1991 - JAMB Mathematics Past Questions and Answers - page 4

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

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

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

= (x² - 1)^{-(\frac{1}{2})}

Given (-2, 0) and (0, -4)

Gradient = (\frac{-4 - 0}{0 - (-2)})

= (\frac{-4}{2})

= -2

For y = ax² - bx + c for minimum y

(\frac{dy}{dx}) = 2x - 2

= (\frac{dy}{dx}) = 0

∴ 2x - 2 = 0

x = 1

BD = (\sqrt{x^2 + x^2})

= (\sqrt{12^2 + 12^2})

= (\sqrt{144 + 144})

= 2(144)

= 12(\sqrt{2cm})

(\frac{x}{4}) = (\frac{\tan 60}{1})

x = 4 tan60

= 4(\sqrt{3})

BD = 2x

= 8(\sqrt{3})

(\frac{BC}{60}) = (\frac{tan 62}{1})

BC = 60 tan 62

(\frac{AC}{60}) = (\frac{tan 62}{1})

AC = 60 tan 64

AB = AC - BC

= 60(tan 64^o - tan 62^o)

cos² x (sec²x + sec² x tan²x)

= (\frac{cos^2x}{cos^2x}) x (\frac{cos^2x sin^2x}{cos^2x cos^2x})

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

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

cosx = (\sqrt{\frac{a}{b}})

y² + (\sqrt{(a)^2}) = (\sqrt{(b)^2}) by pythagoras

y² = b - a

∴ y = b - a

cosec x = (\frac{1}{sin x}) = (\frac{1}{y})

(\frac{b}{y}) = (\frac{\sqrt{b}}{\sqrt{b - a}})

= (\sqrt{\frac{b}{b - a}})