Let `f(x)={{:((1+|sinx|)^((1)/(|sinx|))",",-(pi)/(6)ltxlt0),(e^((tan2x)/(tan3x))",", 0ltxlt(pi)/(6)),(m",",x=0):}` at `x=0`. Then, the value of l and m are

Options

A. `l=-(2)/(3), m=e^((2)/(3))`
B. `l=(2)/(3), m=e^(-(2)/(3))`
C. `l=(2)/(3), m=e^((2)/(3))`
D. None of these

Answer

To solve the problem, we need to find the values of \( l \) and \( m \) such that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined piecewise as follows:

\[ f(x) = \begin{cases} (1 + |\sin x|)^{\frac{l}{|\sin x|}}, & -\frac{\pi}{6} < x < 0 \\ e^{\frac{\tan 2x}{\tan 3x}}, & 0 < x < \frac{\pi}{6} \\ m, & x = 0 \end{cases} \]

### Step 1: Find the Right-Hand Limit (RHL) as \( x \to 0^+ \)

For \( x \) approaching \( 0 \) from the right, we use the second piece of the function:

\[ f(x) = e^{\frac{\tan 2x}{\tan 3x}} \]

We need to evaluate:

\[ \lim_{x \to 0^+} e^{\frac{\tan 2x}{\tan 3x}} \]

Using the fact that \( \tan x \approx x \) as \( x \to 0 \):

\[ \tan 2x \approx 2x \quad \text{and} \quad \tan 3x \approx 3x \]

Thus,

\[ \frac{\tan 2x}{\tan 3x} \approx \frac{2x}{3x} = \frac{2}{3} \]

Now we can evaluate the limit:

\[ \lim_{x \to 0^+} e^{\frac{\tan 2x}{\tan 3x}} = e^{\frac{2}{3}} \]

### Step 2: Find the Left-Hand Limit (LHL) as \( x \to 0^- \)

For \( x \) approaching \( 0 \) from the left, we use the first piece of the function:

\[ f(x) = (1 + |\sin x|)^{\frac{l}{|\sin x|}} \]

As \( x \to 0 \), \( |\sin x| \to 0 \) and \( 1 + |\sin x| \to 1 \). We need to evaluate:

\[ \lim_{x \to 0^-} (1 + |\sin x|)^{\frac{l}{|\sin x|}} \]

This limit is of the form \( 1^\infty \). We can rewrite it using the exponential limit:

\[ \lim_{x \to 0^-} (1 + |\sin x|)^{\frac{l}{|\sin x|}} = e^{\lim_{x \to 0^-} \frac{l \ln(1 + |\sin x|)}{|\sin x|}} \]

Using the approximation \( \ln(1 + u) \approx u \) for small \( u \):

\[ \ln(1 + |\sin x|) \approx |\sin x| \quad \text{as } x \to 0 \]

Thus, we have:

\[ \lim_{x \to 0^-} \frac{l \ln(1 + |\sin x|)}{|\sin x|} = \lim_{x \to 0^-} \frac{l |\sin x|}{|\sin x|} = l \]

### Step 3: Set the Limits Equal for Continuity

For the function to be continuous at \( x = 0 \), we set the left-hand limit (LHL) equal to the right-hand limit (RHL) and to \( f(0) = m \):

\[ e^{\frac{2}{3}} = m \]

And also,

\[ e^{l} = e^{\frac{2}{3}} \]

### Step 4: Solve for \( l \) and \( m \)

From \( e^{l} = e^{\frac{2}{3}} \), we find:

\[ l = \frac{2}{3} \]

From \( m = e^{\frac{2}{3}} \), we have:

\[ m = e^{\frac{2}{3}} \]

### Final Values

Thus, the values of \( l \) and \( m \) are:

\[ \boxed{l = \frac{2}{3}, \quad m = e^{\frac{2}{3}}} \]