1.
evidence of anti-differentiation
eg s =
 6e
3x
(M1)

 4 dx
s = 2e3t + 4t + C
substituting t = 0,
7=2+C
C=5
s = 2e3t + 4t + 5
A2A1
(M1)
A1
A1
N3
[7]
2.
(a)
y
20
10
–1
–2
1
2
x
–10
–20
A1A1A1
N3
A1
N1
A1
N1
R1
N1
f  x  2 dx
A2
N2
(accept 33.3 )
A2
N2
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
(ii)

(iii)
2
0
1
2
(must be an equation)
f ( x ) dx
Valid reason
eg reference to area undefined or discontinuity
Note: GDC reason not acceptable.
(c)

1.5
(i)
V=
(ii)
V = 105
1
1
(d)
f (x) = 2e2x  1  10(2x  1)2
(e)
(i)
x = 1.11
(ii)
p = 0, q = 7.49
(accept (1.11, 7.49))
(accept 0  k < 7.49)
A1A1A1A1
N4
A1
N1
A1A1
N2
[17]
4.
(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
(ii)
A1A1
A1
 3 ( x 2  1)  (3 x) (2 x)
( x 2  1) 2
f ′ (x) =
 3x 2  3  6 x 2
( x 2  1) 2
A1
f ′ (x) =
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
2
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)



d
x

0 ( x 2  1) 2

a
3x 2  3
A1
a
A2
0
Setting up equation (seen anywhere)
(M1)
Correct equation
eg

a
0
a=
1
2
a=
1
2
3x 2  3
( x  1)
2
2
N0
A1

3a 
dx = 2, 2  2   2  0 = 2, 2a2 + 3a  2 = 0
 a  1
a=2
A1
N2
[24]
5.
(a)
A1A1A1
Notes:
N3
Award A1 for both asymptotes shown.
The asymptotes need not be labelled.
Award A1 for the left branch in
approximately correct position,
A1 for the right branch in
approximately correct position.
3
4
(b)
(c)
5
(must be equations)
2
(i)
y = 3, x =
(ii)
x=
14
6
7
 14  
 or 2.33 , also accept , 0  
 6 
3
(iii)
y=
14
6
 y  2.8


