Let `f(x) = {{:((cos x-e^(x^(2)//2))/(x^(3))",",x ne 0),(0",",x = 0):}`, then
Statement I f(x) is continuous at x = 0.
Statement II `lim_(x to 0 )(cos x-e^(x^(2)//2))/(x^(3)) = - (1)/(12)`

Options

A. Statement I is correct, Statement II is also correct, Statement II is the correct explanation of Statement I
B. Statement I is correct, Statement II is also correct, Statement II is not the correct explanation of Statement I
C. Statement I is correct, Statement II is incorrect
D. Statement I is incorrect, Statement II is correct.

Answer

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

\[ f(x) = \begin{cases} \frac{\cos x - e^{\frac{x^2}{2}}}{x^3} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

We need to check two statements:
1. Statement I: \( f(x) \) is continuous at \( x = 0 \).
2. Statement II: \( \lim_{x \to 0} \frac{\cos x - e^{\frac{x^2}{2}}}{x^3} = -\frac{1}{12} \).

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

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

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

Since \( f(0) = 0 \), we need to compute \( \lim_{x \to 0} f(x) \):

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\cos x - e^{\frac{x^2}{2}}}{x^3} \]

### Step 2: Evaluate the Limit

As \( x \to 0 \), both \( \cos x \) and \( e^{\frac{x^2}{2} } \) approach 1, leading to an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and denominator.

#### First Application of L'Hôpital's Rule

Differentiate the numerator and denominator:

- Derivative of the numerator \( \cos x - e^{\frac{x^2}{2}} \):
\[ -\sin x - \left( e^{\frac{x^2}{2}} \cdot x \right) \]

- Derivative of the denominator \( x^3 \):
\[ 3x^2 \]

Now we have:

\[ \lim_{x \to 0} \frac{-\sin x - x e^{\frac{x^2}{2}}}{3x^2} \]

### Step 3: Evaluate the New Limit

As \( x \to 0 \), this limit is still of the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

#### Second Application of L'Hôpital's Rule

Differentiate again:

- Derivative of the new numerator:
\[ -\cos x - \left( e^{\frac{x^2}{2}} + x \cdot x e^{\frac{x^2}{2}} \right) = -\cos x - e^{\frac{x^2}{2}}(1 + x^2) \]

- Derivative of the new denominator:
\[ 6x \]

Now we have:

\[ \lim_{x \to 0} \frac{-\cos x - e^{\frac{x^2}{2}}(1 + x^2)}{6x} \]

### Step 4: Evaluate the Limit Again

This limit is still of the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule one more time.

#### Third Application of L'Hôpital's Rule

Differentiate again:

- Derivative of the numerator:
\[ \sin x - \left( e^{\frac{x^2}{2}}(2x) + e^{\frac{x^2}{2}}(1 + x^2) \right) \]

- Derivative of the denominator:
\[ 6 \]

Now we have:

\[ \lim_{x \to 0} \frac{\sin x - e^{\frac{x^2}{2}}(2x + 1 + x^2)}{6} \]

### Step 5: Final Evaluation

As \( x \to 0 \):

\[ \sin 0 = 0 \quad \text{and} \quad e^{0} = 1 \]

Thus, we have:

\[ \lim_{x \to 0} \frac{0 - 1(0 + 1 + 0)}{6} = \frac{-1}{6} \]

### Conclusion

Since \( \lim_{x \to 0} f(x) = 0 \) and \( f(0) = 0 \), we conclude that \( f(x) \) is continuous at \( x = 0 \).

Now, for Statement II, we found that the limit is \( -\frac{1}{6} \), not \( -\frac{1}{12} \).

Thus, the conclusions are:
- Statement I is **True**: \( f(x) \) is continuous at \( x = 0 \).
- Statement II is **False**: \( \lim_{x \to 0} \frac{\cos x - e^{\frac{x^2}{2}}}{x^3} \neq -\frac{1}{12} \).

### Final Answer
- Statement I is correct.
- Statement II is incorrect.