JAMB Mathematics Past Questions & Answers - Page 307

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

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

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

= -42 + 5(-12)

= -42 - 60

= -102

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

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

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


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

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

= \(\frac{5}{9}\)

-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

\(\frac{1}{2}\)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 > \(\frac{3}{2}\)

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\(\frac{1}{4}\)