Further Mathematics Questions and Answers

\(\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp = 2\)

\(adj^{2} = 2^{2} - (\sqrt{3})^{2} = 1 \implies adj = 1\)

\(\cos \theta = \frac{1}{2}\)

\(\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}\)

\(\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}\)

\(\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\)

= \(- \sqrt{3}\)

Equation of a circle with centre coordinates (a, b) : \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Area of circle = \(\pi r^{2} = 25\pi cm^{2} \implies r^{2} = 25 \)

\(\therefore r = 5cm\)

(a, b) = (-2, 3)

Equation: \((x - (-2))^{2} + (y - 3)^{2} = 5^{2}\)

\(x^{2} + 4x + 4 + y^{2} - 6y + 9 = 25 \implies x^{2} + y^{2} + 4x - 6y + 13 - 25 = 0\)

= \(x^{2} + y^{2} + 4x - 6y - 12 = 0\)

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

\(\frac{\mathrm d y}{\mathrm d x} = -3 + 6x^{2}\)

At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = -3 + 6(1^{2}) = -3 + 6 = 3\)

\(y = mx - 3 \implies \frac{\mathrm d y}{\mathrm d x} = m = 3\) (Tangent with equal gradient)

\(x + 3 = 0 \implies x = -3\)

Using remainder theorem, if x + 3 is a factor, f(-3) = 0.

\(f(-3) = (-3)^{3} + 3(-3)^{2} + n(-3) - 12 = 0\)

\(-27 + 27 - 3n - 12 = 0 \implies -3n = 12\)

\(n = -4\)

No explanation has been provided for this answer.