w
w
ap
eP
m
e
tr
.X
w
CAMBRIDGE INTERNATIONAL EXAMINATIONS
s
er
om
.c
Pre-U Certificate
MARK SCHEME for the May/June 2014 series
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 should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, Pre-U,
GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level
components.
Page 2
1
(i)
(ii)
Mark Scheme
Pre-U – May/June 2014
X – Y ~ N(10, 5)
8 − 10
11 − 10
z1 =
[= –0.894…]; z2 =
[= 0.447…]
5
5
P (8 < Breadth < 11) = 0.487
X 1 + X2 + Y1 + Y2 ~ N(80, 10)
75 − 80
z=
[= –1.58…]
10
Syllabus
9795
B1
2
Either. cc or σ : M1A0
Both. FT on σ
cwo
As above
[e.g. N(80, 20), 0.8681: 2/4]
P (Perimeter > 75) = 0.943
2
(i)
Paper
02
cwo
σ2
)
stated or implied
n
µ + 0.5σ − µ
> 1.96
σ/ n
X ~ N(µ,
z=
µ − 0.1σ − µ
= −0.4
σ /4
1.96
(i)
Let N be the estimated number;
(ii)
98% confidence: z = ±2.326
ps = 0.05
0.05 ± 2.326
FT on their integer n, and
also –0.392: M1A1A0
400 20
=
⇒ N = 8000
N
400
M1
A1√
B1 [1]
B1
B1
0.05 × 0.95
400
400 ÷ 0.07535 = 5309 and
M1A1
A1 [3]
pq
n
0.05 and variance correct
Both, correct to 3 SF
Need
= (0.0246(5), 0.0753(5))
(iii)
B1
M1A1√
A1 [4]
M1A1 [5]
⇒ P( X > µ – 0.1σ) = 0.655
3
A1 [4]
B1
⇒ 0.5 n > 1.96 ⇒ n > 15.355 ⇒ Least n is 16
(ii)
M1A1
400 ÷ 0.02465 = 16228
400 ÷ CV
Integer answers, a.r.t. 5310,
16200
© Cambridge International Examinations 2014
M1A1
A1 [5]
M1
A1 [2]
Page 3
4
Mark Scheme
Pre-U – May/June 2014
(i)
[− 3e ]
(ii)
M X (t ) =
−x
k
0
=1
∫
⇒ –3e–k + 3 = 1
⇒ e–k =
2
[AG]
3
M1A1 [2]
k
M1
A1
e (t −1)k
3
3
−
=
1 − e − k e kt
(t − 1) (t − 1) (t − 1)
3  2 kt 
=
1 − e  [AG]
(t − 1)  3 
(
1
 2

M X (t ) = 3(1 + t + t 2 )1 − 1 + kt + k 2t 2  
2

 3
)
M1
A1 [4]
Both series correct
= 1 + (1 – 2k)t + (1 – 2k – k2)t2 [AG]
(iv)
5
(i)
(ii)
3
E(X) = 1 – 2k = 1–21n  
2
[AG]
Deals with
z2 =
229.5 − 250
250
260.5 − 250
(iii) We are approximating to B (250, 0.1).
Using N (25,22.5)
or
Using Po (25)
30.5 − 25
30.5 − 25
(= 1.1595)
(= 1.1)
z=
z=
22.5
25
P(>30) = 0.123
P(>30) = 0.136
B1
B1
B1 [3]
B1 [1]
In tables
M1
A1
M1
A1
A1 [5]
= −1.297
= 0.664
250
P(230 ≤ X ≤ 260) = 0.649
2
3
Allow longer methods if correct
G(t) = e λ (t – 1)
G′(t) = λe λ (t – 1)
⇒ µ = G′(1) = λ
G′′(t) = λ 2e λ (t – 1)
⇒ G′(1) = λ 2
∴σ 2 = E(X 2) + µ – µ 2 = λ 2 + λ – λ 2 = λ
z1 =
Paper
02
 e (t −1)x 
3e (t −1)x dx = 3

