2005 - JAMB Mathematics Past Questions and Answers - page 3

Sum of interior ∠s = 1800^o

∴(n - 2) 180^o = 1800^o

180n -360^o = 1800^o

180n = 1800^o + 360^o

180n = 2160^o

n = 2160^o/180^o

n = 12 sides

Each exterior ∠ = 360^o/n

= 360^o/12

= 30^o

∠OUT = ∠OTU = a (Base ∠s of 180 Δ)

∴ a + a + 70 = 180^o (sum of ∠s of a Δ)

2a = 180^o - 70^o

2a = 110^o

a = 55^o

But ∠RUT + ∠RBT = 180^o (opposite ∠s of a Cyclic quad)

∴ x + a = 100 = 180

x + 55 + 100 = 180

x = 180 - 155

x = 25

so ∠RUO = x = 25^o

Area of a sector = (\frac{\theta}{360}\times \pi r^2\55=\frac{\theta}{360}\times \frac{22}{7}\times \frac{10\times 10}{1}\theta=\frac{360 \times 55 \times 7}{22 \times 10 \times 10}\theta = 63^{\circ})

L² = 12² + 5²

= 144 + 25

= 169

L = √169

= 13

Curved surface Area = πrL

= ⁵/₁ * ¹³/₁ * π

= 65πcm²

Gradient of the line y + 2x = 5 = -2

Gradient of the line perpendicular to

y + 2x + 5 = ¹/₂

equation of a line perpendicular to

y + 2x = 5 at the point (4, 3)

y - y₁ = m(x - x₁)

y - 3 = ¹/₂(x - 4)

2y - 6 = x -4

2y - x = 2

Sin 30 = x/14

x =14 sin 30

= 14 * ¹/₂

= 7

AB = 2x

= 2 * 7 = 14cm

sin θ = -¹/₂

= -0.5

θ = sin^-1 (0.5)

θ = 30^o

Since θ is negative

θ = 180 + 30 = 210^o

θ = 360 - 30 = 330^o

∴θ = 210^o and 330^o

he diagram x = 44^o (alternate ∠s) the bearing of R from S

= 180 + x

= 180 + 44

= 224^o

y = (1 - 2x)³

let u = 1-2x

du/dx = -2

∴y = u³

dy/du = 3u²

But dy/dx = du/dx * dy/du

= -2 * 3u²

= -6u²

= -6(1-2x)²

At x = -1: dy/dx = -6(1 - 2(-1))²

dy/dx = -6(1 + 2)²

= -6 * 3²

= -6 * 9

= -54