Let $A = \{x \in \mathbb{N} \mid x \text{ is a prime number and } x < 10\}$, $B = \{x \in \mathbb{N} \mid x \text{ is an even number and } x < 9\}$, and $C = \{x \in \mathbb{N} \mid x \text{ is a multiple of } 3 \text{ and } x < 10\}$.
Then $((A \cap B) - C) \times (B - (A \cup C))$ is:
$\{(2, 4), (2, 8)\}$
To solve the problem, we need to carefully determine the sets involved and perform the set operations as instructed. Let's break it down step by step:
- Identify Set Elements:
- Set \(A = \{x \in \mathbb{N} \mid x \text{ is a prime number and } x < 10\}\)
The prime numbers less than 10 are 2, 3, 5, and 7. Hence, \(A = \{2, 3, 5, 7\}\). - Set \(B = \{x \in \mathbb{N} \mid x \text{ is an even number and } x < 9\}\)
The even numbers less than 9 are 2, 4, 6, and 8. Hence, \(B = \{2, 4, 6, 8\}\). - Set \(C = \{x \in \mathbb{N} \mid x \text{ is a multiple of } 3 \text{ and } x < 10\}\)
The multiples of 3 less than 10 are 3, 6, and 9. Hence, \(C = \{3, 6, 9\}\).
- Set \(A = \{x \in \mathbb{N} \mid x \text{ is a prime number and } x < 10\}\)
- Calculate the Intersection and Differences:
- \(A \cap B = \{2\}\)
- \((A \cap B) - C = \{2\} - \{3, 6, 9\} = \{2\}\)
- \(A \cup C = \{2, 3, 5, 7\} \cup \{3, 6, 9\} = \{2, 3, 5, 6, 7, 9\}\)
- \(B - (A \cup C) = \{2, 4, 6, 8\} - \{2, 3, 5, 6, 7, 9\} = \{4, 8\}\)
- Calculate the Cartesian Product:
- The Cartesian product \(((A \cap B) - C) \times (B - (A \cup C))\) is:
- \(\{2\} \times \{4, 8\} = \{(2, 4), (2, 8)\}\)
Hence, the answer is \(\{(2, 4), (2, 8)\}\), which matches the third option provided.
Important Questions from Relations and Functions
If the function of \(f(x)=\dfrac{x}{x-1}\) express f(3x) in terms of f(x)
If \(f(x)-\dfrac{1}{1+2^{1/x}}\) then at x = 0 the function is:
If f(x) is a periodic function and a is a positive real number such that f(x + 2α) + f(x) = 0 for all x ∈ ℝ, then the period of f(x) is:
Let R be the relation in the set N given by R = {(a, b) ∶ a = b − 2, b > 6}, then:
The interval in which y = x2e−x is increasing is:
