If `f'(x) = g(x) (x-a)^2`, where g(a) `!= 0` and g is continuous at x = a, then :

Options

A. f is increasing near a if `g(a) lt 0`
B. f is decreasing near a if ` g(a) gt 0`
C. f is decreasing near a if ` g(a) lt 0`
D. f is increasing near a if ` g(a) gt 0`

Answer

To solve the problem, we start with the given information:

1. We have \( f'(x) = g(x) (x - a)^2 \).
2. It is given that \( g(a) \neq 0 \) and \( g \) is continuous at \( x = a \).

### Step 1: Analyze \( f'(x) \)

Since \( f'(x) = g(x) (x - a)^2 \), we note that \( (x - a)^2 \) is always non-negative for all \( x \). This means that the sign of \( f'(x) \) is determined by the sign of \( g(x) \).

### Step 2: Evaluate \( g(a) \)

Given that \( g(a) \neq 0 \), we have two cases to consider:
- Case 1: \( g(a) > 0 \)
- Case 2: \( g(a) < 0 \)

### Step 3: Case 1 - \( g(a) > 0 \)

If \( g(a) > 0 \), and since \( g \) is continuous at \( x = a \), there exists a neighborhood around \( a \) where \( g(x) \) remains positive. Thus, in this neighborhood, \( f'(x) = g(x) (x - a)^2 > 0 \). This implies that \( f(x) \) is increasing near \( a \).

### Step 4: Case 2 - \( g(a) < 0 \)

If \( g(a) < 0 \), and again using the continuity of \( g \) at \( x = a \), there exists a neighborhood around \( a \) where \( g(x) \) remains negative. Therefore, in this neighborhood, \( f'(x) = g(x) (x - a)^2 < 0 \). This implies that \( f(x) \) is decreasing near \( a \).

### Conclusion

From our analysis, we conclude:
- \( f(x) \) is increasing near \( a \) if \( g(a) > 0 \).
- \( f(x) \) is decreasing near \( a \) if \( g(a) < 0 \).

### Final Answer

Thus, the correct options are:
1. \( f \) is increasing near \( a \) if \( g(a) > 0 \).
2. \( f \) is decreasing near \( a \) if \( g(a) < 0 \).