1993 - JAMB Mathematics Past Questions and Answers - page 2

(\frac{1 + ax}{1 - ax}) = (\frac{p}{q}) by cross multiplication,

q(1 + ax) = p(1 - ax)

q + qax = p - pax

qax + pax = p - q

∴ x = (\frac{p - q}{a(p + q)})

Factorize 15 + 7x - 2x²

(5 - x)(3 + 2x); suppose 15 + 7x = 2x² = 0

∴ (5 - x)(3 + 2x) = 0

x = 5 or x = -(\frac{3}{2})

Since 5 is a root, then (x - 5) is a factor

(x + (\frac{1}{x}) + 1)² - (x + (\frac{1}{x}) + 1)²

= (x + (\frac{1}{x}) + 1 + x + (\frac{-1}{x}) - 1)(x - (\frac{1}{x}) + 1 - x + (\frac{1}{x}) + 1)

= (2x) (2 + (\frac{2}{x})) = 2x x 2(1 + (\frac{1}{x}))

4x (1 + (\frac{1}{x})) = 4x + 4

= 4(1 + x)

x² + y - 5 = 0.....(i)

y - 7x + 3 = 0.........(ii)

y = 7x - 3, substituting the value of y in equation (i)

x² + (7x - 3) - 5 = 0

x² + 7x + 3 = 0

(x + 8)(x - 1) = 0

x = -8 or 1

(3x - 2)(5x - 4) = (3x - 2)² = 5x² - 22x + 6

= 9x² = 12x + 4

6x² - 10x + 4 = 0

6x² - 6x - 4x + 4 = 0

6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0

x = 1 or (\frac{2}{3})

(\frac{1}{p}) - (\frac{1}{q}) + (\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})

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{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