4757
1 (i)
Mark Scheme
June 2006
⎛ − 1⎞ ⎛ − k ⎞ ⎛ 4k − 4 ⎞
⎛ 2⎞
⎜ ⎟ ⎜
⎟ ⎜
⎟
⎜ ⎟
⎜ 4 ⎟ × ⎜ 4 ⎟ = ⎜ 2 − 2k ⎟ [ = 2(k − 1) ⎜ − 1⎟ ]
⎜ 3 ⎟ ⎜ k + 2 ⎟ ⎜ 4k − 4 ⎟
⎜ 2⎟
⎝ ⎠ ⎝
⎠ ⎝
⎠
⎝ ⎠
B1
AB and CD (Condone
M1
A2
BA and DC )
k =1
B1
1
(ii)(A)
(B)
CA × AB
=
45
AB
26
( ≈ 8.825 )
M1
A1
⎛ − 2 − λ − 1 ⎞ ⎛ − 1⎞
⎜
⎟ ⎜ ⎟
OR CP . AB = ⎜ − 3 + 4λ − 5 ⎟ ⋅ ⎜ 4 ⎟ = 0
⎜ 2 + 3λ + 2 ⎟ ⎜ 3 ⎟
⎝
⎠ ⎝ ⎠
⎛ − 95 ⎞
⎟
1 ⎜
CP =
⎜ − 140 ⎟ Distance is
26 ⎜
⎟
⎝ 155 ⎠
M1
M1A1
For appropriate vector product
Evaluation Dependent on
previous M1
M1
M1
A1
M1
⎛ − 3 ⎞ ⎛ − 1⎞ ⎛ − 40 ⎞
⎜ ⎟ ⎜ ⎟ ⎜
⎟
CA × AB = ⎜ − 8 ⎟ × ⎜ 4 ⎟ = ⎜ 5 ⎟
⎜ 4 ⎟ ⎜ 3 ⎟ ⎜ − 20 ⎟
⎝ ⎠ ⎝ ⎠ ⎝
⎠
Distance is
(C)
Evaluating vector product
4 Give A1 ft for one element
correct
Method for finding shortest
distance
Dependent on first M1
Calculating magnitudes
6 Dependent on previous M1
Accept 8.82 to 8.83
M2A1
52650
26
⎛ − 40 ⎞
⎛ 8 ⎞
⎜
⎟
⎜ ⎟
Normal vector is CA × AB = ⎜ 5 ⎟ = −5 ⎜ − 1⎟
⎜ − 20 ⎟
⎜ 4⎟
⎝
⎠
⎝ ⎠
Equation of plane is 8 x − y + 4 z = −16 + 3 + 8
8x − y + 4 z + 5 = 0
Finding CP Dependent on
previous M1
Dependent on previous M1
M1
Dependent on previous M1
Allow −40 x + 5 y − 20 z = 25 etc
M1
A1
3
(iii)
AC . ( AB × CD )
AB × CD
=
⎛ k + 2⎞ ⎛ 2 ⎞
⎜
⎟ ⎜ ⎟
⎜ 8 ⎟ ⋅ ⎜ − 1⎟ (2k − 2)
⎜ −4 ⎟ ⎜ 2 ⎟
⎝
⎠ ⎝ ⎠
Shortest distance is
3(2k − 2)
2k − 12
3
M1
M1
A1 ft
A1
For AC . ( AB × CD )
Fully correct method (evaluation
not required) Dependent on
previous M1
Correct evaluated expression
for distance ft from (i)
4 Simplified answer
Modulus not required
4757
Mark Scheme
(iv) Intersect when k = 6
−2 − λ = 6 − 6 μ
− 3 + 4λ = 5 + 4 μ
2 + 3λ = −2 + 8μ
Solving, λ = 4 , μ = 2
Point of intersection is ( − 6 , 13 , 14 )
June 2006
B1 ft
M1
A1 ft
M1
Forming at least two equations
Two correct equations
Solving to obtain λ or μ
Dependent on previous M1
One value correct
A1
A1
6
−2 − λ = k − kμ
M1
OR − 3 + 4λ = 5 + 4 μ
A1
2 + 3λ = −2 + (k + 2) μ
Solving, k = 6
M1A1
λ = 4, μ = 2
A1
Point of intersection is ( − 6 , 13 , 14 ) A1
Forming three equations
All equations correct
Dependent on previous M1
One value correct
4757
Mark Scheme
M1
A1
2 (i)
⎛ 2x − 4 y ⎞
⎜
⎟
Normal vector is ⎜ − 4 x + 6 y ⎟
⎜ − 4z ⎟
⎝
⎠
(ii)
June 2006
Partial differentiation
⎛ x⎞
⎛ 2x − 4 y ⎞
⎜ ⎟
⎜
⎟
Condone r = ⎜ y ⎟ + λ ⎜ − 4 x + 6 y ⎟
⎜z⎟
⎜ − 4z ⎟
⎝ ⎠
⎝
⎠
A1
A1
⎛ 18 ⎞
⎜
⎟
At Q normal vector is ⎜ − 44 ⎟
⎜ −4 ⎟
⎝
⎠
4 For 4 marks the normal must
appear as a vector (isw)
M1
For 18 x − 44 y − 4 z
Dependent on previous M1
Using Q to find constant
Accept any correct form
M1
Tangent plane is
18 x − 44 y − 4 z = 306 − 176 − 4 = 126
M1
A1
9 x − 22 y − 2 z = 63
4
(iii)
18h − 44 p − 4(−h) ≈ 0
p≈
1
2
For 18δx − 44δy − 4δz
M1
A1 ft
18δx − 44δy − 4δz ≈ 0
M1
h
A1
4
OR 9(17 + h) − 22(4 + p) − 2(1 − h) ≈ 63
If left in terms of x, y, z:
M1A0M1A0
M2A1 ft
1
2
p≈ h
A1
OR (17 + h) 2 − 4(17 + h)(4 + p ) + ... = 0
−44 p + 22h ≈ 0
M2A1
1
2
p≈ h
OR p =
Neglecting second order terms
A1
4h + 44 ± 28h 2 + 88h + 1936
6
M2A1
p ≈ 12 h
A1
(iv) Normal parallel to z-axis requires
2 x − 4 y = 0 and − 4 x + 6 y = 0
2
x = y = 0 ; then − 2 z − 63 = 0
No solutions; hence no such points
M1A1 ft
M1
A1 (ag)
Correctly shown
4
OR 2 x − 4 y = −4 x + 6 y , so y = 53 x
−
8
25
Similarly if only 2 x − 4 y = 0 used
x 2 − 2 z 2 − 63 = 0 , hence no points
M2A2
(v)
2 x − 4 y = 5λ
− 4 x + 6 y = −6λ
M1A1 ft
− 4 z = 2λ
x = − λ , y = −2λ , z = − λ
3
2
1
2
Substituting into equation of surface
9 2
λ − 12λ2 + 12λ2 − 12 λ2 − 63 = 0
4
λ = ±6
M1
Obtaining x, y, z in terms of λ
or x = 3 z , y = 4 z
M1
M1
M1
Obtaining a value of λ (or
equivalent)
4757
Mark Scheme
Point ( − 9 , − 12 , − 3 ) gives k = −45 + 72 − 6 = 21 A1
A1
Point ( 9 , 12 , 3 ) gives k = 45 − 72 + 6 = −21
June 2006
Using a point to find k
If λ = 1 is assumed:
8 M0M1M0M0M1
4757
3 (i)
Mark Scheme
2
June 2006
2
⎛ dx ⎞
⎛ dy ⎞
⎜ ⎟ + ⎜ ⎟ = ( 6t 2 − 6 ) 2 + ( 12t ) 2
⎝ dt ⎠
⎝ dt ⎠
M1A1
= 36t 4 + 72t 2 + 36
= 36( t 2 + 1 ) 2
A1
1
⌠
Arc length is ⎮ 6( t 2 + 1 ) dt
⌡0
⌠ ⎛ dx ⎞ 2 ⎛ dy ⎞ 2
⎟ + ⎜ ⎟ dt
⎝ dt ⎠
⌡ ⎝ dt ⎠
M1
Using ⎮ ⎜
⎮
A1
For 2t 3 + 6t
1
⎡
⎤
= ⎢ 2t 3 + 6t ⎥
⎣
⎦0
=8
A1
6
M1
(ii)
Using ∫ ... y ds (in terms of t)
Curved surface area is
with ‘ds’ the same as in (i)
1
⌠
2
2
∫ 2π y ds = ⎮⌡0 2π ( 6t ) 6( t + 1) dt
⎡
=π ⎢
⎣
=
1
72 5
t
5
192π
5
Any correct integral form in
terms of t
(limits required)
Integration
For π ( 725 t 5 + 24t 3 )
A1
M1
⎤
+ 24t 3 ⎥
⎦0
A1
( ≈ 120.6 )
A1
5
(iii)
dy
12t
= 2
dx 6t − 6
Method of differentiation
M1
A1
2t ⎞
⎛
⎜ = 2
⎟
⎝ t −1 ⎠
Equation of normal is
1 − t2
( x − 2t 3 + 6t )
2t
1 ⎛1 ⎞
y − 6t 2 = ⎜ − t ⎟ x − t 2 (1 − t 2 ) + 3(1 − t 2 )
2⎝t
⎠
y − 6t 2 =
y=
M1
1 ⎛1 ⎞
⎜ − t ⎟ x + 2t 2 + t 4 + 3
2⎝t
⎠
A1 (ag)
4 Correctly obtained
M1
(iv) Differentiating partially with respect to t
1⎛ 1
⎞
0 = ⎜ − 2 − 1⎟ x + 4t + 4t 3
2⎝ t
⎠
1
2t 2
A2
Give A1 if just one error or
omission
( 1 + t 2 ) x = 4t ( 1 + t 2 )
x = 8t 3
t=
At least one intermediate step
required
1
1 3
x
,
2
so y =
1
2
y=
1
x3
−1
(2 x 3
−
2
3 3
x
2
4
3 3
x
16
−
1
2
)x +
+3
1
2
2
x3
+
1
16
4
x3
M1
+3
For obtaining a x = b t 3
M1
Eliminating t
A1
6
4757
(v)
Mark Scheme
P lies on the envelope of the normals
2
Or a fully correct method for
finding the centre of curvature at
a general pt
[ ( 8t 3 , 6t 2 − 3t 4 + 3 ) ]
M1
4
Hence a = 32 × 64 3 − 163 × 64 3 + 3
= −21
June 2006
M1
Or t = 2 and a = 6 × 2 2 − 3 × 2 4 + 3
A1
3
4757
Mark Scheme
June 2006
4 (i)
I
J
K
L
–I –J –K –L
I
I
J
K
L
–I –J –K –L
J
J
–I
L –K –J
K
K –L –I
L
L
–I
–I –J –K –L
–J –J
I
J –K L
–L K
I
K –J –I –L –K J
I
–K –K L
–L K
I
–L –L –K J
–J
Give B5 for 30 (bold) entries
6 correct
Give B4 for 24 (bold) entries
correct
Give B3 for 18 (bold) entries
correct
Give B2 for 12 (bold) entries
correct
Give B1 for 6 (bold) entries
correct
B3
Give B2 for six correct
3 Give B1 for three correct
B3
Give B2 for six correct
3 Give B1 for three correct
I
I
J
K
J
–I
L –K
–J K –L –I
B6
L
J
I
L
K –J –I
L
–I –J –K –L
(ii)
Eleme
nt
I
J
Invers
e
I
–J –K –L –I
Eleme
nt
I
J
K
L
–I –J –K –L
Order
1
4
4
4
2
K
J
K
L
(iii)
4
4
4
(iv) Only two elements of G do not have order 4;
so any subgroup of order 4 must contain an
M1A1
element of order 4
A subgroup of order 4 is cyclic if it contains
B1
an element of order 4
A1
Hence any subgroup of order 4 is cyclic
(may be implied)
For completion
4
OR If a group of order 4 is not cyclic,
it contains three elements of order 2
B1
G has only one element of order 2; so
this cannot occur
M1A1
So any subgroup of order 4 is cyclic A1
(v)
{ I, − I }
{ I, J, − I, − J }
{ I, K, − I, − K }
{ I, L, − I, − L }
B1
B1
B1
B1
B1
For { I , − I } , at least one
5 correct subgroup of order 4, and
no wrong subgroups. This mark
is lost if G or { I } is included
4757
Mark Scheme
June 2006
(vi) The symmetry group has 5 elements of order M1
2
A1
(4 reflections and rotation through 180° )
G has only one element of order 2, hence G
is not isomorphic to the symmetry group
A1
Considering elements of order 2
(or self-inverse elements)
Identification of at least two
elements of order 2 in the
symmetry group
3 For completion
4757
Mark Scheme
Pre-multiplication by transition matrix
5 (i)
0 ⎞
⎛ 0.8 0.1
⎜
⎟
P = ⎜ 0.2 0.6 0.15 ⎟
⎜ 0 0.3 0.85 ⎟
⎝
⎠
(ii)
For the three columns
B1B1B1
3
⎛ 0.6 ⎞ ⎛ 0.3204 0.1545 0.0927 ⎞ ⎛ 0.6 ⎞ ⎛ 0.254 ⎞ M1
⎜ ⎟ ⎜
⎟⎜ ⎟ ⎜
⎟
7
P ⎜ 0.4 ⎟ = ⎜ 0.3089 0.2895 0.2780 ⎟ ⎜ 0.4 ⎟ = ⎜ 0.301 ⎟ M1
⎜ 0 ⎟ ⎜ 0.3706 0.5560 0.6293 ⎟ ⎜ 0 ⎟ ⎜ 0.445 ⎟
⎝ ⎠ ⎝
⎠⎝ ⎠ ⎝
⎠ A1
M1
A1
A1
Division 3 is the most likely
(iii)
June 2006
Considering P 7 (or P 8 or P 6 )
Evaluating a power of P
For P 7 (Allow ±0.001
throughout)
Evaluation of probabilities
One probability correct
6 Correctly determined
⎛ 0.1429 0.1429 0.1429 ⎞
⎜
⎟
P → ⎜ 0.2857 0.2857 0.2857 ⎟
⎜ 0.5714 0.5714 0.5714 ⎟
⎝
⎠
M1
Considering powers of P
M1
Obtaining limit
Equilibrium probabilities are 0.143, 0.286,
0.571
A1
n
Must be accurate to 3 dp if
3 given as decimals
OR
0.8 p + 0.1q = p
⎛ p⎞ ⎛ p⎞
⎜ ⎟ ⎜ ⎟
P⎜ q ⎟ = ⎜ q ⎟ ⇒ 0.2 p + 0.6q + 0.15r = q
⎜r⎟ ⎜r⎟
0.3q + 0.85r = r
⎝ ⎠ ⎝ ⎠
M1
Obtaining at least two equations
q = 2 p , r = 2q = 4 p and p + q + r = 1
M1
Solving (must use p + q + r = 1 )
p=
(iv)
(v)
1
7
, q=
2
7
, r=
0
⎛ 0.8 0.1
⎜
⎜ 0.2 0.6 0.15
Q=⎜
0 0.3 0.75
⎜
⎜ 0
0
0.1
⎝
⎛ 0 ⎞ ⎛ 0.4122
⎜ ⎟ ⎜
5 ⎜ 1 ⎟ ⎜ 0.3131
Q ⎜ ⎟=⎜
0.2369
0
⎜ ⎟ ⎜
⎜ 0 ⎟ ⎜ 0.0378
⎝ ⎠ ⎝
4
7
0⎞
⎟
0⎟
0⎟
⎟
1 ⎟⎠
0.1566 0.0592 0 ⎞ ⎛ 0 ⎞
⎟⎜ ⎟
0.2767 0.2052 0 ⎟ ⎜ 1 ⎟
0.4105 0.4030 0 ⎟ ⎜ 0 ⎟
⎟⎜ ⎟
0.1563 0.3326 1 ⎟⎠ ⎜⎝ 0 ⎟⎠
A1
B1
Third column
B1
