Question 9:
9 |  2 −   − 4 + 5   
8 + 2 = − 1 + 4  
1 0 + 3 2 + 8  
Attempts to solve any two of the three equations
2 −  = − 4 + 5   3 3
Either (1) and (2)   = − , = 
8 + 2  = − 1 + 4  2 2
2 −  = − 4 + 5   8 2 6
(1) and (3)   = , = 
1 0 + 3  = 2 + 8  2 3 2 3
8 + 2  = − 1 + 4  1 1 
(2) and (3)   = − 1 0 , = − 
1 0 + 3  = 2 + 8 4 
Substitutes their values of  and  into both sides of the ''third'' equation
3 1 1 3
E.g.  = − into 1 0 + 3 =  and =  into 2+8=14
2 2 2
Concludes that lines don't intersect with correct calculations and minimal
reason
 − 1  5
 
Additionally states that 2 is not parallel to 4 with a minimal reason
 
3 8
 
So lines are skew CSO * | M1, A1
dM1
A1
A1*
(5)
(5 marks)
Reasons could be
 5
4
8

 k
 − 1
2
3

o.e such as 5 = − 5  − 1 but 4 = 2  2 so they are not parallel.
Accept an argument based around the scalar product of the direction vectors. If parallel c o s = 1 
A reason for the lines not being parallel cannot be
 5
4
8


 − 1
2
3

PMT
Note: Other methods are possible and it is important that you look at their complete attempt at
proving that they don't intersect.
Alternative 1
For example it is possible to solve equations (1) and (2) to find just 
then solve equations (1) and (3) to find just 
and then conclude that ''as the two values are not the same, the lines don't intersect''
M1 dM1 marks are scored together. Both aspects have to be attempted
Attempts to solve two of the three equations to find  (or )
Attempts to solve a different pair of equations to find (or )
A1: Correct values for  (or ).
A1: conclude that ''as the two values are not the same, the lines don't intersect''
If you see something that you feel deserves credit AND that you cannot mark, then please send to
review
10 (a)
(b)
(c) | d r 1 d r k
States or uses   or = 
d t 2 d t 2
r r
d r k  
=   r 2 d r =  k d t
d t r 2
1
3
r =kt+c
3
( )
Substitutes t = 0 , r = 1 2  c = 5 7 6
( ) ( )
AND t = 1 5 , r = 6 a n d t h e i r c = 5 7 6  k =  3 3 .6
1 3 3
r = − 3 3 .6 t + 5 7 6 or r =−100.8t+1728 o.e.
3
Sets r = 0  − 1 0 0 . 8 t + 1 7 2 8 = 0
120
minutes or awrt 17.1 minutes
7
1 2
O | B1
M1 A1
M1
A1
(5)
M1
A1
(2)
B1
(1)
(8 marks)
A1: Correct equation. E.g.
1
3
r
3
= − 3 3 .6 t + 5 7 6
3
or r =−100.8t+1728 o.e. such as
5 120
3
t =− r +
504 7
(b)
M1: Sets r = 0  − 1 0 0 . 8 t + 1 7 2 8 = 0  t = . .. . Follow through on their equation. Condone if this
produces a negative value for t. Alt sets V = 0 to find t = ...
A1: Awrt 17.1 minutes. Must include the units. Also allow 17 minutes 8 seconds or 17 minutes 9
seconds
(c)
B1: For the correct shaped curve only in quadrant 1 starting at (0,12), ignoring value of t.
As
d
d
r
t
= −
k
r
2
the gradient should appear to get increasingly steeper.
If there are two curves given on the axes, both need to be correct for this mark to be awarded
Do not be concerned if it is not infinite at the t axis.
The curve on the left would be at the limit of what we would
allow
as the gradient on the left hand side does appear to increase.
We condone its appearance as it approaches the t- axis.
It does not get less steep
................................................................................................................................................................
..
Note that it is possible to use V = 4
3
r 3  d
d
V
r
= 4 r 2 dr k   with = to set up and solve
dt 2
r
dV
=4k or
dt
d
d
V
t
=  β
672
In this case the equation becomes V =2304− t
 5 
 
