JAMB Mathematics Past Questions & Answers - Page 267

x $\ast$ y = x^y

x $\ast$ 2 = x²

x $\ast$ 2 = x

∴ x² - x = 0

x(x - 1) = 0

x = 0 or 1

$\frac{x + 1}{3}$ > $\frac{1}{X + 3}$ = $\frac{x + 1}{3}$ > $\frac{x + 3}{X + 3}$

= (x + 1)(x + 3)² > 3(x + 3) = (x + 1)[x² + 6x + 9] > 3(x + 3)

x³ + 7x² + 15x + 9 > 3x + 9 = x³ + 7x² + 12x > 0

= x(x + 3)9x + 4) > 0

Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0

= x > -4 (solution only)

Case 2 (+, -, -) = x > 0, x + 4 < 0

= x > 0, x < -3, x < -4 = x < -3(solution only)

Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)

Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0

= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)

combining the solutions -4 < x < -3

y² - 3y > 18 = 3y - 18 > 0

y² - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0

= (y + 3) (y - 6) > 0

Case 1 (+, +) $\to$ (y + 3) > 0, (y - 6) > 0

= y > -3 y > 6

Case 2 (-, -) $\to$ (y + 3) < 0, (y - 6) < 0

= y < -3, y < 6

Combining solution in case 1 and Case 2

= x < -3y < 6

= -3 < y < 6

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

$\frac{q - p}{pq}$ x $\frac{pq}{p^2q^2}$ = $\frac{q - p}{p^2 - q^2}$

$\frac{-(p - q)}{(p + q)(p - q)}$

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

No explanation has been provided for this answer.