Fourth column
B1
Fully correct
3
M1
Considering Q 5 (or Q 6 or Q 4 )
M1
Evaluating a power of Q
A1
For 0.1563 (Allow 0.156 ± 0.001 )
M1
A1 ft
For 1 − a4 , 2
⎛ 0.1566 ⎞
⎜
⎟
⎜ 0.2767 ⎟
=⎜
0.4105 ⎟
⎜
⎟
⎜ 0.1563 ⎟
⎝
⎠
P(still in league) = 1 − 0.1563
= 0.844
5
(vi) P(out of league) is element a4, 2 in Q n
M1
When n = 15 , a4, 2 = 0.4849
M1
When n = 16 , a4, 2 = 0.5094
A1
First year is 2031
A1
ft dependent on M1M1M1
Considering Q n for at least two
more values of n
Considering a4 , 2 Dep on
previous M1
For n = 16
4 SR With no working,
n = 16 stated B3
2031 stated
B4
4757
Mark Scheme
Post-multiplication by transition matrix
5 (i)
0 ⎞
⎛ 0.8 0.2
(ii)
(0.6
⎜
⎟
P = ⎜ 0.1 0.6 0.3 ⎟
⎜ 0 0.15 0.85 ⎟
⎝
⎠
B1B1B1
0.4 0) P 7
M1
M1
A1
For the three rows
3
⎛ 0.3204 0.3089 0.3706 ⎞
⎜
⎟
= (0.6 0.4 0) ⎜ 0.1545 0.2895 0.5560 ⎟
⎜ 0.0927 0.2780 0.6293 ⎟
⎝
⎠
= (0.254 0.301 0.445)
M1
A1
A1
Division 3 is the most likely
(iii)
June 2006
Considering P 7 (or P 8 or P 6 )
Evaluating a power of P
For P 7 (Allow ±0.001
throughout)
Evaluation of probabilities
One probability correct
6 Correctly determined
⎛ 0.1429 0.2857 0.5714 ⎞
⎜
⎟
P → ⎜ 0.1429 0.2857 0.5714 ⎟
⎜ 0.1429 0.2857 0.5714 ⎟
⎝
⎠
M1
Considering powers of P
M1
Obtaining limit
Equilibrium probabilities are 0.143, 0.286,
0.571
A1
n
OR
(p
Must be accurate to 3 dp if
3 given as decimals
q r) P = (p q r)
0.8 p + 0.1q = p
0.2 p + 0.6q + 0.15r = q
M1
Obtaining at least two equations
0.3q + 0.85r = r
q = 2 p , r = 2q = 4 p and p + q + r = 1
p=
(iv)
(v)
1
7
, q=
2
7
, r=
4
7
M1
Solving (must use p + q + r = 1 )
A1
0
0 ⎞
⎛ 0.8 0.2
⎜
⎟
0
.
1
0
.
6
0
.
3
0
⎜
⎟
Q=⎜
0 0.15 0.75 0.1⎟
⎜
⎟
⎜ 0
0
0
1 ⎟⎠
⎝
B1
Third row
B1
Fourth row
B1
Fully correct
(0
M1
Considering Q 5 (or Q 6 or Q 4 )
M1
Evaluating a power of Q
A1
For 0.1563 (Allow 0.156 ± 0.001 )
M1
A1 ft
For 1 − a2 , 4
1 0 0) Q 5
⎛ 0.4122 0.3131 0.2369 0.0378 ⎞
⎜
⎟
⎜ 0.1566 0.2767 0.4105 0.1563 ⎟
= (0 1 0 0) ⎜
0.0592 0.2052 0.4030 0.3326 ⎟
⎜
⎟
⎜ 0
0
0
1 ⎟⎠
⎝
= (0.1566 0.2767 0.4105 0.1563)
P(still in league) = 1 − 0.1563
= 0.844
3
5
ft dependent on M1M1M1
4757
Mark Scheme
(vi) P(out of league) is element a2, 4 in Q n
June 2006
M1
When n = 15 , a2, 4 = 0.4849
M1
When n = 16 , a2, 4 = 0.5094
A1
First year is 2031
A1
Considering Q n for at least two
more values of n
Considering a2 , 4 Dep on
previous M1
For n = 16
4 SR With no working,
n = 16 stated B3
2031 stated
B4
4757
1 (i)
Mark Scheme
⎛ 8 ⎞ ⎛ 6 ⎞ ⎛ 33 ⎞
⎛ 3 ⎞
⎜
⎟ ⎜ ⎟ ⎜
⎟
⎜ ⎟
d K = ⎜ − 1 ⎟ × ⎜ 2 ⎟ = ⎜ − 44 ⎟ [ = 11⎜ − 4 ⎟ ]
⎜ − 14 ⎟ ⎜ − 5 ⎟ ⎜ 22 ⎟
⎜ 2 ⎟
⎝
⎠ ⎝ ⎠ ⎝
⎠
⎝ ⎠
⎛ 8 ⎞ ⎛ 2 ⎞ ⎛ 15 ⎞
⎛ 3 ⎞
⎜
⎟ ⎜ ⎟ ⎜
⎟
⎜ ⎟
d L = ⎜ − 1 ⎟ × ⎜ 1 ⎟ = ⎜ − 20 ⎟ [ = 5 ⎜ − 4 ⎟ ]
⎜ − 14 ⎟ ⎜ − 1⎟ ⎜ 10 ⎟
⎜ 2 ⎟
⎝
⎠ ⎝ ⎠ ⎝
⎠
⎝ ⎠
* These marks can be earned
anywhere in the question
A1
M1*A1*
⎤ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 48 ⎞
⎟
⎟ ⎜
⎥ ⎜ ⎟ ⎜ ⎟ ⎜
⎥ × ⎜ − 4 ⎟ = ⎜ 24 ⎟ × ⎜ − 4 ⎟ = ⎜ − 6 ⎟
⎥ ⎜⎝ 2 ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ − 84 ⎟⎠
⎦
482 + 6 2 + 84 2
32 + 4 2 + 2 2
=
9396
29
= 18
−87 + 29λ = 0 ,
M1
For (b − a) × d
M1
A1
Correct method for finding
distance
9
⎛ 6 + 3λ − 3 ⎞ ⎛ 3 ⎞
⎟ ⎜ ⎟
⎜
OR ⎜ 28 − 4λ − 4 ⎟ ⋅ ⎜ − 4 ⎟ = 0
⎟ ⎜ 2 ⎟
⎜
2λ
⎠ ⎝ ⎠
⎝
For (b + λ d − a) . d = 0
M1
λ =3
Finding λ , and the magnitude
M1
Distance is 12 2 + 12 2 + 6 2 = 18
(ii)
Correctly shown
Finding one point on K or L
or (6 , 0 , 2) or (0 , 8 , − 2) etc
Or (27 , 0 , 14) or (0 , 36 , − 4) etc
A1*
i.e. (6 , 28 , 0)
Distance is
Finding direction of K or L
One direction correct
M1*
A1*
Hence K and L are parallel
For a point on K, z = 0 , x = 3 , y = 4
i.e. (3 , 4 , 0)
For a point on L, z = 0 , x = 6 , y = 28
⎡ ⎛ 6 ⎞ ⎛ 3⎞
⎢⎜ ⎟ ⎜ ⎟
⎢ ⎜ 28 ⎟ − ⎜ 4 ⎟
⎢ ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠
⎣
June 2007
A1
Distance from (3 , 4 , 0) to R is
M1A1 ft
2 × 3 + 4 − 0 − 40
2 2 + 12 + 12
=
(iii)
30
6
=
A1 ag
30 6
=5 6
6
K, M intersect if 1 + 5λ = 3 + 3μ
−4 − 4λ = 4 − 4μ
3λ = 2μ
Solving (2) and (3): λ = 4 , μ = 6
3
M1
(1)
(2)
(3)
A1 ft
M1M1
M1A1
A1
Check in (1): LHS = 1 + 20 = 21 ,
RHS = 3 + 18 = 21
Hence K, M intersect, at (21 , − 20 , 12)
OR M meets P when
M1
8(1 + 5λ ) − (−4 − 4λ ) − 14(3λ ) = 20
A1
M meets Q when
6(1 + 5λ ) + 2(−4 − 4λ ) − 5(3λ ) = 26
A1
Both equations have solution λ = 4
A1
Point is on P, Q and M; hence on K and M
M2
Point of intersection is (21 , − 20 , 12)
A1
35
At least 2 eqns, different
parameters
Two equations correct
Intersection correctly shown
Can be awarded after
7 M1A1M1M0M0
Intersection of M with both P and
Q
4757
(iv)
Mark Scheme
⎡⎛6⎞ ⎛ 1 ⎞
⎟
⎢⎜ ⎟ ⎜
⎢ ⎜ 28 ⎟ − ⎜ − 4 ⎟
⎢ ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠
⎣
Distance is
⎤ ⎡⎛ 3 ⎞ ⎛ 5 ⎞ ⎤ ⎛ 5 ⎞ ⎛ − 4 ⎞
⎟⎥ ⎜ ⎟ ⎜ ⎟
⎥ ⎢⎜ ⎟ ⎜
⎥ ⋅ ⎢⎜ − 4 ⎟ × ⎜ − 4 ⎟ ⎥ = ⎜ 32 ⎟ ⋅ ⎜ 1 ⎟ = 12
⎥ ⎢⎜⎝ 2 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎥ ⎜⎝ 0 ⎟⎠ ⎜⎝ 8 ⎟⎠
⎦
⎦ ⎣
12
12 4
=
=
2
2
2
9 3
4 +1 + 8
36
June 2007
For evaluating d L × d M
For (b − c) . (d L × d M )
M1A1 ft
M1
Numerical expression for distance
A1 ft
A1
5
4757
2 (i)
(ii)
Mark Scheme
∂z
= y 2 − 8 xy − 6 x 2 + 54 x − 36
∂x
∂z
= 2 xy − 4 x 2
∂y
At stationary points,
B1
3
∂z
∂z
= 0 and
=0
∂x
∂y
M1
M1
y = ±6 ; points (0 , 6 , 20) and (0 , − 6 , 20)
2
Give B1 for 3 terms correct
B2
When x = 0 , y 2 − 36 = 0
2
June 2007
2
When y = 2 x , 4 x − 16 x − 6 x + 54 x − 36 = 0
− 18 x 2 + 54 x − 36 = 0
x = 1, 2
A1A1
If A0, give A1 for y = ±6
M1
or y = 2, 4
M1A1
A1
Points (1, 2 , 5) and (2 , 4 , 8)
A0 if any extra points given
8
(iii) When x = 2 , z = 2 y 2 − 16 y + 40
B1
‘Upright’ parabola
B1
(2 , 4 , 8) identified as a minimum
(in the first quadrant)
When y = 4 , z = −2 x 3 + 11x 2 − 20 x + 20
⎛ d2z
⎜
= −12 x + 22 = −2 when x = 2
⎜ dx 2
⎝
⎞
⎟
⎟
⎠
B1
‘Negative cubic’ curve
B1
(2 , 4 , 8) identified as a stationary
point
Fully correct (unambiguous
minimum and maximum)
B1
The point is a minimum on one section and a
maximum on the other; so it is neither a
maximum nor a minimum
(iv)
Require
∂z
∂z
= −36 and
=0
∂x
∂y
B1
6
∂z
= 36 can earn all M marks
∂x
M1
When x = 0 , y 2 − 36 = −36
y = 0 ; point (0 , 0 , 20)
When y = 2 x , 4 x 2 − 16 x 2 − 6 x 2 + 54 x − 36 = −36
− 18 x 2 + 54 x = 0
x = 0, 3
M1
A1
M1
M1
x = 0 gives (0 , 0 , 20) same as above
Solving to obtain x (or y) or
stating ‘no roots’ if appropriate
(e.g. when +36 has been used)
A1
A1
x = 3 gives (3 , 6 , − 7)
7
37
4757
3 (i)
Mark Scheme
2
2
1 ⎞
⎛ dy ⎞
⎛
1+ ⎜ ⎟ =1+ ⎜ x −
⎟
d
4
x
x ⎠
⎝ ⎠
⎝
1
1
1
1
= 1 + x2 − +
= x2 + +
2
2 16 x
2 16 x 2
1 ⎞
⎛
=⎜ x+
⎟
4x ⎠
⎝
M1
2
A1
a
⌠ ⎛
1 ⎞
Arc length is ⎮ ⎜ x +
⎟ dx
4x ⎠
⌡1 ⎝
=
[
1
2
x 2 + 14 ln x
M1
]
= 12 a 2 + 14 ln a −
(ii)
June 2007
a
1
M1
1
2
A1 ag
⌠
2
⎛ dy ⎞
⎟ dx
⎝ dx ⎠
For ⎮ 1 + ⎜
⎮
⌡
5
M1
Curved surface area is ∫ 2π x ds
4
1 ⎞
⎛
⌠
= ⎮ 2π x⎜ x +
⎟ dx
4
x ⎠
⎝
⌡1
4
⎡
= 2π ⎢ 13 x 3 +
⎣
=
87π
2
A1 ft
1
4
Any correct integral form
(including limits)
M1
A1
⎤
x ⎥
⎦1
( ≈ 137 )
for
1
3
x 3 + 14 x
A1
5
3
(iii)
ρ=
⎛ ⎛ dy ⎞ 2 ⎞ 2
⎜1 + ⎜ ⎟ ⎟
⎜ ⎝ dx ⎠ ⎟
⎝
⎠
1 ⎞
⎛
⎜a +
⎟
4
a⎠
⎝
=
1
1+ 2
4a
2
d y
dx 2
3
3
1 ⎞
⎛
a⎜a +
⎟
2
1 ⎞
4a ⎠
⎛
= a⎜a +
= ⎝
⎟
1
4a ⎠
⎝
a+
4a
B1
any form, in terms of x or a
B1
any form, in terms of x or a
M1
Formula for ρ or κ
ρ or κ correct, in any form, in
terms of x or a
A1
A1 ag
5
(iv)
At (1 ,
1
2
dy
=1−
dx
), ρ = (
1
4
⎛1⎞
c = ⎜⎜ 1 ⎟⎟ +
⎝2⎠
=
25
16
3
4
5
4
)2 =
25
16
⎛− 3 ⎞
, so nˆ = ⎜ 4 5 ⎟
⎜
⎟
⎝ 5 ⎠
Finding gradient
Correct normal vector (not
necessarily unit vector); may be in
terms of x
OR M2A1 for obtaining equation
of normal line at a general point
and differentiating partially
M1
A1
⎛− 3 ⎞
⎜ 5⎟
⎜ 4 ⎟
⎝ 5 ⎠
M1
⎛ 1 7 ⎞
,
⎟
⎝ 16 4 ⎠
Centre of curvature is ⎜
A1A1
5
38
4757
(v)
Mark Scheme
M1
A1
Differentiating partially w.r.t. p
0 = x 2 − 2 p ln x
p=
x2
x4
x4
−
and y =
2 ln x
2 ln x 4 ln x
y=
June 2007
M1
4
x
4 ln x
A1
4
39
4757
4 (i)
Mark Scheme
June 2007
M1
A1
M1
A1
By Lagrange’s theorem,
a proper subgroup has order 2 or 5
A group of prime order is cyclic
Hence every proper subgroup is cyclic
Using Lagrange (need not be
mentioned explicitly)
or equivalent
For completion
4
(ii)
M1
A1
e.g. 2 2 = 4 , 23 = 8 , 2 4 = 5 , 25 = 10 ,
26 = 9 , 27 = 7 , 28 = 3 , 29 = 6 , 210 = 1
A1
A1
2 has order 10, hence M is cyclic
(iii)
B1
B2
{ 1 , 10 }
{ 1, 3 , 4 , 5 , 9 }
(iv)
(v)
E is the identity
Considering order of an element
Identifying an element of order 10
(2, 6, 7 or 8)
Fully justified
For conclusion (can be awarded
4 after M1A1A0)
3 Ignore { 1 } and M
Deduct 1 mark (from B1B2) for
each (proper) subgroup given in
excess of 2
