`f(x)` is differentiable function and `(f(x).g(x))` is differentiable at `x = a`. Then (a) `g(x)` must be differentiable at `x=a` (b.) if `g(x)` is discontinuous, then `f(a)=0` (c.) if `f(a)!=0`, then `g(x)` must be differentiable (d.) none of these

Options

A. g(X) must be differentiable at x=a
B. if g(X) is discontinuous , then f(a)=0
C. if `f(a) ne 0, ` then g(x) must be differentiable
D. none of these

Answer

To solve the problem, we will analyze each option step by step based on the properties of differentiable functions and the product rule.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that \( f(x) \) is a differentiable function.
- The product \( f(x)g(x) \) is also differentiable at \( x = a \).

2. **Analyzing Option (a)**:
- **Statement**: \( g(x) \) must be differentiable at \( x = a \).
- **Counterexample**: Let \( f(x) = x^2 \) and \( g(x) = |x| \). At \( x = 0 \), \( f(x)g(x) = x^2|x| \) is differentiable, but \( g(x) \) is not differentiable at \( x = 0 \).
- **Conclusion**: This option is **false**.

3. **Analyzing Option (b)**:
- **Statement**: If \( g(x) \) is discontinuous, then \( f(a) = 0 \).
- **Analysis**: If \( g(x) \) is discontinuous at \( x = a \) and \( f(a) \neq 0 \), then \( g(x) \) must be continuous because \( f(x)g(x) \) is differentiable (and hence continuous) at \( x = a \). Therefore, if \( g(x) \) is discontinuous, \( f(a) \) must be zero.
- **Conclusion**: This option is **true**.

4. **Analyzing Option (c)**:
- **Statement**: If \( f(a) \neq 0 \), then \( g(x) \) must be differentiable.
- **Analysis**: If \( f(a) \neq 0 \), we can define \( h(x) = \frac{1}{f(x)} \), which is continuous and differentiable at \( x = a \). The product \( f(x)g(x) \) being differentiable implies \( g(x) \) must be differentiable as well.
- **Conclusion**: This option is **true**.

5. **Analyzing Option (d)**:
- **Statement**: None of these.
- Since we have found options (b) and (c) to be true, this option is **false**.

### Final Conclusion:
The correct options are (b) and (c). Therefore, the answer is that both statements (b) and (c) are true.