2007 - JAMB Mathematics Past Questions and Answers - page 9

Area of a square = x * x

144 = x²

√144 = x

12 = x

But d² = 12² x 12²

= 144 + 144

= 288

d = √288

= √144 x 2

= 12√2

x = (1, 4, 9, 16, 25, 36}

y = {1, 3, 5, 7, 9, 11, 13, 15}

X ∩ y = {1, 9}

If y = (1 + x)², find (\frac{dy}{dx})

y = (1 + x)²

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

= 2 + 2x

If y = x cosx, find (\frac{dy}{dx}).

Using product rule (\frac{dy}{dx}) = cosx - x sinx

Each exterior angle e = (\frac{360^o}{8})

= 45^o

Integrate (\frac{x^2 - \sqrt{x}}{x}) with respect to x

∫(\frac{x^2 - \sqrt{x}}{x})dx = ∫((\frac{x^2}{x} - \frac {\sqrt{x}}{x}))dx

= ∫(x - x^{(\frac{1}{2})})dx

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

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

Determine the value of (∫^{\frac{\pi}{2}}_{0}) (-2 cosx)dx

(∫^{\frac{\pi}{2}}_{0}) (-2 cosx)dx = -2 sin x(∫^{\frac{\pi}{2}} _{0})

= 2(sin 0 - sin(\frac{\pi}{2}))

= 2(0 - 1)

= -2

f(x) = 2x³ - x² - 4x + 4

f(x) = 6x² - 2x - 4 at turning point, f¹(x) = 0

6x² - 2x - 4 = 0, 3x² - x - 2 = 0, 3x² - 3x + 2x - 2 = 0

(3x + 2)(x - 1) = 0, x = -(\frac{2}{3}) or 1

f¹¹(x) = 12x - 2,

when x = (\frac{2}{3}), f¹¹(x) = 12(-(\frac{2}{3})) - 2 = -10 < 0

(\to) f(x) is maximum @ x = -(\frac{2}{3})

when x = 1, f¹¹(x) = 12(1)- 2 = 10 > 0

(\to) f(x) is maximum @ x = 1

Volume of hemisphere = (\frac{2}{3}\pi r^3)

(\frac{2}{3}\pi r^3) = 718(\frac{2}{3})

(\frac{2}{3}) x (\frac{22}{7} r^3)

= (\frac{2156}{3})

r³ = (\frac{2156}{3}) x (\frac{3}{2}) x (\frac{7}{22})

r³ = 49 x 7

r = 3(\sqrt{343})

= 7cm³

tan 30^o = (\frac{h}{40m})

h = 40 tan 30^o

= 40 x (\frac{1}{\sqrt{30}})m

by rationalizing

h = (\frac{40\sqrt{3}}{3})m