Mathematics 2024 Jamb Examination

To find the roots of the cubic equation \(x^{3} - 19x - 30 = 0\), we can use the factor theorem by testing factors of the constant term (-30).
If we let \(x = 5\):\[ 5^{3} - 19(5) - 30 = 125 - 95 - 30 = 0 \]So, \((x - 5)\) is a factor. Dividing the cubic by \((x - 5)\) gives the quadratic \(x^{2} + 5x + 6\).
Factorising the quadratic:\[ x^{2} + 5x + 6 = (x + 3)(x + 2) \]Setting each factor to zero, we get the roots: \(x = 5, x = -3, x = -2\).

Inverse proportionality is defined by the relationship \(x = \frac{k}{y}\), where \(k\) is a constant.
To find \(k\), substitute the given values \(x = 9\) and \(y = 4\):\[ 9 = \frac{k}{4} \implies k = 9 \times 4 = 36 \]Substituting \(k\) back into the general form gives the law:\[ x = \frac{36}{y} \]

First, calculate \(P^{2}\):\[ P^{2} = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (1\times1 + 2\times2) & (1\times2 + 2\times3) \\ (2\times1 + 3\times2) & (2\times2 + 3\times3) \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} \]Next, calculate \(4P\):\[ 4P = \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix} \]Now, perform the operation \(P^{2} - 4P - I\):\[ \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ 8 & 12 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} (5-4-1) & (8-8-0) \\ (8-8-0) & (13-12-1) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

First, find the transpose of matrix \(A\). \(A^{T}\) is obtained by turning the rows of \(A\) into columns:\[ A^{T} = \begin{pmatrix} 3 & 4 \\ 2 & 2 \\ 1 & -1 \end{pmatrix} \]Now add matrix \(B\) to \(A^{T}\):\[ A^{T} + B = \begin{pmatrix} 3 & 4 \\ 2 & 2 \\ 1 & -1 \end{pmatrix} + \begin{pmatrix} 1 & 4 \\ 0 & 1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} (3+1) & (4+4) \\ (2+0) & (2+1) \\ (1+3) & (-1+2) \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ 2 & 3 \\ 4 & 1 \end{pmatrix} \]

Given \(\tan\theta = 8/15\) (Opposite/Adjacent), we find the Hypotenuse (H):\[ H = \sqrt{8^{2} + 15^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17 \]Thus, \(\sin\theta = 8/17\) and \(\cos\theta = 15/17\).
Substitute these into the expression:\[ \text{Numerator} = \frac{8}{17} - \frac{15}{17} = -\frac{7}{17} \]\[ \text{Denominator} = (\frac{8}{17})^{2} - \frac{8}{17} = \frac{64}{289} - \frac{136}{289} = -\frac{72}{289} \]\[ \text{Result} = -\frac{7}{17} \div -\frac{72}{289} = \frac{7}{17} \times \frac{289}{72} = \frac{7 \times 17}{72} = \frac{119}{72} \]

Based on the operation definition \(x * y = x^{y}\):\[ x * 2 = x^{2} \]The equation given is \(x * 2 = x\), which implies:\[ x^{2} = x \]\[ x^{2} - x = 0 \]\[ x(x - 1) = 0 \]Therefore, the possible values for \(x\) are \(x = 0\) or \(x = 1\).

The mode is the value that appears most frequently in a dataset. Counting the occurrences of each weight:
25: 1 time
30: 2 times
32: 1 time
38: 1 time
40: 1 time
42: 3 times
45: 1 time
48: 1 time
50: 1 time
51: 1 time
52: 1 time

The weight 42 occurs the most frequently (3 times), so it is the mode.

From the Venn diagram data, the total number of students is
\[ 144 = (42 - x) + (30 - x) + (34 + 4x) + (3x) + (6 + x) + (2x)\] \[ 144 = 112 + 8x\] \[ 8x = 144 - 112\] \[ 8x = 32\] \[ x = 4\]  

The number of students that registered for physics and mathematics = 4x + 2x = 4 x 4 + 2 x 4 = 24.

Square both sides to remove the square root:\[ t^{2} = \frac{pq}{r} - r^{2}q \]To clear the fraction, multiply through by \(r\):\[ rt^{2} = pq - r^{3}q \]Factor out \(q\) on the right side:\[ rt^{2} = q(p - r^{3}) \]Divide by \((p - r^{3})\) to isolate \(q\):\[ q = \frac{rt^{2}}{p - r^{3}} \]

First, simplify the term inside the bracket:\[ \sqrt{8a} = \sqrt{4 \times 2a} = 2\sqrt{2a} \]The expression becomes: \((\sqrt{a} + 2\sqrt{2a})^{2}\).
Expand the bracket:\[ (\sqrt{a})^{2} + 2(\sqrt{a})(2\sqrt{2a}) + (2\sqrt{2a})^{2} \]\[ = a + 4\sqrt{2a^{2}} + 4(2a) \]\[ = a + 4a\sqrt{2} + 8a \]\[ = 9a + 4a\sqrt{2} \]Compare this to the given result \(54 + b\sqrt{2}\):\[ 9a = 54 \implies a = 6 \]\[ b\sqrt{2} = 4a\sqrt{2} \implies b = 4a \]\[ b = 4(6) = 24 \]So, \(a = 6\) and \(b = 24\).