Let f be defined as
`f(x)= {((1+|sin x|)^((3a)/(|sin x|))",",-(pi)/(4) lt x lt 0),(b",",x=0),(e^(cot 4x//cot 2x)",",0 lt x lt (pi)/(4)):}`
If f is continuous at x = 0, then the value of `6a+b^(2)` is equal to

Answer

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

### Step 1: Calculate the Left-Hand Limit (LHL)
The left-hand limit is given by:
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1 + |\sin x|)^{\frac{3a}{|\sin x|}} \]
As \( x \) approaches \( 0 \), \( |\sin x| \) approaches \( 0 \). Thus, we can rewrite the limit in the form of \( (1 + u)^{v} \) where \( u = |\sin x| \) and \( v = \frac{3a}{|\sin x|} \).

This limit can be transformed using the exponential limit:
\[ \lim_{u \to 0} (1 + u)^{\frac{1}{u}} = e \]
So we rewrite our limit:
\[ \text{LHL} = e^{\lim_{x \to 0} \frac{3a}{|\sin x|} \cdot |\sin x|} = e^{3a} \]

### Step 2: Calculate the Right-Hand Limit (RHL)
The right-hand limit is given by:
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{\cot(4x)/\cot(2x)} \]
Using the cotangent identity, we can rewrite this limit:
\[ \cot(4x) = \frac{\cos(4x)}{\sin(4x)} \quad \text{and} \quad \cot(2x) = \frac{\cos(2x)}{\sin(2x)} \]
Thus, we have:
\[ \text{RHL} = \lim_{x \to 0^+} e^{\frac{\cos(4x) \sin(2x)}{\sin(4x) \cos(2x)}} \]
As \( x \) approaches \( 0 \), both \( \sin(2x) \) and \( \sin(4x) \) approach \( 0 \), so we can use the small angle approximation:
\[ \sin(kx) \approx kx \quad \text{for small } x \]
This gives us:
\[ \text{RHL} = \lim_{x \to 0^+} e^{\frac{2x}{4x}} = e^{\frac{1}{2}} \]

### Step 3: Set the Limits Equal
For continuity at \( x = 0 \):
\[ e^{3a} = e^{\frac{1}{2}} \]
Taking the natural logarithm of both sides:
\[ 3a = \frac{1}{2} \implies a = \frac{1}{6} \]

### Step 4: Calculate \( f(0) \)
From the definition of \( f \):
\[ f(0) = b \]
Thus, we have:
\[ b = e^{\frac{1}{2}} = \sqrt{e} \]

### Step 5: Calculate \( 6a + b^2 \)
Now we can substitute the values of \( a \) and \( b \):
\[ 6a + b^2 = 6 \cdot \frac{1}{6} + (\sqrt{e})^2 = 1 + e \]

### Final Answer
Thus, the value of \( 6a + b^2 \) is:
\[ \boxed{1 + e} \]