Further Mathematics Paper 2, May/June. 2015

Question 4

Solve: tan (2x – 15)o – 1 = 0, for values of x such that 0o ≤ x ≤ 360o.


This question was reported to be very popular among the candidates. Majority of them attempted it and performed well in it. However, majority of them did not score all the marks allocated the question because they did not give all the four values of x. Candidates were expected to rewrite the given equation as tan (2x – 15)o = 1.

Since 0o ≤ x ≤ 360o, then 0o ≤ 2x ≤ 720o. Hence, possible values of tan-1(1) = 45o, 225o, 405o and 585o. i.e. (2x – 15)o = 45o or 225o or 405o or 585o. Therefore, 2x = 60o or 240o or 420o or 600o. This gave the values of x as 30o or 120o or 210o or 300o.