B1
M1
A, C, G, I are rotations
B, D, F, H, J are reflections
A1
A1
P and M are not isomorphic
M is abelian, P is non-abelian
B1
B1
Considering elements of order 2
(or equivalent) Implied by four
of B, D, F, H, J in the same set
Give A1 if one element is in the
4 wrong set; or if two elements are
interchanged
2
(vi)
Order
A B C D E
F
G H
I
J
5
2
5
5
2 B3
2
5
2
1
2
3
Valid reason
e.g. M has one element of order 2
P has more than one
Give B2 for 7 correct
B1 for 4 correct
Ignore { E } and P
(vii)
M1
{ E , B }, { E , D }, { E , F }, { E , H }, { E , J }
A1 ft
{E, A, C, G, I}
B2 cao
40
Subgroups of order 2
Using elements of order 2
(allow two errors/omissions)
Correct or ft. A0 if any others
given
Subgroups of order greater than 2
4 Deduct 1 mark (from B2) for each
extra subgroup given
4757
Mark Scheme
June 2007
Pre-multiplication by transition matrix
5 (i)
0 0. 4 0. 3 ⎞
⎛ 0
⎟
⎜
0 0. 6 0. 7 ⎟
⎜ 0
P=⎜
0.8 0.1 0
0 ⎟
⎟
⎜
⎜ 0. 2 0. 9 0
0 ⎟⎠
⎝
(ii)
(iii)
B2
Give B1 for two columns correct
2
0
0 ⎞
⎛ 0.3366 0.3317
⎟
⎜
0
.
6634
0
.
6683
0
0 ⎟
⎜
4
P =⎜
0
0
0.3366 0.3317 ⎟
⎟
⎜
⎜ 0
0
0.6634 0.6683 ⎟⎠
⎝
B2
Give B1 for two non-zero elements
correct to at least 2dp
0
0.3334 0.3333 ⎞
⎛ 0
⎟
⎜
0
0.6666 0.6667 ⎟
⎜ 0
P7 = ⎜
0.3335 0.3333
0
0 ⎟
⎟
⎜
⎜ 0.6665 0.6667
0
0 ⎟⎠
⎝
B2
Give B1 for two non-zero elements
4 correct to at least 2dp
M1
Using P 7 (or P 8 ) and initial probs
⎛ 0.4 ⎞ ⎛ 0.1000 ⎞
⎜ ⎟ ⎜
⎟
7 ⎜ 0.3 ⎟ ⎜ 0.2000 ⎟
P ⎜ ⎟=⎜
0.2334 ⎟
0. 2
⎜ ⎟ ⎜
⎟
⎜ 0.1 ⎟ ⎜ 0.4666 ⎟
⎝ ⎠ ⎝
⎠
P(8th letter is C) = 0.233
A1
2
(iv) 0.1000 × 0.3366 + 0.2000 × 0.6683
Using probabilities for 8th letter
Using diagonal elements from P 4
M1
M1
A1 ft
A1
+ 0.2334 × 0.3366 + 0.4666 × 0.6683
= 0.558
4
⎛ 0.4 ⎞ ⎛ 1 3 1 3 0 0 ⎞ ⎛ 0.4 ⎞ ⎛ 0.2333 ⎞
⎟⎜ ⎟ ⎜
⎜ ⎟ ⎜
⎟
(v)(A)
2
2
0 0 ⎟ ⎜ 0.3 ⎟ ⎜ 0.4667 ⎟
3
n ⎜ 0 .3 ⎟ ⎜ 3
=
P ⎜ ⎟≈⎜
0 0 1 3 1 3 ⎟ ⎜ 0.2 ⎟ ⎜ 0.1 ⎟
0. 2
⎟⎜ ⎟ ⎜
⎜ ⎟ ⎜
⎟
⎜ 0.1 ⎟ ⎜ 0 0 2 2 ⎟ ⎜ 0.1 ⎟ ⎜ 0.2 ⎟
⎝ ⎠ ⎝
⎠ ⎝
⎠
3
3⎠ ⎝
P( (n + 1) th letter is A) = 0.233
(B)
Approximating P n when n is large
and even
M1
A1
⎛ 0.4 ⎞ ⎛ 0 0 1 3 1 3 ⎞ ⎛ 0.4 ⎞ ⎛ 0.1 ⎞
⎟⎜ ⎟ ⎜
⎟
⎜ ⎟ ⎜
0 2 3 2 3 ⎟ ⎜ 0 .3 ⎟ ⎜ 0 .2 ⎟
n ⎜ 0 .3 ⎟ ⎜ 0
P ⎜ ⎟≈⎜1 1
=
0 0 ⎟ ⎜ 0.2 ⎟ ⎜ 0.2333 ⎟
0. 2
⎟⎜ ⎟ ⎜
⎟
⎜ ⎟ ⎜ 3 3
⎟
⎜ 0. 1 ⎟ ⎜ 2 2
⎟⎜ ⎟ ⎜
⎝ ⎠ ⎝ 3 3 0 0 ⎠ ⎝ 0.1 ⎠ ⎝ 0.4667 ⎠
P( (n + 1) th letter is A) = 0.1
Approximating P n when n is large
and odd
M1
A1
4
(vi)
0 0.4 0.3 ⎞
⎛ 0
⎟
⎜
0 0.6 0.6 ⎟
⎜ 0
Q=⎜
0.8 0.1 0
0 ⎟
⎟
⎜
⎜ 0.2 0.9 0 0.1 ⎟
⎠
⎝
B1
1
41
4757
(vii)
Mark Scheme
⎛ 0.1721
⎜
⎜ 0.3105
Qn → ⎜
0.1687
⎜
⎜ 0.3487
⎝
0.1721 0.1721 0.1721 ⎞
⎟
0.3105 0.3105 0.3105 ⎟
0.1687 0.1687 0.1687 ⎟
⎟
0.3487 0.3487 0.3487 ⎟⎠
Probabilities are 0.172, 0.310, 0.169, 0.349
June 2007
M1
Considering Q n for large n
OR at least two eqns for equilib
probs
M1
Probabilities from equal columns
OR solving to obtain equilib probs
A2
Give A1 for two correct
4
0.3487 × 0.1 × 0.1
(viii)
M1M1
A1
= 0.0035
Using 0.3487 and 0.1
3
42
4757
Mark Scheme
June 2007
Post-multiplication by transition matrix
5 (i)
0 0. 8 0. 2 ⎞
⎛ 0
⎟
⎜
0 0. 1 0. 9 ⎟
⎜ 0
P=⎜
0.4 0.6 0
0 ⎟
⎟
⎜
⎜ 0. 3 0 . 7 0
0 ⎟⎠
⎝
(ii)
(iii)
B2
Give B1 for two rows correct
2
0
0 ⎞
⎛ 0.3366 0.6634
⎟
⎜
0
.
3317
0
.
6683
0
0 ⎟
⎜
4
P =⎜
0
0
0.3366 0.6634 ⎟
⎟
⎜
⎜ 0
0
0.3317 0.6683 ⎟⎠
⎝
B2
Give B1 for two non-zero elements
correct to at least 2dp
0
0.3335 0.6665 ⎞
⎛ 0
⎟
⎜
0
0.3333 0.6667 ⎟
⎜ 0
P7 = ⎜
0.3334 0.6666
0
0 ⎟
⎟
⎜
⎜ 0.3333 0.6667
0
0 ⎟⎠
⎝
B2
Give B1 for two non-zero elements
4 correct to at least 2dp
(0.4
M1
Using P 7 (or P 8 ) and initial probs
0.3 0.2 0.1) P 7
= (0.1000 0.2000 0.2334 0.4666 )
A1
P(8th letter is C) = 0.233
2
(iv) 0.1000 × 0.3366 + 0.2000 × 0.6683
M1
M1A1 ft
+ 0.2334 × 0.3366 + 0.4666 × 0.6683
Using probabilities for 8th letter
Using diagonal elements from P 4
A1
= 0.558
4
(v)(A)
⎛ 13 2 3
⎜
⎜1 2
n
u P ≈ (0.4 0.3 0.2 0.1) ⎜ 3 3
0 0
⎜
⎜0 0
⎝
= (0.2333 0.4667 0.1 0.2 )
0⎞
⎟
0⎟
2 ⎟
3⎟
2 ⎟
3⎠
0
0
1
1
3
3
P( (n + 1) th letter is A) = 0.233
(B)
⎛0 0
⎜
⎜0 0
n
u P ≈ (0.4 0.3 0.2 0.1) ⎜ 1 2
⎜ 3 3
⎜1 2
⎝ 3 3
= (0.1 0.2 0.2333 0.4667 )
Approximating P n when n is large
and even
M1
A1
1
1
3
3
0
0
2
3⎞
2
3⎟
⎟
Approximating P n when n is large
and odd
M1
0⎟
⎟
0 ⎟⎠
A1
P( (n + 1) th letter is A) = 0.1
4
(vi)
0 0.8 0.2 ⎞
⎛ 0
⎜
⎟
0
0 0.1 0.9 ⎟
Q=⎜
⎜ 0.4 0.6 0
0 ⎟
⎜
⎟
⎝ 0.3 0.6 0 0.1 ⎠
B1
1
43
4757
(vii)
Mark Scheme
⎛ 0.1721
⎜
⎜ 0.1721
n
Q →⎜
0.1721
⎜
⎜ 0.1721
⎝
0.3105
0.3105
0.3105
0.3105
0.1687
0.1687
0.1687
0.1687
0.3487 ⎞
⎟
0.3487 ⎟
0.3487 ⎟
⎟
0.3487 ⎟⎠
Probabilities are 0.172, 0.310, 0.169, 0.349
June 2007
M1
Considering Q n for large n
OR at least two eqns for equilib
probs
M1
Probabilities from equal rows
OR solving to obtain equilib probs
A2
Give A1 for two correct
4
0.3487 × 0.1 × 0.1
(viii)
M1M1
A1
= 0.0035
Using 0.3487 and 0.1
3
44
4757
Mark Scheme
June 2008
4757 (FP3) Further Applications of Advanced Mathematics
1 (i)
⎛ 6 ⎞ ⎛ 10 ⎞ ⎛ 33 ⎞
⎜ ⎟ ⎜ ⎟ ⎜
⎟
AB × AC = ⎜ 8 ⎟ × ⎜ −5 ⎟ = ⎜ 44 ⎟
⎜ 5 ⎟ ⎜ 1 ⎟ ⎜ −110 ⎟
⎝ ⎠ ⎝ ⎠ ⎝
⎠
ABC is 3x + 4 y − 10 z = −9 + 20 − 20
3 x + 4 y − 10 z + 9 = 0
B2
Ignore subsequent working
Give B1 for one element correct
SC1 for minus the correct vector
M1
A1
For 3x + 4 y − 10 z
Accept 33x + 44 y − 110 z = −99 etc
4
(ii)
(iii)
M1
A1 ft
3 × 5 + 4 × 4 − 10 × 8 + 9
Distance is
2
2
2
3 + 4 + 10
40
8
= ( −)
( =
)
125
5
A1
⎛ 6 ⎞ ⎛ −2 ⎞ ⎛ 20 ⎞
⎛1⎞
⎜ ⎟ ⎜ ⎟ ⎜
⎟
⎜ ⎟
AB × CD = ⎜ 8 ⎟ × ⎜ 4 ⎟ = ⎜ −40 ⎟ [ = 20 ⎜ −2 ⎟ ]
⎜ 5 ⎟ ⎜ 5 ⎟ ⎜ 40 ⎟
⎜ 2⎟
⎝ ⎠ ⎝ ⎠ ⎝
⎠
⎝ ⎠
⎛ 10 ⎞ ⎛ 1 ⎞
⎜ ⎟ ⎜ ⎟
⎜ −5 ⎟ . ⎜ −2 ⎟
⎜1⎟ ⎜ 2⎟
Distance is AC . nˆ = ⎝ ⎠ ⎝ ⎠
12 + 22 + 22
22
=
3
Using distance formula (or other
complete method)
Condone negative answer
3 Accept a.r.t. 3.58
M1
Evaluating AB × CD or method
for finding end-points of
common perp PQ
A1
or P ( 3 2 , 11, 23 4 ) &
Q (7118 , 55 9 , 383 36)
or PQ = (22 9 , − 44 9 , 44 9 )
M1
A1
4
M1
(iv)
Volume is
1
6
(AB × AC) . AD
Scalar triple product
A1
⎛ 33 ⎞ ⎛ 8 ⎞
1⎜
⎟ ⎜ ⎟
44 ⎟ . ⎜ −1⎟
6 ⎜⎜
⎟ ⎜ ⎟
⎝ −110 ⎠ ⎝ 6 ⎠
220
= ( −)
3
=
M1
A1
Accept a.r.t. 73.3
4
(v)
E is (−3 + 10λ , 5 − 5λ , 2 + λ )
3(−3 + 10λ ) − 2(2 + λ ) + 5 = 0
M1
2
λ=
7
F is (−3 + 8μ , 5 − μ , 2 + 6μ )
3(−3 + 8μ ) − 2(2 + 6 μ ) + 5 = 0
2
μ=
3
Since 0 < λ < 1 , E is between A and C
Since 0 < μ < 1 , F is between A and D
A1
M1
A1
B1
5
31
4757
(vi)
Mark Scheme
June 2008
M1
VABEF = 16 (AB × AE) . AF
= 16 λμ (AB × AC) . AD
= λμ VABCD
=
Ratio of volumes is
4
17
:
21 21
= 4 : 17
2 (i)
(ii)
⎛ 7 ⎞
⎛4⎞
⎜ 3 ⎟
⎝
⎠
⎜9⎟
⎝ ⎠
4
21
Partial differentiation
Any correct form, ISW
4
M1
A1
Evaluating partial derivatives at
P
All correct
A1 ft
3 Condone omission of ‘ r = ‘
δ g ≈ 16 δ x − 4 δ y + 36 δ z
M1
Alternative:
M3 for substituting x = 7 + 4λ ,
…
into g = 125 + h and neglecting
M1
( = 392λ )
λ2
A1 ft for linear equation in
λ and h
A1 for n correct
A1 ft
M1
⎛4⎞
⎛4⎞
h ⎜ ⎟
1 ⎜ ⎟
−1 , so n =
−1
PQ ≈
392 ⎜⎜ ⎟⎟
392 ⎜⎜ ⎟⎟
⎝9⎠
⎝9⎠
A1
5
∂g ∂g
=
=0
∂x ∂y
M1
−2 x − 4 y = 0 and x + 2 y + 3z = 0
x + 2 y = 0 and z = 0
M1
Require
4 SC1 for 4 : 17 deduced from
without working
A1
h = δ g , so h ≈ 392λ
2
g( x , y , z ) = 0 − 0 = 0 ≠ 125
M1
A1
Hence there is no such point on S
(v)
Finding ratio of volumes of two
parts
A1
∂g
∂g
∂g
= 16,
= −4,
= 36
∂x
∂y
∂z
⎛4⎞
If PQ = λ ⎜⎜ −1⎟⎟ ,
⎜9⎟
⎝ ⎠
δ g ≈ 16(4λ ) − 4(−λ ) + 36(9λ )
(iv)
M1
M1
A1
Normal line is r = ⎜⎜ −7.5 ⎟⎟ + λ ⎜⎜ −1⎟⎟
(iii)
61
) ft if numerical
( 13 63
A1 ag
∂g
= 6 z − 2( x + 2 y + 3 z ) = −2 x − 4 y
∂x
∂g
= −4( x + 2 y + 3z )
∂y
∂g
= 6 x − 6( x + 2 y + 3 z ) = −12 y − 18 z
∂z
At P,
A1
4
V
21 ABCD
∂g
=0
∂z
∂g
∂g
=5
and
∂y
∂x
Require
Useful manipulation using both
eqns
Showing there is no such point
4 on S
Fully correct proof
M1
−4 x − 8 y − 12 z = 5(−2 x − 4 y )
32
∂g
∂g
= λ,
= 5λ
∂x
∂y
M1
Implied by
M1
This M1 can be awarded for
−2 x − 4 y = 1 and − 4 x − 8 y − 12 z = 5
4757
Mark Scheme
y = − 32 z and x = 5 z
June 2008
or z = − 23 y and x = − 103 y
A1
or y = − 103 x and z = 15 x
or x = − 54 λ , y = 83 λ , z = − 14 λ
6(5 z ) z − (5 z ) 2 = 125
M1
z = ±5
M1
Points are (25, − 7.5, 5)
and (−25, 7.5, − 5)
3 (i)
A1
A1 ft
2
ft is minus the other point,
8 provided all M marks have been
earned
B1
x 2 + y 2 = (24t 2 ) 2 + (18t − 8t 3 )2
4
or x : y : z = 10 : − 3 : 2
Substituting into g( x , y , z ) = 125
Obtaining one value of x, y, z or
λ
Dependent on previous M1
4
= 576t + 324t − 288t + 64t
6
= 324t 2 + 288t 4 + 64t 6
M1
= (18t + 8t 3 ) 2
A1 ag
2
⌠
Arc length is ⎮ (18t + 8t 3 ) dt
⌡0
M1
Note
2
= ⎡ 9t 2 + 2t 4 ⎤
⎣
⎦0
A1
= 68
A1
2
2
⌠
3
4
⎮ (18 + 8t ) dt = ⎡⎣ 18t + 2t ⎤⎦ 0 = 68
6 ⌡0
earns M1A0A0
M1
