If the function `f(x) = {{:((1+ |sin x|^(a/(sin x))), -pi//6 lt x lt 0),(b, x=0),(e^((tan 2x)/(tan 3x)), 0 lt x lt pi//6):}`, is continous at x=0, then

Options

A. `a = log_(e)b, a =2//3`
B. `b=log_(e)a, a =2//3`
C. `a = log_(e) b, b=2`
D. none of these

Answer

To determine the values of \( a \) and \( b \) such that the function

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

is continuous at \( x = 0 \), we need to ensure that the left-hand limit \( f(0^-) \), the value at the point \( f(0) \), and the right-hand limit \( f(0^+) \) are all equal.

### Step 1: Find \( f(0^-) \)

We calculate the left-hand limit as \( x \) approaches \( 0 \) from the left:

\[ f(0^-) = \lim_{x \to 0^-} \left( 1 + | \sin x |^{\frac{a}{\sin x}} \right) \]

As \( x \to 0 \), \( \sin x \to 0 \) and \( | \sin x | \to 0 \). This gives us the indeterminate form \( 1 + 0^{\infty} \).

To resolve this, we can rewrite it using the exponential function:

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

As \( x \to 0 \), \( \sin x \approx x \), so:

\[ \frac{\ln(|\sin x|)}{\sin x} \to -\infty \quad \text{(since } \ln(|\sin x|) \to -\infty \text{)} \]

Thus, we have:

\[ f(0^-) = e^{a \cdot (-\infty)} = 0 \quad \text{if } a > 0 \]

So,

\[ f(0^-) = 1 + 0 = 1 \]

### Step 2: Find \( f(0) \)

By definition,

\[ f(0) = b \]

### Step 3: Find \( f(0^+) \)

Now we calculate the right-hand limit as \( x \) approaches \( 0 \) from the right:

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

As \( x \to 0 \), both \( \tan 2x \) and \( \tan 3x \) approach \( 0 \), leading to the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule:

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

Thus,

\[ f(0^+) = e^{\frac{2}{3}} \]

### Step 4: Set the limits equal for continuity

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

\[ f(0^-) = f(0) = f(0^+) \]

This gives us:

\[ 1 = b = e^{\frac{2}{3}} \]

### Step 5: Solve for \( a \) and \( b \)

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

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

And since \( f(0^-) = 1 \):

\[ 1 = e^{\frac{a}{\sin x}} \quad \text{as } x \to 0 \]

This implies:

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

### Conclusion

Thus, the values of \( a \) and \( b \) are:

\[ a = \frac{2}{3}, \quad b = e^{\frac{2}{3}} \]