos α + sin θ sin α
⇒ 3 cos θ cos α = 5 sin θ sin α
3
⇒
= 5 tan θ
tan α
⇒ tan θ = 1.2 ⇒ θ = 50.2 0 (Allow tan −1 1.2 or 0.876 rad.)
(down plane).
© University of Cambridge International Examinations 2011
M1A1√
M1
A1A1
[5]
Page 8
12
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
Paper
02
B1
400000
N
v
M1A1
250000 × 10
20000 +
= 25000 (No penalty for g=9.8
500
Driving force at speed v is
Resistive forces total
here.)
At maximum speed these are equal
400000
= 25000 ⇒ v = 16 ms–1. (Allow 16.1 from g= 9.8)
v
M1A1
[5]
(ii) Apply Newton II
400000
dv
16
dv
− 25000 = 250000v ⇒
− 1 = 10v
v
dx
dx
v
⇒
M1A1
1
v2
1
256 

dx =
dv ⇒
dx =  − v − 16 +
dv
10
16 − v
10
16 − v 

∫
∫
∫
M1A1√
∫
x
v2
= − − 16v − 256 ln 16 − v + c
10
2
v = 6 when x = 0 ⇒ c = 114 + 256 ln 10
M1
⇒
(703.5)
(acf) (Or use of M1A1
limits)
⇒
x
5
= 114 + 256 ln 10 − 72 − 192 − 256 ln 4 = 256 ln  − 150 when
10
2
M1
v = 12.
⇒ x = 846 (AWRT)
© University of Cambridge International Examinations 2011
A1
[9]