JAMB Mathematics Past Questions & Answers - Page 198

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