Mathematics 2023 Jamb Examination

To find the number of arrangements of a word with repeating letters, we use the formula \(\dfrac{n!}{p!q!...}\), where \(n\) is the total number of letters and \(p, q...\) are the frequencies of repeating letters.
The word SYLLABUS has 8 letters.
The letter 'L' repeats 2 times.
The letter 'S' repeats 2 times.
Therefore, the number of possible words is \(\dfrac{8!}{2! 2!}\).

First, combine the terms with base 16 using the law of indices \(a^m \times a^n = a^{m+n}\):\[16^{0.16 + 0.04} \times 2^{0.2} = 16^{0.2} \times 2^{0.2}\]Next, convert 16 to base 2:\[(2^4)^{0.2} \times 2^{0.2} = 2^{0.8} \times 2^{0.2}\]Combine the exponents:\[2^{0.8 + 0.2} = 2^1 = 2\]

First, calculate the values inside the brackets using the '*' operation:\[(-4 * 2) = (-4)^2 \times 2 = 16 \times 2 = 32\]\[(7 * -1) = 7^2 \times (-1) = 49 \times (-1) = -49\]Now, apply the '^' operation to the results:\[32 \text{ ^ } (-49) = 2(32) + (-49)\]\[= 64 - 49 = 15\]

First, rewrite the radical as a power: \(\sqrt[3]{x^2} = x^{2/3}\).
Integrate the expression:\[\int (4x - 6x^{2/3})dx = \left[ \frac{4x^2}{2} - \frac{6x^{5/3}}{5/3} \right]_0^1\]\[= \left[ 2x^2 - \frac{18}{5}x^{5/3} \right]_0^1\]Substitute the limits 1 and 0:\[\left(2(1)^2 - \frac{18}{5}(1)^{5/3}\right) - (0) = 2 - 3.6 = -1.6\]Converting to a fraction: \(-1.6 = -\frac{8}{5}\).

First, find the decrease in population:\[1,230 - 1,040 = 190\]Now, calculate the percentage decrease based on the original population:\[\text{Percentage Decrease} = \frac{\text{Decrease}}{\text{Original Population}} \times 100\]\[= \frac{190}{1230} \times 100 = \frac{1900}{123} \approx 15.447...\]Rounding to two decimal places gives 15.45%.

Let the exterior angle be \(x\). Then the interior angle is \(5x\).
The sum of the interior and exterior angles is 180°:\[x + 5x = 180 \implies 6x = 180 \implies x = 30^o\]To find the number of sides (n), use the sum of exterior angles (360°):\[n = \frac{360^o}{30^o} = 12\]A polygon with 12 sides is a dodecagon.

Given \(\dfrac{dA}{dt} = 1.5\). We need to find \(\frac{dr}{dt}\).
From \(A = \pi r^2\), we find \(r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{2}{\pi}}\).
Differentiating \(A = \pi r^2\) with respect to \(t\):\[\frac{dA}{dt} = 2\pi r \frac{dr}{dt} \implies 1.5 = 2\pi \left(\sqrt{\frac{2}{\pi}}\right) \frac{dr}{dt}\]\[\frac{dr}{dt} = \frac{1.5}{2\sqrt{2\pi}} \approx \frac{1.5}{5.013} \approx 0.299 \text{ cms}^{-1}\]

Cross-multiply to clear the fraction:\[y(4x + 5d) = 3x - 9c\]\[4xy + 5dy = 3x - 9c\]Collect terms with x on one side and the rest on the other:\[4xy - 3x = -9c - 5dy\]Factor out x:\[x(4y - 3) = -(9c + 5dy)\]\[x = \frac{-(9c + 5dy)}{4y - 3}\]

Expand the brackets on both sides:\[3x - 3 \le 2x - 6\]Collect terms with x on the left and constants on the right:\[3x - 2x \le -6 + 3\]\[x \le -3\]

From ∆PSR
|PS| = |SR| (If two tangents are drawn from an external point of the circle, then they are of equal lengths)
∴ ∆PSR is isosceles
∠PSR + ∠SRP + ∠SPR = 180^o (sum of angles in a triangle)
Since |PS| = |SR|; ∠SRP = ∠SPR
⇒ ∠PSR + ∠SRP + ∠SRP = 180^o
∠PSR + 2∠SRP = 180^o
36^o + 2∠SRP = 180^o
2∠SRP = 180^o - 36^o
2∠SRP = 144^o
∠SRP = 144/2=72^o
∠SRP = ∠PQR (angle formed by a tangent and chord is equal to the angle in the alternate segment)
∴ ∠PQR = 720