WAEC Further Mathematics Past Questions & Answers - Page 109

The upper quartile occurs at the $\frac{3N}{4}$ position.

= $\frac{3 \times 50}{4} = 37.5$

This occurs in the class 25 - 29.

$\text{Total frequency} = 50$

Median occurs at $\frac{50}{2} = \text{25th position}$

= Class 20 - 24 with classmark $\frac{20 + 24}{2} = 22$

Surface area of sphere, $A = 4\pi r^{2}$

$\frac{\mathrm d A}{\mathrm d r} = 8\pi r$

The rate of change of radius with time $\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}$

$\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})$

= $8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}$

$h : x \to 2 - \frac{1}{2x - 3}$

$h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}$

Let x = h(y)

$x = \frac{4y - 7}{2y - 3}$

$x(2y - 3) = 4y - 7  \implies 2xy - 4y = 3x - 7$

$y = \frac{3x - 7}{2x - 4}$

$h^{-1}(x) = \frac{3x - 7}{2x - 4}$

$\therefore  h^{-1}(\frac{1}{2}) = \frac{3(\frac{1}{2}) - 7}{2(\frac{1}{2}) - 4}$

= $\frac{\frac{-11}{2}}{-3} = \frac{11}{6}$

$h : x \to 2 - \frac{1}{2x - 3}$

$h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}$

Let x = h(y)

$x = \frac{4y - 7}{2y - 3}$

$x(2y - 3) = 4y - 7  \implies 2xy - 4y = 3x - 7$

$y = \frac{3x - 7}{2x - 4}$

$h^{-1}(x) = \frac{3x - 7}{2x - 4}$