Mathematics 2008 Jamb Examination

Let \(e\) be the identity element, so:\[m \Delta e = m + e + me = m\]Rearranging:\[m + e + me = m\]\[e(1 + m) = 0\]Since the operation excludes \(m = -1\), \(1 + m \neq 0\) for all allowed values. Therefore:\[e = 0\]Thus, zero acts as the identity element.

The line PQ intersects the upper transversal at an angle of 50°.
By applying the property of angles on a straight line:
180° − 50° = 130°
This gives the adjacent exterior angle.
Using angles on a straight line again to determine the angle inside the larger triangle:
180° − 130° = 50°
Recall that the sum of angles in a triangle is 180°.
Therefore, the remaining angle in the larger triangle is:
180° − 50° − 30° = 100°
Finally, using angles on a straight line at the point where x is located:
x = 180° − 100° = 80°

The exterior angle of any regular polygon is given by:\[ \text{Exterior angle} = \dfrac{360°}{n} \]For \(n = 12\):\[ \dfrac{360°}{12} = 30° \]The exterior angle is therefore 30°.

Volume of a cylinder is:\[ V = \pi r^{2}h \]Substituting values:\[ V = \dfrac{22}{7} \times 1^{2} \times 14 = 44\; \text{m}^3 \]Since \(1\;\text{m}^3 = 1000\;\text{litres}\):\[ 44 \times 1000 = 44,000\; \text{litres} \]The capacity is 44,000 litres.

The midpoint of PR is:\[ M = \left( \dfrac{6 + 4}{2}, \dfrac{2 + 2}{2} \right) = (5, 2) \]The perpendicular bisector of a line segment always passes through its midpoint.
Hence the locus passes through (5,2).

Rewrite the line in slope-intercept form:\[2y = -3x - 1\]\[y = -\dfrac{3}{2}x - \dfrac{1}{2}\]The gradient is \(-\dfrac{3}{2}\).
The gradient of a perpendicular line is the negative reciprocal:\[ m = \dfrac{2}{3} \]Thus, the required gradient is \(\dfrac{2}{3}\).

To check closure, substitute any positive integers x and y into \(2x - 3y + 2\). For some values, such as x = 1 and y = 2, the result is \(2(1) - 3(2) + 2 = -2\), which is not a positive integer. Hence, it is not closed.
The operation is commutative only if \(2x - 3y + 2 = 2y - 3x + 2\), which simplifies to \(5x = 5y\). This is only true when x = y, so it is not generally commutative.
Therefore, the operation is neither commutative nor closed.

The identity element e must satisfy:
\(m \Delta e = m\)
Substitute into the operation:
\(m + e + me = m\)
Rearranging gives:
\(e(1 + m) = 0\)
Since m ≠ −1 by definition, \(1 + m ≠ 0\). Thus e = 0 is the identity element.
Zero leaves every element unchanged under the operation.

PQ is parallel to RS; therefore alternate interior, co-interior, and corresponding angles apply depending on the transversal shown.
The given adjacent angle is 100°, and angle x forms a co-interior pair with it. Co-interior angles sum to 180°.
Hence:\[x + 100^{\circ} = 180^{\circ}\]\[x = 80^{\circ}\]Therefore, the angle marked x is 80°.

Capacity = volume of cylinder.\[ V = \pi r^{2}h \]With r = 1 m and h = 14 m:\[ V = \dfrac{22}{7} \times 1^{2} \times 14 = 44 \, \text{m}^{3} \]1 cubic metre = 1000 litres:\[ 44 \times 1000 = 44,000 \, \text{litres} \]The capacity is 44,000 litres.