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

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

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