2009 - JAMB Mathematics Past Questions and Answers - page 4

x² + 3³ = 5²
x² + 9 = 25
x² = 25 – 9
x² = 16

x = √ 16 

= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0

Gradient =	y2 - y1
x2 - x1

2 3

A(4,2) and B(1,r), AB = 3 units

3 = √ (x₂- x₁)² + (y₂-y₁)²


3 = √ (4-1)² + (2-r)²


3 = √ 3² + (2-r)²


3 = √ 9 + 4 – 4r + r²


3 = √ r² - 4r + 13

9 = r² - 4r + 13 By squaring both sides
r² - 4r + 4 = 0
(r-2)(r-2) = 0
r = 2

y = 3 cos 4x
dy/dx = 3x -4 sin 4x
= -12sin 4x

\(Sin 45 - cos 30\\ \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\\ =\frac{\sqrt{2}-\sqrt{3}}{2}\)

Tan 60 =	300
x

x =	300
Tan 60

x =	300
√3

x = (300/√3) x (√3/√3)

x =	300√3
.....3

x = 100√3

x = (2+3t)(5t-4)
Let u = 2+3t ∴^du/_dt = 3
and v = 5t-4 ∴^dv/_dt = 5
^dx/_dt = V^du/_dt + U^dv/_dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x⁴/₅ - 2
= 24 - 2
= 22

x(4-x) = 4x-x^2
dy/_dx = 4-2x
As ^dy/_dx = 0
4-2x = 0
2x = 4
x = 2

S = t³ - t² - t + 5
^ds/_dt = 3t² - 2t - 1
As ds = 0
3t² - 2t - 1 = 0
(3t+1)(t-1) = 0
∴ t = 1 or -¹/₃
At min pt t = -¹/₃
S = t³ - t² - t + 5 put t = -¹/₃
= (-¹/₃)³ - (-¹/₃)² - (-¹/₃) + 5
= -¹/₂₇ - ¹/₉ + ¹/₃ + 5
= 140 / 27
= 5.2

∫sec²θ dθ = ∫ 1/cos² dθ
∫(cos)^-2 dθ,
let u = cos θ
∴∫u^-2 = ¹/_u + c
∫cos θ = sin θ + c
∫sec^-2θ = ¹/_u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c