
Question 12

(a) Given that ∫^{m}_{1}X²  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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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 ∫^{m}_{1}(x²  2x + 1 )dx = (x²/3  X² + x)^{m}_{1} =[ (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
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