JAMB Mathematics Past Questions & Answers - Page 337

x * y = x^y
x * 2 = 12 - x

Thus by comparison,

x = x, y = 2

But x * y = x * 2

x^y = 12 - x

x² = 12 - x

x² + x - 12 = 0

x² + 4x - 3x - 12 = 0

x(x + 4) - 3(x + 4) = 0

(x - 3)(x + 4) = 0

x - 3 = 0 or x + 4 = 0

So x = 3 or x = -4

Common ratio r of the G.P is

\(r = \frac{T_n + 1}{T_n} = \frac{T_2}{T_1}\)

\(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}}\)

\(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}} \)

\( = \frac{(\sqrt{10})(\sqrt{10}) + (\sqrt{10})(-\sqrt{5}) + (2\sqrt{5})(\sqrt{10}) + (2\sqrt{5})(-\sqrt{5})}{(\sqrt{10})^2 - (\sqrt{5})^2}\)

\(\frac{10 - \sqrt{50} + 2\sqrt{50} - 10}{10 - 5}\)

\(\frac{\sqrt{50}}{5}\)

\(\frac{\sqrt{25 \times 2}}{5}\)

\(\frac{5\sqrt{2}}{5}\)

\(\sqrt{2}\)

a + 3d = 13 .......... (1)
a + 9d = 31 .......... (2)

(2) - (1): 6d = 18

d = 18/6 = 3

From (1), a + 3(3) = 13

a + 9 = 13

a = 13 - 9 = 4

Hence,
T24 = a + 23d
T24 = 4 + 23(3)
T24 = 4 + 69
T24 = 73

\(\frac{x}{2} + \frac{3}{4} \leq \frac{5x}{6} - \frac{7}{12}\)

\(12\frac{x}{2} + 12\frac{3}{4} \leq 12\frac{5x}{6} - 12\frac{7}{12}\)

6x + 9 \(\leq\) 10x - 7

6x - 10x \(\leq\) - 7 - 9

-4x \(\leq\) -16

-4x/-4 \(\geq\) -16/-4

x \(\geq\) 4

Consider the range -3 < x < -1

= { -2, -1, 0}, for instance

When x = -2,

\(\frac{-2 - 5}{-2 + 3} < -1\)

\(\frac{-7}{1} < -1\)

When x = -1,

\(\frac{-1 - 5}{-1 + 3} < -1\)

\(\frac{-6}{2} < -1\)

= -3 < -1

When x = 0,

\(\frac{0 - 5}{0 + 3} < -1\)

\(\frac{- 5}{3} < -1\)

Hence -3 < x < 1