Question 6:
--- 6(a) ---
6(a) | 2sin2sin
Gradient m
3cos3cos | Correct attempt at chord gradient –
do not allow slips unless a correct
method is clear | M1
y2sinm(x3cos)
or y2sinm(x3cos)
or ymxcand attempts to find c
using P or Q | A correct straight line method
using their chord gradient and the
point P or the point Q | M1
2sin2sin
y2sin (x3cos)
3cos3cos
2sin2sin
y2sin (x3cos)
3cos3cos
4cossin
y2sin 2 2 (x3cos)
6sinsin
2 2
2sin2sin 3cos2sin2sin
y x2sin
3cos3cos 3cos3cos
2cos 2coscos
y 2 x2sin 2
3sin sin
2 2 | A1
A correct equation for the chord in any form.
1 1 1 1
3ysin ()2xcos ()6(coscos ()sinsin ())
2 2 2 2
or
1 1 1 1
3ysin ()2xcos ()6(coscos ()sinsin ())
2 2 2 2
x 1 y 1 1
cos () sin ()cos ()**ag**
3 2 2 2 2 | A1cso
This is cso – there must no errors in applying the factor formulae and
sufficient working must be shown to justify the printed answer but allow
1 1 
coscos ()sinsin ()cos
2 2 2
(4)
(b) | 3cos3cos 2sin2sin
,
 
 2 2 
or
1 1 1 1
(3cos ()cos (),2sin ()cos ())
2 2 2 2
or
1 1 1 1
(3cos ()cos (),2sin ()cos ())
2 2 2 2 | B1
Correct coordinates of mid-point in any form
Coordinates must be in this order but condone outer brackets missing
(1)
(c) | Centre of chord is
1 1 1 1
(3cos ()cos (),2sin ()cos ())
2 2 2 2
Attempt factor formulae on both coordinates of mid-point at any stage in
(c)
y
May be implied by their  below
x | M1
1 1  1 
2sin ()cos () 2sin ()
y 2 2  2 
   
x 1 1 1
3cos ()cos () 3cos ()
2 2  2 
Or
1 1  1 
3cos ()cos () 3cos ()
x 2 2  2 
   
y 1 1 1
2sin ()cos () 2sin ()
2 2  2  | dM1A1
1 1
M1: Obtains an expression for k or –k or or 
k k
Dependent on the previous M1 (factor formulae must have been
used)
A1: Correct expression in any form
1
2cos ()
2
m
1
3sin ()
2 | Must be seen or used in (c) | B1
1
sin ()
2 2 y 2 2  4
 So   k 
 
1 3m x 3 3m 9m
cos ()
2 | A1cso
(5)
Total 10
7.(a) | dy  1
y r2x2  x(r2 x2) 2
dx
Or
dy
2x2y 0
dx | dy  1
 px(r2 x2) 2
dx
Or
dy
pxqy 0
dx | M1
Attempts to differentiate explicitly or implicitly to give one of the given forms
dy 2 x2 x2
1 1 or1
 
dx r2x2 y2 | Substitutes their derivative into
2
dy
1
 
dx | M1
r2 x2 x2 r2
  *
r2 x2 r2 x2 | cso | A1*
dy x
This is cso and so there must be no errors e.g.  could give the correct
dx y
answer but loses the A1but allow to show equivalence of lhs and rhs
(3)
(b) | r2
S 2y dx
r2 x2 | 2
dy
M1: Use of y 1 dx
 
dx
using their answer to part (a)
(must be y and not y2)
2not required here | M1A1
A1: Correct expression including
2(may be implied by later work
but must appear before any
integration)
r2
2
r2 x2 dx
r2 x2 | Substitutes for y in terms of x.
Dependent on first M. | dM1
2rxr 2rxr
or
r 0 | Substitutes the limits r and –r or 0
and r into an expression of the
form krxand subtracts. The use
of the 0 limit can be taken on trust
if omitted. Dependent on both
previous method marks. | ddM1
If they reach 2r2correctly then double, then some justification is
needed e.g. some mention of symmetry
4r2 * | cso | A1
(5)
r
 r2
Note that S 22 y dxfollowed by correct work could
r2 x2
0
score full marks as could the correct use of S 2y r2dx
y2
(c) | arc length 
2 | Ignore any working | B1
(1)
Total 9
8.(a) |  1  1 0 1 1
         
OA 1 ,OB 1 ,OCAB 2 ,BC= 1 ,AC= 3
         
         
0 1 1 0 1
         
ABACij2k
Or e.g.
BABCij2k | M1: Attempt vector product for two
sides of the triangle. If the method is
unclear, at least 2 components must be
correct. | M1A1
A1: Correct vector
1
Area ABC = 1212 22
2 | 1
Attempts theirABAC
2
Dependent on the first M | dM1
1
6
2 | Accept equivalents or awrt 1.22 | A1
Note that triangles OAB and OBC have the same area but score 0/4
It must be triangle ABC
(4)
(b) | bc(ijk)(2jk)ij2k | Attempt bc. If the method is
unclear, at least 2 components must
be correct. | M1
1 1
  (ij).(-i-j+2k)  (11)0
6 6 | M1: Attempt scalar product of a with
their bc to obtain a number not a
vector. | M1A1
A1: Obtains = 0 with no errors
(allow omission of 1for all 3 marks)
6
Just a.(-i-j+2k)0 would lose
the A1
(3)
Alternative
a.bc | 1 1 0
1 1 1
0 2 1 | Writes this statement (allow other
brackets provided the determinant is
implied later) | M1
121100 | M1: Clear attempt at determinant | M1A1
A1: Obtains = 0 with no errors
(allow omission of 1for all 3 marks)
6
(c) | Volume of tetrahedron (OABC) = 0
abcoe or cbaoe
b x c is perpendicular to a or a is parallel to CB
All vectors/points lie in the same plane
OABC is a parallelogram
a, b and c are linearly dependent
Do not isw – if there are contradictory or wrong statements award B0 | B1
(1)
Total 8