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Cambridge International Examinations
Cambridge Pre-U Certificate
FURTHER MATHEMATICS (PRINCIPAL)
Paper 2 Further Application of Mathematics
9795/02
For Examination from 2016
SPECIMEN MARK SCHEME
3 hours
MAXIMUM MARK: 120
The syllabus is approved for use in England, Wales and Northern Ireland as a Cambridge International Level 3 Pre-U Certificate.
This document consists of 7 printed pages and 1 blank page.
© UCLES 2013
[Turn over
2
Mark Scheme Notes
Marks are of the following three types:
M
Method mark, awarded for a valid method applied to the problem. Method marks are not lost for
numerical errors, algebraic slips or errors in units. However it is not usually sufficient for a candidate just
to indicate an intention of using some method or just to quote a formula; the formula or idea must be
applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula.
Correct application of a formula without the formula being quoted obviously earns the M mark and in
some cases an M mark can be implied from a correct answer.
A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks
cannot be given unless the associated method mark is earned (or implied).
B
Mark for a correct result or statement independent of method marks.
The following abbreviations may be used in a mark scheme:
AG
Answer Given on the question paper (so extra checking is needed to ensure that the detailed working
leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
aef Any equivalent form
art
Answers rounding to
cwo Correct working only (emphasising that there must be no incorrect working in the solution)
ft
Follow through from previous error is allowed
o.e. Or equivalent
© UCLES 2013
9795/02/SM/16
3
1
(i)
M X (t ) =
=k
=
(ii)
∫
∞
0
∫
∞
0
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
1
k
′
⇒ E ( X ) = M X ( 0) =
k
(k − t ) 2
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 CAO)
2
(i)
(ii)
(iii)
M1 A1
A1
M1 A1
A1
2
2
2 1
1
⇒ Var ( X ) = 2 −   = 2
2
k
k
k
k
M1 A1
E (aX + bY ) = µ
M1
E (aX + bY ) = aE ( X ) + bE (Y )
⇒ aµ + 3bµ = µ ⇒ a + 3b = 1
M1
A1
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
⇒ Varmin ( aX + bY ) =
© UCLES 2013
M1 A1
−1
t t2
1
2
 t
M X (t ) = 1 −  = 1 + + 2 + ... = 1 + t + 2 t 2 + ...
k k
k
k
 k
1
E( X ) =
k
E( X 2 ) =
M1
A1
AG
′
M X (t ) =
Alternatively:
M1
σ2 
3
9  4σ 2
1 − 6 × + 13 ×
=
n 
13
169  13n
9795/02/SM/16
M1
M1 A1
M1 A1
A1
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4
3
x = 1.675
(i)
B1
0.1
(ft on wrong mean)
6
99% confidence interval is (1.57, 1.78) art
99% confidence limits are 1.675 ± 2.576 ×
M1
A1ft
A1
(ii)
sn = 0.09215 or sn–1 = 0.1009
v = 5 ⇒ t5 (0.99) = 4.032
B1
B1
0.1009
6
99% confidence interval is (1.51, 1.84) art
99% confidence limits are 1.675 ± 4.032 ×
(iii)
4
or 1.675 ± 4.032 ×
0.09215
5
A1
Sensible comment referring to the fact that 1.8 is outside the 1st interval but inside
2nd. (ft on their confidence intervals)
P( X = r ) = e
(i)
G X (t ) =
−λ
∞
λr
r!
0
r
∞
0
B1ft
(may be implied by next line)
B1
(λt ) r
= e −λ e λt = e λ (t −1)
r!
M1 A1
∑ p t = ∑e
r
M1 A1
−λ
AG
(ii)
G X +Y (t ) = e λ ( t −1) .e µ ( t −1) = e ( λ + µ )(t −1)
(iii)
P( [0, 4] or [1, 3] or [2, 2] ) = 0.2231 × 0.1336 + 0.3347 × 0.2138 + 0.2510 × 0.2565
(B1 for any two correct)
0.1657
P(X Y 2 | X + Y = 4 ) =
0.1954
= 0.848 art
⇒ ( X + Y ) ~ Po(λ + µ)
M1 A1
B2,1,0
M1
A1ft
A1
5
1
4
np = 100 × = 20 and npq = 20 × = 16
5
5
Standardisation in either (a) or (b)
(i)
(a)
(b)
(ii)
© UCLES 2013
B1
M1
14.5 − 20
= −1.375 ⇒ P([ 15) = 0.915 (within range [0.915,0.916])
4
(ft on variance of 20)
z=
z=
12.5 − 20
= −1.875 ⇒ P(<12) = 1 – 0.9696 = 0.0304 (within range
4
[0.0303,0.0304])
mean = variance = 36 ⇒ S.D. = 6
z = 1.645
1

 N +  − 36
2

> 1.645 (Allow working with equality, but must be
6
⇒ N > 45.37 ∴ least N = 46
9795/02/SM/16
A1ft A1
A1 A1
B1
B1
1

