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