2006 - JAMB Mathematics Past Questions and Answers - page 5

Cost of removing a 6m² room = N540
Cost of renovating a 1m² room = 540 / 6
Cost of removing a 9m² room
= (540 / 6) * 9
= N810

Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = ⁿ/₂(2a + (n-1)d)
= ⁿ/₂ (2 + n – 1)
= ¹/₂n(n + 1)

T = 2 π √¹/_g
^T/_2π = √¹/_g
(^T/_2π)² = (√¹/_g)^2
T2/_4π² = ¹/_g
T² x g = 4π² x l
G = (4π² x l) / T²

y = x² - x - 12
= (x - 4)(x - 3)
∴ x = 4 or x = -3
-3 < x < 4
But y \( \geq \) 0
∴ -3 < x \( \leq \) 4

3, -6, +12, -24
a = 3, r = -2
\(8n = \frac{a(1-r^n)}{1-r}\\ ∴1-2^8 = \frac{3(1-(-2^{n-1}))}{1-(-2)}\\ 1-2^8 = \frac{3(1-(-2^{n-1}))}{3}\)
1-2⁸ = 1-(-2)^n-1
-2⁸ = -2^n-1
8 = n-1
n = 9

(tan \(\theta\) = \(\frac{opp}{adj}\))

|AB|² = 5² + 4² \(\to\) |AB|² = 41

\(\to\) AB = \(\sqrt{41}\) sin²\(\theta\) - cos²\(\theta\)

\(\to\) \(\frac{5^2}{\sqrt{41}}\) - (\(\frac{4}{\sqrt{41}}^2\)) = \(\frac{25}{41}\) - \(\frac{16}{41}\)

= (\(\frac{9}{41}\))

PN = NQ

1 + q + 5 + 2q = 3 + 2 + 3 + 4

6 + 3q = 12 \(\to\) 3q = 12 - 6 = 6

q = 2

2p x 2 + 8 x 1 = 24

\(\to\) 4p = 24 - 8 = 16, p = 4

3 x 1 + -5q x 2 = -17

\(\to\) -10q = -17 - 3

-10q = -20

q = 2