If `f(x)={{:((1+|sinx|)^((p)/(|sinx|)),",",-(pi)/(6)ltxlt0),(q,":",x=0),(e^(tan3x.cot5x),":",0ltxlt(pi)/(6)):}` is continuous at x = 0, then the value of `2p+10lnq` is equal to

Answer

To solve the problem, we need to ensure the function \( f(x) \) is continuous at \( x = 0 \). This means that the left-hand limit as \( x \) approaches 0 from the left must equal the right-hand limit as \( x \) approaches 0 from the right, and both must equal \( f(0) \).

1. **Define the function**:
\[ f(x) = \begin{cases} \frac{(1 + |\sin x|)^p}{|\sin x|} & \text{for } -\frac{\pi}{6} < x < 0 \\ q & \text{for } x = 0 \\ e^{\tan(3x) \cot(5x)} & \text{for } 0 < x < \frac{\pi}{6} \end{cases} \]

2. **Calculate the right-hand limit as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{\tan(3x) \cot(5x)} \]
We know that \( \tan(3x) \approx 3x \) and \( \cot(5x) \approx \frac{1}{5x} \) as \( x \to 0 \).
Thus,
\[ \tan(3x) \cot(5x) \approx 3x \cdot \frac{1}{5x} = \frac{3}{5} \]
Therefore,
\[ \lim_{x \to 0^+} f(x) = e^{\frac{3}{5}} \]

3. **Calculate the left-hand limit as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{(1 + |\sin x|)^p}{|\sin x|} = \lim_{x \to 0^-} \frac{(1 + \sin x)^p}{-\sin x} \]
As \( x \to 0 \), \( \sin x \to 0 \), and we can use the expansion \( \sin x \approx x \):
\[ \lim_{x \to 0^-} \frac{(1 + \sin x)^p}{-\sin x} = \lim_{x \to 0^-} \frac{(1 + x)^p}{-x} \]
This limit is of the form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule:
\[ = \lim_{x \to 0^-} \frac{p(1 + x)^{p-1}}{-1} = -p \]

4. **Set the limits equal**:
Since the function is continuous at \( x = 0 \), we have:
\[ q = e^{\frac{3}{5}} \quad \text{and} \quad -p = e^{\frac{3}{5}} \]
Thus, we find:
\[ p = -\frac{3}{5} \]

5. **Calculate \( 2p + 10 \ln q \)**:
First, substitute \( p \) and \( q \):
\[ 2p + 10 \ln q = 2\left(-\frac{3}{5}\right) + 10 \ln\left(e^{\frac{3}{5}}\right) \]
Simplifying:
\[ = -\frac{6}{5} + 10 \cdot \frac{3}{5} = -\frac{6}{5} + 6 = \frac{24}{5} \]

6. **Final answer**:
\[ 2p + 10 \ln q = \frac{24}{5} \]