This equation was reported to have been attempted by majority of the candidates and their performance was regarded as fair.  A few candidates were reported not to have expressed the given fraction in partial fraction correctly. 

In part(a), candidates were expected to respond as follows:   ≡  +  +                    i.e  3 + 1 = A(+1)(-2) + B(-2) + C( + 1)2.  When  = 2, 9C = 7.  This implied that C = . When x = -1, -3B = -2.  This implied that B = .  When x = 0, 1 = -2A – 2B + C i.e. 1 = -2A -  + .  Solving this gave A .  Thus    =  +  - .
In part(b), candidates were expected to show that if   = 6, then 1(-1-0) -2(1-0) + (2 + 1) = 6, i.e. -1-2 + 3 = 6.  Solving this equation gave  = 3.