Further Mathematics Questions and Answers

\(f(4) = 3(4) - 2 = 12 - 2 = 10\)

\(f(-3) = 5(-3) - 3 = -15 - 3 = -18\)

\(f(4) - f(-3) = 10 - (-18) = 28\)

The angle subtended by the minor arc = \(\frac{\pi}{3} radians\)

The angle subtended by the major arc = \(2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)

Perimeter of the major arc = \(r\theta + 2r\)

= \(12 \times \frac{5\pi}{3} + 2(12) = 20\pi + 24\)

= \(4(5\pi + 6)\)

\((x = \frac{nx_{1} + mx_{2}}{n + m}, y = \frac{ny_{1} + my_{2}}{n + m})\)

Given P(-2, 3) and Q(4, 9),

\((\frac{2(4) + 3(-2)}{2 + 3}, \frac{2(9) + 3(3)}{2 + 3})\)

= \((\frac{2}{5}, \frac{27}{5})\)

= \((\frac{2}{5}, 5\frac{2}{5})\)

\(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x  = (\frac{8x^{1 + 1}}{2} - \frac{4x^{2+1}}{3})|_{0}^{2}\)

= \((4x^{2} - \frac{4x^{3}}{3}) |_{0}^{2}\)

= \((4(2^2) - \frac{4(2^3)}{3})\)

= \(16 - \frac{32}{3} = \frac{16}{3}\)

\(s = ut + \frac{at^{2}}{2}\)

This movement is against gravity, so it is negative.

\(s = ut - \frac{gt^{2}}{2}\)

\(s = 20m, u = 25ms^{-1}\)

\(20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}\)

\(5t^{2} - 25t + 20 = 0 \)

\(5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0\)

\(5t - 20 = \text{0 or t - 1 = 0}\)

\(t = \text{1 or 4}\)