Mathematics 2024 Jamb Examination

The total number of subsets of a set with \(n\) elements is given by the formula \(2^{n}\). This total includes all proper subsets and the one improper subset (the set itself).

For set K, there are \(n = 5\) elements.\[ \text{Total subsets} = 2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32 \]This number 32 represents the sum of all proper and improper subsets.

In calculus, the derivative of a constant value is always zero. The expression \(\cos 25^{\circ} - \sin 25^{\circ}\) consists only of constant numerical values (there is no variable \(x\) or \(\theta\)).

Therefore:\[ \frac{d}{dx}(\cos 25^{\circ} - \sin 25^{\circ}) = 0 - 0 = 0 \]
However, if the question implies differentiating the trigonometric functions \(\cos x - \sin x\) and then evaluating at 25, the derivatives are:\[ \frac{d}{dx}(\cos x) = -\sin x \]\[ \frac{d}{dx}(\sin x) = \cos x \]\[ \text{Result} = -\sin 25^{\circ} - \cos 25^{\circ} = -(\sin 25^{\circ} + \cos 25^{\circ}) \]

The mean is the sum of the numbers divided by the count of the numbers. There are 7 numbers in total.\[ \text{Mean} = \frac{13 + 16 + x + 18 + 21 + 2x + 35}{7} = 22 \]Summing the constants and variables:\[ 103 + 3x = 22 \times 7 \]\[ 103 + 3x = 154 \]\[ 3x = 154 - 103 \]\[ 3x = 51 \]\[ x = \frac{51}{3} = 17 \]

[Image illustrating the mathematical relationship between variance and standard deviation]
Variance is the square of the standard deviation (\(\sigma^{2}\)).
Given Standard Deviation (\(\sigma\)) = 12.34:\[ \text{Variance} = (12.34)^{2} \]\[ \text{Variance} = 152.2756 \]Rounding to the nearest whole number, we get 152.

[Image showing step-by-step standard deviation calculation table]
1. Find the mean (\(\bar{x}\)):\[ \bar{x} = \frac{5+4+6+7+8}{5} = \frac{30}{5} = 6 \]2. Find the squares of the deviations from the mean \((x - \bar{x})^{2}\):
(5-6)\(^{2}\) = 1; (4-6)\(^{2}\) = 4; (6-6)\(^{2}\) = 0; (7-6)\(^{2}\) = 1; (8-6)\(^{2}\) = 4.
3. Sum the squares:\[ \sum(x - \bar{x})^{2} = 1 + 4 + 0 + 1 + 4 = 10 \]4. Calculate variance (\(\sigma^{2}\)):\[ \sigma^{2} = \frac{10}{5} = 2 \]5. Standard deviation (\(\sigma\)) = \(\sqrt{2}\).

This problem requires combinations because the order of selection does not matter.
Ways to choose 2 girls from 5: \(^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10\)
Ways to choose 3 boys from 7: \(^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)

Total ways = \(10 \times 35 = 350\) ways.

[Image showing selection from two groups with a specific constraint]
The committee must have exactly one female. This means we choose 1 female from 3 AND 4 males from 7 (to make a total committee of 5).
\[ \text{Ways} = ^{3}C_{1} \times ^{7}C_{4} \]\[ ^{3}C_{1} = 3 \]\[ ^{7}C_{4} = ^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]\[ \text{Total ways} = 3 \times 35 = 105 \text{ ways} \]

Total balls = 7 + 4 = 11. Since the balls are replaced, the probabilities remain the same for both picks.
P(Red) = \(7/11\), P(Black) = \(4/11\).
Two balls of the same colour means (Red and Red) OR (Black and Black):\[ P(\text{Same}) = P(R,R) + P(B,B) \]\[ P(\text{Same}) = \left( \frac{7}{11} \times \frac{7}{11} \right) + \left( \frac{4}{11} \times \frac{4}{11} \right) \]\[ P(\text{Same}) = \frac{49}{121} + \frac{16}{121} = \frac{65}{121} \]

The variation is expressed as \(U = k\sqrt{V}\).
First, find the constant \(k\):\[ 24 = k\sqrt{9} \implies 24 = 3k \implies k = 8 \]Now, find \(V\) when \(U = 16\):\[ 16 = 8\sqrt{V} \]\[ \sqrt{V} = \frac{16}{8} = 2 \]\[ V = 2^{2} = 4 \]

The variation is expressed as:\[ B = \frac{K}{C^{1/3}} \text{ or } K = B \times C^{1/3} \]Substitute the given values \(B = 2\) and \(C = 27\):\[ K = 2 \times 27^{1/3} \]\[ K = 2 \times 3 = 6 \]The constant of proportionality is 6.