(ii) Curved surface area is ∫ 2π y ds
=
=
2
⌠
⎮ 2π (9t 2 − 2t 4 )(18t + 8t 3 ) dt
⎮
⌡0
2
⌠
⎮ π (324t 3 + 72t 5 − 32t 7 ) dt
⎮
⌡0
2
4
6
8 ⎤
⎡
=π
⎣
81t + 12t − 4t
= 1040π
Using ds = (18t + 8t 3 ) dt
Correct integral expression
including limits (may be implied
by later work)
M1
A1
M1
⎦0
M1
( ≈ 3267 )
A1
6
(iii)
κ=
=
=
xy − xy
3
2 2
( x2 + y )
2
=
(24t 2 )(18 − 24t 2 ) − (48t )(18t − 8t 3 )
2
3 3
(18t + 8t )
2
48t (9 − 12t − 18 + 8t )
8t 3 (9 + 4t 2 )3
−6
=
M1
A1A1
Using formula for κ (or ρ )
For numerator and denominator
M1
Simplifying the numerator
−48t 2 (9 + 4t 2 )
8t 3 (9 + 4t 2 )3
A1 ag
t (4t 2 + 9) 2
5
33
4757
(iv)
Mark Scheme
June 2008
6
When t = 1, x = 8, y = 7 , κ = − 169
ρ = ( −)
169
6
B1
3
dy y 18t − 8t
10
= =
=
dx x
24
24t 2
⎛ 513 ⎞
nˆ = ⎜
12 ⎟
⎝ − 13 ⎠
M1
Finding gradient (or tangent
vector)
M1
A1
⎛ 513 ⎞
⎛8⎞
c = ⎜ ⎟ + 169
6 ⎜ 12 ⎟
7
⎝ ⎠
⎝ − 13 ⎠
Finding direction of the normal
Correct unit normal (either
direction)
M1
Centre of curvature is ( 18 56 , − 19 )
A1A1
7
B1
4 (i) Commutative: x ∗ y = y ∗ x (for all x, y)
Associative: ( x ∗ y ) ∗ z = x ∗ ( y ∗ z )
(for all x, y, z)
(ii)
3 Give B1 for a partial
explanation, e.g.
‘Position of brackets does not
matter’
2( x + 12 )( y + 12 ) − 12 = 2 xy + x + y + 12 − 12
B1 ag
= 2 xy + x + y = x ∗ y
x+
2( x +
> 0 and y +
1
2
1
)( y + 12 ) − 12
2
> − , so x ∗ y ∈ S
> 0, so 2( x +
1
)( y + 12 )
2
Intermediate step required
1
M1
(iii)(A) If x , y ∈ S then x > − 12 and y > − 12
1
2
Accept e.g. ‘Order does not
matter’
B2
>0
1
2
A1
A1
3
(B) 0 is the identity
since 0 ∗ x = 0 + x + 0 = x
B1
B1
If x ∈ S and x ∗ y = 0 then
2 xy + x + y = 0
y=
y + 12 =
−x
2x + 1
1
>0
2(2 x + 1)
(since x > − 12 )
M1
or 2( x + 12 )( y + 12 ) − 12 = 0
A1
or y + 12 =
M1
so y ∈ S
Dependent on M1A1M1
A1
6
S is closed and associative; there is an
identity; and every element of S has an
inverse in S
(iv)
1
4( x + 12 )
M1
If x ∗ x = 0, 2 x 2 + x + x = 0
x = 0 or − 1
0 is the identity (and has order 1)
−1 is not in S
A1
A1
3
34
4757
(v)
Mark Scheme
4 ∗ 6 = 48 + 4 + 6 = 58
June 2008
B1
= 56 + 2 = 7 × 8 + 2
So 4 6 = 2
B1 ag
2
(vi)
(vii)
Element
0
1 2
4
5
6
Order
1
6 6
3
3
2
B3
3
B1
B1
B1
{0}, G
{ 0, 6 }
{ 0, 4, 5 }
35
Give B2 for 4 correct
B1 for 2 correct
Condone omission of G
If more than 2 non-trivial
subgroups are given, deduct 1
3
mark (from final B1B1) for each
non-trivial subgroup in excess
of 2
4757
Mark Scheme
June 2008
Pre-multiplication by transition matrix
⎛ 0.1 0.7 0.1 ⎞
⎜
⎟
P = ⎜ 0.4 0.2 0.6 ⎟
⎜ 0.5 0.1 0.3 ⎟
⎝
⎠
5 (i)
(ii)
B2
Give B1 for two columns correct
2
M1
⎛ 1 3 ⎞ ⎛ 0.328864 ⎞
⎜ ⎟ ⎜
⎟
P ⎜ 1 3 ⎟ = ⎜ 0.381536 ⎟
⎜ 1 ⎟ ⎜ 0.2896 ⎟
⎠
⎝ 3⎠ ⎝
6
Using P 6 (or P 7 )
For matrix of initial probabilities
For evaluating matrix product
Accept 0.381 to 0.382
M1
M1
A1
P(B used on 7th day) = 0.3815
4
M1
M1
A1
(iii)
0.328864 × 0.1 + 0.381536 × 0.2 + 0.2896 × 0.3
= 0.1961
Using diagonal elements from P
Correct method
Accept a.r.t. 0.196
3
(iv)
⎛ 0.352 0.328 0.304 ⎞
⎜
⎟
P 3 = ⎜ 0.364 0.404 0.372 ⎟
⎜ 0.284 0.268 0.324 ⎟
⎝
⎠
0.328864 × 0.352 + 0.381536 × 0.404 + 0.2896 × 0.324
= 0.3637
(v)
M1
For evaluating P3
M1
M1
A1
Using diagonal elements from
P3
Correct method
4 Accept a.r.t. 0.364
Deduct 1 if not given as a (3×3)
matrix
Deduct 1 if not 4 dp
B1
⎛ 0.3289 0.3289 0.3289 ⎞
⎜
⎟
Q = ⎜ 0.3816 0.3816 0.3816 ⎟
⎜ 0.2895 0.2895 0.2895 ⎟
⎝
⎠
B1
B1
0.3289, 0.3816, 0.2895 are the long-run
probabilities for the routes A, B, C
B1
4 Accept ‘equilibrium probabilities’
(vi) ⎛ 0.1 0.7 a ⎞ ⎛ 0.4 ⎞ ⎛ 0.4 ⎞
⎜
⎟⎜ ⎟ ⎜ ⎟
⎜ 0.4 0.2 b ⎟ ⎜ 0.2 ⎟ = ⎜ 0.2 ⎟
⎜ 0.5 0.1 c ⎟ ⎜ 0.4 ⎟ ⎜ 0.4 ⎟
⎝
⎠⎝ ⎠ ⎝ ⎠
M1
0.04 + 0.14 + 0.4a = 0.4, so a = 0.55
0.16 + 0.04 + 0.4b = 0.2, so b = 0
0.2 + 0.02 + 0.4c = 0.4, so c = 0.45
M1
Obtaining a value for a, b or c
A2
Give A1 for one correct
4
After C, routes A, B, C will be used with
probabilities 0.55, 0, 0.45
M1
M1
A1
(vii)
0.4 × 0.1 + 0.2 × 0.2 + 0.4 × 0.45
= 0.26
36
Using long-run probs 0.4, 0.2,
0.4
Using diag elements from new
3 matrix
4757
Mark Scheme
June 2008
Post-multiplication by transition matrix
⎛ 0.1 0.4 0.5 ⎞
⎜
⎟
P = ⎜ 0.7 0.2 0.1 ⎟
⎜ 0.1 0.6 0.3 ⎟
⎝
⎠
5 (i)
B2
Give B1 for two rows correct
2
M1
(ii)
( 13
1
3
1
3
) P6 = ( 0.328864
0.381536 0.2896 )
P(B used on 7th day) = 0.3815
Using P 6 (or P 7 )
For matrix of initial probabilities
For evaluating matrix product
Accept 0.381 to 0.382
M1
M1
A1
4
M1
M1
A1
(iii)
0.328864 × 0.1 + 0.381536 × 0.2 + 0.2896 × 0.3
= 0.1961
Using diagonal elements from P
Correct method
Accept a.r.t. 0.196
3
(iv)
⎛ 0.352 0.364 0.284 ⎞
⎜
⎟
P = ⎜ 0.328 0.404 0.268 ⎟
⎜ 0.304 0.372 0.324 ⎟
⎝
⎠
M1
For evaluating P3
M1
M1
A1
Using diagonal elements from
0.328864 × 0.352 + 0.381536 × 0.404 + 0.2896 × 0.324
= 0.3637
⎛ 0.3289 0.3816 0.2895 ⎞
⎜
⎟
Q = ⎜ 0.3289 0.3816 0.2895 ⎟
⎜ 0.3289 0.3816 0.2895 ⎟
⎝
⎠
B1B1B1
0.3289, 0.3816, 0.2895 are the long-run
probabilities for the routes A, B, C
B1
3
(v)
(vi)
( 0.4
P3
Correct method
4 Accept a.r.t. 0.364
Deduct 1 if not given as a (3×3)
matrix
Deduct 1 if not 4 dp
4 Accept ‘equilibrium probabilities’
⎛ 0.1 0.4 0.5 ⎞
⎜
⎟
0.2 0.4 ) ⎜ 0.7 0.2 0.1 ⎟ = ( 0.4 0.2 0.4 )
⎜ a
b
c ⎟⎠
⎝
0.04 + 0.14 + 0.4a = 0.4, so a = 0.55
M1
M1
0.16 + 0.04 + 0.4b = 0.2, so b = 0
0.2 + 0.02 + 0.4c = 0.4, so c = 0.45
Obtaining a value for a, b or c
A2
Give A1 for one correct
4
After C, routes A, B, C will be used with
probabilities 0.55, 0, 0.45
M1
M1
A1
(vii)
0.4 × 0.1 + 0.2 × 0.2 + 0.4 × 0.45
= 0.26
37
Using long-run probs 0.4, 0.2,
0.4
Using diag elements from new
3 matrix
4757
Mark Scheme
June 2009
4757 Further Pure 3
1 (i) Putting x = 0 ,
M1
A1
−3 y + 10 z = 6, − 4 y − 2 z = 8
y = −2, z = 0
Finding coords of a point on the
line
or (2, 0, − 1) , (1, − 1, − 1 2 ) etc
⎛8⎞ ⎛ 3⎞
Direction is given by ⎜⎜ −3 ⎟⎟ × ⎜⎜ −4 ⎟⎟
M1
⎜ 10 ⎟ ⎜ −2 ⎟
⎝ ⎠ ⎝ ⎠
⎛ 46 ⎞
⎜
⎟
= ⎜ 46 ⎟
⎜ −23 ⎟
⎝
⎠
⎛ 0⎞
⎛2⎞
⎜ 0⎟
⎝ ⎠
⎜ −1 ⎟
⎝ ⎠
Equation of L is r = ⎜⎜ −2 ⎟⎟ + λ ⎜⎜ 2 ⎟⎟
(ii)
or finding a second point
A1
A1 ft
5 Dependent on M1M1
Accept any form
Condone omission of ‘r =’
JJJG
⎛ 7 ⎞ ⎛2⎞ ⎛6⎞
JJJG
⎜
⎟ ⎜ ⎟ ⎜ ⎟
AB × d = ⎜ −14 ⎟ × ⎜ 2 ⎟ = ⎜ 15 ⎟
⎜ 4 ⎟ ⎜ −1⎟ ⎜ 42 ⎟
⎝
⎠ ⎝ ⎠ ⎝ ⎠
M1
A2 ft
Evaluating AB× d
Give A1 ft if just one error
⎡
⎢
Distance is ⎢
⎢
⎣
M1
Appropriate scalar product
A1 ft
Fully correct expression
⎛2⎞
⎜ ⎟
[ =3⎜ 5 ⎟ ]
⎜ 14 ⎟
⎝ ⎠
⎛ −1⎞ ⎛ 2 ⎞
⎜ ⎟ ⎜ ⎟
⎜ 14 ⎟ . ⎜ 5 ⎟
⎛ −1 ⎞ ⎛ 0 ⎞ ⎤
⎜ 5 ⎟ ⎜14 ⎟
⎜ ⎟ ⎜ ⎟⎥ ˆ
⎝ ⎠ ⎝ ⎠
⎜ 12 ⎟ − ⎜ −2 ⎟ ⎥ . n =
22 + 52 + 142
⎜5⎟ ⎜ 0⎟⎥
⎝ ⎠ ⎝ ⎠⎦
138 46
=
=
= 9.2
15
5
A1
6
(iii)
⎛6⎞
JJJG
⎜ ⎟
AB× d = ⎜ 15 ⎟ = 62 + 152 + 422
⎜ 42 ⎟
⎝ ⎠
JJJG
AB × d
62 + 152 + 422
Distance is
=
d
22 + 22 + 12
45
=
= 15
3
JJJG
M1
For AB× d
M1
Evaluating magnitude
In this part, M marks are
dependent on previous M marks
M1A1 ft
A1
5
32
4757
(iv)
Mark Scheme
⎛ −1⎞
⎛ k + 1⎞
⎛6⎞
⎛2⎞
⎜5⎟
⎝ ⎠
⎜ −3 ⎟
⎝
⎠
⎜9⎟
⎝ ⎠
⎜ −1⎟
⎝ ⎠
At D, ⎜⎜ 12 ⎟⎟ + λ ⎜⎜ −12 ⎟⎟ = ⎜⎜ −2 ⎟⎟ + μ ⎜⎜ 2 ⎟⎟
M1
Condone use of same
parameter on both sides
A1 ft
Two equations for λ and μ
M1M1
Obtaining λ and μ
(numerically)
Give M1 for λ and μ in terms of
k
Equation for k
12 − 12λ = −2 + 2 μ
5 − 3λ = 9 − μ
λ = , μ =5
1
3
−1 + 13 (k + 1) = 6 + 10
M1
k = 50
A1
D is (6 + 2μ , − 2 + 2μ , 9 − μ )
i.e. (16, 8, 4)
M1
A1
33
June 2009
8 Obtaining coordinates of D
4757
Mark Scheme
Alternative solutions for Q1
1 e.g. 23x − 23 y = 46
(i)
x = t, y = t −2
June 2009
M1A1
3t − 4(t − 2) − 2 z = 8
M1A1 ft
x = t , y = t − 2, z = − 12 t
A1
Eliminating
one of x, y, z
5
⎛ −1 + 7 μ ⎞ ⎛ 2λ ⎞
JJJG
JJJG
JJJG JJJG
(ii) PQ = ⎜12 − 14 μ ⎟ − ⎜ −2 + 2λ ⎟ PQ . d = PQ . AB = 0
⎜
⎟ ⎜
⎟
⎜ 5 + 4 μ ⎟ ⎜ −λ ⎟
⎝
⎠ ⎝
⎠
M1
2(−1 + 7 μ − 2λ ) + 2(14 − 14μ − 2λ ) − (5 + 4 μ + λ ) = 0
A1 ft
7(−1 + 7 μ − 2λ ) − 14(14 − 14μ − 2λ ) + 4(5 + 4 μ + λ ) = 0 λ = 27 / 25 , μ = 47 / 75
A1 ft
JJJG
PQ = (92 / 75) 2 + (230 / 75) 2 + (644 / 75) 2 = 9.2
M1A1 ft
A1
⎛ 6 + 2λ + 1 ⎞ ⎛ 2 ⎞
JJJG
(iii) AX ⋅ d = ⎜ −2 + 2λ − 12 ⎟ ⋅ ⎜ 2 ⎟ = 0
⎜
⎟ ⎜ ⎟
⎜ 9 − λ − 5 ⎟ ⎜ −1 ⎟
⎝
⎠ ⎝ ⎠
Two
equations for
λ and μ
6 Expression
for shortest
distance
M1
2(7 + 2λ ) + 2(2λ − 14) − (4 − λ ) = 0
λ=2
11
⎛
⎞
JJJG ⎜
⎟
AX = ⎜ −10 ⎟
⎜ 2 ⎟
⎝
⎠
A1 ft
M1
AX = 112 + 102 + 22
M1
A1
= 15
5
34
4757
Mark Scheme
June 2009
⎛ 7 ⎞ ⎡ ⎛ k + 1⎞ ⎛ 2 ⎞ ⎤
(iv) ⎜ −14 ⎟ ⋅ ⎢ ⎜ −12 ⎟ × ⎜ 2 ⎟ ⎥ = 0
⎜
⎟ ⎢⎜
⎟ ⎜ ⎟⎥
Appropriate
scalar triple
product
equated to
zero
M1
⎜ 4 ⎟ ⎢ ⎜ −3 ⎟ ⎜ − 1 ⎟ ⎥
⎝
⎠ ⎣⎝
⎠ ⎝ ⎠⎦
⎛ 7 ⎞ ⎛ 18 ⎞
⎜
⎟ ⎜
⎟
⎜ −14 ⎟ ⋅ ⎜ k − 5 ⎟ = 0
⎜ 4 ⎟ ⎜ 2k + 26 ⎟
⎝
⎠ ⎝
⎠
126 − 14k + 70 + 8k + 104 = 0
k = 50
M1
A1
⎛ −1⎞
⎛ 51 ⎞ ⎛ 6 ⎞
⎛2⎞
At D, ⎜⎜ 12 ⎟⎟ + λ ⎜⎜ −12 ⎟⎟ = ⎜⎜ −2 ⎟⎟ + μ ⎜⎜ 2 ⎟⎟
