Further Mathematics Paper 2, Nov/Dec. 2011
 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
1. Solve for x and y in the equations:

3x + 5y – 4z = -5
6x + 3y – 5z = 26
-2x + 2y + z = -11
(b)  A function g is defined by g(x) = . Express g(x) in partial fractions.

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Observation

Candidates were reported to have performed very well in this question. In part (a), two methods were popular among the candidates. They were the Matrix method (Crammer’s rule) and elimination method. Majority of those who used the matrix method were able to obtain the matrix of coefficients as   . Its determinant was Δ =   = -13. Substituting the column matrix   for the first column of the matrix of coefficients and calculating the determinant of the resulting matrix gave   Δx =   =-260. Similarly, by substituting   in the second and third columns of the matrix of coefficients and finding the determinant of each resulting matrix gave Δy = -51 and Δz =  -275 respectively. Hence, x =  =  = 20. In the same way, y and z were obtained as 3.92 and 21.15 respectively.
The partial fraction in part (b) was also reported to be very well handled by majority of the candidates. Candidates were expected to factorize the denominator to obtain (x – 2)(x + 3). Hence,  =   +  . Solving this expression using appropriate methods gave A = -1 and B = -3