If `f (x) = {{:(( x log cos x )/( log (1 + x ^(2)))",", x ne 0), ( 0"," , x = 0):}` then f(x) is

Options

A. continuous as well as differentiable at x = 0
B. continuous but not differentiable at x = 0
C. continuous but not differentiable at x = 0
D. neither continuous nor differentiable a x =0

Answer

To determine the nature of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{x \log(\cos x)}{\log(1 + x^2)} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

we need to check if \( f(x) \) is continuous and differentiable at \( x = 0 \).

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

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

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

For \( x \neq 0 \):

\[ f(x) = \frac{x \log(\cos x)}{\log(1 + x^2)} \]

We will find the limit as \( x \) approaches 0.

### Step 2: Evaluate the Limit

Using L'Hôpital's Rule since both the numerator and denominator approach 0 as \( x \to 0 \):

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

This is in the form \( \frac{0}{0} \). We differentiate the numerator and denominator:

- Derivative of the numerator \( x \log(\cos x) \):
- Using the product rule: \( \log(\cos x) + x \cdot \frac{-\sin x}{\cos x} \)

- Derivative of the denominator \( \log(1 + x^2) \):
- Derivative is \( \frac{2x}{1 + x^2} \)

So we apply L'Hôpital's Rule:

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

### Step 3: Simplify the Limit

Now we simplify the limit:

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

As \( x \to 0 \), \( \log(\cos x) \to 0 \) and \( x \tan x \to 0 \). Thus, we can apply L'Hôpital's Rule again if needed.

### Step 4: Final Evaluation

Evaluating the limit gives:

\[ \lim_{x \to 0} \frac{\log(\cos x)}{\log(1 + x^2)} = \lim_{x \to 0} \frac{-\frac{\sin x}{\cos x}}{\frac{2x}{1 + x^2}} = \lim_{x \to 0} \frac{-\tan x}{\frac{2x}{1 + x^2}} = \lim_{x \to 0} \frac{-1}{2} = 0 \]

Thus,

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

So, \( f(x) \) is continuous at \( x = 0 \).

### Step 5: Check Differentiability at \( x = 0 \)

To check differentiability, we calculate the derivative at \( x = 0 \):

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{h \log(\cos h)}{\log(1 + h^2)} \]

Using L'Hôpital's Rule again since this is also \( \frac{0}{0} \):

After applying L'Hôpital's Rule, we find:

\[ f'(0) = -\frac{1}{2} \]

### Conclusion

Since \( f(x) \) is continuous and differentiable at \( x = 0 \), we conclude that:

**The function \( f(x) \) is continuous and differentiable at \( x = 0 \).**