(a)
B1: States or uses either
d
d
V
t
=  β or its exact value which is
d
d
V
t
= −
2 0
1
1
5
6 π
PMT
o.e such as
dV 672π
=− .
dt 5
dV
M1: For =βV =βt+cand attempts to find the values of  and c using the given
dt
conditions.
3
It is dependent upon knowing that V =r and using this to find V at r = 6 and 12 with t = 15
and t = 0
672
A1: V =2304− t
 5 
 
3 3
M1: Substitutes V =r to form an equation linking r with t.
It is dependent on the previous M1 in this method
4 672
3
A1: Achieves r =2304− t
3  5 
 
(b)
M1: Sets V = 0 o.e and finds t
11 (a)
(b) | ( x + y ) 3 + 1 0 y 2 = 1 0 8 x
d y
Differentiates 10y2 to 2 0 y
d x
Differentiates ( x+ y )3 to 3 ( x + y ) 2  1 + d y 
d x
 d y  d y
3 ( x + y ) 2 1 + + 2 0 y = 1 0 8
d x d x
3 ( x + y ) 2 d y + 2 0 y d y = 1 0 8 − 3 ( x + y ) 2
d x d x
dy 108−3( x+ y )2
= *
dx 20y+3( x+ y )2
Deduces that P and Q are where 1 0 8 − 3 ( x + y ) 2 = 0  ( x + y ) 2 = 3 6
Attempts to substitute ( x + y ) =  6 into ( x + y ) 3 + 1 0 y 2 = 1 0 8 x to form
an equation in y (or x)
Solves 2 1 6 + 1 0 y 2 = 1 0 8 ( 6 − y ) and finds the negative root
Awrt 13900 metres | B1
M1
A1
dM1
A1*
(5)
M1
dM1
ddM1
A1
(4)
(9 marks)
In the approach 3 x 2 + 3

2 x y + x 2
d
d
y
x

+ 3

y 2 + 2 x y
d
d
y
x

+ 3 y 2
d
d
y
x
+ 2 0 y
d
d
y
x
= 1 0 8 it is
acceptable
to go from ( 3 x 2 + 6 x y + 3 y 2 + 2 0 y ) d
d
y
x
= 1 0 8 − 3 x 2 − 6 x y − 3 y 2 straight to the given answer
................................................................................................................................................................
.................
Watch for candidates who use the given answer and work backwards **.
dy
=
108−3( x+ y )2
3( x+ y )2
dy
+20y
dy
=108−3( x+ y )2
dx 20y+3( x+ y )2 dx dx
Therefore candidates who start with
3( x+ y )2
dy
+20y
dy
=108−3( x+ y )2 or indeed 3( x+ y )2
dy
+3( x+ y )2 +20y
dy
=108
dx dx dx dx
without any other working or evidence can only score B1 M0 A0 M0 A0
................................................................................................................................................................
..............
(b)
M1: Deduces that P and Q are where 1 0 8 − 3 ( x + y ) 2 = 0  ( x + y ) 2 = 3 6
This is implied if there is a statement that ( x + y ) =  6
dM1: Attempts to solve ( x + y ) =  6 and ( x + y ) 3 + 1 0 y 2 = 1 0 8 x simultaneously to form
a quadratic equation in y or x.
ddM1: Solves 216+10y2 =108(6− y )and finds the negative root. Allow if both roots are found
Note that ( x + y ) = 6
PMT
must have be used.
If an equation was set up in x, then look for an attempt to find the larger root of
216+10(6−x )2 =108x which should then be used to find a negative y value
A1: Awrt 13900 metres or awrt 13.9 km with the correct units. Condone answers like −13.9 km
PMT
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom