Further Mathematics Paper 2, Nov/Dec. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 11

(a) Find the area enclosed by the x - axis and curve y = 3x² u + 2x - 1.
(b) If 2X² - x - 3 = P(x + Q)² + R where P, Q and R are constants, find the:
(i) values of P, Q and R;
(ii) minimum value of 2X² - x - 3.

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Observation

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.