Further Mathematics Questions and Answers

\(4x^{2} + 5kx + 10 = (2x + \sqrt{10})^{2}\)

Expanding the right hand side equation, we have

\(4x^{2} + 4x\sqrt{10} + 10\)

Comparing with the left hand side, we have

\(5k = 4\sqrt{10}  \implies k = \frac{4}{5}\sqrt{10}\)

\((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4  \implies  \frac{\sqrt{x} + 1}{\sqrt{x} - 1} + \frac{\sqrt{x} - 1}{\sqrt{x} + 1} = 4\)

\(\frac{(\sqrt{x} + 1)(\sqrt{x} + 1) + (\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\)

= \(\frac{x + 2\sqrt{x} + 1 + x - 2\sqrt{x} + 1}{x - 1} \implies \frac{2x + 2}{x - 1} = 4\)

\(2x + 2 = 4x - 4  \therefore 4x - 2x = 2x = 2 + 4= 6\)

\(x = 3\)

\(f(x) = 3x^{2} - 12x + 12\) and \(f(x) = 3\)

\(\therefore f(x) = 3 = 3x^{2} - 12x + 12 \implies 3x^{2} - 12x + 12 - 3 = 0\)

\(3x^{2} - 12x + 9 = 0; 3x^{2} - 9x - 3x + 9 = 0\)

\(3x(x - 3) - 3(x - 3) = (3x - 3)(x - 3) = 0\)

3x - 3 = 0 or x - 3 = 0

\(x = 1 or 3\)

The domain of a function refers to the regions where the function is defined or has a value on a particular region.

\(\frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\) has a domain defined on all set of real numbers because the function is defined on the set of real numbers when the denominator \(\sqrt{9x^{2} + 1} \geq 0\).

\(\sqrt{9x^{2} + 1} \geq 0 \implies 9x^{2} + 1 \geq 0\) which because of the square sign has a value for all values of x, be it negative or positive.

\(\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)

= \(\frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)

= \(\frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3} - \sqrt{3} - 1}\)

= \(\frac{6}{2} = 3\)