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