Further Mathematics Questions and Answers

$^{n}C_{3} = \frac{n!}{(n - 3)! 3!}$

$^{n}P_{2} = \frac{n!}{(n - 2)!}$

$\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}$

$\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}$

Note that $(n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!$

$\frac{(n - 2)(n - 3)!}{(n - 3)! 3!} = 1$

$\frac{n - 2}{3!} = 1 \implies n - 2 = 6$

$n = 2 + 6 = 8$

Given line: $3x - 2y + 4 = 0 \implies 2y = 3x + 4$

$y = \frac{3}{2}x + 2$

$Gradient (\frac{\mathrm d y}{\mathrm d x}) = \frac{3}{2}$

Gradient of perpendicular line = $\frac{-1}{\frac{3}{2}} = \frac{-2}{3}$

$\implies \frac{y - (-3)}{x - 2} = \frac{-2}{3}$

$\frac{y + 3}{x - 2} = \frac{-2}{3}$

$3(y + 3) = -2(x - 2) \implies 3y + 2x + 9 - 4 = 0$

= $3y + 2x + 5 = 0$

 $V = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ and $U = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$

$U + V = \begin{pmatrix} -1 - 2 \\ 5 + 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 9 \end{pmatrix}$

$|U + V| = \sqrt{(-3)^{2} + 9^{2}} = \sqrt{9 + 81} = \sqrt{90}$

= $sqrt{9 \times 10} = 3\sqrt{10}$

Mean ($\bar{x}$) = $\frac{1+2+3+4+5+5+6+7+8+9}{10} = \frac{50}{10} = 5$

$MD = \frac{\sum |x - \bar{x}|}{n} = \frac{20}{10} = 2$

$g(x) = \frac{x + 1}{x + 2}, x \neq 2$

Let y = x, then $g(y) = \frac{y + 1}{y + 2}$

Let x = g(y), so that $x = \frac{y + 1}{y + 2}$

$x(y + 2) = y + 1$

$xy + 2x = y + 1 \implies xy - y = 1 - 2x$

$y(x - 1) = 1 - 2x \implies y = \frac{1 - 2x}{x - 1}$

$y = g^{-1}(x) = \frac{1 - 2x}{x - 1}$

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