1990 - JAMB Mathematics Past Questions and Answers - page 2

Simplify ((\frac{1}{x^{-1}} + \frac{1}{y^{-1}}))^-1 = ((\frac{1}{x^{-1}} + \frac{1}{y^{-1}}))^-1

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

= (\frac{x}{y})

(ab² - bc²)(a²c - abc)

[2(2)² - (- 2x(\frac{1}{2}))] [2²(-(\frac{1}{2})) - 2(-2)(-(\frac{1}{2}))]

[8 = (\frac{1}{2})][-2 - 2] = (\frac{17}{2}) x 4²

= -34

f(x - 4) = x² + 2x + 3

To find f(2) = f(x - 4)

= f(2)

x - 4 = 2

x = 6

f(2) = 6² + 2(6) + 3

= 36 + 12 + 3

= 51

9(x + y)² - 4(x - y)²

Using difference of two squares which says

a² - b² = (a + b)(a - b) = 9(x + y)² - 4(x - y)²

= [3(x + y)]² - [2(x - y)]^-2

= [3(x + y) + 2(x - y) + 2(x - y)][3(x + y) - 2(x - y)]

= [3x +3y + 2x - 2y][3x + 3y - 2x + 2y]

= (5x + y)(x + 5y)

a² + b² = 16 and 2ab = 7

To find all possible values = (a - b)² + b² - 2ab

Substituting the given values = (a - b)²

= 16 - 7

= 9

(a - b) = (\pm)9

= (\pm)3

OR a - b = 3, -3

x + (\frac{1}{x}) = 4, find x² + (\frac{1}{x^2})

= (x + (\frac{1}{x}))² = x² + (\frac{1}{x^2}) + 2

x² + (\frac{1}{x^2}) = ( x + (\frac{1}{x^2}))² - 2

= (4)² - 2

= 16 - 2

= 14

(2x (\frac{-1}{4})² = 4x² + (\frac{1}{x^2}) - 4

what must be added is +(\frac{1}{x^2})

x - 8(\sqrt{x}) + 15 = 0

x + 15 = 8(\sqrt{x})

square both sides = (x + 15)² = (8 (\sqrt{x})²

x² + 225 + 30x = 64x

x² + 225 + 30x - 64x = 0

x² - 34x + 225 = 0

(x - 9)(x - 25) = 0

x = 9 or 25