Mathematics 2024 Jamb Examination

When people are arranged in a circle, the number of ways to sit them is calculated using the circular permutation formula, which is \((n - 1)!\). This is because one position is fixed to account for the rotational symmetry of the circle.

For \(n = 6\):\[ (6 - 1)! = 5! \]\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]Therefore, there are 120 ways for 6 people to sit around the table.

The region shown on the graph is defined by the inequalities
\[ 3x + 4y < 12, \quad x \ge 0, \quad y \ge 0\]
The inequality \(3x + 4y < 12\) represents all the points lying below the straight line \(3x + 4y = 12\).  

The conditions \(x \ge 0\) and \(y \ge 0\) restrict the solution to the first quadrant of the coordinate plane.

Hence, the required region is the set of points in the first quadrant that lie below the line \(3x + 4y = 12\).

First, find \(2B\) by multiplying every element in matrix \(B\) by 2:\[ 2B = 2 \times \begin{pmatrix} 1 & 4 & -2 \\ -3 & 3 & -1 \end{pmatrix} = \begin{pmatrix} 2 & 8 & -4 \\ -6 & 6 & -2 \end{pmatrix} \]Now, add matrix \(A\) to \(2B\):\[ A + 2B = \begin{pmatrix} 2 & -4 & 3 \\ 5 & 1 & 0 \end{pmatrix} + \begin{pmatrix} 2 & 8 & -4 \\ -6 & 6 & -2 \end{pmatrix} \]\[ = \begin{pmatrix} (2+2) & (-4+8) & (3-4) \\ (5-6) & (1+6) & (0-2) \end{pmatrix} = \begin{pmatrix} 4 & 4 & -1 \\ -1 & 7 & -2 \end{pmatrix} \]

The formula for the sum to infinity of a geometric progression is:\[ S_{\infty} = \frac{a}{1 - r} \]Given \(S_{\infty} = 100\) and \(r = -1/2\):\[ 100 = \frac{a}{1 - (-1/2)} \]\[ 100 = \frac{a}{1 + 1/2} \]\[ 100 = \frac{a}{3/2} \]Multiply both sides by \(3/2\):\[ a = 100 \times \frac{3}{2} = 150 \]The first term is 150.

In \(\Delta\) PQR, since \(|PQ| = |QR|\), it is an isosceles triangle, so \(\angle QPR = \angle QRP = 60^{\circ}\). The third angle \(\angle PQR = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}\). Since all angles are 60º, the triangle is equilateral with side \(s = 8\text{ cm}\).
The area of an equilateral triangle is given by:\[ \text{Area} = \frac{\sqrt{3}}{4}s^{2} \]\[ \text{Area} = \frac{\sqrt{3}}{4}(8)^{2} = \frac{\sqrt{3}}{4}(64) \]\[ \text{Area} = 16\sqrt{3}\text{ cm}^{2} \]

Let \(x\) represent the number of students offering both subjects.

Using the principle of inclusion for the total number of students:
\[ 65 = 30 + 45 - x\] \[ 65 = 75 - x\] \[ 65 - 75 = -x\] \[ -10 = -x\] \[ x = 10\]
Hence, \(10\) students offer both subjects.

The probability that a randomly selected student offers both subjects is
\[ P(\text{students offering both subjects}) = \frac{10}{65}\] \[ = \frac{2}{13}\]

Let the horizontal distance be \(x\). The building height is 10m and the tree height is 25m. The difference in height between the fruit and the top of the building is:\[ \text{Opposite side} = 25 - 10 = 15\text{m} \]Using tangent for the right-angled triangle formed:\[ \tan(30^{\circ}) = \frac{\text{Opposite}}{\text{Adjacent}} \]\[ \frac{1}{\sqrt{3}} = \frac{15}{x} \]\[ x = 15\sqrt{3}\text{m} \]

First, find the cost price (CP) and selling price (SP) per orange.\[ \text{CP per orange} = \frac{24}{5} = 4.80 \]\[ \text{SP per orange} = \frac{30}{4} = 7.50 \]The profit per orange is:\[ \text{Profit} = 7.50 - 4.80 = 2.70 \]Assuming the question refers to the profit made on the initial batch (4 or 5) or a specific set, if we calculate based on the lowest common multiple (20 oranges):
CP of 20 = \(20 \times 4.80 = 96\)
SP of 20 = \(20 \times 7.50 = 150\)
Profit = 54. However, looking at the options provided in the JAMB source, the question usually asks for a specific quantity. Based on a standard 4-orange batch sale profit:\[ \text{Total profit on 4 sold} = (7.50 \times 4) - (4.80 \times 4) = 30 - 19.2 = 10.80 \]

To differentiate \(y = (5x + 1)^{4}\), we use the chain rule (function of a function):
Let \(u = 5x + 1\), then \(y = u^{4}\).\[ \frac{dy}{du} = 4u^{3} \]\[ \frac{du}{dx} = 5 \]Applying the chain rule:\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]\[ \frac{dy}{dx} = 4(5x + 1)^{3} \times 5 = 20(5x + 1)^{3} \]

To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \(3 - \sqrt{7}\):\[ \frac{5 + \sqrt{7}}{3 + \sqrt{7}} \times \frac{3 - \sqrt{7}}{3 - \sqrt{7}} \]\[ = \frac{5(3) - 5\sqrt{7} + 3\sqrt{7} - (\sqrt{7})^{2}}{3^{2} - (\sqrt{7})^{2}} \]\[ = \frac{15 - 2\sqrt{7} - 7}{9 - 7} \]\[ = \frac{8 - 2\sqrt{7}}{2} \]Divide both terms in the numerator by 2:\[ = 4 - \sqrt{7} \]