Further Mathematics Questions and Answers

\((3x + 4)^{4} = ^{4}C_{0}(3x)^{0}(4)^{4} + ^{4}C_{1}(3x)^{1}(4)^{3} + ^{4}C_{2}(3x)^{2}(4)^{2} + ^{4}C_{3}(3x)^{3}(4)^{1} + ^{4}C_{4}(3x)^{4}(4)^{0}\)

\(x^{3} = ^{4}C_{3}(3x)^{3}(4) = \frac{4!}{3!1!} \times 3^{3} \times 4\)

= \(432x^{3}\)

\(a . b = |a||b|\cos \theta\)

\(\cos \theta = \frac{a . b}{|a||b|}\)

= \( \frac{(5i + 3j).(3i - 5j)}{(\sqrt{5^2 + 3^2})(\sqrt{3^{2} + (-5)^{2}})}\)

= \(\frac{0}{34} = 0\)

\(\theta = \cos^{-1} 0 = 90°\)

\(BC = BA + AC\)

Given, \(AB\), then \(BA = - AB\)

= \(AB = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \implies BA = \begin{pmatrix} -4 \\ -3 \end{pmatrix}\)

\(\therefore BC = \begin{pmatrix} -4 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}\)

= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\)

\(|BC| = \sqrt{(-2)^{2} + (-6)^{2}} = \sqrt{40} \)

= \(2\sqrt{10}\)

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

\(\int (x^2 + \frac{1}{x^2} - 2) \mathrm {d} x\)

= \(\int (x^2 + x^{-2} - 2) \mathrm {d} x\)

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

For equal roots, \(b^{2} - 4ac = 0\)

From the equation, \(a = P, b = (P+1), c = P\)

\((P+1)^{2} - 4(P)(P) = P^{2} + 2P + 1 - 4P^{2} = 0\)

\(-3P^{2} + 2P + 1 = 0 \implies 3P^{2} - 2P - 1 = 0\)

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

\(3P(P - 1) + 1(P - 1) = 0\)

\(P = \text{1 or }\frac{-1}{3}\)