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 Further Mathematics Paper 2, Nov/Dec. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 2

(a)        Evaluate .

(b)        Given that f(x – 1) = 2 - 3 + 5 + 2, find f (2).

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Observation

It was reported that this question was not as popular as question 1 and majority of the candidates who attempted this question did not perform well.  Their performance in part(b) was however reported to be better than what it was in part (a).
In part (a), candidates were expected to rationalize the fraction before taking the limits.  Taking the limit of the fraction as originally given, gave  which is undefined.  Therefore the fraction should be rationalized to remove the surd in its denominator.  The conjugate of the denominator was 1 +  Therefore rationalizing gave  x     =
=  .Therefore,  =
=  = 2.

In part (b), some candidates were reported to have substituted 2 for x.  This was wrong.  Candidates were expected to show that if f(x-1) = 2x3 – 3x2 + 5x + 2, then f(2) meant that x – 1
= 2 i.e. x = 3.  Hence, they should substitute 3 for x instead of 2.  This would give
f(2)3 – 3(3)2 + 5(3) + 2 = 44.