Question - Intersection and Union of Sets, Trigonometric Functions, Complex Numbers, and Permutations

1. Let A be the set of those who teach physics, B the set of those who teach mathematics, and C the set of those who teach both. We can use the principle of inclusion and exclusion to find the number of teachers who teach either physics or mathematics.

Let \( n(A) \) denote the number of physics teachers, \( n(B) \) the number of math teachers, and \( n(C) \) the number of teachers who teach both physics and mathematics.

We are given \( n(A \cup B \cup C) = 12 \).

We are also given that \( n(A \cup B) = n(A \cup C) \).

To find \( n(A \cup B) \), we will calculate using inclusion-exclusion principle:

\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)

But since \( A \cap B = C \), we have:

\( n(A \cup B) = n(A) + n(B) - n(C) \)

Similarly, for \( n(A \cup C) \), assuming no one teaches only math, we have:

\( n(A \cup C) = n(A) + n(C) - n(A \cap C) \)

And since \( A \cap C = C \), we have:

\( n(A \cup C) = n(A) \)

Given that \( n(A \cup B) = n(A \cup C) \), we can set the equations equal to each other and solve:

\( n(A) + n(B) - n(C) = n(A) \)

\( n(B) - n(C) = 0 \)

\( n(B) = n(C) \)

Now substituting the values we have:

\( 7 + n(B) - 4 = 12 \)

\( n(B) - 4 = 5 \)

\( n(B) = 9 \)

\( n(A) = 12 - n(B) = 12 - 9 = 3 \)

2. To find \( B \) and \( R \) we use the following definitions:

\( B = \{y | y = 2x + 7, x \in \mathbb{R} \text{ and } -5 \leq x \leq 5 \} \)

The range of \( B \) is:

When \( x = -5 \), \( y = 2(-5) + 7 = -3 \)

When \( x = 5 \), \( y = 2(5) + 7 = 17 \)

So, \( R(B) = [-3, 17] \)

3. If \( \sin(\theta) = -\frac{4}{5} \) and \( \theta \) is in the third quadrant, both \( \sin(\theta) \) and \( \cos(\theta) \) are negative in the third quadrant. To find \( \cos(\theta) \), we use the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):

\( \cos(\theta) = -\sqrt{1 - \sin^2(\theta)} = -\sqrt{1 - \left(-\frac{4}{5}\right)^2} = -\frac{3}{5} \)

So, the real numbers \( x' \) and \( y' \) for \( \frac{\sin(\theta)}{\cos(\theta)} \) are:

\( \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} \)

Therefore, \( x' = 4 \) and \( y' = 3 \).

4. The word "INVOLUTE" has 3 vowels and 5 consonants. We can form different sets of 3 vowels and 2 consonants, and treat each set as a separate case for permutation.

The number of permutations of 3 vowels (I, O, U) and 2 consonants out of 5 is given by:

\( P(vowels) = \frac{3!}{(3-3)!} = 3! = 6 \)

\( P(consonants) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 20 \)

The total number of permutations is the product of the two:

\( P(total) = P(vowels) \times P(consonants) = 6 \times 20 = 120 \)

Each of these permutations can be arranged in 5! ways because the order of letters within the set matters.

So, the total number of words is \( P(total) \times 5! = 120 \times 5! = 120 \times 120 = 14400 \).