Let `f` be a function with continuous second derivative and `f(0)=f^(prime)(0)=0.` Determine a function `g` by `g(x)={(f(x))/x ,x!=0 0,x=0` Then which of the following statements is correct? `g` has a continuous first derivative `g` has a first derivative `g` is continuous but `g` fails to have a derivative `g` has a first derivative but the first derivative is not continuous

Options

A. g has a continuous first derivative
B. g has a first derivative
C. g is continuous but g fails to have a derivative
D. g has a first derivative but the first derivative is not continuous

Explanation

To solve the problem, we need to analyze the function \( g(x) \) defined as:

\[ g(x) = \begin{cases} \frac{f(x)}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

Given that \( f(0) = 0 \) and \( f'(0) = 0 \), we want to determine the continuity and differentiability of \( g(x) \).

### Step 1: Check Continuity of \( g(x) \) at \( x = 0 \)

To check if \( g(x) \) is continuous at \( x = 0 \), we need to evaluate:

\[ \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{f(x)}{x} \]

Since both \( f(0) = 0 \) and \( f'(0) = 0 \), we have an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f'(x)}{1} = f'(0) = 0 \]

Thus,

\[ \lim_{x \to 0} g(x) = 0 = g(0) \]

This shows that \( g(x) \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability of \( g(x) \) at \( x = 0 \)

To check if \( g(x) \) is differentiable at \( x = 0 \), we need to evaluate the limit:

\[ g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{g(h)}{h} \]

For \( h \neq 0 \):

\[ g(h) = \frac{f(h)}{h} \]

Thus,

\[ g'(0) = \lim_{h \to 0} \frac{f(h)}{h^2} \]

Again, since \( f(0) = 0 \), we have another indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

\[ g'(0) = \lim_{h \to 0} \frac{f'(h)}{2h} \]

This limit is also of the form \( \frac{0}{0} \) since \( f'(0) = 0 \). We apply L'Hôpital's Rule one more time:

\[ g'(0) = \lim_{h \to 0} \frac{f''(h)}{2} = \frac{f''(0)}{2} \]

Since \( f''(0) \) exists (as \( f \) has a continuous second derivative), we conclude that \( g'(0) \) exists.

### Conclusion

Since \( g(x) \) is continuous at \( x = 0 \) and \( g'(0) \) exists, we can confirm that:

1. \( g(x) \) has a continuous first derivative.
2. \( g(x) \) has a first derivative.

Thus, the correct statements are:

- \( g \) has a continuous first derivative.
- \( g \) has a first derivative.

### Final Answer

The correct options are:
- \( g \) has a continuous first derivative.
- \( g \) has a first derivative.