Let f (x)= `{{:(((e ^(1/(x-2))-3))/((1)/(3 ^(x-2))+1), x gt 2), ((b sin{-x})/({-x}), x lt 2),(c, x=2):},` where {.} denotes fraction part function, is continuous at `x=2,` then `b+c=`

Options

A. 0
B. 1
C. 2
D. 4

Answer

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

Given the function:

\[ f(x) = \begin{cases} \frac{e^{\frac{1}{x-2}} - 3}{\frac{1}{3^{x-2}} + 1} & \text{if } x > 2 \\ \frac{b \sin(-x)}{-x} & \text{if } x < 2 \\ c & \text{if } x = 2 \end{cases} \]

### Step 1: Find the right-hand limit as \( x \) approaches 2

We need to calculate:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{e^{\frac{1}{x-2}} - 3}{\frac{1}{3^{x-2}} + 1} \]

As \( x \to 2^+ \), \( x - 2 \to 0^+ \). Therefore, \( \frac{1}{x-2} \to +\infty \) and \( e^{\frac{1}{x-2}} \to +\infty \).

Thus, we can simplify:

\[ \lim_{x \to 2^+} f(x) = \frac{+\infty - 3}{\frac{1}{3^{0}} + 1} = \frac{+\infty}{1 + 1} = +\infty \]

### Step 2: Find the left-hand limit as \( x \) approaches 2

Next, we calculate:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{b \sin(-x)}{-x} \]

As \( x \to 2^- \), \( -x \to -2 \). Thus:

\[ \lim_{x \to 2^-} f(x) = \frac{b \sin(-2)}{-2} = -\frac{b \sin(2)}{2} \]

### Step 3: Set the limits equal to ensure continuity

For the function to be continuous at \( x = 2 \), we need:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \]

This gives us:

\[ -\frac{b \sin(2)}{2} = +\infty = c \]

Since \( c \) must also equal \( +\infty \), we cannot have a finite value for \( b \) that satisfies this condition unless \( b = 0 \).

### Step 4: Set \( b \) and \( c \)

From the above analysis, we conclude:

1. \( b = 0 \)
2. \( c = 0 \)

### Step 5: Calculate \( b + c \)

Thus:

\[ b + c = 0 + 0 = 0 \]

### Final Answer

The value of \( b + c \) is:

\[ \boxed{0} \]