0
 (t − 1)  0
k
=3
(iii)
Syllabus
9795
One cc correct
Both ccs, √250
M1A1
A1
Correct handling of tails
M1
Answer in range [0.649, 0.650]
A1 [5]
Distribution stated or implied
Correct CC and √
M1
A1
Exact binomial 0.125: M0
Exact Poisson 0.136691: 3
A1 [3]
© Cambridge International Examinations 2014
Page 4
6
(i)
Mark Scheme
Pre-U – May/June 2014
[
]
[
(ii)
(iii)
]
0.41
0.41
4
4
tan −1 x 0 = 0.495K, tan −1 x 0 = 0.506K
π
π
⇒ 0.41 < median < 0.42 [AG]
m
π
4
tan −1 x 0 = 0.5 ; m = = 0.414…
Or:
π
8
[
]
4 1 x
dx
π 0 1 + x2
1
2
2
= ln 1 + x 2 0 = ln 2
π
π
∫
E(X) =
E(X 2) =
=
[
4
π
∫
4
x2
dx
2
0 1+ x
π
[
4
x − tan −1 x
π
]
1
0
=
Paper
02
Correct integral
M1
One correct output
A1
Conclusion fully justified (need
some comment if 0.41 and 0.42
used)
A1 [3]
∫ x f(x) dx attempted
]
1
Syllabus
9795
[AG]
1
0
A1 [2]
1
∫ 1− 1+ x
2
dx
4 π 4
1 −  = − 1
π  4 π
∫ x2 f(x) dx seen
Method for integration
4

 − 1 or 0.2732
π

M1
M1
Answers from calculator can
get full marks
A1 [5]
Quotient of relevant
probabilities
One correct
M1
2

4  2
Var (X) =  − 1 −  1n 2  = 0.0785
π
π


 
(iv)
[
[
]
]
4
tan −1 x
π
4
tan −1 x
π
1
2 −1
1
3/ 3
1 1
=
2 2
2 1
= 1− =
3 3
=1−
P(X > 1/3 3 | X >
7
π π
−
1 1
2
2
or 4 6 =
2 – 1) = ÷ =
π π
3 2
3
3
−
4 8
Recognising (3, 4, 5) triangle
Resolving vertically for P:
T= 0.5g
Resolving vertically for Q:
3
T = mg
5
Resolving horizontally for Q:
⇒
4
T = 4 mω 2
5
A1
M1
A1
A1 [3]
[can award B1 if answer
correctly obtained without
use of T = 5]
Needs resolving with trig (if
resolve || QR must have
cos or sin × mrω 2)
m = 0.3
ω =
M1
g
or 1.83
3
© Cambridge International Examinations 2014
B1
B1
M1
A1
M1
A1
A1
A1 [8]
Page 5
Mark Scheme
Pre-U – May/June 2014
Components of speed || to AB are 12 and 3 ms–1.
CLM: 2 × 12 – 3 × 3 = 2vA + 3vB
8
NEL: − 2 (12 + 3) = v A − v B
3
vA = –3, vB = 7
Components of velocity ⊥ to AB are 5 and 4 ms–1.
Speed A =
Speed B =
9
(i)
(ii)
750 =
2
–1
2
7 + 4 = 65 or 8.06 ms (OE)
∫
t
0
dt =
∫
15
5
vd v
15
Also
Consistent signs on RHS for
A1
(both √ on x-components)
Both
Pt  v 2 
=   (= 100 )
⇒
m  2 5
15
P
P 750
dv
= mv
⇒ ∫ dx = ∫ v 2 dv
5
v
m 0
dx
B1
M1A1√
M1A1√
A1
B1
either
M1
Both answers, correct to 3SF if
necessary
A1 [9]
2
] found, need second
15
equation for M1
Or:
1
Energy: 750F = m(152 – 52)
2
10m
F
=75
15 = 5 + t ⇒ t =
F
m
M1
A1
If a [=
Let P denote the constant power of the engine.
P
dv
P
=m
⇒
v
dt
m
Paper
02
Any signs on RHS for A1
32 + 52 = 34 or 5.83 ms–1,
1
(5 + 15)t ⇒ t = 75
2
Newton II:
Syllabus
9795
P
seen
v
If ∆KE used, must equate to
Pt
15
750 P  v 3   3250 
=   =

m
3 
 3 5 
P 13
900
= t and t =
or 69.2 seconds
⇒
m 9
13
⇒
© Cambridge International Examinations 2014
For F =
(B1)
(B1) [2]
B1
M1
A1
A1
A1
M1
A1
A1
A1 [9]
Page 6
10
Mark Scheme
Pre-U – May/June 2014
Syllabus
9795
0.05λ
= 2.5g ⇒ λ = 250 N [AG]
0.5
(i)
Hooke’s Law:
(ii)
Let e m be the extension in the new equilibrium position.
M1A1 [2]
250e
⇒ e = 0.08 m
0.5
250(0.08 + x )
⇒ &x&= −125 x
Newton II: 4 &x& = 4 g −
0.5
Find new e
4g =
(iii)
(a)
Amplitude of new motion is 0.08 – 0.05 = 0.03 m
(b)
Maximum speed = 125 × 0.03 = 0.335 ms–1.
(c)
T=
(d)
11
(i)
Paper
02
Needs 0.08√
SR: &x& = 10 − 125 x and then
x = y – 0.08 : B2
A1 [4]
cwo
B1 [1]
√ on a or ω
B1√ [1]
√ on ω
B1√ [1]
2π
2π
= 0.562 seconds.
=
ω
125
1
π
−1 2 
 + sin
 = 0.206 seconds
3
125  2
2
2
”) /ω or cos–1(“ ”)/ω
3
3
Deal with quadrants
sin–1(“
gx2 tan 2α – 2V 2x tan α + (gx2 + 2V 2y) = 0 or equiv
For boundary of accessible points B 2 – 4AC = 0
(*)
⇒ 4V 4x2 – 4gx2(gx2 + 2V 2y) = 0
1
⇒ 2gV 2y = V 4 – g2x2 ⇒ y =
[AG]
V 4 − g 2 x2
2 gV 2
(
y must be part of quadratic
Allow ∆ ≥ 0
)
gx 2
dy
= x sec 2 α − 2 tan α sec 2 α
dα
2V
2
1
V
V 4 − g 2 x2
⇒ y=
=0⇒ x=
g tan α
2 gV 2
(
⇒x=
(
“Quadratic equation” required
in question, so B0
)
1
g 2h2 − g 2 x 2
2 g 2h
B1
M1
A1
B0M1
A1
A1
[max 3]
)
y = h: M1A0
M1
A1
A1 [3]
3h
(b) Substituting for x in (*) gives
3gh2 tan 2α – 2 3 gh2 tan α + 3gh2 – 2gh2 = 0
⇒ 3 tan 2α – 2 3 tan α + 1 = 0
π
1
⇒ α = 30° or
⇒ tan α =
6
3
M1
M1
A1 [3]
A1 [4]
Or:
(ii) (a) Putting y = – h and V 2 = gh ⇒ − h =
M1
A1
M1
Or quote tan α =
© Cambridge International Examinations 2014
−B
2A
M1
A1
A1 [3]
Page 7
Mark Scheme
Pre-U – May/June 2014
Syllabus
9795
Paper
02
Notation: Speed of wind relative to: ground, w; cyclist at 15 ms–1, u; cyclist at 10 ms–1, v
Angle of wind velocity with due west, θ (see diagram in method α)
12
Method α:
D
10
B
θ 135°
∠ABC = 45°
∠BAC = 120°
∠BCA = 15°
5
w
A
120°
v
u
Diagram, with all three
C
v
5
5 sin 120°
=
⇒v=
= 16.73
sin 120° sin 15°
sin 15°
u
5
=
o
sin 135
sin 15o
5 sin 135o
⇒u =
= 13.66 or 5 + 5
sin 15o
Or :
w2 = 102 + 16.732 – 2 × 10 × 16.73 × cos 135o
⇒ w = 24.8(3)
sin θ sin 135°
=
16.73
24.83
⇒ θ = 28.45°.
⇒ Wind blows from bearing 118°.
u = 5 + 5 3 = 13.66
or v =
)
(
3
= w cos θ
10 + w sin θ = w cos θ
tan θ =
)
5
2 3 + 3 = 9.66
2
)
w sin θ
M1A1
M1A1
M1
(
Method γ: 15 +
M1A1
A1
M1
 − 5 7 + 3  − 21.83

