WAEC Further Mathematics Past Questions & Answers - Page 136

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

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

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

$2x + 2 = 4x - 4  \therefore 4x - 2x = 2x = 2 + 4= 6$

$x = 3$

$f(x) = 3x^{2} - 12x + 12$ 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.

$\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}$

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

= $\frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3} - \sqrt{3} - 1}$

= $\frac{6}{2} = 3$

Divide $(2x^{3} - 5x^{2} - x + 6)$ by $(2x^{2} - x - 3)$ to get the other factor. Be careful of sign conventions.