2007 - JAMB Mathematics Past Questions and Answers - page 6

x = -2, x = -1 and x = 1
then the factors; x+2, x+1 and x-1
Product of the factors; (x+2)(x+1)(x-1)
= y = (x + 2)(x² - x + x - 1)
= y = (x+2)(x²-1)
x³ - x + 2x² - 2 = y
x³ + 2x² - x - 2 = y

\(\sqrt{\frac{42w}{5l}}\)
square both side of the equation
\(d^2 = \left(\sqrt{\frac{42W}{5l}}\right)^2\\ d^2 = \frac{42W}{5l}\\ 5ld^2=42W\\ l = \frac{42W}{5d^2}\)

(x³ + x - 12) ≥ 0
(x + 4)(x - 3) ≥ 0
Either x + 4 ≥ 0 implies x ≥ -4
Or x - 3 ≥ 0 implies x ≥ 3
∴ x ≥ 3 or x ≥ -4

2t² + t - 15 = (2t-5)(t+3)

-3(x-3) < -2(x+3) = -3x + 6 < -2x - 6
-3x + 2x < -6 - 6
-x < - 12
x > ^-12/_-1
x > 12

W ∝ L²
W = KL²
K = ^W/_L²
K = ⁶/_4²
K = ⁶/₁₆ = ³/₈
W = ³/₈ L²
W = ³/₈(√17)²
W = ³/₈ x 17
W = ⁵¹/₈ = 6³/₈

a*e = a + e + 1 = a
implies e+ 1 = 0
∴ e = -1
7 * e = -1
∴ a + 7 + 1 = -1
a + 8 = -1
a+8 = -1
a = -1-8
a = -9

³/₂, 3, 7, 16, 35, 74, ....
Using the method of substitution
When n = 1, 5 . 2^n-2 - 1 = 5 . 2^1-2 - 1
= 5 x 2^-1 - 1
= 5 x ¹/₂ - 1
= ⁵/₂ - 1 = ³/₂
When n = 1, 5 . 2^n-2 - n
= 5 . 2^2-2 - 2
= 5 x 2⁰ - 2
= 5 x 1 – 2 = 3
When n = 3, 5 . 2^n-2 - n
= 5 . 2^3-2 - 3
= 5 x 2¹ - 3
= 5 x 2 – 3
= 10 -3 = 7
When n = 4, 5 . 2^n-2 - n
= 5 . 2^4-2 - 4
= 5 x 2² - 4
= 5 x 4 – 4
= 20 – 4 = 16
When n = 5, 5 . 2^n-2 - n
= 5 . 2^5-2 - 5
= 5 x 2³ - 5
= 5 x 8 – 5
= 40 – 5
= 35
When n = 6, 5 . 2^n-2 - n
= 5 . 2^6-2 - 6
= 5 x 2⁴ - 6
= 5 x 16 – 6
= 80 – 6 = 74
∴ 5 . 2^n-2 - 1 = ³/₂, 3, 7, 16, 35, 74

\(a=2\\ r = \frac{3}{4}\\ S = \frac{a}{1-r}\\ S= \frac{2}{1-\frac{3}{4}}\\ = \frac{2}{\frac{1}{4}}\\ S = \frac{2}{1}\times \frac{4}{1}\\ = 8\)

\(\sqrt{12}-\sqrt{147}+y\sqrt{3} = 0\\ \sqrt{4\times 3}-\sqrt{49\times 3}+y\sqrt{3} = 0\\ 2\sqrt{3}-7\sqrt{3}+y\sqrt{3} = 0\\ y\sqrt{3} = 7\sqrt{3} - 2\sqrt{3}\\ y=\frac{5\sqrt{3}}{\sqrt{3}}\\ y = 5\)