WAEC Further Mathematics Past Questions & Answers - Page 103

Since f(3) = 0, then (x - 3) is a factor of f(x).

Dividing f(x) by (x - 3), we get $2x^{2} + 3x - 2$.

$2x^{2} + 3x - 2 = 2x^{2} - x + 4x - 2$

$x(2x - 1) + 2(2x - 1) = (x + 2)(2x - 1)$

Therefore, $f(x) = (x - 3)(x + 2)(2x -1)$

$2x^{2} - 6x + 5 = 0 \implies a = 2, b = -6, c = 5$

$\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3$

$\alpha\beta = \frac{c}{a} = \frac{5}{2}$

$\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^{2} + \alpha^{2}}{\alpha\beta}$

$\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{\alpha\beta} = \frac{3^{2} - 2(\frac{5}{2})}{\frac{5}{2}}$

= $\frac{4}{\frac{5}{2}} = \frac{8}{5}$

$\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}$

Squaring both sides, we have

$(\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}$

$x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1$

$2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0$

$(2\sqrt{x(x + 1)})^{2}= 0^{2}  \implies 4(x(x + 1)) = 0$

$\therefore x(x + 1) = 0$

$x = \text{0 or -1}$

$(a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...$

Third term = $^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)$

= $15a^{4}b^{2}$

$f(x) = x^{2}, g(x) = \sin x$

$g \circ f = g(x^{2}) = \sin x^{2}$