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)