Find k, if possible, so that
`f (x)= [{:((ln (2- cos 2x))/(ln ^(2) (1+ sin 3x)),,, x lt 0),(k,,, x=0),((e ^(sin 2x )-1)/(ln (1+ tan 9x)),,, x gt 0):}` is continious at `x=0.`

Options

A. `2/3`
B. `1/9`
C. `2/9`
D. not possible

Answer

To find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{\ln(2 - \cos(2x))}{\ln^2(1 + \sin(3x)) & \text{if } x < 0 \\ k & \text{if } x = 0 \\ \frac{e^{\sin(2x)} - 1}{\ln(1 + \tan(9x)) & \text{if } x > 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the left-hand limit (LHL) and right-hand limit (RHL) at \( x = 0 \) are equal to \( k \).

### Step 1: Calculate the Left-Hand Limit (LHL) as \( x \to 0^- \)

For \( x < 0 \):

\[ f(x) = \frac{\ln(2 - \cos(2x))}{\ln^2(1 + \sin(3x))} \]

We need to find:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\ln(2 - \cos(2x))}{\ln^2(1 + \sin(3x))} \]

As \( x \to 0 \), \( \cos(2x) \to 1 \) and \( \sin(3x) \to 0 \). Thus:

\[ \ln(2 - \cos(2x)) \to \ln(2 - 1) = \ln(1) = 0 \]

and

\[ \ln(1 + \sin(3x)) \to \ln(1 + 0) = \ln(1) = 0 \]

This gives us a \( \frac{0}{0} \) form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Using L'Hôpital's Rule:

\[ \lim_{x \to 0^-} \frac{\ln(2 - \cos(2x))}{\ln^2(1 + \sin(3x))} = \lim_{x \to 0^-} \frac{\frac{d}{dx}[\ln(2 - \cos(2x))]}{\frac{d}{dx}[\ln^2(1 + \sin(3x))]} \]

Calculating the derivatives:

1. **Numerator**:

\[ \frac{d}{dx}[\ln(2 - \cos(2x))] = \frac{1}{2 - \cos(2x)} \cdot (2\sin(2x)) = \frac{2\sin(2x)}{2 - \cos(2x)} \]

2. **Denominator**:

Using the chain rule:

\[ \frac{d}{dx}[\ln^2(1 + \sin(3x))] = 2\ln(1 + \sin(3x)) \cdot \frac{3\cos(3x)}{1 + \sin(3x)} \]

### Step 3: Evaluate the Limit Again

Now we evaluate:

\[ \lim_{x \to 0^-} \frac{\frac{2\sin(2x)}{2 - \cos(2x)}}{2\ln(1 + \sin(3x)) \cdot \frac{3\cos(3x)}{1 + \sin(3x)}} \]

As \( x \to 0 \):

- \( \sin(2x) \to 2x \)
- \( \cos(2x) \to 1 \)
- \( \ln(1 + \sin(3x)) \to 3x \)

Thus, the limit simplifies to:

\[ \lim_{x \to 0^-} \frac{\frac{2(2x)}{2 - 1}}{2(3x) \cdot \frac{3}{1}} = \lim_{x \to 0^-} \frac{4x}{6x} = \frac{4}{6} = \frac{2}{3} \]

### Step 4: Calculate the Right-Hand Limit (RHL) as \( x \to 0^+ \)

For \( x > 0 \):

\[ f(x) = \frac{e^{\sin(2x)} - 1}{\ln(1 + \tan(9x))} \]

We need to find:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{\sin(2x)} - 1}{\ln(1 + \tan(9x))} \]

As \( x \to 0 \):

- \( e^{\sin(2x)} - 1 \to \sin(2x) \to 2x \)
- \( \tan(9x) \to 9x \)

Thus:

\[ \ln(1 + \tan(9x)) \to \ln(1 + 9x) \approx 9x \]

This gives us another \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again:

### Step 5: Apply L'Hôpital's Rule Again

Calculating the derivatives:

1. **Numerator**:

\[ \frac{d}{dx}[e^{\sin(2x)} - 1] = e^{\sin(2x)} \cdot 2\cos(2x) \]

2. **Denominator**:

\[ \frac{d}{dx}[\ln(1 + \tan(9x))] = \frac{9\sec^2(9x)}{1 + \tan(9x)} \]

Now we evaluate:

\[ \lim_{x \to 0^+} \frac{2e^{\sin(2x)}\cos(2x)}{\frac{9\sec^2(9x)}{1 + \tan(9x)}} \]

As \( x \to 0 \):

\[ = \frac{2 \cdot 1 \cdot 1}{9} = \frac{2}{9} \]

### Step 6: Set Limits Equal for Continuity

For \( f(x) \) to be continuous at \( x = 0 \):

\[ \text{LHL} = \text{RHL} = k \]

Thus:

\[ \frac{2}{3} = k \quad \text{and} \quad \frac{2}{9} = k \]

Since we need \( k \) to be equal for both limits, we find that:

\[ k = \frac{2}{9} \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{\frac{2}{9}} \]