Question 11:
--- 11(a) ---
11(a) | 1 
Uses  r2 d or  r2 d OE
2 0 | AO1.1a | M1 | 1
 4+2cos2 d
2
1 π
 (16+16cos+4cos2) d
2 π
π
 (8+8cos+(1+cos2)) d
π
π
 1 
88sin sin2
 
 2 
π
=
π
 1 
 98sin sin2
 
 2 

(9π+0+0)(9π00)
= 18π
Rewrites cos2 in terms of
cos2 | AO1.1a | M1
Correctly integrates ‘their’
expression, ft non-zero coefficients. | AO1.1b | A1F
Obtains required answer from fully
correct mathematical argument | AO2.1 | R1
(b) | Selects appropriate method to
determine polar equation by
equating OA and OB to find θ | AO3.1a | M1 | Let A  r ,   and Br ,  
1 2 2
1
OAOBr r
1 2
42cos θ 42cos θ
1 2
 
1 2
π π
Angle AOB    
3 1 2 3
π π
  and θ  
1 6 2 6
OAOB4+ 3
AB is perpendicular to the initial
line
Polar equation of AB is
  3  
rcos 4+ 3 for  
2 6 6
Uses the above to find two values
of θ and hence deduce the lengths
of OA and OB
Award this mark for correct
deduction using ‘their’ values of θ | AO2.2a | R1
Uses the correct polar equation for
a perpendicular line r = dsec θ | AO3.1a | M1
Obtains a correct equation for AB
(including correct specified range)
CAO | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution