Further Mathematics Questions and Answers

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

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

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

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

\(xy + 2x = y + 1 \implies xy - y = 1 - 2x\)

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

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

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

\(PQ = \begin{pmatrix} 7 - 3 \\ 4 - 1 \end{pmatrix}\)

\(= \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

\(\hat{n} = \frac{\overrightarrow{PQ}}{|PQ|} \)

\(|PQ| = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\)

\(\hat{n} = \frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\) 

P(winning) = \(\frac{2}{10}\)

P(both tickets winning) = \(\frac{2}{10} \times \frac{1}{9} = \frac{1}{45}\)

\(P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}\) 

PQ = \(\begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2 - 2x & 6 + 4x \end{pmatrix} = R\)

\(\implies 2 - 2x = -8; -2x = -8 - 2 = -10\)

\(6 + 4x = 26 \implies 4x = 26 - 6 = 20\)

\(\implies x = 5\)

Arranging the scores in ascending order, we have: 2, 5, 12, 17, 21, 29, 33, 41, 43, 45.

The upper quartile = 41.