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$\frac{1}{4}$