Candidates' performance in this question was described as poor.

In part (a), candidates were expected to recognize the problem as an application of integration.
In solving this problem, candidates should obtain the values of x when y = 0 (on the x - axis) as
they will serve as the limits of integration i.e 3x² + 2x - 1 = 0 ⇒ 3x² + 3x - x - 1 = O. Factorising,
we have (3x -1)(x + 1) = 0 ~ x = -1, 1/3

Area = &int ^{1/3} -1 (3x² + 2x -1)dx = [x³ + x² - x]^{1/3}-1 = [(1/27 + 1/9 - 1/3) - (-1 + 1 + 1)] = 22/27sq. unit

In part (b), candidates expected responses were:

2x² - x -3 = 2(x² -1/2x) - 3 = 2(x² - 1/2x + 1/16 - 1/16) - 3 = 2(x - 1/4)² - 1/8 - 3 = 2(x - 1/4)² - 25/8.Comparing 2(x -1/4)² -25/8 with p(x + Q)² +R, we have P =2, Q =-1/4 and R = -25/8.

The minimum valve of y = R =-25/8.