Question 1:
1 | (a) | (i) | B1
B1
B1
[3] | 1.1b
1.1b
1.1b | Shape, position of cusps in correct
quadrants or on axes, points on
axes.
Shape, position of cusps on axes.
Shape, position of cusps in correct
quadrants or on axes, points on
axes.
(ii) | e.g. bounded
e.g. closed
e.g. reflective symmetry in the x-axis.
e.g. rotational symmetry
e.g. cusps | B1
[1] | 1.2 | B1 for one common feature
FT provided at least 1 graph
correct in 1(a)(i)
(iii) | Reflective symmetry in the y axis OE | B1
[1] | 1.2 | B1 for one feature unique to m = 4
(b) | (i) | dx
sin((m1)t)sin(t)
dt
dy
cos(t)cos((m1)t)
dt
dx
0
dt
sin((m1)t)sin(t)0
2k (2l1)
t  or t  where k,l are integers.
m m2
2 4 2(m1)
t 0, , ,...
m m m
 3 5 (2m3)
or , , ,...,
m2 m2 m2 m2
dy
0
dt
cos(t)cos((m1)t)0
2k 2l
t  or t  where k,l are integers.
m m2
2 4 2(m1)
t 0, , ,...
m m m
2 4 6 2(m1)
or , , ,...,
m2 m2 m2 m2
dx dy
Values of t for which 0 and 0are
dt dt
 3 5 (2m3)
, , ,...,
m2 m2 m2 m2 | M1
M1
M1
A1
[4] | 1.1b
1.1b
2.2b | dx
For evidence of solving 0
dt
At least one from each group.
At least one from each group
CAO, must include all values in
correct range.
(ii) | The tangent to the curve is vertical at points corresponding
to such values of t. | B1
[1] | 1.2
(c) | (i) | x(t) cos(t) 1 cos((m1)t)
    
y(t) sin(t) m1sin((m1)t)
cos(t) 1 cos((m1)t) 1
  1,    .
sin(t) m1sin((m1)t) m1
Considering vector addition and these magnitudes,
x(t) 1
  is at most a distance of 1 from the origin
y(t) m1
1
and at least a distance of 1 from the origin.
m1
Alternative version
[r] x(t)2y(t)2
 cos((m1)t) 2  sin((m1)t) 2
 cos(t)  sin(t) 
 m1   m1 
m22mcos(mt)2m2cos(mt)2

m22m1
m2 1
 1 when cos(mt)1
(m1)2 m1
and
(m2)2 1
 1 when cos(mt)1
(m1)2 m1 | M1
E1
M1
A1 | 1.1b
2.4
1.1b
2.4 | Consideration of the magnitude of
the vector(s) involved or
equivalent words
Any reasonable argument
regarding the specific distance
bounds.
1 2cos(mt)
Or 1 
(m1)2 m1
Find expression for distance to the
orgin and simplify to a suitable
form. Square roots may be
omitted.
Considering range of cos or use of

suitable values of t (e.g. 0 and
m
). SC1 for max or min correctly
found.
(ii) | 1 m
Area of circles with radius r where r 1 
1 1 m1 m1
1 m2
and radius r where r 1  .
2 2 m1 m1 | 1B1w1 2s
[1] | ws1w.11b1
(iii) | m2 (m2)2
Both and tend to π as m tends to
(m1)2 (m1)2
infinity. Hence the result by a sandwich of limits. | B1
[1] | 1.1b | No need for a formal argument
(d) | 2 2
dx dy

   
dt   dt 
 (sin((m1)t)sin(t))2(cos(t)cos((m1)t)2
 22cos(mt)
mt
2sin

 2 
The length of the curve is therefore
2
mt
 2sin dt
 2 
0
2
mt
2 sin dt
 2 
0
2
m mt
2m  sin dt (by looking at the
 2 
0
graph of the function on previous line)
2
2 mt m
2m cos 
m  2 
0
8
This is independent of m. | M1
M1
M1
A1
A1
B1
[6] | 1.1a
1.1b
3.1a
3.1a
1.1b
3.2a | Evidence of correct application of
formula involving expressions for
dx dy
and
dt dt
Suitable intermediate step
For obtaining an expression
suitable for integration (previous
M1 implied), condone lack of
modulus symbol
Setting out and splitting the
integral into m equal parts
Correct value for integral.
Conclusion about this being
Independent of m.