⎜5⎟
⎜ −3 ⎟ ⎜ 9 ⎟
⎜ −1⎟
⎝ ⎠
⎝
⎠ ⎝ ⎠
⎝ ⎠
Equation for
k
M1
−1 + 51λ = 6 + 2 μ
12 − 12λ = −2 + 2μ
5 − 3λ = 9 − μ
A1 ft
Condone use
of same
parameter on
both sides
λ = , μ =5
1
3
M1
D is (6 + 2μ , − 2 + 2μ , 9 − μ )
i.e. (16, 8, 4)
M1
A1
8
Two
equations for
λ and μ
Obtaining
λ or μ
Obtaining
coordinates
of D
35
4757
Mark Scheme
M1
A2
2 (i) ∂z
= 3( x + y )3 + 9 x( x + y ) 2 − 6 x 2 + 24
∂x
∂z
= 9 x( x + y )2
∂y
(ii)
At stationary points,
June 2009
Partial differentiation
Give A1 if just one minor error
A1
4
∂z
∂z
= 0 and
=0
∂x
∂y
M1
2
9 x( x + y ) = 0 ⇒ x = 0 or y = − x
If x = 0 then 3 y 3 + 24 = 0
M1
y = −2 ; one stationary point is (0, − 2, 0)
A1A1
If y = − x then −6 x 2 + 24 = 0
M1
x = ±2 ; stationary points are (2, − 2, 32)
A1
and
A1
(−2, 2, − 32)
(iii)
At P (1, − 2, 19) ,
∂z
∂z
= 24 ,
=9
∂x
∂y
⎛1⎞
⎛ 24 ⎞
⎜ 19 ⎟
⎝ ⎠
⎜ −1 ⎟
⎝ ⎠
B1
M1
A1 ft
Normal line is r = ⎜⎜ −2 ⎟⎟ + λ ⎜⎜ 9 ⎟⎟
(iv)
If A0A0, give A1 for x = ±2
7
∂z
∂z
δx+ δ y
∂x
∂y
= 24 δ x + 9 δ y
δz ≈
For normal vector (allow sign
error)
3 Condone omission of ‘r =’
M1
A1 ft
3h ≈ 24k + 9h
k ≈ − 14 h
M1
A1
4
OR Tangent plane is 24 x + 9 y − z = −13
24(1 + k ) + 9(−2 + h) − (19 + 3h) ≈ −13 M2A1 ft
A1
k ≈ − 14 h
(v) ∂z
∂x
= 27 and
∂z
=0
∂y
(Allow M1 for
M1
9 x( x + y ) 2 = 0 ⇒ x = 0 or y = − x
If x = 0 then 3 y 3 + 24 = 27
y = 1, z = 0 ; point is (0, 1, 0)
M1
A1
d =0
If y = − x then −6 x 2 + 24 = 27
A1
M1
x 2 = − 12 ; there are no other points
A1
6
36
∂z
= −27 )
∂x
4757
Mark Scheme
June 2009
3 (i)
2
Forming
M1
A1
2
⎛ dx ⎞ ⎛ dy ⎞
2
2
⎜ dθ ⎟ + ⎜ dθ ⎟ = [a (1 + cos θ )] + (a sin θ )
⎝
⎠ ⎝
⎠
2
⎛ dx ⎞ ⎛ dy ⎞
⎜ dθ ⎟ + ⎜ dθ ⎟
⎝
⎠ ⎝
⎠
2
= a 2 (2 + 2 cos θ )
= 4a 2 cos 2 12 θ
s = ∫ 2a cos
= 4a sin
1
θ
2
1
θ
2
dθ
+C
s = 0 when θ = 0 ⇒ C = 0
(ii) dy
dx
=
=
2sin
cos
Using half-angle formula
M1
A1
Integrating to obtain k sin 12 θ
Correctly obtained (+C not
needed)
A1 ag
a sin θ
a(1 + cos θ )
1θ
2
M1
6 Dependent on all previous
marks
M1
1θ
2
2a cos 2 12 θ
M1
A1
= tan 12 θ
ψ = 12 θ , and so s = 4a sinψ
Using half-angle formulae
A1
4
(iii)
ρ=
M1
A1 ft
ds
= 4a cosψ
dψ
= 4a cos 12 θ
Differentiating intrinsic equation
A1 ag
3
OR
ρ=
=
( 4a
2
cos 2 12 θ
)
3
2
a(1 + cos θ )(a cos θ ) − (−a sin θ )(a sin θ )
8a 3 cos3 12 θ
a 2 (1 + cos θ )
=
8a 3 cos3 12 θ
2a 2 cos 2 12 θ
M1A1 ft
= 4a cos 12 θ
A1 ag
(iv) When
θ = 23 π , ψ = 13 π , x = a( 23 π + 12 3) , y = 23 a
ρ = 2a
B1
M1
A1
⎛ − sinψ ⎞ ⎛ − 12 3 ⎞
nˆ = ⎜
⎟=⎜ 1 ⎟
⎝ cosψ ⎠ ⎜⎝ 2 ⎟⎠
⎛ a( 2 π + 1 3) ⎞
⎛− 1 3⎞
2
⎟ + 2a ⎜ 21 ⎟
c=⎜ 3
3
⎜
⎟
⎜
⎟
a
⎝ 2 ⎠
2
⎝
⎠
Centre of curvature is ( a ( 23 π − 12 3) ,
Correct expression for ρ or κ
Obtaining a normal vector
Correct unit normal
(possibly in terms of θ )
M1
5
2
a)
Accept (1.23a , 2.5a)
A1A1
6
37
4757
Mark Scheme
M1
(v) Curved surface area is ∫ 2π y ds
⌠
π
= ⎮⎮ 2π a(1 − cos θ ) 2a cos 12 θ dθ
π
= ⎮⎮ 8π a 2 sin 2 12 θ cos 12 θ dθ
M1
⌡0
=⎡
⎣
Correct integral expression in
any form
(including limits; may be implied
by later working)
Obtaining an integrable form
A1 ft
⌡0
⌠
June 2009
16
π a2
3
π
sin 3 12 θ ⎤
⎦0
M1
= 163 π a 2
Obtaining k sin 3 12 θ
equivalent
A1
5
38
or
4757
Mark Scheme
3
4
5
All powers of an element of
order 6
M1
4 (i) In G, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1
2
June 2009
6
[ or 5 = 4, 5 = 6 , 5 = 2, 5 = 3, 5 = 1 ]
In H, 52 = 7 , 53 = 17 , 54 = 13, 55 = 11, 56 = 1
[ or 112 = 13, 113 = 17 , 114 = 7 , 115 = 5, 116 = 1
]
(iii)
4
{ 1, 6 }
{ 1, 2, 4 }
G
Ignore { 1 } and G
Deduct 1 mark (from B1B2) for
3 each proper subgroup in excess
of two
B1
B2
H
G
1 ↔ 1
2 ↔ 7
H
1 ↔ 1
2 ↔ 13
OR
3 ↔ 5
4 ↔ 13
All powers correct in both
groups
B1
B1
G has an element 3 (or 5) of order 6
H has an element 5 (or 11) of order 6
(ii)
A1
B4
Give B3 for 4 correct, B2 for 3
4 correct,
B1 for 2 correct
3 ↔ 11
4 ↔ 7
5 ↔ 11
6 ↔ 17
5 ↔ 5
6 ↔ 17
M1
(iv) ad(1) = a(3) = 1
Evaluating e.g. ad(1) (one case
sufficient; intermediate value
must be shown)
For ad = c correctly shown
Evaluating e.g. da(1) (one case
sufficient; no need for any
working)
ad(2) = a(2) = 3
ad(3) = a(1) = 2 , so ad = c
A1
da(1) = d(2) = 2
da(2) = d(3) = 1
M1
da(3) = d(1) = 3 , so da = f
A1
4
(v) S is not abelian; G is abelian
B1
1
or S has 3 elements of order 2;
G has 1 element of order 2
or S is not cyclic etc
(vi) Eleme
nt
a
b
c
d
e
f
Order
3
3
2
2
1
2
B4
Give B3 for 5 correct, B2 for 3
4 correct,
B1 for 1 correct
B1B1B1
B1
(vii) { e , c }, { e , d }, { e, f }
{ e, a , b }
39
Ignore { e } and S
If more than 4 proper subgroups
4 are given, deduct 1 mark for
each proper subgroup in excess
of 4
4757
Mark Scheme
June 2009
Pre-multiplication by transition matrix
5 (i)
⎛ 0 0.1 0 0.3 ⎞
⎜
⎟
0.7 0.8 0 0.6 ⎟
P=⎜
⎜ 0.1 0 1 0.1 ⎟
⎜
⎟
⎝ 0.2 0.1 0 0 ⎠
(ii)
B2
Give B1 for two columns correct
2
M1
⎛ 0.6 ⎞ ⎛ 0.0810 ⎞
⎜ ⎟ ⎜
⎟
13 ⎜ 0.4 ⎟ ⎜ 0.5684 ⎟
P
=
⎜ 0 ⎟ ⎜ 0.2760 ⎟
⎜ ⎟ ⎜
⎟
⎝ 0 ⎠ ⎝ 0.0746 ⎠
A2
(iii) 0.5684 × 0.8 + 0.2760
Using P13 (or P14 )
Give A1 for 2 probabilities
3 correct
(Max A1 if not at least 3dp)
Tolerance ±0.0001
M1M1
A1 ft
= 0.731
For 0.5684 × 0.8 and 0.2760
Accept 0.73 to 0.7312
3
(iv)
⎛ 0.6 ⎞ ⎛ . ⎞
⎜ ⎟ ⎜
⎟
0.4
. ⎟
P 30 ⎜ ⎟ = ⎜
,
⎜ 0 ⎟ ⎜ 0.4996 ⎟
⎜ ⎟ ⎜
⎟
⎝ 0 ⎠ ⎝ . ⎠
⎛ 0.6 ⎞ ⎛ . ⎞
⎜ ⎟ ⎜
⎟
0.4
. ⎟
P 31 ⎜ ⎟ = ⎜
⎜ 0 ⎟ ⎜ 0.5103 ⎟
⎜ ⎟ ⎜
⎟
⎝ 0 ⎠ ⎝ . ⎠
Finding P(C) for some powers of
P
For identifying P31
M1
A1
Level 32
A1
3
(v) Expected number of levels including the next M1
1
=5
change of location is
0.2
A1
Expected number of further levels in B is 4
A1
For 1/(1 − 0.8) or 0.8 / (1 − 0.8)
For 5 or 4
For 4 as final answer
3
(vi)
⎛ 0 0.1 0 0.3 ⎞
⎜
⎟
0.7 0.8 0 0.6 ⎟
Q=⎜
⎜ 0.1 0 0.9 0.1 ⎟
⎜
⎟
⎝ 0.2 0.1 0.1 0 ⎠
⎛ 0.0916 0.0916 0.0916
⎜
0.6183 0.6183 0.6183
Qn → ⎜
⎜ 0.1908 0.1908 0.1908
⎜
⎝ 0.0992 0.0992 0.0992
B1
0.0916 ⎞
⎟
0.6183 ⎟
0.1908 ⎟
⎟
0.0992 ⎠
A: 0.0916 B: 0.6183 C: 0.1908 D:
0.0992
M1
M1
A2
Can be implied
Evaluating powers of Q
or Obtaining (at least) 3
equations
from Qp = p
Limiting matrix with equal
columns
or Solving to obtain one equilib
5 prob
or M2 for other complete
method
Give A1 for two correct
(Max A1 if not at least 3dp)
Tolerance ±0.0001
40
4757
Mark Scheme
0.1 a 0.3 ⎞ ⎛ 0.11 ⎞ ⎛ 0.11 ⎞
⎜
⎟⎜
⎟ ⎜
⎟
b 0.6 ⎟ ⎜ 0.75 ⎟ ⎜ 0.75 ⎟
0.7
0.8
⎜
=
⎜ 0.1 0 c 0.1 ⎟ ⎜ 0.04 ⎟ ⎜ 0.04 ⎟
⎜
⎟⎜
⎟ ⎜
⎟
⎝ 0.2 0.1 d 0 ⎠ ⎝ 0.1 ⎠ ⎝ 0.1 ⎠
June 2009
(vii) ⎛ 0
⎛ 0.11 ⎞
⎜
⎟
0.75 ⎟
Transition matrix and ⎜
⎜ 0.04 ⎟
⎜
⎟
⎝ 0.1 ⎠
M1
A1
0.075 + 0.04a + 0.03 = 0.11
Forming at least one equation
M1
0.077 + 0.6 + 0.04b + 0.06 = 0.75
0.011 + 0.04c + 0.01 = 0.04
0.022 + 0.075 + 0.04d = 0.1
or a + b + c + d = 1
a = 0.125, b = 0.325, c = 0.475, d = 0.075
Give A1 for two correct
A2
5
41
4757
Mark Scheme
June 2009
Post-multiplication by transition matrix
5 (i)
⎛ 0 0.7 0.1 0.2 ⎞
⎜
⎟
0.1 0.8 0 0.1 ⎟
P=⎜
⎜ 0
0
1
0 ⎟
⎜
⎟
⎝ 0.3 0.6 0.1 0 ⎠
(ii)
( 0.6
B2
Give B1 for two rows correct
2
0.4 0 0 ) P13
M1
= ( 0.0810 0.5684 0.2760 0.0746 )
A2
(iii) 0.5684 × 0.8 + 0.2760
Using P13 (or P14 )
Give A1 for 2 probabilities
3 correct
(Max A1 if not at least 3dp)
Tolerance ±0.0001
M1M1
A1 ft
= 0.731
For 0.5684 × 0.8 and 0.2760
Accept 0.73 to 0.7312
3
(iv)
( 0.6
( 0.6
0.4 0 0 ) P
30
= (. . 0.4996 .)
0.4 0 0 ) P
31
= (. . 0.5103 .)
Finding P(C) for some powers of
P
For identifying P31
M1
A1
Level 32
A1
3
(v) Expected number of levels including the next M1
1
change of location is
=5
0.2
A1
Expected number of further levels in B is 4
A1
For 1/(1 − 0.8) or 0.8 / (1 − 0.8)
For 5 or 4
For 4 as final answer
3
(vi)
⎛ 0 0.7 0.1 0.2 ⎞
⎜
⎟
0.1 0.8 0 0.1 ⎟
Q=⎜
⎜ 0
0 0.9 0.1 ⎟
⎜
⎟
0.3
0.6
0.1 0 ⎠
⎝
⎛ 0.0916 0.6183 0.1908
⎜
0.0916 0.6183 0.1908
n
Q →⎜
⎜ 0.0916 0.6183 0.1908
⎜
⎝ 0.0916 0.6183 0.1908
B1
0.0992 ⎞
⎟
0.0992 ⎟
0.0992 ⎟
⎟
0.0992 ⎠
A: 0.0916 B: 0.6183 C: 0.1908 D:
0.0992
M1
M1
A2
Can be implied
Evaluating powers of Q
or Obtaining (at least) 3
equations
from pQ = p
Limiting matrix with equal rows
or Solving to obtain one equilib
prob
5 or M2 for other complete
method
Give A1 for two correct
(Max A1 if not at least 3dp)
Tolerance ±0.0001
42
4757
(vii)
Mark Scheme
⎛ 0 0.7 0.1 0.2 ⎞
⎜
⎟
0.1 0.8 0 0.1 ⎟
( 0.11 0.75 0.04 0.1) ⎜⎜
a
b
c
d ⎟
⎜
⎟
0.3
0.6
0.1
0 ⎠
⎝
= ( 0.11 0.75 0.04 0.1)
June 2009
M1
Transition matrix and
( 0.11
0.75 0.04 0.1)
A1
0.075 + 0.04a + 0.03 = 0.11
Forming at least one equation
M1
0.077 + 0.6 + 0.04b + 0.06 = 0.75
0.011 + 0.04c + 0.01 = 0.04
0.022 + 0.075 + 0.04d = 0.1
or a + b + c + d = 1
a = 0.125, b = 0.325, c = 0.475, d = 0.075
Give A1 for two correct
A2
5
43
4757
1 (i)
Mark Scheme
 5   2   42 
  
