Further Mathematics Questions and Answers

Let the sum of the men's ages be f, so that

\(\frac{f}{n} = 50 .... (1)\)

Also, \(\frac{f - (55 + 63)}{n - 2} = 50 - 1 = 49 .... (2)\)

From (1), \(f = 50n\)

From (2), \(f - 118 = 49(n - 2) = 49n - 98\)

\(f = 49n - 98 + 118 = 49n + 20\)

\(\therefore f = 50n = 49n + 20\)

\(50n - 49n = n = 20\)

\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)

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

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

\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} \frac{-1}{x - 2}\)

= \(\frac{-1}{1 - 2} = \frac{-1}{-1} = 1\)

\(\log_{0.25} 8 = x\)

\(8 = 0.25^{x}\)

\(2^{3} = (2^{-2})^{x} \implies 3 = -2x\)

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

\(S_{\infty} = \frac{a}{1 - r}\) (for an exponential series)

\(r = \frac{24}{96} = \frac{6}{24} = \frac{1}{4}\)

\(S_{\infty} = \frac{96}{1 - \frac{1}{4}} = \frac{96}{\frac{3}{4}}\)

= \(\frac{96 \times 4}{3} = 128\)

\(2x^{2} + kx + 5 = 0\)

\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)

\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)

\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)

\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)

\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)

\(k^{2} = 16 \therefore k = \pm 4\)