2014 - JAMB Mathematics Past Questions and Answers - page 3

\(\begin{vmatrix}
-x & 12 \\ -1 & 4
\end{vmatrix} = - 12\)

-4x - (-1)12 = -12

-4x + 12 = -12

-4x = -12 - 12

-4x = - 24

x = 6

\(0 \begin{vmatrix}
7 & 8 \\ 5 & 4\end{vmatrix}-3 \begin{vmatrix}
1 & 8 \\ 0 & 4\end{vmatrix}+2 \begin{vmatrix}
1 & 7 \\ 0 & 5\end{vmatrix}\)

= 0(28 - 40) - 3(4 - 0) + 2(5 - 0)

= 0(-12) - 3(4) + 2(5)

= 0 - 12 + 10

= -2

If each interior angle of the polygon is 135^o, then each exterior angle is 180^o - 135^o = 45^o. Hence, number of sides =

\(\frac{360^o}{\text{one exterior angle}}\)

\(\frac{360^o}{45^o}\)

= 8

Using \(V = \pi r^2 h\)

6160 = 22/7 x 28 x 28 x h

\(h = \frac{6160}{22 \times 4 \times 28} \)

\(h = 2.5m \)

Mid point of S(-5, 4) and T(-3, -2) is

\([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + 2)]\)

\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)

\([\frac{1}{2}(-8), \frac{1}{2}(2)]\)

\([-4, 1]\)

\(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}\)

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

1 - x = 2(2 - 4)

1 - x = 4 - 8

1 - x = -4

-x = -4 - 1

x = 5

y - 4x + 3 = 0

When y = 0, 0 - 4x + 3 = 0

Then -4x = -3

x = 3/4

So the line cuts the x-axis at point (3/4, 0).

When x = 0, y - 4(0) + 3 = 0

Then y + 3 = 0

y = -3

So the line cuts the y-axis at the point (0, 3)

Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)

\([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\)

\([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\)

\([\frac{3}{8}, \frac{-3}{2}]\)

4x + 3y - 5 = 0 (given)

The equation of the line perpendicular to the given line takes the form 3x - 4y = k

Thus, substitution x = -2 and y = 3 in 3x - 4y = k gives;

3(-2) - 4(3) = k

-6 - 12 = k

k = -18

Hence the required equation is 3x - 4y = -18

3x - 4y + 18 = 0