   

AC  AB   8    1    42 
 26   2   21 

   

 
AC  AB
Perpendicular distance is

AB

B2
Give B1 for one component correct
M1
422  422  212
2 1  2
2
June 2010
2
2

63
3
M1
 21
Calculating magnitude of a vector
product
A1
www
5
  3  2   8    2 

    
OR   8      0   .  1   0
  27  2   1    2 


2(2  5)  (  8)  2(2  26)  0
  6 [ F is (15, 14, 15) ]
CF  7 2  142  142  21
(ii)
M1
A1 ft
M1A1
 2   3   3 p 1 
    
 

AB  CD   1    p    2 p  4 
 2   p  1  2 p  3 
  
 

 5   3 p 1 
   
 

AC . (AB  CD)   8  .  2 p  4 
 26   2 p  3 

 

 5(3 p  1)  8(2 p  4)  26(2 p  3) [  21 p  105 ]
 
AB  CD  (3 p  1) 2  (2 p  4) 2  (2 p  3) 2
 17 p 2  2 p  26
  
AC . (AB  CD)
21 p  5
Distance is

 
AB  CD
17 p 2  2 p  26
(iii)
 42   8 
  


1
V  (  ) (AC  AB).AD  (  ) 6  42  .  p  8 
 21   p  27 



7
 (  ) 56  7( p  8)  2 ( p  27)
1
6
 ()

(iv)
7
2
35
2
Appropriate scalar product
A1
 72 p
B1
B1
B1
M1
A1 ft
B1 ft
M1A1 (ag) Correctly obtained
8
M1
Appropriate scalar triple product
A1 ft
In any form
M1
A1
4
p 5
Intersect when p  5
B1
3
 2  8
 3
 
   
 
 8     1   0   5
 27 
 2   1 
 4
 
   
 
3  2  8  3
8    5
27  2  1  4 
Evaluation of scalar triple product
Dependent on previous M1
1
(105  21 p ) or better
6
[ 8    p ]
[ 27  2  1  ( p  1)  ]
  7,   3
Point of intersection is (17 , 15, 13)
B1 ft
Equations of both lines (may involve p)
M1
Equation for intersection (must have
different parameters)
A1 ft
Equation involving  and 
Second equation involving  and 
or Two equations in  ,  , p
A1 ft
Obtaining  or 
M1
A1
7
1
4757
2 (i)
(ii)
Mark Scheme
g
 ( y  xy  z 2 ) e x  2 y
x
g
 ( x  2 xy  2 z 2 ) e x  2 y
y
g
 2 z e x2 y
z
At (2, 1,  1) ,
June 2010
Partial differentiation
M1
A1
A1
A1
4
g
g
g
4,
 4,
 2
x
y
z
M1
A1
 4
 
Normal has direction  4 
 2 
 
L passes through (2, 1,  1) and has this direction
M1
A1 (ag)
4
(iii)
When g  0, xy  z 2  0
(2  2 )(1  2 )  (1   ) 2  0
M1
3  3 2  0
  1
  1 gives P (0, 3, 0)
M1
A1 (ag)
  1 gives Q (4,  1,  2)
A1
Obtaining a value of 
Or B1 for verifying g(0, 3, 0)  0
and showing that P is on L
4
(iv)
At P,
g
g
g
 3e 6 ,
 0,
0
x
y
z
OR give M2 A1 www for
g(2 , 3  2 ,  )
M1
 (3 2  6  ) e6  6  6  e 6
g
g
g
 g  x  y z
x
y
z
M1
 3e 6 (2  )  0  0  6  e 6
A1 (ag)
3
(v)
When 6 e6  h ,    16 e6 h
M1
Point (2  , 3  2  ,  ) is approximately
( 13 e6 h , 3  13 e6 h ,  16 e6 h )
(vi)
A1 (ag)
2
g
g
g
 e 6 ,
 4e6 ,
 4e6
x
y
z
When x  4  2  , y  1  2  , z  2  
M1
 g  (e6 )(2  )  (4e6 )(2 )  (4e6 )(  )
M1A1
At Q,
M1
OR give M1 M2 A1 www for
g(4  2 ,  1  2 ,  2   )
 6  e6
If 6  e6  h , then   16 e6 h
M1
Point is approximately
( 4  13 e 6 h ,  1  13 e 6 h ,  2  16 e 6 h )
A2
 (3 2  6 ) e 6   6  6 e6
Give A1 for one coordinate correct
7
2
If partial derivatives are not evaluated
at Q, max mark is M0M1M0M0
4757
3 (i)
Mark Scheme
dy 1  1 2 1 1 2
 x 2x
dx 2
June 2010
B1
2
1
 dy 
1
1     1  ( 12 x 2  12 x 2 )2
 dx 
M1
 1  14 x 1  12  14 x  14 x 1  12  14 x
 ( 12 x
 12
 12 x
1
)2
2
A1
a
1

1
Arc length is  ( 12 x 2  12 x 2 ) dx
0
 x

a
(ii)
1
1
2
2
 13 x
 13 a
3
3
a
2

 0
M1
A1 (ag)
2
5
 2 y ds
Curved surface area is
 y ds
M1
For
A1
Correct integral form including limits
M1A1
For
3
3
1
1

1
  2 ( x 2  13 x 2 )( 12 x 2  12 x 2 ) dx
0
3

 2  ( 12  13 x  16 x 2 ) dx
0
3

 2 

1
2
1 3 
x  16 x 2  18
x 
0
 3
1
2
x  16 x 2  181 x3
A1
5
(iii)
dy
3

dx
4
 3 
Unit normal vector is  54 
 
 5
When x  4,
d2 y
dx
2
  14 x
 32
 14 x
 1  ( ) 

3 2
4
5
() 32
3
2
(
 4  25  
c   2   
  3  2  
(iv)
 12
3
5
4
5
B1
M1
A1 ft
(
5
)
32
125
25

)
2
5
64
32
B1




M1
A1
A1
Differentiating partially w.r.t. p
M1
0  2p x
2

Applying formula for  or 
M1
A1 ft
 3 12 

 10 2 
3

1
Finding a normal vector
Correct unit normal (either direction)
9
3
p2 x 2
A1
2
p
x
Envelope is y 
4
2
x
1
2
x
4 32
y x
3

M1
A1
1 8 32
x
3 x3
A1
5
3
4757
4 (i)
Mark Scheme
x
1
 x  x 1
s t( x)  s 


x
 x 1 
x 1
x  ( x  1) 1

  r( x)
x
x
M1
A1 (ag)
x 1
 x 1 
x
t s( x)  t 

 x  x 1 1
x
x 1

 1  x  q( x)
( x  1)  x
(ii)
(iii)
M1
A1
4
p
q
r
s
t
u
p
p
q
r
s
t
u
q
q
p
s
r
u
t
r
r
u
p
t
s
q
s
s
t
q
u
r
p
t
t
s
u
q
p
r
u
u
r
June 2010
t
p
q
Element
p
q
r
s
t
u
Inverse
p
q
r
u
t
s
B3
Give B2 for 4 correct, B1 for 2 correct
3
s
B3
(iv)
(v)
(vi)
{ p }, F
{ p , q }, { p , r }, { p , t }
{ p, s, u }
 j
Element
1
1
e3
Order
1
2
6
Give B2 for 4 correct, B1 for 2 correct
3
Ignore these in the marking
Deduct one mark for each non-trivial
subgroup in excess of four
B1B1B1
B1
4
e
 j
3
6
e
2 j
3
e
 2 j
3
3
3
21  2, 22  4, 23  8, 24  16, 25  13, 26  7
B4
4
Finding (at least two) powers of 2
M1
For 26  7 and 29  18
27  14, 28  9, 29  18 , 210  17 , 211  15, 212  11 A1
(vii)
213  3, 214  6, 215  12, 216  5, 217  10, 218  1
Hence 2 has order 18
A1
G is abelian (so all its subgroups are abelian)
F is not abelian
B1
Give B3 for 4 correct, B2 for 3 correct
B1 for 2 correct
Correctly shown
3 All powers listed implies final A1
Can have ‘cyclic’ instead of ‘abelian’
1
(viii) Subgroup of order 6 is {1, 23 , 26 , 29 , 212 , 215 }
i.e. {1, 7 , 8, 11, 12, 18 }
M1
or B2
A1
2
4
4757
Mark Scheme
Pre-multiplication by transition matrix
5 (i)
 0.16 0.28 0.43

0.84
0
0
P
 0
0.72
0

0
0.57
 0
(ii)
(iii)
(iv)
 1   0.3349 
  

0
0.3243 
P9    
 0   0.2231 
  

 0   0.1177 
June 2010
Allow tolerance of  0.0001 in
probabilities throughout this question
1

0
0

0
Give B1 for two columns correct
B2
2
M2
Prob(C )  0.2231
Using P9
Give M1 for using P10
A1
3
Week 5
 1   0.5020 
  

0
0.2851 
P4    
 0   0.1577 
  

 0   0.0552 
B1
M1
First column of a power of P
A1
3
. . .
.
 .
.



0.2869 . . . 
.
.
P7  
P8  
 .


. . .
. 0.2262



.
.
.
.
.
.



Probability is 0.2869  0.2262
 0.0649
. .

. .
. .

