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Further Mathematics Paper 2, May/June. 2009  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
General Comments
Weakness/Remedies
Strength











Question 1

(a)      Solve 2x² + x – 6 < 0.
                        5 – 2 √10
(b)     Express   3√5 + √2   in the form m√2 + n√5,
        where m and n are rational numbers.

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Observation

The part (a) of the question was reportedly attempted by majority of the candidates. However, most of the candidates who attempted the question could not investigate properly the range of values for which the inequality was valid. They were expected to know that if 2x²+x-6 is factorized to get (2x - 3)(x + 2), then 2x²+x-6 < 0 becomes (2x - 3)(x + 2) < 0 which implies that either (2x-3) < 0 and (x+2) > 0 or (2x -3) > 0 and (x+2) < 0 i.e. either -2 < x < 3/2  or  -2 > x > 3/2 (which is impossible). Hence -2 < x < 3/2.

Many of the candidates who attempted this question were able to get   (2x - 3)(x + 2) < 0 but reasoned that the inequality means (2x – 3)< 0 and (x + 2)< 0 which was wrong.

In part (b), candidates were expected to multiply the numerator and the denominator by the conjugate of the denominator (3√5 - √2) and simplify the results to get -35√2/43 + 19√5/43 or -0.81 √2 + 0.44√5.
The report stated that majority of the candidates who attempted this question performed well.  However, many of them did not express the final answer in the required form.  Others did not multiply by the correct conjugate of the denominator.

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