Mathematics 2011 Jamb Examination

When objects are arranged in a circle, the number of unique arrangements is different from a linear arrangement because the circle has no fixed starting or ending point. To account for rotational symmetry, we fix one person in a position and arrange the others relative to them.

The formula for circular permutations of \(n\) objects is:\[(n - 1)!\]For 5 people sitting around a table, the number of ways is:\[(5 - 1)! = 4!\]Calculating the factorial:\[4! = 4 \times 3 \times 2 \times 1 = 24\]

First, we determine the total number of items in the set. The set starts at 43 and ends at 60, so the total count of numbers is:\[60 - 43 + 1 = 18\]Next, we identify the prime numbers within this range. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. The primes in this set are: 43, 47, 53, and 59. There are 4 prime numbers in total.
The probability \(P\) is given by the ratio of the number of successful outcomes to the total number of outcomes:\[P = \frac{\text{Number of primes}}{\text{Total numbers}} = \frac{4}{18}\]Reducing the fraction to its lowest terms by dividing both the numerator and denominator by 2:\[P = \frac{2}{9}\]

To find the derivative of \(\tan \theta\), we can use the quotient rule by expressing the tangent function as the ratio of sine to cosine:\[y = \tan \theta = \frac{\sin \theta}{\cos \theta}\]Using the quotient rule \(\frac{d}{d\theta}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}\), where \(u = \sin \theta\) and \(v = \cos \theta\):\[\frac{dy}{d\theta} = \frac{\cos \theta (\cos \theta) - \sin \theta (-\sin \theta)}{\cos^2 \theta}\]\[\frac{dy}{d\theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\]Applying the fundamental trigonometric identity \(\cos^2 \theta + \sin^2 \theta = 1\):\[\frac{dy}{d\theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta\]Therefore, the derivative of \(\tan \theta\) is \(\sec^2 \theta\).

In set theory, the intersection \( P \cap Q \) represents the elements that are common to both set \( P \) and set \( Q \). From the diagram, the intersection contains the element \( \{c\} \).

The complement of a set, denoted as \( (P \cap Q)' \), consists of all elements in the universal set that are NOT in the intersection. Looking at the diagram, the elements outside the intersection are \( a \) (only in \( P \)), \( e \) (only in \( Q \)), and \( b, d \) (outside both sets but within the universal set). Combining these, the complement is \( \{a, b, d, e\} \).

A pie chart represents the total distribution as \( 360^\circ \).
To find the percentage for English, we determine the angle occupied by the English sector. From the data provided in the chart:
English: \( 90^\circ \)\[ \text{Percentage} = \left( \frac{90^\circ}{360^\circ} \right) \times 100 = \frac{1}{4} \times 100 = 25\% \]

To find the total number of students, we read the frequency (number of students) for each class from the bar chart and sum them together:
- SS2A: 45 students
- SS2B: 30 students
- SS2C: 40 students
- SS2D: 35 students
- SS2E: 30 students

Summing these values gives:\[\text{Total} = 45 + 30 + 40 + 35 + 30 = 180\]