The Chief Examiner reported that this question was attempted by majority of the candidates and they performed poorly in it. However, their performance was also reported to be better in part (b) than in part (a). 
In part (a), candidates were expected to express  as + +  in partial fractions. This was not done by majority of those who attempted this question.  If  =  + +  , then x2 + 1 = A(x + 2)2 + B(x + 2) + C. i.e x2 + 1 = Ax2 + 4Ax + 4A + Bx + 2B + C.   By comparing coefficients we have, A = 1 ...eqn.(1); 4A + B = 0.... eqn.(2) and 4A+2B + C = 1....eqn3.
Substituting eqn(1) in to eqn.(2) gave 4(1) + B = 0. Which implied that B = -4.  Substituting 1 for A and -4 for B in eqn(3) gave 4(1) + 2(-4) + C = 1 which implied that C=5.  Therefore, the partial fractions of  =   -  + .

In part (b), candidates were expected to integrate (2x - 3) with respect to x to have
y = x2 - 3x + c.  The minimum value of y would occur at the point where the gradient function (2x – 3) = 0 i.e. when x = .  Substituting this into the equation y = x2 – 3 x + c gave 3 =  - 3() + c. Simplifying this gave c = .  Therefore the required curve was 4y = 4x2 – 12x + 21.