Further Mathematics Questions and Answers

\(h(x) = x^{3} - \frac{1}{x^{3}}\)

\(h(a) = a^{3} - \frac{1}{a^{3}}\)

\(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\)

\(h(a) - h(\frac{1}{a}) = (a^{3} - \frac{1}{a^{3}}) - (\frac{1}{a^{3}} - a^{3}) = 2a^{3} - \frac{2}{a^{3}}\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4\)

\(y = \int (3x^{2} - 4) \mathrm {d} x = x^{3} - 4x + c\)

y = 6 when x = 3

\(6 = 3^{3} - 4(3) + c \implies 6 = 27 - 12 + c\)

\(c = 6 - 15 = -9\)

\(y = x^{3} - 4x - 9\)

\(h = 45t - 9t^{2}\)

\(\frac{\mathrm d h}{\mathrm d t} = 45 - 18t = 0\)

\(45 = 18t \implies t = 2.5 s\)

\(h(2.5) = 45(2.5) - 9(2.5)^{2} = 112.5 - 56.25\)

= \(56.25 m.\)

\(F_{1} = (7i + 8j)N ; F_{2} = (3i + 4j)N\)

\(|F_{1} - F_{2}| = |(7i + 8j) - (3i + 4j)| = |4i + 4j|\)

\(|4i + 4j| = \sqrt{4^{2} + 4^{2}} = \sqrt{32} = 4\sqrt{2}\)

\(\tan \theta = \frac{y}{x} = \frac{4}{4} = 1\)

\(\theta = \tan^{-1} 1 = 045°\)

= \((4\sqrt{2} N, 045°)\)

Given lines \(OA\) and \(OB\) inclined at angle \(\theta\), the line \(AB\) is gotten using cosine rule.

\(|AB|^{2} = |OA|^{2} + |OB|^{2} - 2|OA||OB|\cos \theta\)

\(|AB|^{2} = 4^{2} + 5^{2} - 2(4)(5)\cos 60\)

= \(16 + 25 - 20\)

\(|AB|^{2} = 21 \implies |AB| = \sqrt{21}\)

\(\implies \text{The two particles are} \sqrt{21} m \text{apart in 1 sec}\)

In two seconds, the particles will be \(2\sqrt{21} m\) apart.