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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Pre-U Certificate
MARK SCHEME for the May/June 2012 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 2012 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 2012
M X (t ) =
=k
=
(ii)
∫
∞
0
∫
∞
0
Syllabus
9795
e tx ke −kx dx (Limits Required)
e ( t − k ) x dx = k
[
− k −(k −t ) x
e
k −t
]
∞
0
=
∫
∞
0
e − ( k − t ) x dx (Limits not required)
k
k −t
(i)
M1A1
M1A1
A1
[5]
−1
1
2
t t2
 t
M X (t ) = 1 −  = 1 + + 2 + ... = 1 + t + 2 t 2 + ...
k
k k
k
 k
1
E( X ) =
k
2
M1
[3]
k
1
′
⇒ E ( X ) = M X (0) =
2
k
(k − t )
2k
2
″
″
M X (t ) =
⇒ E ( X 2 ) = M X (0) = 2
3
(k − t )
k
(A1 ft if double sign error when differentiating twice, but CA0)
E( X 2 ) =
M1
A1
(AG)
′
M X (t ) =
Alternatively:
Paper
02
M1A1
A1
2
2
2 1
1
⇒ Var( X ) = 2 −   = 2
2
k
k
k
k
E (aX + bY ) = µ
E (aX + bY ) = aE ( X ) + bE (Y )
⇒ aµ + 3bµ = µ ⇒ a + 3b = 1
M1A1
[5]
[8]
M1
M1
A1
[3]
(ii)
(iii)
σ2
σ2
+ 4b 2
n
n
2
2
2
σ
σ
σ
=
(a 2 + 4b 2 ) =
(1 − 6b + 9b 2 + 4b 2 ) =
(1 − 6b + 13b 2 ) (AG)
n
n
n
Var (aX + bY ) = a 2 Var ( X ) + b 2 Var (Y ) = a 2
d
3
Var (aX + bY ) = −6 + 26b = 0 ⇒ b =
db
13
2
σ 
3
9  4σ 2
⇒ Varmin (aX + bY ) =
1 − 6 × + 13 ×  =
n 
13
13  13n
© University of Cambridge International Examinations 2012
M1
M1A1
[3]
M1A1
A1
[3]
[9]
Page 3
3
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
9795
Paper
02
x = 1.675
B1
99% confidence limits are 1.675 ± 2.576 ×
M1A1
0.1
(ft on wrong mean)
6
99% confidence interval is (1.57, 1.78) (AWRT)
A1
[4]
(ii)
sn = 0.09215 or sn–1 = 0.1009
v = 5 ⇒ t5 (0.99) = 4.032
B1
B1
99% confidence limits are 1.675 ± 4.032 ×
M1A1
0.1009
0.09215
or 1.675 ± 4.032 ×
6
5
99% confidence interval is (1.51, 1.84) (AWRT)
A1
[5]
(iii)
4
(i)
Sensible comment referring to the fact that 1.8 is outside the 1st interval but
inside 2nd. (ft on their CI’s.)
P( X = r ) = e −λ
∞
λr
r!
∞
0
(ii)
(iii)
5
(i)
(a)
−λ
0
(λt ) r
= e −λ e λt = e λ (t -1)
r!
G X +Y (t ) = e λ ( t −1) .e µ ( t −1) = e ( λ + µ )(t −1)
(AG)
⇒ ( X + Y ) ~ Po(λ + µ)
1
4
np = 100 × = 20 and npq = 20 × = 16
5
5
Standardisation in either (a) or (b)
14.5 − 20
= −1.375 ⇒ P(≥ 15) = 0.915
4
[10]
M1A1
[3]
M1A1
[2]
P( [0, 4] or [1, 3] or [2, 2] ) = 0.2231 × 0.1336 + 0.3347 × 0.2138 + 0.2510 ×
0.2565
0.1657
P(X≤2 | X + Y = 4 ) =
0.1954
= 0.848 (AWRT)
(S.C. If no marks earned in (iii), award B1 for P(X ≤ 2 ) = 0.1657 if seen.)
z=
[1]
B1
(may be implied by next line.)
G X (t ) = ∑ p r t = ∑ e
r
B1
B2,1,0
M1A1
A1
[5]
[10]
B1
M1
(AWFW [0.915,0.916])
A1 A1
(ft on variance of 20)
(b)
z=
12.5 − 20
= −1.875 ⇒ P(<12) = 1 – 0.9696 = 0.0304
4
A1A1
[6]
(AWFW [0.0303,0.0304])
(ii)
mean = variance = 36 ⇒ S.D. = 6
z = 1.645
1

