If `f and g` are one-one functions, then a. `f+g` is one one b. `fg` is one one c. `fog` is one one d. `none of these`

Options

A. `f+g` is one-one
B. `fg` is one-one
C. `fog` is one-one
D. None of these

Explanation

To determine which of the given options is correct, we need to analyze the properties of one-one functions (injective functions) and how they behave under addition, multiplication, and composition.

### Step-by-Step Solution:

1. **Understanding One-One Functions**:
- A function \( f \) is called one-one (or injective) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). This means that different inputs lead to different outputs.

2. **Analyzing \( f + g \)**:
- Let \( f \) and \( g \) be one-one functions. Consider \( f(x) = \sin(x) \) and \( g(x) = \cos(x) \) for \( x \) in the interval \( [0, \frac{\pi}{2}] \).
- We find \( f + g = \sin(x) + \cos(x) \).
- Evaluating at \( x = 0 \): \( f(0) + g(0) = \sin(0) + \cos(0) = 0 + 1 = 1 \).
- Evaluating at \( x = \frac{\pi}{2} \): \( f(\frac{\pi}{2}) + g(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1 \).
- Here, \( f + g \) gives the same output (1) for two different inputs (0 and \( \frac{\pi}{2} \)), thus \( f + g \) is **not one-one**.

3. **Analyzing \( f \cdot g \)**:
- Now consider \( f \cdot g = \sin(x) \cdot \cos(x) \).
- Evaluating at \( x = 0 \): \( f(0) \cdot g(0) = \sin(0) \cdot \cos(0) = 0 \cdot 1 = 0 \).
- Evaluating at \( x = \frac{\pi}{2} \): \( f(\frac{\pi}{2}) \cdot g(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) \cdot \cos(\frac{\pi}{2}) = 1 \cdot 0 = 0 \).
- Again, \( f \cdot g \) gives the same output (0) for two different inputs (0 and \( \frac{\pi}{2} \)), thus \( f \cdot g \) is **not one-one**.

4. **Analyzing \( f \circ g \)**:
- Now consider the composition \( f(g(x)) \).
- Since both \( f \) and \( g \) are one-one, the composition \( f \circ g \) is also one-one. This is a fundamental property of one-one functions.

5. **Conclusion**:
- From the analysis:
- \( f + g \) is not one-one.
- \( f \cdot g \) is not one-one.
- \( f \circ g \) is one-one.
- Therefore, the correct answer is option **c. \( f \circ g \) is one-one**.