. .
SC Give B0M1A1 for Week 9 and
0.3860 0.3098 0.2066 0.0976
M1M1
Elements from P 7 and P8
M1
A1
Multiplying appropriate probabilities
4
(v)
Expected run length is
(vi)
Pn
 0.3585

0.3011
 
 0.2168

 0.1236
1
 1.19 (3 sf)
1  0.16
M1
A1
0.3585 0.3585 0.3585 

0.3011 0.3011 0.3011 
0.2168 0.2168 0.2168 

0.1236 0.1236 0.1236 
M1
Allow 1.2
2
A: 0.3585 B: 0.3011 C: 0.2168 D: 0.1236
Evaluating P n with n  10
or Obtaining (at least) 3 equations
from P p  p
Limiting matrix with equal columns
or Solving to obtain one equilib prob
Give A1 for two correct
M1
A2
4
(vii)
Expected number is 145  0.3585
M1
A1 ft
 52
2
(viii)  a
b
c

0
0
1  a
 0 1 b
0

0 1 c
 0
1   0.4   0.4 

 

0   0.25   0.25 

0   0.2   0.2 

 

0   0.15   0.15 
 0.4 


0.25 
Transition matrix and 
 0.2 


 0.15 
M1
A1
0.4a  0.25b  0.2c  0.15  0.4
Forming at least one equation
Dependent on previous M1
M1
0.4(1  a )  0.25
0.25(1  b)  0.2
0.2(1  c)  0.15
a  0.375, b  0.2, c  0.25
A1
4
5
4757
Mark Scheme
Post-multiplication by transition matrix
5 (i)
0
0 
 0.16 0.84


0.28
0
0.72
0 
P
 0.43
0
0
0.57 


0
0
0 
 1
(ii)
1
June 2010
Allow tolerance of  0.0001 in
probabilities throughout this question
Give B1 for two rows correct
B2
2
0 0 0  P9
M2
Using P9
Give M1 for using P10
  0.3349 0.3243 0.2231 0.1177 
Prob(C )  0.2231
A1
3
(iii)
Week 5
1 0 0 0  P 4
B1
M1
  0.5020 0.2851 0.1577 0.0552 
First row of a power of P
A1
3
(iv)
.
 . 0.2869 . . 
. .



.
.
. .
. . 0.2262
P7  
P8  
.


.
. .
. .
.



.
.
.
.
.
.
.



Probability is 0.2869  0.2262
 0.0649
.

.
.

.
SC Give B0M1A1 for Week 9 and
0.3860 0.3098 0.2066 0.0976
M1M1
Elements from P 7 and P8
M1
A1
Multiplying appropriate probabilities
4
(v)
Expected run length is
(vi)
Pn
 0.3585

0.3585
 
 0.3585

 0.3585
1
 1.19 (3 sf)
1  0.16
M1
A1
0.3011 0.2168 0.1236 

0.3011 0.2168 0.1236 
0.3011 0.2168 0.1236 

0.3011 0.2168 0.1236 
M1
Allow 1.2
2
A: 0.3585 B: 0.3011 C: 0.2168 D: 0.1236
Evaluating P n with n  10
or Obtaining (at least) 3 equations
from p P  p
Limiting matrix with equal rows
or Solving to obtain one equilib prob
Give A1 for two correct
M1
A2
4
(vii)
Expected number is 145  0.3585
M1
A1 ft
 52
2
(viii)
0
0 
 a 1 a


b
0
1

b
0 

0.4
0.25
0.2
0.15


c
0
0 1 c 


0
0
0 
1
  0.4 0.25 0.2 0.15 
M1
Transition matrix and
 0.4 0.25 0.2 0.15
A1
0.4a  0.25b  0.2c  0.15  0.4
Forming at least one equation
Dependent on previous M1
M1
0.4(1  a )  0.25
0.25(1  b)  0.2
0.2(1  c)  0.15
a  0.375, b  0.2, c  0.25
A1
4
6
4757
1 (i)
Mark Scheme
Distance is
2(2)  (7)  2(7)  11
2 1  2
2
2
June 2011
Formula, or other complete method
Numerical expression for distance
M1
A1
2
2
A1
3
(ii)
(iii)
 1   2   15 
     
 4    1   12 
 7   2   9 
     
(v)
Vector product of normals, or finding a
point on AD, e.g. (0,  2.6, 4.2) ,
(3.25, 0, 2.25) , (7 , 3, 0)
Correct direction
2
5
 
 
Equation of AD is r   1    4 
3
 3 
 
 
A1
 5   5   10 

     
CA  d   6    4    5 
 2   3   10 
     

CA  d
102  52  102
Distance is


d
52  42  32
M1
Appropriate vector product
A2 ft
Give A1 if just one error
M1
M1
Formula for distance
Finding magnitude
Both dependent on first M1
 4.5 
(iv)
M1
3
2

3
225
50
3 2
 2.12
2
A1
6
 5   9   12 
 4
      
 
d  BC   4    12    3   3  1 
 3   6   24 
8
     
 
 4   4 
   
 6  .  1 
 4  8
10
Distance is     
2
2
2
9
4 1  8
  4   5 
  
  
1
1 



V 6 (AB AC).AD 6   6    6 
  4   2 

Accept any form
A1 ft
M1
Vector product of directions
A1 ft
Appropriate scalar product
Evaluation of scalar product
For denominator
M1
A1 ft
M1
A1
6
 5
  
.  4 
  3 

 12   5 

  
1
 6   12  .  4   5
 6   3 

  
V  20    4
D is (22, 15,  9) or (18,  17 , 15)
M1
Appropriate scalar triple product
M1
Evaluation of scalar triple product
A1 ft
or 2a  2b  c  3 (simplified)
for D (a , b , c)
M1
A1A1
Obtain a value of λ , or one of a, b, c
6
2
4757
Mark Scheme
June 2011
Alternative methods for Question 1
1 (ii)
Eliminating x
3y  4z  9
x  7  53 t , y  3  43 t , z  t
1 (iii)





Eliminating one of x, y, z
or 3x  5 z  21 or 4 x  5 y  13
M1
A1A1
3
2
 5  7   5 
 
     
 1     4    5   .  4   0
3
 3   1    3 
 
     
25  25  16  24  9  6  0
  1.1 , F is (7.5, 3.4,  0.3)
M1
Appropriate scalar product
A1 ft
CF  (0.5) 2  (1.6) 2  (1.3) 2
 4.5
A1 ft
M1
Obtaining a value for λ
M1
Finding magnitude
A1
6
1 (iv)





 2 
9 2
5
 
   
 
7
12
1







 
   
4
 7
 6   3 
 3 
 
   
 
 5
  
 . 4   0
  3 

M1
A1 ft
 5  9  4   9 

  
and  4  12   6  .  12   0
 3  6  4   6 

  
4
128
 , 
81
243
2
2
 40   10   80 
 81    81    81 
     
Distance is

Two appropriate scalar products
A1 ft
M1
Obtaining values for λ and µ
M1
Obtaining distance
2
10
9
A1
6
3
4757
2 (i)
Mark Scheme
When y  3, z  3x 2  36 x  216
B1
When x  6, z  8 y 3  216 y  756
B1
Correct shape (parabola) and position
B1
For (6,  324)
B1
B1
Correct shape and position
B1
For (3,  324)
B1
(ii)
(iii)
(6,  3,  324) is a SP on both sections;
hence it is a SP on S
B1
Saddle point
B1
z
z
 12 xy  30 x  36,
 24 y 2  6 x 2
x
y
B1B1
At a SP,
June 2011
For (3,  1188)
7 If B0B0 then give B1 for x  3
2
z z

0
x y
M1
24 y 2  6 x 2  0  y   12 x
y  12 x   6 x 2  30 x  36  0
 x  6, 1;
M1
SPs are (6,  3,  324)
(1, 0.5, 19)
A1
y   x  6 x  30 x  36  0
M1
 x  2, 3;
A1
A1
1
2
2
SPs are (2,  1, 28)
(3,  1.5, 27)
8
(iv)
z
z
 120 and
0
x
y
( Allow M1 for
M1
z
 120 )
x
y  12 x   6 x 2  30 x  84  0; D  302  4  6  84 M1
D ( 1116)  0 ; so this has no roots
A1
y   x  6 x  30 x  84  0  x  7 ,  2
M1
M1
A1
A1
1
2
2
When x  7 , y  3.5, z  203; so k  637
When x  2, y  1, z  148; so k  92
Obtaining at least one value of x
Obtaining a value of k
7
4
4757
Mark Scheme
3(a)(i) dy 1 12 x 1  12 x
 e 2e
dx 2
June 2011
B1
2
 dy 
1 x
1
1 x
 dx   4 e  2  4 e
 
M1
2
1x
1 x
 dy 
1     14 e x  12  14 e x  ( 12 e 2  12 e 2 ) 2
 dx 
(ii)
Correct completion
A1 (ag)
3
ln a
1x
1 x

Length is  ( 12 e 2  12 e 2 ) dx
0
M1
1x
1 x

For  ( 12 e 2  12 e 2 ) dx

 1 x 1 x 
  e2  e 2 

0
A1
For e 2  e



A1 (ag)
ln a
1 
a 1
  (1  1) 
a 
a
a
1x
1 x
2
Correctly shown
3
 2 y ds
M1
For
1x
1x
1 x
1 x

  2 ( e 2  e 2 )( 12 e 2  12 e 2 ) dx
0
A1
Correct integral form including limits
M1
Obtaining integrable expression
   e x  2 x  e x 

0
A1
For e x  2 x  e x
1 

   a  2 ln a  
a


A1
(iii) Curved surface area is
 y ds
ln a
ln a

   ( e x  2  e  x ) dx
0
ln a
(b)(i)
5
dy
cos 

dx 2sin 
B1
2sin 
(  2 tan  )
cos 
Normal is y  sin   2 tan  ( x  2 cos  )
y  2 x tan   3sin 
M1
Gradient of normal is
(ii)
A1 (ag)
Correctly shown
3
Differentiating partially w.r.t. 
M1
0  2 x sec   3cos 
2
x  32 cos3 
y  3cos  tan   3sin 
3
 3sin  (cos 2   1)  3sin 3 
2
2
2
A1
M1
A1
Obtaining an expression for y
Any correct form
M1
Using 1  cos 2   sin 2 
2
(2 x) 3  y 3  (3cos3  ) 3  (3sin 3  ) 3
2
2
 3 3 (cos 2   sin 2  )  3 3
Correctly shown
A1 (ag)
6
(iii) (2, 0) has   0
(A)
Centre of curvature is (

3
2
Using param eqn with   0
(or other method for  or cc)
M1
, 0)
1
2
A1
(B) (0, 1) has   1 
2
Using param eqn with   12 
M1
Centre of curvature is ( 0 ,  3 )
 4
(or other method for  or cc)
A1
4
5
4757
Mark Scheme
4 (i)
1
3
4
5
9
1
1
3
4
5
9
3
3
9
1
4
5
4
4
1
5
9
3
5
5
4
9
3
1
9
9
5
3
1
4
Composition table shows closure
Identity is 1
Element
1
3
4
5
9
Inverse
1
4
3
9
5
June 2011
B2
Give B1 if not more than 4 errors
B1
B1
Dependent on B2 for table
Give B1 for 3 correct
B2
6
So every element has an inverse
(ii)
Since 5 is prime,
a group of order 5 must be cyclic
Two cyclic groups of the same order must be
isomorphic
B1
B1
B1
3
(iii)
(iv)
2 j
4 j
6 j
8 j
H
1
e5
e5
e5
e5
G
1
3
9
5
4
or
1
4
5
9
3
or
1
5
3
4
9
or
1
9
4
3
5
For 1  1
B1
B2
For non-identity elements
3
B1
Identity is (1, 1)
Inverse of (9, 3) is (5 , 4)
B1
2
(v)
( x , y )5  ( x 5 , y 5 )
M1
Since G has order 5, x5  1 and y 5  1
M1
Hence ( x , y )  (1, 1)
A1 (ag)
5
3
(vi)
(vii)
Order of ( x , y ) is a factor of 5 (so must be 1 or 5)
Only identity (1, 1) can have order 1
Hence all other elements have order 5
M1
{ (1, 1) , (9, 3) , (4, 9) , (3, 5) , (5, 4) }
B2
B1
A1 (ag)
3
2
(viii) An element of order 5 generates a subgroup, and so
can be in only one subgroup of order 5
Number is 24  4  6
Or for 24 ÷ 4
Or listing at least 2 other subgroups
Give B1 for unsupported answer 6
M1
A1
2
6
Give B1 ft for 5 elements including
(1, 1) , (9, 3) , (5, 4)
4757
Mark Scheme
Pre-multiplication by transition matrix
5 (i)
0
0
 1 0.07


0 0.8 0.15 0 
P
 0 0.13 0.75 0 


0
0.1 1 
0
(ii)
June 2011
Allow tolerance of  0.0001 in
probabilities throughout this question
Give B1 for two columns correct
B2
2
If system enters an absorbing state, it remains in that
state
A and D are absorbing states
B1
B1
2
(iii)
(iv)
 0   0.2236 
  

0.4
0.2505 
P9    
 0.6   0.1998 
  

 0   0.3261 
Prob(owned by B) = 0.2505
For P9
M1M1
A1
 0 
 
0.4
(or P10 ) and  
 0.6 
 
 0 
3
...
...
... 
1


... 0.4818
...
... 
P4  
 ...
...
0.3856 ... 


...
...
1
 ...
0.2236  0.2505  0.4818  0.1998  0.3856  0.3261
= 0.7474
M1
Using diagonal elements from P 4
M1
A1
Using probabilities for 10th day
3
(v)
 0 
 
0.4
1 0 0 1 P n  
0.6
 
 0 
 ( 0.8971 ) when n  26
 ( 0.9057 ) when n  27
i.e. on the 28th day
M1
Considering P n for some n  20
M1
Evaluating Prob(A or D) for some
values of n
A1 ft
Identifying n  26 or n  27
( Implies M1M1 )
A1
4
(vi)
Pn
(vii)
(viii)
 1 0.5738 0.3443

0
0
0
 
0
0
0

0
0.4262
0.6557

0

0
Q
0

1
B2
 0   0.4361 
  

0.4
0 
Q 
 0.6   0 
  

 0   0.5639 
Prob(eventually owned by A) = 0.4361
Give B1 for two bold elements correct
2 (to 3 dp)
 0 
 
0.4
Using Q and  
 0.6 
 
 0 
M1M1
A1
3
 0   0.5 
   
0
p
(where q  1  p )
Q  
q  0 
   
 0   0.5 
0.5738 p  0.3443q  0.5
p  0.6786, q  0.3214
M1M1
For LHS and RHS
A1 ft
Or 0.4262 p  0.6557 q  0.5
M1
A1
Solving to obtain a value of p
Allow 0.678 - 0.679, 0.321 – 0.322
5
7
4757
1
Mark Scheme
Question
(i)
Answer
June 2013
Marks
 1
 1  6   6 
 2 
        
 
 4   BC   4    18    3  [  3  1  ]
1
1  3  6
 2
 
     
 
 8   2 

  
 2  .  1 

AB . d  13   2 

Shortest distance is
d
22  12  22
40
Shortest distance is
3
Guidance
Vector product of directions
Intention sufficient
M1*
Appropriate scalar product
Dep *
M1
Evaluation of d
Dep **
M1*
A1
A1
[5]
OR
  11  6   3   

 

  18    2  4  
  3  3   10   

  1
  
 . 4  0
  1 

M1* Two appropriate scalar products
 8  6     6 

  
and  2  18  4  .  18   0
 13  3     3 

  
81  18  29 , 123  27   41
 80 
9 

5
82   40 
   ,    , XY 
9 

3
9
 80 

9

2
Shortest distance is
Shortest distance is
2
 80   40   80 
     
 9   9   9 
A1 Two correct equations

M1* Obtaining XY
Dep *
M1
Dep **
2
40
3
A1
5
4757
1
Mark Scheme
Question
(ii)
Answer
 8   6   228 
  
