Which of these functions defines one-to-one correspondence from (0, 2) to (3, ∞)?

A) 2/𝑥 + 2 B) 3/x + 3

C) 1/x + 2

D) 1/x + 3

E) None of the mentioned

To determine which of these functions defines a one-to-one correspondence (bijection) from the domain \((0, 2)\) to the range \((3, \infty)\), we analyze the behavior of each function.

### Step 1: Criteria for One-to-One Correspondence
- **Injective (One-to-One)**: Each \(x\) in the domain maps to a unique \(y\) in the range.
- **Surjective (Onto)**: Every value in the range \((3, \infty)\) must be reachable from some \(x\) in the domain \((0, 2)\).

### Step 2: Analyze Each Function
#### A) \(f(x) = \frac{2}{x} + 2\)
1. **Domain Check**: For \(x \in (0, 2)\), \(f(x)\) is defined.
2. **Range Check**:
   - As \(x \to 0^+\), \(\frac{2}{x} \to \infty\), so \(f(x) \to \infty\).
   - As \(x \to 2^-\), \(\frac{2}{x} \to 1\), so \(f(x) \to 3\).
   - Range: \((3, \infty)\).
3. \(f(x)\) is **strictly decreasing** (derivative is negative), ensuring injectivity.

This function is a bijection.

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#### B) \(f(x) = \frac{3}{x} + 3\)
1. **Domain Check**: For \(x \in (0, 2)\), \(f(x)\) is defined.
2. **Range Check**:
   - As \(x \to 0^+\), \(\frac{3}{x} \to \infty\), so \(f(x) \to \infty\).
   - As \(x \to 2^-\), \(\frac{3}{x} \to \frac{3}{2}\), so \(f(x) \to \frac{3}{2} + 3 = 4.5\).
   - Range: \((4.5, \infty)\).
3. Since \((4.5, \infty) \neq (3, \infty)\), this is **not surjective**.

Not a bijection.

---

#### C) \(f(x) = \frac{1}{x} + 2\)
1. **Domain Check**: For \(x \in (0, 2)\), \(f(x)\) is defined.
2. **Range Check**:
   - As \(x \to 0^+\), \(\frac{1}{x} \to \infty\), so \(f(x) \to \infty\).
   - As \(x \to 2^-\), \(\frac{1}{x} \to \frac{1}{2}\), so \(f(x) \to 2.5\).
   - Range: \((2.5, \infty)\).
3. Since \((2.5, \infty) \neq (3, \infty)\), this is **not surjective**.

Not a bijection.

---

#### D) \(f(x) = \frac{1}{x} + 3\)
1. **Domain Check**: For \(x \in (0, 2)\), \(f(x)\) is defined.
2. **Range Check**:
   - As \(x \to 0^+\), \(\frac{1}{x} \to \infty\), so \(f(x) \to \infty\).
   - As \(x \to 2^-\), \(\frac{1}{x} \to \frac{1}{2}\), so \(f(x) \to 3.5\).
   - Range: \((3.5, \infty)\).
3. Since \((3.5, \infty) \neq (3, \infty)\), this is **not surjective**.

Not a bijection.

---

#### E) None of the mentioned
Since \(f(x) = \frac{2}{x} + 2\) satisfies the conditions for a bijection from \((0, 2)\) to \((3, \infty)\), **this option is incorrect**.

---

### Final Answer:
**A) \(f(x) = \frac{2}{x} + 2\)**

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Do you want a detailed explanation for any specific step? 

Here are 5 related questions to explore:  
1. How do you verify if a function is strictly increasing or decreasing?  
2. What are the conditions for a function to be bijective?  
3. Can you find the derivative of \(f(x) = \frac{2}{x} + 2\) and interpret it?  
4. Why do functions like \(f(x) = \frac{1}{x} + 3\) fail to map onto \((3, \infty)\)?  
5. What changes in the function domain or codomain would make \(f(x) = \frac{1}{x} + 3\) bijective?  

**Tip:** Always test the limits of a function at the boundaries of its domain to verify its range.

Which of these functions defines one-to-one correspondence from (0, 2) to (3, ∞)?
A) 2/x + 2
B) 3/x + 3
C) 1/x + 2
D) 1/x + 3
E) None of the mentioned

Analyze which function defines a bijection from the domain (0, 2) to the range (3, ∞). Explore key concepts like injective and surjective functions, domain, and range with step-by-step reasoning. Suitable for Grades 11-12.

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