`f(x)={{:(((3)/(2))^((cot 3x)/(cot 2x)), 0 le xlt (pi)/(2)),(b+3, x =(pi)/(2)),( (1+|cotx|)^((a tan x|)/b), (pi)/(2) lt xlt pi):}` is continuous at `x= pi/2` , then

Options

A. a=0
B. a=2
C. b=-2
D. b=2

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). This means that the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at that point must all be equal.

### Step 1: Calculate the Left-Hand Limit (LHL) as \( x \to \frac{\pi}{2}^- \)

The function for \( 0 \leq x < \frac{\pi}{2} \) is given by:
\[ f(x) = \left( \frac{3}{2} \right)^{\frac{\cot(3x)}{\cot(2x)}} \]

We need to find:
\[ \text{LHL} = \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} - h\right) \]

Substituting \( x = \frac{\pi}{2} - h \):
\[ \text{LHL} = \lim_{h \to 0} \left( \frac{3}{2} \right)^{\frac{\cot(3(\frac{\pi}{2} - h))}{\cot(2(\frac{\pi}{2} - h))}} \]

Calculating the cotangent values:
\[ \cot(3(\frac{\pi}{2} - h)) = \cot\left(\frac{3\pi}{2} - 3h\right) = \tan(3h) \]
\[ \cot(2(\frac{\pi}{2} - h)) = \cot(\pi - 2h) = -\cot(2h) \]

Thus, we have:
\[ \text{LHL} = \lim_{h \to 0} \left( \frac{3}{2} \right)^{\frac{\tan(3h)}{-\cot(2h)}} \]

Using the fact that \( \cot(2h) = \frac{1}{\tan(2h)} \):
\[ \text{LHL} = \lim_{h \to 0} \left( \frac{3}{2} \right)^{-\tan(3h) \tan(2h)} \]

As \( h \to 0 \), both \( \tan(3h) \) and \( \tan(2h) \) approach 0. Therefore:
\[ \text{LHL} = \left( \frac{3}{2} \right)^{0} = 1 \]

### Step 2: Calculate the Right-Hand Limit (RHL) as \( x \to \frac{\pi}{2}^+ \)

For \( \frac{\pi}{2} < x < \pi \), the function is given by:
\[ f(x) = (1 + |\cot x|)^{\frac{a \tan x}{b}} \]

We need to find:
\[ \text{RHL} = \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} + h\right) \]

Substituting \( x = \frac{\pi}{2} + h \):
\[ \text{RHL} = \lim_{h \to 0} \left( 1 + |\cot(\frac{\pi}{2} + h)| \right)^{\frac{a \tan(\frac{\pi}{2} + h)}{b}} \]

Calculating the cotangent and tangent:
\[ |\cot(\frac{\pi}{2} + h)| = |\tan(h)| \quad \text{and} \quad \tan(\frac{\pi}{2} + h) = -\cot(h) \]

Thus:
\[ \text{RHL} = \lim_{h \to 0} \left( 1 + \tan(h) \right)^{-\frac{a \cot(h)}{b}} \]

As \( h \to 0 \):
\[ \text{RHL} = \lim_{h \to 0} \left( 1 + h \right)^{-\frac{a}{b} \cdot \frac{1}{h}} \text{ (using } \tan(h) \approx h \text{ and } \cot(h) \approx \frac{1}{h}) \]

This is of the form \( 1^{\infty} \). We can use the exponential limit:
\[ \text{RHL} = e^{\lim_{h \to 0} -\frac{a}{b} \cdot \frac{1}{h} \ln(1 + h)} = e^{\lim_{h \to 0} -\frac{a}{b} \cdot \frac{h}{h}} = e^{-\frac{a}{b}} \]

### Step 3: Set Limits Equal for Continuity

Since \( f\left(\frac{\pi}{2}\right) = b + 3 \), we have:
1. \( b + 3 = 1 \) (from LHL)
2. \( e^{-\frac{a}{b}} = 1 \) (from RHL)

From the first equation:
\[ b = -2 \]

From the second equation:
\[ -\frac{a}{b} = 0 \implies a = 0 \text{ (since } b \neq 0\text{)} \]

### Final Result

Thus, the values of \( a \) and \( b \) are:
\[ \boxed{a = 0, b = -2} \]