1998 - JAMB Mathematics Past Questions and Answers - page 2

d² = (y - y)² + (x - x)²

5 = 4x² + 1 = 25= 4x² + 1

= 4x² = 25 - 1= 24

x² = (\frac{24}{4})

x = (\sqrt{6})

y = 4x + 3

when x = 0, y = 3 (\to) (0, 3)

when y = 0, x = -(\frac{3}{4}) (\to) ((\frac{3}{4}), 0)

mid-point (\frac{0 + (-{\frac{3}{4}})}{2}), (\frac{3 + 0}{4})

-(\frac{3}{8}), (\frac{3}{2})

cos x + sin x (\frac{1}{cos x - sinx})

= (cosx + sinx)(cosx - sinx) = 1

= cos²x + sin²x = 1

cos²x - (1 - cos²x) = 1

= 2cos²x = 2

cos²x = 1

= cosx = (\pm)1 = x

= cos^-1x ((\pm), 1)

= 0, (\pi) (\frac{3}{2}\pi), 2(\pi)

(possible solution)

(\frac{150}{Z}) = tan 60^o,

Z = (\frac{150}{tan 60^o})

= (\frac{150}{3})

= 50(\sqrt{3})cm

(\frac{150}{X x Z}) = tan45^o = 1

X + Z = 150

X = 150 - Z

= 150 - 50(\sqrt{3})

= 50( (\sqrt{3}) - (\sqrt{3}))m

y = 243(4x + 5)^-2, find (\frac{dy}{dx})

= -1944(4x + 5)_-3

= 1944(9)_-3

(\frac{dy}{dx}) when x = 1

= -(\frac{1944}{9^3})

= -(\frac{1944}{729})

= (\frac{-8}{3})

let y = (\frac{x}{cosx}) = x sec x

y = u(x) v (x0

(\frac{dy}{dx}) = U(\frac{dy}{dx}) + V(\frac{du}{dx})

dy x [secx tanx] + secx

x = x secx tanx + secx

∫(^{\pi}_{2})(sec² x - tan²x)dx

∫(^{\pi}{2}) dx = [X](^{\pi}{2})

= (\pi) - 2 + c

when c is an arbitrary constant of integration

m = (\frac{dy}{dv}) = 6x - 5

∫dy = ∫(6x - 5)dx

y = 3x² - 5x + C

when x = 2, y = 5

∴ 5 = 3(2)² - 5(2) +C

C = 3

∴ y = 3x² - 5x + 3

mean(x) = (\frac{\sum x}{N})

= (\frac{48}{8})

= 5.875

re-arranging the numbers;

2, 3, 5, 6, 2, 7, 8, 9

median = (\frac{6 + 7}{2})

= (\frac{1}{2})

= 6.5

m + 2n = 5.875 + (6.5)²

= 13 + 5.875

= 18.875

= (\approx) = 19