Solution

Solution

Explanation

• Total possible ordered pairs on a set of 10 elements = 10² = 100.
• A reflexive relation must include all pairs of the form (a,a) for all a ∈ A.
• Thus, at least 10 pairs must be in R ⇒ m ≥ 10.
• The maximum relation is the universal relation with all 100 pairs ⇒ m ≤ 100.

✅ Final Answer

The number of ordered pairs m can take any value in the range:
10 ≤ m ≤ 100

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Let \(|S|=m\).

Count incidences via the \(A_i\): There are 30 sets each of size 5, so total memberships \(=30\times 5=150\). Each element of \(S\) lies in exactly 10 of the \(A_i\), so also \(= m\times 10\). Hence \(m=\dfrac{150}{10}=15\).

Count incidences via the \(B_j\): There are \(n\) sets each of size 3, so total memberships \(=n\times 3\). Each element of \(S\) lies in exactly 9 of the \(B_j\), so also \(= m\times 9 = 15\times 9=135\).

Thus \(n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.\)

Solution

We are given:

\[ \sin(4x) = \frac{1}{2}, \quad x \in (-9\pi,\ 3\pi) \]

General solutions for \( \sin(θ) = \frac{1}{2} \)

\[ θ = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad θ = \frac{5\pi}{6} + 2n\pi \]

Let \( θ = 4x \), so we get:

• \( x = \frac{\pi}{24} + \frac{n\pi}{2} \)
• \( x = \frac{5\pi}{24} + \frac{n\pi}{2} \)

Count how many such \( x \) fall in the interval \( (-9\pi, 3\pi) \)

By checking all possible \( n \) values, we find:

• For \( x = \frac{\pi}{24} + \frac{n\pi}{2} \): 24 valid values
• For \( x = \frac{5\pi}{24} + \frac{n\pi}{2} \): 24 valid values

Total distinct values = 24 + 24 = 48

✅ Final Answer: $\boxed{48}$

Solution

Solution

Given: $$x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}$$

Solutions to the equation are: $$\{(1,1), (1,-1), (-1,1), (-1,-1)\}$$

Among them, only \( (1, -1) \) satisfies \( x > y \).

Answer: $$\boxed{1}$$

Solution

Maximum Students Passing All Three Exams

Given:

• Total students = 50
• \( |M| = 37 \), \( |P| = 24 \), \( |C| = 43 \)
• \( |M \cap P| \leq 19 \), \( |M \cap C| \leq 29 \), \( |P \cap C| \leq 20 \)

We use the inclusion-exclusion principle:

\[ |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C| \]

Let \( x = |M \cap P \cap C| \). Then:

\[ 50 \geq 37 + 24 + 43 - 19 - 29 - 20 + x \Rightarrow 50 \geq 36 + x \Rightarrow x \leq 14 \]

✅ Final Answer: \(\boxed{14}\)

Solution

Solution

Solution

Solution

Solution

Given relation: $R = \{(x,y) : y = 2x - 7\}$ where $A = \{1,2,3,\dots,20\}$.

We need both $x, y \in A$.

So, $1 \le y = 2x - 7 \le 20$

⇒ $1 \le 2x - 7 \le 20$

Add 7: $8 \le 2x \le 27$

Divide by 2: $4 \le x \le 13.5$

Hence, integer values of $x$ are $4,5,6,7,8,9,10,11,12,13$ → total $10$ values.

✅ Number of elements in R = 10

Solution

Solution

Given totals: \(|M|=32,\ |B|=38,\ |L|=30\)

Intersections (inclusive): \(|M\cap L|=7,\ |M\cap B|=10,\ |B\cap L|=8,\ |M\cap B\cap L|=5\).

Compute “only” counts:

\(\displaystyle |M\ \text{only}|=|M|-|M\cap B|-|M\cap L|+|M\cap B\cap L|=32-10-7+5=20\)
\(\displaystyle |B\ \text{only}|=|B|-|M\cap B|-|B\cap L|+|M\cap B\cap L|=38-10-8+5=25\)
\(\displaystyle |L\ \text{only}|=|L|-|M\cap L|-|B\cap L|+|M\cap B\cap L|=30-7-8+5=20\)

Exactly one subject: \(20+25+20=\boxed{65}\).

Solution

Given: \(A=\{1,2,3,4\}\), \(B=\{3,4,5\}\)

\(A\cup B=\{1,2,3,4,5\}\Rightarrow |A\cup B|=5\)
\(A\cap B=\{3,4\}\Rightarrow |A\cap B|=2\)
\(A\triangle B=(A\cup B)\setminus(A\cap B)=\{1,2,5\}\Rightarrow |A\triangle B|=3\)

Size of Cartesian product: \(|(A\cup B)\times(A\cap B)\times(A\triangle B)| = 5\times 2\times 3 = \mathbf{30}\).

Solution

Let \(|S|=m\).

Count incidences via the \(A_i\): There are 30 sets each of size 5, so total memberships \(=30\times 5=150\). Each element of \(S\) lies in exactly 10 of the \(A_i\), so also \(= m\times 10\). Hence \(m=\dfrac{150}{10}=15\).

Count incidences via the \(B_j\): There are \(n\) sets each of size 3, so total memberships \(=n\times 3\). Each element of \(S\) lies in exactly 9 of the \(B_j\), so also \(= m\times 9 = 15\times 9=135\).

Thus \(n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.\)

Solution

Given: \( S = \{2, (1,4)\} \)

Here, the set \(S\) has two elements:

• The number \(2\)
• The ordered pair \((1,4)\)

Hence, \(|S| = 2\).

Formula: The number of elements in the power set of a set having \(n\) elements is \(2^n\).

\[ |P(S)| = 2^{|S|} = 2^2 = 4 \]

✅ Therefore, the number of elements in \(P(S)\) is 4.

Solution

Solution

Solution

Solution

Given: \(A = \{x, y, z\}\)

Step 1: Find the number of subsets of \(A\):

If a set has \(n\) elements, its power set has \(2^n\) subsets.

\(|A| = 3 \Rightarrow |P(A)| = 2^3 = 8.\)

Step 2: Now we need the number of subsets of \(P(A)\), i.e. the power set of the power set.

So, \(|P(P(A))| = 2^{|P(A)|} = 2^8 = \boxed{256}.\)

✅ Final Answer: 256

Solution

Solution

Sets: \(A=\{(x,y)\mid y=\tfrac{1}{x},\ x\in\mathbb{R}\setminus\{0\}\}\), \(B=\{(x,y)\mid y=-x,\ x\in\mathbb{R}\}\).

Intersection: Solve \( \frac{1}{x} = -x \) with \(x\neq 0\). \[ \frac{1}{x} = -x \;\Longrightarrow\; 1 = -x^2 \;\Longrightarrow\; x^2 = -1, \] which has no real solution.

Conclusion: \(A \cap B = \varnothing\) (they are disjoint in \(\mathbb{R}^2\)).

Note: Over complex numbers, the intersection would be at \(x=\pm i\), but for real \(x\), there is none.