 N +  − 36
1
2


> 1.645 (Allow working with equality, but must be  N +  .)
6
2

⇒ N > 45.37 ∴ least N = 46
B1
B1
M1A1
A1
[5]
© University of Cambridge International Examinations 2012
[11]
Page 4
6
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
9795
Above x-axis between (0, 0) to (3, 0)
Correct concavity. (Do not condone parabolas.)
Paper
02
B1
B1
[2]
(ii)
µ=
4
27
∫ (3x
3
0
3
)
− x 4 dx
(Limits required.)
M1
3
4  x4 x5 
=
3 −  = 1.8
27  4
5 0
4
f ′( x) =
(6 x − 3 x 2 ) = 0
27
A1A1
M1A1
⇒ x = 0, 2 ∴ Mode = 2
A1
[6]
(iii)
Mean less than mode in (ii) matches negative skew in sketch.
B1
[1]
(iv)
P( X − µ < σ ) =
4
27
∫ (3x
2.4
2
1.2
)
− x 3 dx
(i)
(ii)
= 0.64
A1A1
[3]
Tractive force = Resistance at maximum speed ⇒
75 3
dv
− v = 90
v 4
dt
F = ma ⇒
M1
2.4
4  3 x4 
=
x − 
27 
4  1.2
7
(Limits required.)
⇒
75
3
= 10k ⇒ k = . (AG)
10
4
25 1
dv
− v = 30
(AG) (3 terms
v 4
dt
B1
[1]
M1A1
[2]
req. for M1)
(iii)
t
120v
dv
3 100 − v 2
7
− 2v
t = −60
dv = − 60 ln 100 − v 2
3 100 − v 2
∫ dt = ∫
0
7
∫
[
 91 
= −60 ln 51 + 60 ln 91 = 60 ln 
 51 
[12]
M1
]
7
3
(Limits not required)
( = 34.7) seconds.
© University of Cambridge International Examinations 2012
M1A1
M1A1
[5]
[8]
Page 5
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
8
3
Distances
Syllabus
9795
x 2 = 32 + 82 − 2 × 3 × 8 cos135° ⇒ x = 10.341
sin θ sin 135°
=
⇒ θ = 11.837°
3
8
(θ as in distance diagram.)
Paper
02
M1A1
M1A1
135º
8
x
θ
180º – (45º + 11.837º) = 123.163º
sin α sin 123.163°
=
⇒ α = 33.923°
40
60
(α as in velocities diagram.)
Velocities
60
B1
M1A1
40
α
123.163º
8
Course is : (45º + 11.837º) – 33.923º = 22.914º
Course is 023º (Bearing notation not required.)
A1
[8]
[8]
[8]
[8]
Alternative solution:
Equate easterly and northerly displacements at time t hours, where θ is required
bearing:
Attempt at either equation
60 cos θ t = 4 2 + 40t
60 sin θ t = 4 2 + 3
Attempt to eliminate t.
2
(4 2 + 3)
3
8.656... cos θ − 5.656... sin θ = 5.7712....
M1
A1
A1
M1
(4 2 + 3) cos θ − 4 2 sinθ =
A1
Rearrange to obtain cos(θ + 33.16) = 0.5580…
A1
Solve: θ + 33.16 = 56.076… ⇒ θ = 22.9
Course is 023º (Bearing notation not required.)
M1
A1
Obtain e.g.
or
© University of Cambridge International Examinations 2012
Page 6
9
(i)
(ii)
(iii)
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
Syllabus
9795
Paper
02
T cos θ = mg (Must be seen to score in part (i).)
θ = 90° ⇒ mg = 0 ∴ AP can never be horizontal.
 mg 
∴ 0 < cos θ < 1 ⇒ T  =
 > mg
 cos θ 
B1
B1
T sin θ = ml sin θϖ 2 ⇒ T = ml ϖ 2
M1A1
B1
[3]
[2]
h
mgl
= mg ⇒ T =
l
h
mgl
⇒ mlϖ 2 =
h
2
⇒ω h = g.
T
B1
B1
B1
h = 0.5 ⇒ ϖ = 20 = 4.47 Angular speed is 4.47 rad/s.
B1
(Allow 4.43 from g = 9.8.)
[4]
10 (i)
(ii)
Let u denote speed of sphere Q before impact, v1 and v2 the speeds of spheres Q
and P, respectively, after impact and α the angle between Q’s initial direction of
motion and the line of centres.
After impact, if moving perpendicularly, Q moves perpendicular to line of
centres and P moves along line of centres. (Stated or implied.)
B1
CLM: mu cos α = 0 + 3mv2 or mux = 3mv
NEL: eu cos α = v2 or eux = v.
M1A1
A1
∴e =
A1
1
3
[9]
[5]
v1 = u sin α and v2 =
1
u cos α.
3
B1 (both)
Loss in KE is
1
1
1
u 2 cos 2 α
mu 2 − mu 2 sin 2 α − .3m
2
2
2
9
1
= mu 2 (Or remaining KE is 5/6 of initial KE etc.)
12
But cos2 α + sin2 α = 1 (used)
⇒ ... ⇒ sin 2 α =
3
4
⇒ sin α =
M1A1
A1
M1
3
⇒ α = 60º
2
© University of Cambridge International Examinations 2012
M1A1
[7]
[12]
Page 7
11 (i)
T=
Mark Scheme: Teachers’ version
Pre-U – May/June 2012
(
6mg 9a 2 + x 2 − a
a
Syllabus
9795
)
M1A1
Let θ be the angle between each string and line of motion of particle.
( 9a
12mg
mx = −2T cos θ = −
a
⇒ x = −
(ii)
12 gx 
a
1−