N + )
2

M1 A1
A1
5
6
(i)
Above x-axis between (0, 0) to (3, 0)
Correct concavity. (Do not condone parabolas)
(ii)
µ=
4
27
∫ (3x
3
0
3
)
− x 4 dx
B1
B1
(Limits required)
M1
3
4  x 4 x5 
=
3 −  = 1.8
27  4
5 0
4
f ′( x) =
(6 x − 3 x 2 ) = 0
27
⇒ x = 0, 2 ∴ Mode = 2
A1 A1
M1 A1
A1
(iii)
Mean less than mode in (ii) matches negative skew in sketch.
(iv)
P( X − µ < σ ) =
4
27
∫ (3x
2.4
1.2
2
)
− x 3 dx
B1
25 1
dv
AG (3 terms required for M1)
− v = 30
v 4
dt
M1 A1
(ii)
F = ma ⇒
∫
0
75 3
dv
− v = 90
v 4
dt
⇒
A1 A1
75
3
= 10k ⇒ k =
AG
10
4
Tractive force = Resistance at maximum speed ⇒
t
120v
dv
100 − v 2
7
7
− 2v
(Limits not required)
t = −60
dv = − 60 ln 100 − v 2
3
3 100 − v 2
 91 
= −60 ln 51 + 60 ln 91 = 60 ln  ( = 34.7) seconds.
 51 
dt =
∫
7
3
∫
M1
= 0.64
(i)
(iii)
8
(Limits required)
2.4
4  3 x4 
=
x − 
27 
4  1.2
7
B1
[
]
M1
M1 A1
M1 A1
(i)
Resolve vertically NB = 2 × 10
Take moments about e.g. intersection of normals: NB × 0.2 cos60 = F × 0.4 sin60
moments equation correct (if in equilibrium)
F = 5.77, NB = 20
F > µNB
Correctly deduce not in equilibrium therefore rod does slip
M1
M1
A1
A1
M1
A1
(ii)
Consider forces horizontally NA non-zero
Correctly deduce impossibility hence leftwards force required for equilibrium
M1
A1
© UCLES 2013
9795/02/SM/16
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6
9
(i)
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 
(ii)
T sin θ = ml sin θϖ 2 ⇒ T = ml ϖ 2
(iii)
T
10
(i)
B1
mgl
h
B1
⇒ ω 2h = g .
B1
h = 0.5 ⇒ ϖ = 20 = 4.47 Angular speed is 4.47 rad/s.
B1
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
Conservation of linear motion: mu cos α = 0 + 3mv2 or mux = 3mv
Newtons experimental law: eu cos α = v2 or eux = v.
1
∴e =
3
(ii)
© UCLES 2013
B1
M1 A1
h
mgl
= mg ⇒ T =
l
h
⇒ mlϖ 2 =
B1
B1
1
u cos α.
3
Loss in kinetic energy is
u 2 cos 2 α
1
1
1
mu 2 − mu 2 sin 2 α − ⋅ 3m
2
2
2
9
1
2
= mu (Or remaining kinetic energy is 5/6 of initial kinetic energy etc.)
12
But cos2 α + sin2 α = 1 (used)
3
3
⇒ K ⇒ sin 2 α =
⇒ sin α =
⇒ α = 60°
4
2
v1 = u sin α and v2 =
9795/02/SM/16
M1 A1
A1
A1
B1
(both)
M1 A1
A1
M1
M1 A1
7
11
(i)
6mg  9a 2 + x 2 − a 


T=
a
Let θ be the angle between each string and line of motion of particle.
12mg 
x
m&x& = −2T cos θ = −
 9a 2 + x 2 − a  ×
2

a 
9a + x 2
⇒ &x& = −
(ii)
12 gx 
a
1−
2
a 
9a + x 2
∴ &x& ≈ (−12 g + 4 g )




12
(i)
x
8g
=−
x
a
a
(ii)
(iii)
a
8g
or π
a
.
2g
ga 8 g 2
a2
a
=
A ⇒ A2 =
⇒ A=
200 a
1600
40
where A is the amplitude.
y = 20 sin α ⋅
M1 A1
2
x
x
x


2
− 5
 = x tan α − (1 + tan α ) AG
20 cos α
80
 20 cos α 
M1 A1
B1
M1 A1
A1
x = R cos β and y = R sin β ⇒ y = x tan β .
∴
x2
R 2 (1 − sin 2 β )
⇒ R sin β = 20 −
80
80
R 2 (1 − sin 2 β ) + 80 R sin β − 1600 = 0 ⇒ ( R[1 − sin β ] + 40)( R[1 + sin β ] − 40) = 0
40
− 40
⇒R=
(up)
or
(down)
1 + sin β
1 − sin β
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 =
=
=
1 + sin β
cos 2 β
1 − sin 2 β
⇒ R down =
© UCLES 2013
A1
B1 B1
2
x2 tan2 α – 80x tan α + x2 + 80y = 0 (Can be implied by what follows.)
Real roots ⇒ 6400x2 – 4x2(x2 + 80y) [ 0
x2
⇒ 1600 – x2 – 80y [ 0 ⇒ y Y 20 –
, (x ≠ 0).
80
In y = 20 −
(iii)
A1 A1
M1 A1
vmax = ϖa ⇒
x = 20 cos αt y = 20 sin αt – 5t2
M1
A1
Which is simple harmonic motion of period 2π
(iii)
M1 A1
− 40 sin β − 40 40(1 + sin β )
40
=
=
2
2
1 + sin β
cos β
1 − sin β
9795/01/SM/16
M1 A1
M1
A1
B1
B1
B1
B1
8
BLANK PAGE
© UCLES 2013
9795/02/SM/16