Question 6:
6 | (a) | r = 0 when = 0 and when  = /2 (and the
function is continuous so there is a loop).
Must be in 1st quadrant since  only takes
values between 0 and ½.
There are only two values of 𝜃 for which r = 0,
therefore there is only one loop (two loops
require three values of 𝜃/tangents at the pole). | B1
B1
B1
[3] | 2.4
2.1
2.2a | Must link domain of with
quadrant.
Convincingly explaining why it
is a single loop.  values/r = 0
must be linked to loop. | Condone use of “origin” rather than
“pole” in explanations.
or “There are no other solutions
tor = 0 (in the domain)”.
1

6 | 1

4 | 5

1 2 | 1

2
0.912 | 1.589 | 1.351 | 0
(c) | d r 1 d d 1  
6 s i n h ( s i n 2 ) 6 s i n 2 s i n h     =  + 
d 3 d d 3   
d r 1 1
1 2 c o s 2 s i n h 2 s i n 2 c o s h     = +
d 3 3 
1 1
A a t , 1 2 c o s 2 s i n h 2 s i n 2 c o s h 0      + =
3 3
1 1
1 2 c o s 2 s i n h 2 s i n 2 c o s h      = −
3 3
1
6 t a n h t a n 2    = −
3
1 1
 =tanh−1(− tan2)
3 6
 1 
1+(− tan2)
1  6 
= ln 
2 1−(− 1 tan2)
 6 
3 6−tan2
= ln 
2 6+tan2 | M1
A1
M1
A1
[4] | 3.1a
1.1
1.1
2.2a | Attempting to differentiate r()
using the product rule.
All correct (any form)
Setting their derivative to 0 and
rearranging to a form containing
tan and tanh only.
AG so intermediate step must be
shown.
Condone use of  rather than
but final answer must be in
terms of  (or explicitly stating
“ = ”). | 1 1
asinh  cos2+bsin2cosh 
3 3
where a, b are non-zero constants
If using exponential form, must
reach tan and reduce to one
exponential term.
2
eg ⅇ3 𝜙(6+tan2𝜙) = 6−tan2𝜙
(d) | r( = 6sin(21.0207)sinh(1.0207/3))
= 1.85 (3 sf) cao | B1
[1] | 1.1 | 0.0741 comes from using degrees.