If `f(x)={((x-2)2^(-(1/(|x-2|)+1/(x-2))),x!=2),(0,x=2):}` then `f(x)` at `x=2` is

Options

A. