1.
(a)
y
20
10
–2
–1
1
x
2
–10
–20
A1A1A1
N3
A1
N1
Note: Award A1 for the left branch asymptotic
to the x-axis and crossing the y-axis,
A1 for the right branch approximately
the correct shape,
A1 for a vertical asymptote at
1
approximately x = .
2
(b)
(i)
x
2
1
2
(must be an equation)
(ii)

f ( x ) dx
A1
N1
(iii)
Valid reason
R1
N1
A2
N2
A2
N2
A1A1A1A1
N4
A1
N1
0
eg reference to area undefined or discontinuity
Note: GDC reason not acceptable.
(c)

1.5
(i)
V=
(ii)
V = 105
1
f x  2 dx
(accept 33.3 )
(d)
f (x) = 2e2x  1  10(2x  1)2
(e)
(i)
x = 1.11
(accept (1.11, 7.49))
1
(ii)
(accept 0  k < 7.49)
p = 0, q = 7.49
A1A1
N2
[17]
Note: Accept exact answers given in terms of .
2.
(a)
Evidence of using l = r
arc AB = 7.85 (m)
(b)
A1
1
Evidence of using A  r 2 θ
2
Area of sector AOB = 58.9 (m2)
(c)
(M1)
N2
(M1)
A1
N2
METHOD 1
π6
2 π
3
angle =

30
6
(A1)

6
attempt to find 15 sin
height = 15 + 15 sin
M1

6
= 22.5 (m)
A1
N2
METHOD 2
π
angle =
3

60
3
attempt to find 15 cos
height = 15 + 15 cos
= 22.5 (m)
(A1)

3
M1

3
A1
N2
2
(d)
(i)
 
  
h  15 15 cos   
4
2 4
(M1)
= 25.6 (m)
(ii)
A1
 
h(0) = 15  15 cos  0  
 4
(M1)
= 4.39(m)
(iii)
N2
A1
N2
METHOD 1
Highest point when h = 30
R1


30 = 15  15 cos  2t  
4

M1


cos  2t   = 1
4

(A1)
3 

t = 1.18  accept 
8 

A1
N2
METHOD 2
h
30
2π
Sketch of graph of h
Correct maximum indicated
t = 1.18
t
M2
(A1)
A1
N2
METHOD 3
Evidence of setting h(t) = 0


sin  2t    0
4

Justification of maximum
M1
(A1)
R1
eg reasoning from diagram, first derivative test, second
derivative test
3 

t = 1.18  accept

8 

(e)


h(t) = 30 sin  2t   (may be seen in part (d))
4

(f)
(i)
A1
N2
A1A1
N2
3
h(t)
30
π
2
π
t
–30
A1A1A1
N3
(M1)
A1
N2
(M1)
A1
N2
Notes: Award A1 for range 30 to 30, A1
for two zeros.
Award A1 for approximate correct
sinusoidal shape.
(ii)
METHOD 1
Maximum on graph of h
t = 0.393
METHOD 2
Minimum on graph of h
t = 1.96
METHOD 3
Solving h(t) = 0
One or both correct answers
(M1)
A1
t = 0.393, t = 1.96
N2
[22]
3.
(a)
evidence of using area of a triangle
(M1)
1
eg A   2  2  sin θ
2
A = 2 sin 
(b)
A1
N2
METHOD 1
PÔA =   
1
2  2  sin   θ  (= 2 sin (  ))
2
since sin (  ) = sin 
then both triangles have the same area
area OPA =
(A1)
A1
R1
AG
N0
R3
AG
N0
METHOD 2
triangle OPA has the same height and the same base as triangle OPB
then both triangles have the same area
(c)
1
2
 2  2
2
area  APB = 2 sin  + 2 sin  (= 4 sin )
S = area of semicircle  area APB (= 2  4 sin )
S = 2( − 2 sin )
area semi-circle =
A1
A1
M1
AG
N0
4
(d)
METHOD 1
attempt to differentiate
(M1)
ds
  4 cos θ
dθ
setting derivative equal to 0
(M1)
eg
correct equation
A1
eg 4 cos  = 0, cos  = 0, 4 cos  = 0
=

2
A1
N3
EITHER
evidence of using second derivative
(M1)
S() = 4 sin 
A1
 
S    4
2
A1

it is a minimum because S    0
2
R1
N0
OR
evidence of using first derivative
(M1)
for  <

, S () < 0 (may use diagram)
2
A1
for  >

, S () > 0 (may use diagram)
2
A1
it is a minimum since the derivative goes from negative
to positive
R1
N0
METHOD 2
2  4 sin  is minimum when 4 sin  is a maximum
4 sin  is a maximum when sin  = 1
=
(e)

