WAEC Further Mathematics Past Questions & Answers - Page 112

$(x * y) = \frac{x+y}{2}$

$(3 * b) = \frac{3+b}{2}$

$x \circ y = \frac{x^{2}}{y}$

$(\frac{3+b}{2}) \circ 48 = \frac{(\frac{3+b}{2})^{2}}{48} = \frac{1}{3}$

$\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}$

$(3 + b)^{2} = \frac{48 \times 4}{3} = 64$

$b^{2} + 6b + 9 = 64 \implies b^{2} + 6b + 9  - 64 = 0$

$b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0$

$b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (} b + 11) = 0$

Since b > 0, b - 5 = 0 

b = 5.

When you have two lines, $y_{1}, y_{2}$, perpendicular to each other, the product of their slopes = -1.

$3x + 4y + 6 = 0 \implies 4y = -6 - 3x$

$\therefore y = \frac{-6}{4} - \frac{3}{4}x$

$\frac{\mathrm d y}{\mathrm d x} = \frac{-3}{4}$

Also, $4x - by + 3 = 0  \implies by = 4x + 3$

$y = \frac{4}{b}x + \frac{3}{b}$ 

$\frac{\mathrm d y}{\mathrm d x} = \frac{4}{b}$

$\frac{-3}{4} \times \frac{4}{b} = -1 \implies \frac{4}{b} = \frac{4}{3}$

$b = 3$

$(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin \theta + \sin \theta - \sin^{2} \theta$

$= 1 - \sin^{2} \theta$

Recall, $\cos^{2} \theta + \sin^{2} \theta = 1$

$\therefore 1 - \sin^{2} \theta = \cos^{2} \theta$.

$\frac{1}{5^{-y}} = 25(5^{4-2y})$

$\implies 5^{y} = (5^{2})(5^{4-2y})$

$5^{y} = 5^{2+4-2y}$

Comparing bases, we have

$y = 6 - 2y$

$3y = 6 \implies y = 2$

No explanation has been provided for this answer.