Further Mathematics Questions and Answers

\(\frac{^{8}P_{x}}{^{8}C_{x}} = \frac{8!}{(8 - x)!} ÷ \frac{8!}{(8 - x)! x!} = 6\)

\(\frac{8!}{(8 - x)!} \times \frac{(8 - x)! x!}{8!} = x! = 6\)

\(x = 3\)

\(\log_{3}(x - y) = 1 \implies x - y = 3^{1} = 3 .... (1)\)

\(\log_{3}(2x + y) = 2 \implies 2x + y = 3^{2} = 9 ..... (2)\)

From (1), y = x - 3

From (2), y = 9 - 2x

\(\implies 9 - 2x = x - 3\)

\(9 + 3 = x + 2x = 3x\)

\(x = 4\)

\((216)^{-\frac{2}{3}} = (\frac{1}{216})^{\frac{2}{3}} = (\sqrt[3]{\frac{1}{216}})^{2} = (\frac{1}{6})^{2} = \frac{1}{36}\)

\((0.16)^{-\frac{3}{2}} = (\frac{100}{16})^{\frac{3}{2}} = (\sqrt{100}{16})^{3} = \frac{1000}{64}\)

\(\frac{1}{36} \times \frac{1000}{64} = \frac{125}{288}\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - \frac{4}{x^5} = 3x^{2} - 4x^{-5}\)

\(y = \int (3x^{2} - 4x^{-5}) \mathrm {d} x \)

\(y = x^{3} + \frac{1}{x^{4}} + c\)

f(1) = 4; \(4 = 1^{3} + \frac{1}{1^{4}} + c \implies 4 = 2 + c\)

\(c = 2\)

\(f(x) = x^{3} + \frac{1}{x^{4}} + 2\)

Given the roots of an equation such that you can find the sum and product of the roots, the equation can be given as:

\(x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0 \)

\(\alpha + \beta = -3 + 1 = -2\)

\(\alpha \beta = -3 \times 1 = -3\)

Equation: \(x^{2} - (-2)x + (-3) = 0 \implies x^{2} + 2x - 3 = 0\)