a 
9a 2 + x 2
∴ x ≈ (− 12 g + 4 g )
12 (i)
2
)
+ x2 − a ×




A1
[6]
a
8g
M1A1
or π
a
.
2g
A1
[3]
ga 8 g 2
a2
a
=
A ⇒ A2 =
⇒ A=
200 a
1600
40
where A is the amplitude.
vmax =ϖa ⇒
M1A1
[2]
x = 20 cos αt y = 20 sin αt – 5t2 (Allow g or 9.8 here for B mark).
y = 20 sin α ⋅
[11]
B1B1
2
x
x
x


− 5
(1 + tan 2 α )
 = x tan α −
20 cos α
20
cos
α
80


(AG)
x2 tan2 α – 80x tan α + x2 + 80y = 0 (Can be implied by what follows.)
Real roots ⇒ 6400 x 2 − 4 x 2 ( x 2 + 80 y ) ≥ 0
⇒ 1600 − x 2 − 80 y ≥ 0 ⇒ y ≤ 20 −
(iii)
A1A1
9a 2 + x 2
2
(ii)
M1
x
x
8g
=−
x
a
a
Which is SHM of period 2π
(iii)
Paper
02
x2
, (x ≠ 0).
80
M1A1
[4]
B1
M1A1
A1
[4]
x = R cos β and y = R sin β ⇒ y = x tan β .
In y = 20 −
∴
x2
R 2 (1 − sin 2 β )
⇒ R sin β = 20 −
80
80
M1A1
R 2 (1 − sin 2 β ) + 80 R sin β − 1600 = 0 ⇒ (R[1 − sin β ] + 40)(R[1 + sin β ] − 40) M1
⇒R=
12 (iii)
40
1 + sin β
(up)
or
− 40
1 − sin β
(down)
Alternative solution:
x2 + 80 tan βx – 1600 = 0
⇒ x = −40 tan β ± 40 sec β
− 40 sin β + 40 40(1 − sin β )
40
=
⇒ R up =
=
2
2
1 + sin β
cos β
1 − sin β
⇒ R down =
− 40 sin β − 40 40(1 + sin β )
40
=
=
2
2
+
1
sin β
cos β
1 − sin β
A1
[4]
B1
B1
B1
B1
[4]
© University of Cambridge International Examinations 2012
[12]