`f+g` may be a continuous function, if (a) f is continuous and g is discontinuous

Options

A. f is continuous and g is discontinuous (B) f is discontinuous and g is continuous
B. f is discontinuous and g is continuous
C. f and g are discontinuous
D. None of these

Answer

To determine if \( f + g \) can be a continuous function when \( f \) is continuous and \( g \) is discontinuous, we can analyze the properties of continuous and discontinuous functions.

### Step-by-Step Solution:

1. **Understanding Continuity**:
- A function \( f(x) \) is continuous at a point \( c \) if:
\[ \lim_{x \to c} f(x) = f(c) \]
- This means that as \( x \) approaches \( c \), \( f(x) \) approaches the value of \( f(c) \).

2. **Properties of Continuous Functions**:
- If \( f \) and \( g \) are both continuous functions, then \( f + g \) is also continuous.
- If either \( f \) or \( g \) is discontinuous, we need to analyze the behavior of their sum.

3. **Assuming \( f \) is Continuous**:
- Let’s assume \( f(x) \) is continuous at \( x = c \). Therefore:
\[ \lim_{x \to c} f(x) = f(c) \]

4. **Assuming \( g \) is Discontinuous**:
- Let \( g(x) \) be discontinuous at \( x = c \). This means:
\[ \lim_{x \to c} g(x) \neq g(c) \]
- There could be different types of discontinuities (jump, infinite, removable, etc.).

5. **Analyzing \( f + g \)**:
- We need to evaluate \( \lim_{x \to c} (f(x) + g(x)) \):
\[ \lim_{x \to c} (f(x) + g(x)) = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) \]
- Since \( f(x) \) is continuous, we have:
\[ \lim_{x \to c} f(x) = f(c) \]
- However, since \( g(x) \) is discontinuous, \( \lim_{x \to c} g(x) \) does not equal \( g(c) \).

6. **Conclusion**:
- Therefore, \( \lim_{x \to c} (f(x) + g(x)) \) does not equal \( f(c) + g(c) \).
- This implies that \( f + g \) is discontinuous at \( x = c \).

### Final Conclusion:
- Hence, \( f + g \) cannot be a continuous function if \( f \) is continuous and \( g \) is discontinuous. Therefore, the statement that \( f + g \) may be a continuous function under these conditions is false.