   

AB  BC   2    18    54 
 13   3   132 

   

 
AB  BC  2282  542  1322

BC  62  182  32
 
AB  BC
72324
Shortest distance is


369
BC
Shortest distance is 14
Marks
M1*
June 2013
Guidance
Appropriate vector product
A2
Give A1 if one error
M1*
Dep *
M1
Dep **
A1
Sign error in vector product can earn
M1A1M1M1A1
[6]
OR
  11  6   3 

  
  18    2 
  3  3   10 

  6 
  
 .  18   0
  3 

M1* Allow one error
A1
M1* Obtaining a value of 
1

3
Shortest distance is
Dep *
A1
(6) 2  (4) 2  (12) 2
M1
Shortest distance is 14
A1
6
Dep **
4757
1
Mark Scheme
Question
(iii)
Answer
Marks
 11 
 6   3 
 1
 

  
 
 0     18    2     4 
 3 
 k  3  10 
1
 

  
 
11  6  3  
18  2  4 
  5 ,   22
3   (k  3)  10  
k 4
3
 1
 
 
Point of intersection is  2   22  4 
10 
1
 
 
Point of intersection is (19, 90, 32)
1
(iv)
 1
 1
 1

12  
 
 
 4   18 , so AD  () 18  4   2 2  4 
1
1
1
 
 
 
  
Volume is 16 (AB  AC) . AD
 8   2 
1
 

   2    16 
6 
  13   10 

 228   1
2
  

54  .  4  

3 
  
 132   1 
 40 2
June 2013

 1

 
 . (2 2)  4 
1

 

2
(120)
3
M1
A1
Guidance
Allow one error
Two correct equations
Must use different parameters
Obtaining a value of k
Other methods possible
(e.g. distance between lines is 0)
A1
M1
A1
M1
A1
[7]

M1*
A1
Obtaining AD or D
M1*
Appropriate scalar triple product
A1 ft
Correct expression
Can be implied
M1
Evaluating scalar triple product
Dep **
A1
[6]
Accept 56.6
7
4757
2
Mark Scheme
Question
(i)
Answer
z
 6 x 2  6 x  12 y
x
z
 6 y 2  6 y  12 x
y
z z

, 6 x 2  6 x  12 y  6 y 2  6 y  12 x
If
x y
x2  y 2  x  y  0
( x  y )( x  y  1)  0
y  x or y  1  x
June 2013
Marks
Guidance
B1
B1
Identifying factor ( x  y )
M1
E1E1
SC If M0, then give
B1 for verifying y  x
B1 for verifying y  1  x
[5]
2
(ii)
z z

0
x y
M1
If y  x then 6 x 2  6 x  12 x  0
M1
Obtaining quadratic in x (or y)
M1
B1A1
Obtaining a non-zero value of x
Condone (0, 0) for B1
x  0,  3
Stationary points (0, 0, 0) and ( 3,  3, 54)
Can be implied
Or quartic, and factorising as
x(linear)(quadratic)
If y  1  x then 6 x 2  6 x  12(1  x)  0
x2  x  2  0
Which has no real roots ( D  7  0 )
2
(iii)
At P,
z 21 z 21
,


x 2
y 2
z
z
x y
x
y
21
21
w h h
2
2
w
h
21
z 
M1
A1
[7]
Obtaining quadratic with no real roots
Correctly shown
M1
A1
Substituting into
M1
A1 ft
A1
8
z
z
or
x
y
Just stating ’No real roots’ M1A0
Correct value, or substitution seen
4757
Mark Scheme
Question
2
(iv)
Answer
June 2013
Marks
[5]
Guidance
z z

 24
x y
M1
Allow sign error
24λ is M0 unless   1 appears
later
If y  x then 6 x 2  6 x  12 x  24
M1
Obtaining quadratic in x (or y)
Or quartic, and one linear factor
x  1,  4
Points (1, 1, 22) and (4,  4, 32)
If y  1  x then 6 x 2  6 x  12(1  x)  24
If neither correct, give A1 for x  1,  4
A1A1
Obtaining quadratic in x (or y)
M1
Or third linear factor of quartic
x  2,  1
Points (2,  1, 5) and (1, 2, 5)
3
(a)
If neither correct, give A1 for x  2,  1
A1A1
[7]
2
 dr 
2
2
2
r2  
  a (1  cos  )  (a sin  )

d


Condone ...  ( a sin  )2
B1
or 4a 2 cos 4 12   4a 2 sin 2 12  cos 2 12 
 a 2 (1  2cos  cos 2   sin 2  )  2a 2 (1  cos )
Using 1  cos   2cos 2 12 
M1
 4a 2 cos 2 12 
A1

2

2
 dr 
1
d

Arc  r 2  

 2a cos 2  d


d



0

1


1
2


  4a sin 12  

0
2 2 a
For 4a sin 12 
A1
A1
[6]
9
2
 dr 
 d in terms of θ
 d 
For  r 2  

M1
Limits not required
4757
3
Question
(b)
(i)
Mark Scheme
Answer
2
 x2
1
 dy 
1   1 
 2

d
x
2
2x
 





Marks




B1
M1
2
A1
2
 dy 
 dx
 dx 
Area is  2 y 1  


M1*
2

 x3 1  x 2
1
  2  


 6 2 x 
 2 2 x 2
1


2

 2 

1
Guidance
2
x4 1
1
  4
4 2 4x
 x2
1

 2
 2 2x

 x5 x
1 
   3  dx
 12 3 4 x 

 dx

A1 ft
Integral expression including limits
M1
Obtaining integrable form
A1
Allow one error
2
 x6 x 2
1 
 2 

 2 
 72 6 8 x  1

47
16
June 2013
A1
[8]
10
Dep *
4757
3
Mark Scheme
Question
(b)
(ii)
Answer
2
d y
dx
2
 x
1
x
3
 x2
1
 2

2
2x

1
x 3
x
(



17
)
8
Marks
June 2013
Guidance
B1
3
Using formula for ρ or κ
M1
A1 ft
Correct expression for ρ or κ
A1 ft
Correct numerical expression for ρ
3
  15 2  2
3
 17 
1    
 8 
 
  8 

1
17
2
8
8
289

64
E1
Correctly shown
[5]
3
(b)
(iii)
1  15 
dy 15
 , so unit normal is


17  8 
dx 8
M1
A1
Obtaining a normal vector
Correct unit normal
Allow M1 for 
 2  289  1517 


c  19  


 64  8
 12 
 17 
M1
Allow sign errors
Must use a unit vector
 127 89 
,

 64 24 
Centre of curvature is  
A1A1
[5]
11
 8   15 
 or 

 15   8 
4757
4
4
Mark Scheme
Question
(a)
(i)
(a)
Answer
Identity is e
Element
Inverse
a
b
b
a
c
c
June 2013
Marks
B1
d
g
e
e
f
h
g
d
h
f
(ii)
d2  a, d4  c
Guidance
B2
Give B1 for four correct
[3]
M1
A1
Finding powers of an element
Identifying d (or f or g or h) as a generator
A1
Or f 2  b ,
At least fourth power
Implies previous M1
f4 c
Or g 2  b , g 4  c
Hence d has order 8, and G is cyclic
4
(a)
(iii)
H
G
or
or
or
0
e
e
e
e
2
d
f
g
h
4
a
b
b
a
6
f
d
h
g
8
c
c
c
c
Or h 2  a , h 4  c
Correctly shown
E1
[4]
10
h
g
f
d
12
b
a
a
b
14
g
h
d
f
B1
B1
B1
For e  0 and c  8
For { d , f , g , h }  { 2, 6,10,14 }
For a fully correct isomorphism
In any order
Correct statement about rotations and/or
reflections which implies non-IM
Or (4) reflections (and 180°
rotation) have order 2
Or composition of reflections (or
90° rotation and reflection) is
not commutative
[3]
4
(a)
(iv)
Rotations have order 2 or 4
Reflections have order 2
B1
Or More than one element of order 2
Or Not commutative
Fully correct explanation
There is no element of order 8
Hence not isomorphic
E1
[2]
12
Dependent on previous B1
4757
4
Mark Scheme
Question
(b)
(i)
Answer
x
1  nx
f m f n ( x) 
 x 
1 m

 1  nx 
x
x


 f m n ( x)
1  nx  mx 1  (m  n) x
June 2013
Marks
Guidance
M1
Composition of functions
In either order
E1
Correctly shown
E0 if in wrong order
[2]
4
(b)
(ii)
(f m f n ) f p  f m  n f p  f m  n  p
M1
Combining three functions
Hence S is associative
E1
Correctly shown
For any f m , f n in S, f m f n  f m  n
[2]
M1
Referring to this in context
f m (f n f p )  f m f n  p  f m  n  p
4
4
(b)
(b)
(iii)
(iv)
f m f n is in S (so S is closed)
Identity is f 0
Inverse of f n is f  n
since f n f  n  f n  n  f 0
A1
S is also associative, and hence is a group
E1
{ f 2 n } for all integers n
[6]
B2
B0 for x
B1
M1E1 bod for
(f m f n )f p  f m  n  p  f m (f n f p )
B1 for n  0
B1
B1
Closure, associativity, identity and inverses
must all be mentioned in (iii)
Or { f3n } , etc
Give B1 for multiples of 2 (or 3, etc) but not
completely correctly described
[2]
13
Dependent on previous 5 marks
e.g. { f 0 , f 2 , f 4 , f 6 , ...}
4757
Mark Scheme
Question
5
(i)
Answer
Pre-multiplication by transition matrix
0
0
0
 1 0.5


0
0
 0 0.05 0.5
P   0 0.45 0.05 0.5 0 


0
0.45 0.05 0 
0
0
0
0
0.45 1 

Marks
June 2013
Guidance
Allow tolerance of  0.0001 in
probabilities throughout this question
Give B2 for four columns correct
Give B1 for two columns correct
B3
[3]
5
(ii)
 0   0.5042 
1  

 3   0.0230 
P8  13    0.0278 
1  

 3   0.02071 
 0   0.4242 
  

M1
For P8 (allow P 7 or P9 ) and initial
column matrix
E1
Correctly shown
P(3 lives) = 0.0207 (4 dp)
[2]
5
(iii)
Let q(n) = P(not yet ended after n tasks)
0
1 
 3
  0 1 1 1 0  P n  13 
1 
 3
0
 
q(10)  0.0371
M1
Obtaining probabilities after 10 tasks
M1
Adding probabilities of 1, 2, 3 lives
A1
[3]
14
Allow M1 for using P9 or P11
4757
5
Mark Scheme
Question
(iv)
Answer
q(9)  q(10)
 0.05072  0.03709
 0.0136
OR
5
(v)
.
0 

1  

 3   0.01506 

P9  13   
.
1  

 3   0.01355 
0 

.
  

0.01506  0.5  0.01355  0.45
 0.0136
q(13)  0.01374
q(14)  0.00998
Smallest N is 14
5
(vi)
 1 0.7880 0.5525 0.2908 0 


0
0
0
0
0
Pn   0
0
0
0
0  L


0
0
0
0
0
 0 0.2120 0.4475 0.7092 1 


Marks
M1
M1
A1
[3]
Guidance
Using q(9) and q(10)
Evaluating q(9)
M1 Probs of 1 and 3 lives after 9 tasks
M1
A1
M1
M1
Evaluating q(n) for some n  10
Consecutive values each side of 0.01
A1
[3]
Must be clear that their answer is 14
Give B1 for any element correct to 3 dp
(other than 0 or 1)
B2
[2]
5
(vii)
 0   0.5438 
1  

 3  0 
L  13    0 
1  

 3  0 
 0   0.4562 
  

P(wins a prize) = 0.4562
June 2013
M1M1
A1
[3]
15
Using L and the initial column matrix
Just N  14 www earns B3
4757
Mark Scheme
5
Question
(viii)
5
(ix)
Answer
Maximum probability is 0.7092
Always start with 3 lives
Marks
B1 ft
B1
[2]
 0   0.4 
   
 0.1  0 
L p  0 
   
 q   0 
 0   0.6 
   
0.7880  0.1  0.5525 p  0.2908(0.9  p)  0.4
5
Or 0.0212  0.4475 p  0.7092(0.9  p )  0.6
Obtaining a value for p or q
Accept values rounding to 0.227, 0.673
M1
A1
[3]
Allow tolerance of  0.0001 in
probabilities throughout this question
Post-multiplication by transition matrix
(i)
Guidance
M1
P(2 lives) = 0.2273 , P(3 lives) = 0.6727
5
June 2013
0
0
0
0 
 1


0
0 
 0.5 0.05 0.45
P 0
0.5 0.05 0.45
0 


0
0.5 0.05 0.45 
 0
 0
0
0
0
1 

Give B2 for four rows correct
Give B1 for two rows correct
B3
[3]
5
(ii)
0
1
3
1
3
1
3
0  P8
  0.5042 0.0230 0.0278 0.02071 0.4242
P(3 lives) = 0.0207 (4 dp)
For P8 (allow P 7 or P9 ) and initial row
matrix
M1

E1
[2]
Correctly shown
16
Allow use of p  q  1
4757
5
Mark Scheme
Question
(iii)
Answer
Let q(n) = P(not yet ended after n tasks)
 0
1
3
1
3
q(10)  0.0371
5
(v)
3
q(9)  q(10)
 0.05072  0.03709
 0.0136
(iv)
OR
5
1
 0
 
1
n  
0 P 1
 
1
 0
 
0
1
3
1
3
1
3
0  P9
  . 0.01506 . 0.01355 . 
0.01506  0.5  0.01355  0.45
 0.0136
q(13)  0.01374
q(14)  0.00998
Smallest N is 14
5
(vi)
 1

 0.7880
n
P   0.5525

 0.2908
 0

0 0 0
0 0 0
0 0 0
0 0 0
0 0 0


0.2120 
0.4475   L

0.7092 
1 
Marks
June 2013
Guidance
M1
Obtaining probabilities after 10 tasks
M1
Adding probabilities of 1, 2, 3 lives
A1
[3]
M1
M1
A1
[3]
Using q(9) and q(10)
Evaluating q(9)
Allow M1 for using P9 or P11
M1 Probs of 1 and 3 lives after 9 tasks
M1
A1
M1
M1
Evaluating q(n) for some n  10
Consecutive values each side of 0.01
A1
[3]
Must be clear that their answer is 14
0
Give B1 for any element correct to 3 dp
(other than 0 or 1)
B2
[2]
17
Just N  14 www earns B3
4757
5
Mark Scheme
Question
(vii)
0
Answer
1
3
1
3
1
3
Marks
0 L
  0.5438 0 0 0 0.4562
M1M1
(viii)
Maximum probability is 0.7092
Always start with 3 lives
5
(ix)
0
Guidance
Using L and the initial row matrix

P(wins a prize) = 0.4562
5
June 2013
0.1 p q 0  L
A1
[3]
B1 ft
B1
[2]
  0.4 0 0 0 0.6 
0.7880  0.1  0.5525 p  0.2908(0.9  p)  0.4
M1
P(2 lives) = 0.2273 , P(3 lives) = 0.6727
M1
A1
[3]
Or 0.0212  0.4475 p  0.7092(0.9  p )  0.6
Allow use of p  q  1
Accept values rounding to 0.227, 0.673
18