2010 - JAMB Mathematics Past Questions and Answers - page 1

27x³ - 8y³ = (3x - 2y)³

But 9x² + 6xy + 4y² = (3x +2y)²

So, 27x³ - 8y³ = (3x - 2y)(3x - 2y)²

Hence the other factor is 3x - 2y

(\frac{x^3 + 3x^2 - 10x}{2x^2 - 8}) = (\frac {x(x^2 + 3x - 10)}{2(x^2 - 4)})

= (\frac {x(x^2 + 5x - 2x - 10)}{2(x + 2)(x - 2)})

= (\frac {x(x - 2)(x + 5)}{2(x + 2)(x - 2)})

= (\frac {x(x + 5)}{2(x + 2)})

x - y = 2 ...........(1)

x² - y² = 8 ........... (2)

x - 2 = y ............ (3)

Put y = x -2 in (2)

x² - (x - 2)² = 8

x² - (x² - 4x + 4) = 8

x² - x² + 4x - 4 = 8

4x = 8 + 4 = 12

x = (\frac{12}{4})

= 3

from (3), y = 3 - 2 = 1

therefore, x = 3, y = 1

y = (\alpha)√x ........(1)

y = k√x ........(2)

When y = 3, x = 16,

(2) becomes 3 = k√16 or 3 = k x 4

giving k = (\frac{3}{4})

from (2), y = (\frac{3}{4})√x

When x = 64, y = (\frac{3}{4})√64

y = (\frac{3}{4}) x 8

= 6

x (\alpha) (\frac{1}{y}) .........(1)

x = k x (\frac{1}{y}) .........(2)

When x = 2(\frac{1}{2})

= (\frac{5}{2}), y = 2

(2) becomes (\frac{5}{2}) = k x (\frac{1}{2})

giving k = 5

from (2), x = (\frac{5}{y})

so when y =4, x = (\frac{5}{y}) = 1(\frac{1}{4})

(\frac{1}{2})x + (\frac{1}{4}) > (\frac{1}{3})x + (\frac{1}{2})

Multiply through by through by the LCM of 2, 3 and 4

12 x (\frac{1}{2})x + 12 x (\frac{1}{4}) > 12 x (\frac{1}{3})x + 12 x (\frac{1}{2})

6x + 3 > 4x + 6

6x - 4x > 6 - 3

2x > 3

(\frac{2x}{2}) > (\frac{3}{2})

x > (\frac{3}{2})

-6 (\leq) 4 - 2x < 5 - x

split inequalities into two and solve each part as follows:

-6 (\leq) 4 - 2x = -6 - 4 (\leq) -2x

-10 (\leq) -2x

(\frac{-10}{-2}) (\geq) (\frac{-2x}{-2})

giving 5 (\geq) x or x (\leq) 5

4 - 2x < 5 - x

-2x + x < 5 - 4

-x < 1

(\frac{-x}{-1}) > (\frac{1}{-1})

giving x > -1 or -1 < x

Combining the two results, gives -1 < x (\leq) 5

Using S(\infty) = (\frac{a}{1 - r})

r = (\frac{0.05}{0.5}) = (\frac{1}{10})

S(\infty) = (\frac{0.5}{{\frac{1}{10}}})

= (\frac{0.5}{({\frac{9}{10}})})

= (\frac{0.5 \times 10}{9})

= (\frac{5}{9})

If (\begin{vmatrix} x & 3 \ 2 & 7 \end{vmatrix}) = 15

7x - 2 x 3 = 15

7x - 6 = 15

7x = 15 + 6 = 21

therefore x = (\frac{21}{7})

= 3

(\begin{vmatrix} 2 & 0 & 5 \ 4 & 6 & 3 \ 8 & 9 & 1 \end{vmatrix})

= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)

= 2(-21) - 0 + 5(-12)

= -42 + 5(-12)

= -42 - 60

= -102