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

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})

x = (300/√3) x (√3/√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