WAEC Further Mathematics Past Questions & Answers - Page 96

In order to do this, simply find the option in the range where only the cos is +ve. This occurs in the range \(270° \leq x \leq 360°\).

Check: \(\sin 300 = - \sin 60 = \frac{-\sqrt{3}}{2}\)

\(\cos 300 = \cos 60 = \frac{1}{2}\)

\(T_{n} = ar^{n - 1}\) (Geometric progression)

\(a = \sqrt{6}, r = \frac{T_{2}}{T_{1}} = \frac{3\sqrt{2}}{\sqrt{6}} \)

\(r = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}\)

\(\therefore T_{8} = (\sqrt{6})(\sqrt{3})^{8 - 1} \)

= \((\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}\)

\(x^{2} - 2x \geq 3 \implies x^{2} - 2x - 3 \geq 0\)

\(x^{2} + x - 3x - 3 = (x + 1)(x - 3) \geq 0\) 

\(x = -1 ; x = 3\)

Check: \(x = -1  : (-1)^{2} - 2(-1)  = 1 + 2 \geq 3\)  (satisfied)

\(-1 < x < 3 : 0^{2} - 2(0) = 0 \geq 3\) (not satisfied)

\(x < -1 : (-2)^{2} - 2(-2) = 4 + 4 = 8 \geq 3\) (satisfied)

\(x = 3 : 3^{2} - 2(3) = 9 - 6 = 3 \geq 3\) (satisfied)

\(x > 3 : 4^{2} - 2(4) = 16 - 8 = 8 \geq 3\) (satisfied)

\(\therefore x^{2} - 2x \geq \text{3 is satisfied in the region x} \leq \text{-1 and x} \geq 3\) 

\(\frac{\cos 2\theta - 1}{\sin 2\theta}\)

\(\cos (x + y) = \cos x \cos y - \sin x \sin y \implies \cos 2\theta = \cos^{2} \theta - \sin^{2} \theta\)

\(\cos^{2} \theta = 1 - \sin^{2} \theta \implies \cos 2\theta = 1 - 2\sin^{2} \theta\)

\(\sin 2\theta = 2\sin \theta \cos \theta\)

\(\therefore \frac{\cos 2\theta - 1}{\sin 2\theta} = \frac{1 - 2\sin^{2}\theta - 1}{2\sin \theta \cos \theta}\)

= \(\frac{-2 \sin^{2} \theta}{2\sin \theta \cos \theta} = \frac{- \sin \theta}{\cos \theta}\)

= \(-\tan \theta\)

No explanation has been provided for this answer.