2
S is greatest when 4 sin  is smallest (or equivalent)
 = 0 (or )
R3
(A2)
A3
N3
(R1)
A1
N2
[18]
4.
(a)
(i)
n=5
(A1)
T = 280  1.125
T = 493
(ii)
evidence of doubling
eg 560
setting up equation
eg 280  1.12n = 560, 1.12n = 2
n = 6.116...
in the year 2007
A1
N2
(A1)
A1
(A1)
A1
N3
5
(b)
(i)
2560000
10  90 e 0.15
(A1)
P = 39 635.993...
(A1)
P
P = 39 636
(ii)
(c)
(i)
A1
N3
P = 46 806.997...
not doubled
valid reason for their answer
eg P < 51200
A1
A1
R1
N0
correct value
A2
N2
P
e.g.
(ii)
2560000
10  90 e 0.17 
25600
, 91.4 , 640 : 7
280
setting up an inequality (accept an equation, or reversed
inequality)
e.g.
M1
P
2560000
 70 ,
 70
T
10  90e 0.1n 280 1.12 n


finding the value 9.31....
after 10 years
(A1)
A1
N2
[17]
5.
(a)
P(F  S) = 1  0.14 (= 0.86)
(A1)
Choosing an appropriate formula
(M1)
eg P(A  B) = P(A) + P(B)  P(A  B)
Correct substitution
eg P(F  S) = 0.93  0.86
A1
P(F  S) = 0.07
Notes: There are several valid approaches. Award
(A1)(M1)A1 for relevant working using any
appropriate strategy eg formula, Venn
Diagram, or table.
AG
N0
Award no marks for the incorrect solution
P(F  S) = 1  P(F) + P(S) = 1  0.93 = 0.07
(b)
Using conditional probability
(M1)
 P (F  S ) 

eg P(F  S)  
P ( S ) 

P(F  S) =
0.07
0.62
= 0.113
(A1)
A1
N3
6
(c)
F and S are not independent
A1
N1
R1R1
N2
If independent P(F  S) = P(F) P(S), 0.07  0.31  0.62 (= 0.1922) R1R1
N2
EITHER
If independent P(F  S) = P(F), 0.113  0.31
OR
(d)
Let P(F) = x
P(S) = 2P(F) (= 2x)
(A1)
For independence P(F  S) = P(F)P(S) (= 2x2)
(R1)
Attempt to set up a quadratic equation
(M1)
eg P(F  S) = P(F)P(S)  P(F)P(S), 0.86 = x + 2x  2x2
2x2  3x + 0.86 = 0
A2
x = 0.386, x = 1.11
(A1)
P(F) = 0.386
(A1)
N5
[16]
6.
(a)
(b)
(i)
p=2
A1
N1
(ii)
q=1
A1
N1
(i)
f (x) = 0
2
(M1)
3x
=0
x 2 1
x= 
(2x2  3x  2 = 0)
1
x=2
2
 1 
  , 0
 2 
(ii)
A1
Using V =
A1

b
a
y2dx (limits not required)
N2
(M1)
2
0
V=
∫
1
2

3x 
  2  2  dx
x 1 

V = 2.52
(c)
(i)
Evidence of appropriate method
A2
A1
N2
M1
eg Product or quotient rule
Correct derivatives of 3x and x2  1
Correct substitution
eg
A1A1
A1
 3 ( x 2  1)  (3 x) (2 x)
f ′ (x) =
( x 2  1) 2
 3x 2  3  6 x 2
( x 2  1) 2
A1
7
f ′ (x) =
ii)
3x 2  3
3( x 2  1)
=
( x 2 1) 2
( x 2 1) 2
AG
N0
METHOD 1
Evidence of using f ′(x) = 0 at max/min
(M1)
3 (x2 + 1) = 0 (3x2 + 3 = 0)
A1
no (real) solution
R1
Therefore, no maximum or minimum.
AG
N0
METHOD 2
Evidence of using f ′(x) = 0 at max/min
(M1)
Sketch of f ′(x) with good asymptotic behaviour
A1
Never crosses the x-axis
R1
Therefore, no maximum or minimum.
AG
N0
METHOD 3
Evidence of using f ′ (x) = 0 at max/min
(d)
(M1)
Evidence of considering the sign of f ′ (x)
A1
f ′ (x) is an increasing function (f ′ (x)  0, always)
R1
Therefore, no maximum or minimum.
AG
For using integral
Area =

(M1)

g ( x) dx  or
0

a
Recognizing that

a
0

a
0
f  ( x) dx or
g ( x) dx  f ( x)

a
0

dx 
( x  1)

3x 2  3
2
2
A2
0
Correct equation

a
3x 2  3
0
( x 2  1) 2
a=
1
2
a=
1
2
A1
a
Setting up equation (seen anywhere)
eg
N0
(M1)
A1

3a 
dx = 2, 2  2   2  0 = 2, 2a2 + 3a  2 = 0
 a  1
a=2
A1
N2
[24]
8