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