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 12

(a) Given that m1X² - 2x + 1 )dx = 1/3,m > 0, determine the value of m.
(b) If (x - y)² = 3xy + 1, find the gradient at point (1, 0).

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Observation

The report stated that more candidates attempted this question than question 11. It further stated that candidates' performance on this question was poor.
In part (a) candidates were expected to integrate the expression with respect to x, substitute the limits of integration and equate the result to 1/3. Candidates were reported not to have followed this procedure. A few others were not able to manipulate the fractions involved. Evaluating the
definite integral, we obtained m1(x² - 2x + 1 )dx = (x²/3 - X² + x)m1 =[ (m³/3 - m² + m - (1/3 - 1 + 1) ]
=m³/3 - m² + m - 1/3. This implied that m³/3 - m² + m - 1/3 = 1/3.→(m³/3 - m² + m - 2/3 = 0
i.e. m³ - 3m² +3m - 2 = 0. =(m - 2)(m² - m + 1) = 0. Hence, m = 2.
In part (b), majority of the candidates who attempted this part of the question performed well in it. Candidates were expected to differentiate the equation with respect to x and substitute 1 for x and 0 for y in the resulting differential coefficient as follows:
(x - y)² = 3xy + 1 = X² -2xy + y² = 3xy + 1
2x - (2y +2x dy) +2y dy = 3y + 3x dy. By bring like term together, and simplifying, we obtain
dx         dx                  dx
dy =    5y - 2x , When x = 1 and y = 0,
dx
2y - 5x²

dy =    5(0) - 2(1)
dx
2(0) - 5(1)

2 or 0.4

5