Which of the following functions is an injective (one-one) function in its respective domain? (A) `f(x) = 2x + sin 3x ` (B) `x. [x]` , (where `[.]` denotes the G.I.F) (C) `f(x)=(2^x-1)/(4^x+1)` (D) `f(x)=(2^x+1)/(4^x-1)`

Options

A. `f(x)=2x+sin 3x`
B. `f(x)=x*[x],` (where [.] denotes the G.I.F)
C. `f(x)=(2^(x)-1)/(4^(x)+1)`
D. `f(x)=(2^(x)+1)/(4^(x)-1)`

Explanation

To determine which of the given functions is injective (one-one) in its respective domain, we will analyze each function step by step.

### Step 1: Analyze Option A - \( f(x) = 2x + \sin(3x) \)

1. **Find the derivative**:
\[ f'(x) = 2 + 3\cos(3x) \]
2. **Determine the range of the derivative**:
- Since \( \cos(3x) \) varies between -1 and 1, we have:
\[ -3 \leq 3\cos(3x) \leq 3 \]
- Adding 2 to the entire inequality:
\[ -1 \leq f'(x) \leq 5 \]
3. **Conclusion**:
- Since \( f'(x) \) can be negative (e.g., when \( \cos(3x) = -1 \)), the function is not injective.

### Step 2: Analyze Option B - \( f(x) = x \cdot [x] \) (where \([.]\) denotes the greatest integer function)

1. **Evaluate the function**:
- For \( x = 0.5 \), \( f(0.5) = 0.5 \cdot [0.5] = 0.5 \cdot 0 = 0 \).
- For \( x = 1/3 \), \( f(1/3) = (1/3) \cdot [1/3] = (1/3) \cdot 0 = 0 \).
2. **Conclusion**:
- Since \( f(0.5) = f(1/3) = 0 \), the function is not injective.

### Step 3: Analyze Option C - \( f(x) = \frac{2^x - 1}{4^x + 1} \)

1. **Set up the equation for injectivity**:
- Assume \( f(x_1) = f(x_2) \):
\[ \frac{2^{x_1} - 1}{4^{x_1} + 1} = \frac{2^{x_2} - 1}{4^{x_2} + 1} \]
2. **Cross-multiply**:
\[ (2^{x_1} - 1)(4^{x_2} + 1) = (2^{x_2} - 1)(4^{x_1} + 1) \]
3. **Simplify**:
- This leads to a complex equation, but we can analyze the behavior of the function.
- As \( x \) increases, \( 2^x \) and \( 4^x \) both increase, and the function is strictly increasing.
4. **Conclusion**:
- Since the function is strictly increasing, it is injective.

### Step 4: Analyze Option D - \( f(x) = \frac{2^x + 1}{4^x - 1} \)

1. **Set up the equation for injectivity**:
- Assume \( f(x_1) = f(x_2) \):
\[ \frac{2^{x_1} + 1}{4^{x_1} - 1} = \frac{2^{x_2} + 1}{4^{x_2} - 1} \]
2. **Cross-multiply**:
\[ (2^{x_1} + 1)(4^{x_2} - 1) = (2^{x_2} + 1)(4^{x_1} - 1) \]
3. **Analyze the behavior of the function**:
- The function is more complicated, but we can differentiate it to check for monotonicity.
4. **Find the derivative**:
- After differentiating and analyzing, we find that the derivative changes signs, indicating that the function is not injective.

### Final Conclusion:
The only injective function among the options is **Option C: \( f(x) = \frac{2^x - 1}{4^x + 1} \)**.

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