Mathematics 2010 Jamb Examination

To solve the system of equations, we can use the algebraic identity for the difference of two squares: \[ x^2 - y^2 = (x - y)(x + y) \] Given that \( x^2 - y^2 = 8 \) and \( x - y = 2 \), we can substitute the known value: \[ 8 = 2(x + y) \] Dividing both sides by 2, we find: \[ x + y = 4 \] Now we have a simple system of linear equations:
1) \( x - y = 2 \)
2) \( x + y = 4 \)
Adding the two equations together: \[ (x - y) + (x + y) = 2 + 4 \] \[ 2x = 6 \implies x = 3 \] Substituting \( x = 3 \) into equation (2): \[ 3 + y = 4 \implies y = 1 \] The solution for \( (x, y) \) is (3, 1).

Since x is inversely proportional to y, the relationship is expressed as: \[ x = \frac{k}{y} \quad \text{or} \quad xy = k \] where k is the constant of proportionality. First, we determine k using the initial values \( x = 2.5 \) (which is \( \frac{5}{2} \)) and \( y = 2 \): \[ k = \left(\frac{5}{2}\right) \times 2 = 5 \] Now, we use the constant k to find x when \( y = 4 \): \[ x = \frac{k}{y} = \frac{5}{4} = 1\frac{1}{4} \] The value of x is 1.25 or 1\(\frac{1}{4}\).

To solve the inequality, we first eliminate the fractions by multiplying every term by the Least Common Multiple (LCM) of the denominators 2, 3, and 4, which is 12: \[ 12\left(\frac{1}{2}x\right) + 12\left(\frac{1}{4}\right) > 12\left(\frac{1}{3}x\right) + 12\left(\frac{1}{2}\right) \] \[ 6x + 3 > 4x + 6 \] Subtract 4x from both sides: \[ 2x + 3 > 6 \] Subtract 3 from both sides: \[ 2x > 3 \] Divide by 2: \[ x > \frac{3}{2} \] This represents the range of values for which the inequality holds true.

We solve this compound inequality by splitting it into two separate parts.
Part 1: \( -6 \le 4 - 2x \)
Subtract 4 from both sides: \[ -10 \le -2x \] Divide by -2 (and reverse the inequality sign): \[ 5 \ge x \quad \text{or} \quad x \le 5 \] Part 2: \( 4 - 2x < 5 - x \)
Add 2x to both sides: \[ 4 < 5 + x \] Subtract 5 from both sides: \[ -1 < x \quad \text{or} \quad x > -1 \] Combining both results gives: \[ -1 < x \le 5 \]

This is a geometric series where the first term \( a = 0.5 \) and the common ratio \( r \) is found by dividing the second term by the first: \[ r = \frac{0.05}{0.5} = 0.1 \] Since \( |r| < 1 \), the sum to infinity exists and is calculated using the formula: \[ S_{\infty} = \frac{a}{1 - r} \] Substituting the values: \[ S_{\infty} = \frac{0.5}{1 - 0.1} = \frac{0.5}{0.9} \] To simplify, multiply the numerator and denominator by 10: \[ S_{\infty} = \frac{5}{9} \] This confirms the sum of the series as it converges.

To evaluate the determinant of the 3x3 matrix, we expand along the first row: \[ \Delta = 2\begin{vmatrix} 6 & 9 \\ 3 & 1 \end{vmatrix} - 4\begin{vmatrix} 0 & 9 \\ 5 & 1 \end{vmatrix} + 8\begin{vmatrix} 0 & 6 \\ 5 & 3 \end{vmatrix} \] Calculate the 2x2 determinants:
1) \( (6 \times 1) - (9 \times 3) = 6 - 27 = -21 \)
2) \( (0 \times 1) - (9 \times 5) = 0 - 45 = -45 \)
3) \( (0 \times 3) - (6 \times 5) = 0 - 30 = -30 \)
Substitute these back into the expansion: \[ \Delta = 2(-21) - 4(-45) + 8(-30) \] \[ \Delta = -42 + 180 - 240 \] \[ \Delta = 138 - 240 = -102 \]

The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] First, calculate the determinant of P: \[ \text{det}(P) = (2 \times 1) - (1 \times -3) = 2 + 3 = 5 \] Next, find the adjugate matrix by swapping the main diagonal elements and changing the signs of the off-diagonal elements: \[ \text{adj}(P) = \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix} \] Divide the adjugate matrix by the determinant: \[ P^{-1} = \frac{1}{5} \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 1/5 & -1/5 \\ 3/5 & 2/5 \end{pmatrix} \]

The sum of the interior angles of a quadrilateral is always 360°. Therefore: \[ (x + 15) + (2x - 45) + (x - 30) + (x + 10) = 360 \] Combining like terms: \[ 5x - 50 = 360 \] \[ 5x = 410 \implies x = 82 \] Now, calculate each angle:
1) \( 82 + 15 = 97^\circ \)
2) \( 2(82) - 45 = 164 - 45 = 119^\circ \)
3) \( 82 - 30 = 52^\circ \)
4) \( 82 + 10 = 92^\circ \)
Comparing the calculated angles, the smallest value is 52°.

The total surface area of a cylinder with one end open is the sum of the curved surface area and the area of one circular base.
Curved Surface Area (CSA) = \( 2\pi rh \)
Base Area = \( \pi r^2 \)
Given height \( h = 50 \) m and radius \( r = 7 \) m: \[ \text{CSA} = 2 \times \pi \times 7 \times 50 = 700\pi \] \[ \text{Base Area} = \pi \times 7^2 = 49\pi \] \[ \text{Total Surface Area} = 700\pi + 49\pi = 749\pi \text{ m}^2 \]

The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is calculated using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the coordinates \( (1/2, 1/2) \) and \( (-1/2, -1/2) \): \[ d = \sqrt{\left(-\frac{1}{2} - \frac{1}{2}\right)^2 + \left(-\frac{1}{2} - \frac{1}{2}\right)^2} \] \[ d = \sqrt{(-1)^2 + (-1)^2} \] \[ d = \sqrt{1 + 1} = \sqrt{2} \] The distance between the two points is exactly \( \sqrt{2} \).