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