Mathematics 2025 Jamb Examination

This is an Arithmetic Progression (AP) where the first term \(a = 3\) and common difference \(d = 5 - 3 = 2\).
The general formula for the \(k^{th}\) term is \(T_{k} = a + (k - 1)d\).
For the \((n - 2)^{th}\) term, substitute \(k = n - 2\):\[ T_{n-2} = 3 + ((n - 2) - 1)2 \]\[ T_{n-2} = 3 + (n - 3)2 \]\[ T_{n-2} = 3 + 2n - 6 = 2n - 3 \]

To differentiate \(y = \sin(4x)\), we use the chain rule.
The derivative of \(\sin(u)\) is \(\cos(u) \cdot \frac{du}{dx}\).
Here, \(u = 4x\) and \(\frac{du}{dx} = 4\).\[ \frac{dy}{dx} = \cos(4x) \cdot 4 = 4\cos 4x \]

1. Start with \(A = \displaystyle \frac{\theta \pi r^{2}}{360}\).
2. Multiply both sides by 360: \(360A = \theta \pi r^{2}\).
3. Divide both sides by \(\pi r^{2}\) to isolate \(\theta\):\[ \theta = \displaystyle \frac{360A}{\pi r^{2}} \]

Since the coefficient of \(x^{2}\) is negative (-2), the parabola opens downwards and has a maximum value (the question likely intended maximum).
The x-coordinate of the vertex is \(x = \displaystyle \frac{-b}{2a} = \displaystyle \frac{-(-4)}{2(-2)} = \displaystyle \frac{4}{-4} = -1\).
Substitute \(x = -1\) into the equation:\[ y = 2 - 4(-1) - 2(-1)^{2} \]\[ y = 2 + 4 - 2(1) = 6 - 2 = 4 \]

For two persons, the possible outcomes are:
1. Survival (p) and Survival (p) = \(p^{2}\)
2. Death (q) and Death (q) = \(q^{2}\)
3. Death (q) and Survival (p) = \(qp\)
4. Survival (p) and Death (q) = \(pq\)

The probability of exactly one death and one survival is \(qp + pq = 2pq\). 

The diagram shows a construction starting with a 60° arc and a 90° perpendicular. The point Z is located by bisecting the space between the 60° and 90° marks.\[ \text{Angle} = 60^{\circ} + \displaystyle \frac{90^{\circ} - 60^{\circ}}{2} = 60^{\circ} + 15^{\circ} = 75^{\circ} \]

Divide 137 by 5 repeatedly and record the remainders:\[ \begin{array}{r|l}
\text{Divide} & \text{Remainder} \\ \hline
137 \div 5 = 27 & 2 \\
27 \div 5 = 5 & 2 \\
5 \div 5 = 1 & 0 \\
1 \div 5 = 0 & 1 \\
\end{array}
\quad \uparrow\] \[ \text{Reading the remainders from bottom to top: } 1022_5\]

Equation: \(P = a + bQ\).
1. \(32 = a + 16b\)
2. \(20 = a + 12b\)
Subtract (2) from (1): \(12 = 4b \implies b = 3\).
Substitute \(b = 3\) into (2): \(20 = a + 12(3) \implies a = 20 - 36 = -16\).
Law: \(P = -16 + 3Q\).
When \(Q = 28\): \(P = -16 + 3(28) = -16 + 84 = 68\).

The word HANDIER contains 7 letters, and all letters are distinct (H, A, N, D, I, E, R).
The number of arrangements is \(7!\) (7 factorial):\[ 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]

Modal class is 30-39 (highest frequency 12).
\(L = 29.5\) (lower boundary), \(d_{1} = 12 - 6 = 6\), \(d_{2} = 12 - 8 = 4\), \(w = 10\).\[ \text{Mode} = L + \left( \displaystyle \frac{d_{1}}{d_{1} + d_{2}} \right)w \]\[ \text{Mode} = 29.5 + \left( \displaystyle \frac{6}{6 + 4} \right)10 = 29.5 + 6 = 35.5 \]Note: Using standard boundaries, result is 35.5. In JAMB multiple choice, 34.5 or similar values may appear if boundaries are treated differently, but 35.5 is the calculated estimate.