2008 - JAMB Mathematics Past Questions and Answers - page 6

Simplify 16(\frac{1}{2}) x 4(\frac{1}{2}) x 27(\frac{1}{3})

= 16(\frac{1}{2}) x 4(\frac{1}{2}) x 27(\frac{1}{3})

= (\frac{1}{\sqrt{16}^1} \times \frac{1}{\sqrt{4}^1}) x (3(\sqrt{27}))¹

= (\frac{1}{4}) x (\frac{1}{2}) x (\frac{3}{1})

= (\frac{3}{8})

(logX^{\frac{1}{2}}) 64 = 3, find the value of X

recall that X = 10^p

log₁₀X = P

64 = (X^{\frac{1}{2}})

4³ = (X^{\frac{1}{2}})

(4^{3 \times \frac{1}{2}}) = (X^{\frac{1}{2} \times \frac{1}{3}})

4 = (X^{\frac{1}{2}})

(4)² = (X^{\frac{1}{2} \times 2})

x = 16

(\frac{1 + \sqrt{2}}{1 - \sqrt{2}}) x (\frac{1 + \sqrt{2}}{1 + \sqrt{2}})

= (\frac{(1 + \sqrt{2}) (1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})})

= (\frac{1 + \sqrt{2} + \sqrt{2} + (\sqrt{2})^2}{1 + \sqrt{2} - \sqrt{2} - (\sqrt{2})^2})

= (\frac{1 + 2\sqrt{2} + \sqrt{2}}{1^2 - 2})

= (\frac{3 + 2\sqrt{2}}{-1})

-3 - 2(\sqrt{2}) = x + y(\sqrt{2})

x = -3, y = -2

= (-3, -2)

X = {n² + 1 : n = 0, 2, 3}

y = {n + 1 : n = 2, 3, 4, 5}

find x ∩ y

x = (n² + 1)(2² + 1)(3² + 1)

y = (2 + 1) (3 + 1)(5 + 1)

x = {1, 5, 10}

y = {3, 4, 6}

x ∩ y = (\Phi)

L = (\frac {4}{3}) m (\sqrt{PQ})

make Q the subject

(\frac {L}{1}) = (\frac {4}{3}) m (\sqrt{PQ})

3L = 4m (\sqrt{PQ})

Taking the square of both sides

(3L)² = (4m (\sqrt{PQ}))

(\frac {9L^2}{16m^2P}) = (\frac {16m^2PQ}{16m^2P})

Q = (\frac {9L^2}{16m^2P})

(4x + 3y)² - (3x - 2y)²

= 16x² + 24xy + 9y - (9x² - 12xy + 4y²)

= 7x² + 36xy + 5y

= 7x² + xy + 35xy + 5y²

= (7x² + xy) + (35xy + 5y²)

= x(7x + y) + 5y(7x + y)

= (7x + y)(x + 5y)

x - 3 x y²

x - 3 = ky²

when x = 5, y = 2

5 - 3 = k(2)²

2 = 4k

k = (\frac{2}{4})

k = (\frac{1}{2})

Find x when y = 6

from x - 3 = ky²

x - 3 = (\frac{1}{2})ky²

x - 3 = (\frac{1}{2}) x 36

x - 3 = 18

x = 18 + 3

x = 21

P (\alpha) (\frac{1}{q^2})

p = (\frac{k}{q^2})

k = pq²

k = 8 x 4²

k = 8 x 16 = 128

find k = pq²

q² = (\frac{k}{p})

q = (\frac{k}{p})

q = (\frac{128}{32})

q = √4

q = (\pm)2