Further Mathematics Paper 2, Nov/Dec. 2014
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 9

(a)   Resolve x2 + 1 into partial fractions.
3

(b)   The gradient of a curve at point (x, y) is (2x – 3). If the minimum value on the curve
is 3, find the equation of the curve.

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Observation

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.