 14 
 accept 0 ,  or 0 , 2.8
5






1
9  6 
 dx  9 x 
2 

2
x

5


2
x

5


1
3 ln 2 x  5 
C
22 x  5
A1A1
N2
A1
N1
A1
N1

(i)
A1A1A1
A1A1
(ii)
Evidence of using V =

b
a
y 2 dx
(M1)
Correct expression
eg

a
3
N5
A1
2

1 
 dx , 
 3 
2 x  5 


a
3
1
9  6 

2 x  5 2 x  52


 dx,


a


1
9 x  3 ln 2 x  5 

22 x  5 3


 
1
1
   27  3ln 1  
Substituting  9a  3 ln 2a  5 
22a  5  
2

Setting up an equation
9a 
A1
(M1)
1
1
 28

 27   3 ln 2a  5  3 ln 1    3 ln 3 
22a  5
2
 3

Solving gives a = 4
A1
N2
[17]
8.
Evidence of integration
s = 0.5 e2t + 6t2 + c
Substituting t = 0, s = 2
(M1)
A1A1
(M1)
eg 2 = 0.5 + c
c = 2.5
(A1)
s = 0.5 e2t + 6t2 + 2.5
A1
N4
[6]
11.
(a)
METHOD 1
Attempting to interchange x and y
(M1)
Correct expression x = 3y  5
(A1)
f
1
( x) 
x5
3
A1
N3
METHOD 2
Attempting to solve for x in terms of y
(M1)
5
Correct expression x 
f
(b)
1
( x) 
y5
3
(A1)
x5
3
A1
For correct composition (g1◦ f) (x) = (3x  5) + 2
(A1)
(g1◦ f) (x) = 3x  3
(c)
(d)
A1
x3
 3x  3 x  3  9 x  9
3
x
N3
N2
(A1)
12
8
A1
N2
A1A1A1
N3
A1A1
N2
A1A1
N2
(i)
y=3
x=2
Note:
(ii)
Award A1 for approximately correct x
and y intervals, A1 for two branches of
correct shape, A1 for both asymptotes.
(Vertical asymptote) x = 2, (Horizontal asymptote) y = 3
(Must be equations)
(e)
(f)
(i)
3x + ln (x  2) + C
(ii)
3x  ln x  253
(3x + ln x  2 + C)
(M1)
= (15 + ln 3)  (9 + ln1)
A1
= 6 + ln 3
A1
N2
A1
N1
Correct shading (see graph).
[18]
6
12.
(a)
3
x 1
2
(i)
f (x) = 
(ii)
For using the derivative to find the gradient of the tangent
(M1)
f (2) =  2
(A1)
A1A1
N2
1
Using negative reciprocal to find the gradient of the normal   M1
2
1
1


y  3  ( x  2)  or y  x  2 
2
2


(iii)
Equating 
A1
3 2
1
x  x  4  x  2 (or sketch of graph)
4
2
N3
M1
3x2  2x  8 = 0
(A1)
(3x + 4)(x  2) = 0
4
  1.33
3
x= 
(b)
(i)

2
45
11.25
4
Area =
(iii)
Attempting to use the formula for the volume
eg

1

N2
A2
 3 2
  1 3 1 2

  x  x  4  dx ,  x  x  4 x
1  4
4
2
 
 1
2
(ii)
k
A1
Any completely correct expression (accept absence of dx)
eg
(c)
4
 4 4
(accept  ,  or x   , x  2)
3
 3 3
N2
(accept 11.3)
 3

   x 2  x  4  dx , 
1 
4

2

A1
N1
(M1)
2
 3 2

  x  x  4  dx
1 
4

2
A2
N3
k
1
 1

f ( x) dx   x3  x 2  4 x
2
 4
1
Note:
Award A1 for 
A1A1A1
1
1 3
x , A1 for x 2 , A1 for 4x.
2
4
1
 1
  1 1 
Substituting   k 3  k 2  4k       4 
2
 4
  4 2 
1
1
=  k 3  k 2  4k  4.25
4
2
(M1)(A1)
A1
N3
[21]
7
14.
Note:
Area =

k
0
sin 2 xdx
Using area = 0.85
There are many approaches possible.
However, there must be some evidence
of their method.
(must be seen somewhere)
(A1)
(must be seen somewhere)
(M1)
EITHER
k
 1

Integrating  cos 2 x 
2
0
1
 1

cos 2k  cos 0 

2
 2

Simplifying
Equation
(A1)
1
cos 2k  0.5
2
1
cos 2k  0.5 = 0.85
2
(A1)
(cos 2k =  0.7)
OR
Evidence of using trial and error on a GDC
Eg


2 sin
0
2 xdx = 0.5 ,
(M1)(A1)

too small etc
2
OR
Using GDC and solver, starting with

k
0
sin 2 xdx  0.85 = 0(M1)(A1)
THEN
k = 1.17
(A2) (N3)
[6]
15.
(a)
s = 25t 
4 3
t c
3
Note: Award no further marks if “c” is
missing.
Substituting s = 10 and t = 3
10 = 25  3 
(M1)(A1)(A1)
(M1)
4 3
(3)  c
3
10 = 75  36 + c
c =  29
(A1)
4
s = 25t  t 3  29
3
(b)
(A1) (N3)
METHOD 1
s is a maximum when v =
ds
 0 (may be implied)
dt
(M1)
8
25  4t2 = 0
25
4
t2 =
t=
(A1)
5
2
(A1) (N2)
METHOD 2
Using maximum of s ( 12
25t 
2
, may be implied)
3
4 3
2
t  29 12
3
3
(A1)
t = 2.5
(c)
25t 
(M1)
(A1) (N2)
4 3
t  29  0
3
(accept equation)
m = 1.27, n = 3.55
(M1)
(A1)(A1) (N3)
[12]
17.
Attempting to integrate.
y  x3  5 x  c
substitute (2, 6) to find c  6  23  5(2)  c 
c 8
(M1)
(A1)(A1)(A1)
(M1)
(A1)
y  x3  5x  8 (Accept x3  5 x  8 )
(C6)
[6]
18. (a)
y0
2 x
(1  x2 )2
(A1)
1
(A1)(A1)(A1)
3
(b)
f ( x) 
(c)
6 x2  2
 0 (or sketch of f ( x) showing the maximum)
(1  x 2 )3
(M1)
6 x2  2  0
(A1)
x
x
(d)

