JAMB Mathematics Past Questions & Answers - Page 204

C = a + k

$\frac{1}{N}$ = c

= $\frac{aN + k}{N}$

CN = aN + K

30(100) = a(100) + k

3000 = 100a + k.......(i)

60(40) = a(40) + k

2400 = 40a + k.......(ii)

eqn (i) - eqn (ii)

600 = 60a

a = 10

subt. for a in eqn (i) 3000 = 100(10) + K

3000 - 1000 = k

k = 2000

CN = 10N + 2000. when N = 50,

50C = 10(50) + 2000

50C = 500 + 2000

C = $\frac{2500}{50}$

= 50k

$(\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2})(\sqrt{3} + \sqrt{2})$

$(\sqrt[4]{3} + \sqrt[4]{2})(\sqrt[4]{3} - \sqrt[4]{2}) = \sqrt{3} - \sqrt[4]{6} + \sqrt[4]{6} - \sqrt{2}$

= $\sqrt{3} - \sqrt{2}$

$(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) = 3 + \sqrt{6} - \sqrt{6} - 2$

= $3 - 2 = 1$

$e^{x} = 1 + x + \frac{x^{2}}{1.2} + \frac{x^{3}}{1.2.3} + ...$

$\frac{1}{e^{\frac{1}{2}}} = e^{-\frac{1}{2}}$

$e^{-\frac{1}{2}} = 1 - \frac{x}{2} + \frac{x^{2}}{1.2^{3}} - \frac{x^{3}}{1.2^{4}.3} + ...$

Equal roots implies b² - 4ac = 0

a = 1b = - (p - 2), c = 2p + 1

[-(p - 2)]² - 4 x 1 x (2p + 1) = 0

p² - 4p + 4 - 4(2p + 1) = 0

p² - 4p = 4 - 8p - 4 = 0

p² - 12p = 0

p(p - 12) = 0

p = 0 or 12

N = 22$\frac{1}{2}$%, M = 17$\frac{1}{10}$%

M = $\frac{171}{10}$%, N = $\frac{45}{2}$

$\frac{45}{2}$ x $\frac{10}{171}$

= $\frac{225}{171}$

= 1 $\frac{54}{171}$

= 1 $\frac{18}{57}$