JAMB Mathematics Past Questions & Answers - Page 195

0.0000152 x 0.042 = A x 10⁸

1 \(\leq\) A < 10, it means values of A includes 1 - 9

0.0000152 = 1.52 x 10^-5

0.00042 = 4.2 x 10^-4

1.52 x 4.2 = 6.384

10^-5 x 10^-4

= 10^-5-4

= 10^-9

= 6.38 x 10^-9

A = 6.38, B = -9

Given Cos z = L, z is an acute angle

\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z

= \(\frac{\text{cos z}}{\text{sin z}}\)

cosec z = \(\frac{1}{\text{sin z}}\)

cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\)

cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\)

sec z = \(\frac{1}{\text{cos z}}\)

tan z = \(\frac{\text{sin z}}{\text{cos z}}\)

sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\)

= \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)

the original eqn. becomes

\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\)

= \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\)

= \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\)

= sin z + 1

= 1 + \(\sqrt{1 - L^2}\)

= \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)

From the diagram, sin 60^o = \(\frac{PR}{8}\)

PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\)

= 4\(\sqrt{3}\)

Cos 45^o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\)

PS Cos45^o = 4\(\sqrt{3}\)

PS = 4\(\sqrt{3}\) x 2

= 4\(\sqrt{6}\)

4x - 3 = 3x + y = x - y = 3.......(i)

3x + y = 2y + 5x - 12.........(ii)

eqn(ii) + eqn(i) 3x = 15

x = 5

substitute for x in equation (i)

5 - y = 3

y = 2

\(x \propto y^{3} z^{4}\)

\(x = ky^{3} z^{4}\)

If y is halved and z is doubled, we have

\(x = k (\frac{1}{2} y)^{3} (2z)^{4}\)

\(x = k (\frac{y^{3}}{8}) (16z^{4})\)\)

\(x = 2k y^{3} z^{4}\)

\(\therefore \text{x is increased in the ratio 2 : 1}\)