WAEC Further Mathematics Past Questions & Answers - Page 56

$f(x) = 2x - 1; g(x) = x^{2} + 1$

$f(x) = 2x - 1$

$f(y) = 2y - 1$

$x = 2y - 1 \implies 2y = x + 1$

$y = \frac{x + 1}{2}$

$g(3) = 3^{2} + 1 = 10$

$f^{-1}(10) = \frac{10 + 1}{2} = \frac{11}{2}$

$S_{n} = \frac{n}{2}(2a + (n - 1)d)$ ( for an arithmetic progression)

$S_{3} = 18 = \frac{3}{2}(2(4) + (3 - 1) d)$

$18 = \frac{3}{2} (8 + 2d)$

$18 = 12 + 3d \implies 3d = 6$

$d = 2$

$\therefore T_{1} = 4 \implies T_{2} = 4 + 2 = 6; T_{3} = 6 + 2 = 8$

Their product = $4 \times 6 \times 8 = 192$

No explanation has been provided for this answer.

Given $ax^{2} + bx + c = 0 (\text{general form of a quadratic equation})$

We have $x^{2} + \frac{b}{a}x + \frac{c}{a} = 0$

$x^{2} - (-\frac{b}{a})x + \frac{c}{a} = 0$

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

$\therefore \text{The equation is} x^{2} - 3x + 2 = 0$ 

Two lines are parallel if and only if their slopes are equal.

$kx - 5y + 6 = 0 \implies 5y = kx + 6$

$y = \frac{k}{5}x + \frac{6}{5}$

$Slope = \frac{k}{5}$

$mx + ny - 1 = 0 \implies ny = 1 - mx$

$y = \frac{1}{n} - \frac{m}{n}x$

$Slope = -\frac{m}{n}$

$Parallel \implies \frac{k}{5} = -\frac{m}{n}$

$-5m = kn \implies 5m + kn = 0$