Let `f: (-(pi)/(4), (pi)/(4)) rarr R` 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

Options

A. `1-e`
B. `e-1`
C. `1+e`
D. e

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: Find the Left-Hand Limit (LHL)
The left-hand limit as \( x \) approaches 0 is given by the expression for \( f(x) \) when \( x < 0 \):
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1 + |\sin x|)^{\frac{3a}{|\sin x|}} \]
As \( x \) approaches 0, \( |\sin x| \) approaches 0. Thus, we can rewrite the limit:
\[ \text{LHL} = \lim_{x \to 0} (1 + |\sin x|)^{\frac{3a}{|\sin x|}} \]
This limit is of the form \( 1^\infty \). We can use the exponential limit property:
\[ \lim_{x \to 0} (1 + u)^{v} = e^{\lim_{x \to 0} u \cdot v} \quad \text{where } u = |\sin x| \text{ and } v = \frac{3a}{|\sin x|} \]
Thus,
\[ \text{LHL} = e^{\lim_{x \to 0} |\sin x| \cdot \frac{3a}{|\sin x|}} = e^{3a} \]

### Step 2: Find the Right-Hand Limit (RHL)
The right-hand limit as \( x \) approaches 0 is given by the expression for \( f(x) \) when \( x > 0 \):
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{\cot(4x)/\cot(2x)} \]
Using the cotangent limit, we can rewrite:
\[ \cot(4x) = \frac{\cos(4x)}{\sin(4x)} \quad \text{and} \quad \cot(2x) = \frac{\cos(2x)}{\sin(2x)} \]
Thus,
\[ \text{RHL} = \lim_{x \to 0} e^{\frac{\cot(4x)}{\cot(2x)}} = e^{\lim_{x \to 0} \frac{\sin(2x)}{\sin(4x)}} \]
Using the small-angle approximation \( \sin(kx) \approx kx \) as \( x \to 0 \):
\[ \text{RHL} = e^{\lim_{x \to 0} \frac{2x}{4x}} = e^{\frac{1}{2}} \]

### Step 3: Set the Limits Equal
Since \( f \) is continuous at \( x = 0 \):
\[ \text{LHL} = \text{RHL} = f(0) \]
This gives us:
\[ e^{3a} = e^{\frac{1}{2}} \quad \text{and} \quad f(0) = b \]
From \( e^{3a} = e^{\frac{1}{2}} \), we equate the exponents:
\[ 3a = \frac{1}{2} \implies a = \frac{1}{6} \]

### Step 4: Determine \( b \)
Since \( b \) must also equal the right-hand limit:
\[ b = e^{\frac{1}{2}} = \sqrt{e} \]

### Step 5: Calculate \( 6a + b^2 \)
Now we can calculate:
\[ 6a + b^2 = 6 \left(\frac{1}{6}\right) + \left(\sqrt{e}\right)^2 = 1 + e \]

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