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.