1
3
1
(  0.577)
3
(A1)
0.5
0
1
1
1


dx  = 2 
dx = 2 
dx 
2
2
0.5 1  x
0 1 x
0.5 1  x 2


0.5
(A1)
(N4)
4
(A1)(A1)
2
9
[10]
19.
(a)
(b)
(i)
a  1 
 accept (1   , 0) 
(A1)
(ii)
b  1 
 accept (1   , 0) 
(A1)
(i)

1
h ( x)dx   h ( x)dx
2
 2.14
2
(M1)(A1)(A1)
1
OR

1
 2.14
h ( x) dx 

2
h ( x) dx
(M1)(A1)(A1)
h ( x)dx   h ( x)dx
(M1)(A1)(A1)
1
OR

(ii)
(c)
1
 2.14
1
2
5.141...  (0.1585...)
= 5.30
(A2)
(i)
y = 0.973
(A1)
(ii)
0.240  k  0.973
(A3)
5
4
[11
20.
(a)
(b)
(c)
At A, x = 0 => y = sin (e0) = sin (1)
=> coordinates of A = (0,0.841)
OR
A(0, 0.841)
(M1)
(A1)
sin (ex) = 0 => ex = 
=> x = ln  (or k = π)
OR
x = ln  (or k = π)
(M1)
(A1)
(i)
Maximum value of sin function = 1
(A1)
(ii)
dy
= ex cos (ex)
dx
Note:
(iii)
(G2)
(A2)
Award (A1) for cos (ex) and (A1) for ex.
dy
= 0 at a maximum
dx
=> ex =
(i)
Area =

ln 
0
Note:
2
(A1)(A1)
(R1)
ex cos (ex) = 0
=> ex = 0 (impossible) or cos (ex) = 0
(d)
2
sin (e x ) dx
π
2
=> x = ln
(M1)
π
2
(A1)(AG)
6
(A1)(A1)(A1)
Award (A1) for 0, (A1) for ln π, (A1) for sin (ex).
10
(ii)
Integral = 0.90585 = 0.906 (3 sf)
(G2)
5
(e)
y = x3
p
(M1)
At P, x = 0.87656 = 0.877 (3 sf)
(G2)
3
[18]
22.
(a)
(b)
1
 1 -kx 
-kx
0 e dx   k e  0
1
= – (e–k – e0)
k
1
= – (e–k – 1)
k
1
= – (1 – e–k)
k
1
(A1)
(A1)
(A1)
(AG)
3
k = 0.5
(i)
y
(0,1)
1
–1 0
1
2
3
x
(A2)
Note:
Award (A1) for shape, and (A1) for the point (0,1).
11
(ii)
Shading (see graph)
(iii)
Area =
1
e
0
=
-kx dx for k = 0.5
(i)
dy
= –ke–kx
dx
(ii)
x=1
(iii)
At x = 1
(M1)
1
(1 – e0.5)
0 .5
= 0.787 (3 sf)
OR
Area = 0.787 (3 sf)
(c)
(A1)
(A1)
(G2)
5
(A1)
y = 0.8  0.8 = e –k
ln 0.8 = –k
k = 0.223
dy
= –0.223e–0.223
dx
= –0.179 (accept –0.178)
(A1)
(A1)
(M1)
(A1)
OR
dy
= –0.178 or – 0.179
dx
(G2)
5
[13]
12