If A ⊆ B and B ⊆ C, then find the cardinality of A ∪ B ∪ C.

Explanation

• Since A is a subset of B, everything in A is already in B.
• Since B is a subset of C, everything in B (and hence A) is already in C.
• Therefore, A ∪ B ∪ C = C.

✅ Final Answer

|A ∪ B ∪ C| = |C|

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

Number of real solutions

Equation: $$\sin\!\left(e^x\right) \;=\; 5^x + 5^{-x}$$

Reasoning

• For all real \(x\), \(\sin(e^x)\in[-1,1]\).
• By AM–GM, \(5^x + 5^{-x} \ge 2\) (with equality only when \(5^x=5^{-x}\Rightarrow x=0\)).
• At \(x=0\): LHS \(=\sin(1)\approx 0.84\), RHS \(=2\) ⇒ not equal.
• Hence RHS is always \(\ge 2\) while LHS is always \(\le 1\) → they can never match.

✅ Conclusion

No real solution (number of real solutions = 0).

Multiply numerator and denominator by \(10^x\): \( y = \dfrac{10^{2x} - 1}{10^{2x} + 1} \).

Put \( t = 10^{2x} \), then \( y = \dfrac{t-1}{t+1} \). Solving, \( t = \dfrac{1+y}{1-y} \).

Hence, \( 10^{2x} = \dfrac{1+y}{1-y} \). Taking log base 10: \( 2x = \log_{10}\!\Big(\dfrac{1+y}{1-y}\Big) \).

✅ Final Answer

The inverse function is:
\[ f^{-1}(y) = \tfrac{1}{2}\,\log_{10}\!\left(\dfrac{1+y}{1-y}\right), \quad |y|<1 \]

We are given:

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

Step 1: 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:

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

✅ Step 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: $\boxed{48}$

Using the set formula: \[ |T \cup C| = |T| + |C| - |T \cap C| \] \[ 200 = 120 + 144 - x \quad \Rightarrow \quad x = 64 \]

Minimum possible intersection: \[ x \geq |T| + |C| - 200 = 64 \] Maximum possible intersection: \[ x \leq \min(120,144) = 120 \]

✅ Final Answer

The range of \(x\) is: 64 ≤ x ≤ 120 Hence, \(m = 64, \; n = 120\).

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: $$\boxed{1}$$

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: \(\boxed{14}\)

Newspaper Reading Problem

In a class of 50 students:
- 30 students read Hitava
- 35 students read Hindustan
- 10 students read neither

We need to find the number of students who read both newspapers.

Total students who read at least one newspaper = Total students − Students reading neither = 50 − 10 = 40

Using the formula for union of two sets:
$$|H \cup I| = |H| + |I| - |H \cap I|$$
where
\( |H \cup I| = 40 \),
\( |H| = 30 \),
\( |I| = 35 \),
and \( |H \cap I| = ? \)

Rearranging:
$$|H \cap I| = |H| + |I| - |H \cup I|$$
$$= 30 + 35 - 40 = 65 - 40 = 25$$

Number of students reading both newspapers = 25