JAMB Mathematics Past Questions & Answers - Page 193

Pq + 1 = q²......(i)

t = $\frac{1}{p}$ - $\frac{1}{pq}$.........(ii)

p = $\frac{q^2 - 1}{q}$

Sub for p in equation (ii)

t = $\frac{1}{q^2 - \frac{1}{q}}$ - $\frac{1}{\frac{q^2 - 1}{q} \times q}$

t = $\frac{q}{q^2 - 1}$ - $\frac{1}{q^2 - 1}$

t = $\frac{q - 1}{q^2 - 1}$

= $\frac{q - 1}{(q + 1)(q - 1)}$

= $\frac{1}{q + 1}$

$(1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)$

$(x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1$

$(1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)$

a2 - b2 = ($\frac{2x}{1 - x}$)2 - ($\frac{1 + x}{1 - x}$)2

= ($\frac{2x}{1 - x} + \frac{1 + x}{1 - x}$)($\frac{2x}{1 - x} - \frac{1 + x}{1 - X}$)

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

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

< B is the largest since the side facing it is the largest, i.e. (x + 4)cm

Cosine B = $\frac{1}{5}$

= 0.2 given

b² - a² + c² - 2a Cos B

Cos B = $\frac{a^2 + c^2 - b^2}{2ac}$

$\frac{1}{5}$ = $\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}$

$\frac{1}{5}$= $\frac{x(x - 16)}{2x(x - 4)}$

$\frac{1}{5}$ = $\frac{x - 16}{2x - 8}$

= 5(x - 16)

= 2x - 8

3x = 72

x = $\frac{72}{3}$

= 24

If diameter of the circle = 12cm; radius of the circle(L) = $\frac{12}{2}$

= 6cm

$\frac{\theta}{360}$ = $\frac{r}{L}$ where $\theta$ = 210$\theta$, L = 6cm

$\frac{210}{360}$ = $\frac{r}{6}$

where r = radius of the base of the cone

V = $\frac{1260}{360}$

= 3.50cm