 


2
=
Either side = 


 
 5

3 + 3   11.83 


 2
(
M1A1
A1
M1
A1
A1 [10]
 − sin 45°   − 10 
 − sin 30°   − 15 

 + 
 = w
 + 
Method β: u 
 cos 45°   0 
 cos 30°   0 
u
w
+ 15 =
+ 10
2
2
3u w
=
2
2
B1
9+2 3
⇒ θ = 28.45°.
23
A1
(N.B. Accept this vector answer
from any method)
M1A1
A1 [10]
[Note: If a and b for wsinθ and
wcosθ in the above equations,
each equation gets M1A1,
followed by M1A1A1 for v and
M1A1A1 for θ.]
M1A1A1
Wind blows from bearing 118°.
v = 24.8(3). Speed of wind is 24.8 km h–1.
© Cambridge International Examinations 2014
M1A1A1
M1A1
M1A1[10]
Page 8
Mark Scheme
Pre-U – May/June 2014
Syllabus
9795
Method δ:
w
15
15 sin 120°
=
⇒w=
sin 120° sin (60° − θ )
sin 60° cos θ − cos 60° sin θ
⇒w=
15 3
3 cos θ − sin θ
w
10
10 sin 135°
=
⇒w=
sin 135° sin (45° − θ )
sin 45° cos θ − cos 45° sin θ
Paper
02
Two equations for θ and w
M1A1
Compound angle formulae used
M1A1
Correct values of sin and cos
M1A1
Solve
Correct θ; correct bearing
Correct w
M1
A1A1
A1 [10]
Any two
All four
M1
A1
M1
A1
A1
10
cosθ − sin θ
Equate: 15 3 cosθ – 15 3 sinθ = 10 3 cosθ –10 sinθ
⇒w
5 3
⇒ tan θ =
=
9+2 3
23
[= 0.5419]
15 3 − 10
⇒ θ = 28.45 ⇒ Wind blows from bearing 118°
w = 24.83
and
Method ε : Components:
= ux + 15 = vx + 10
wx
wy
= uy = v y
wy
wy
= tan (60°) = 3 ,
= tan (45°) = 1
wx − 15
wx − 10
wx – 10 =
wy =
5
3 (wx – 15) ⇒ wx = (7 + 3) ,
2
M1A1
5
(3 + 3 )
2
⇒ speed = 24.8(3) ms–1
Angle is tan–1 (0.5419) = 28.45°
⇒ wind blows from bearing 118°.
A1
M1
A1 [10]
(
)
− 5 7 + 3 

  − 21.83 

2
=
Summary of answers: w = 
, w = 2 16 + 5 3 = 24.83
 5
 
3 + 3   11.83 

 2

 3+ 3 
 18 + 4 3 
 or tan −1 
 = 28.45° ⇒ wind blows from bearing 118(.45)°
θ = tan −1 
7+ 3
 46 
(
(
)
)
)
 − 5 1+ 3 

  − 6.83 
u=  2
=
, u = 13.66
 5 3 + 3   11.83 
 2

(
(
)
)
 − 5 1+ 3 

  − 11.83 
v=  2
=
, v = 16.73
 5 3 + 3   11.83 
 2

(
© Cambridge International Examinations 2014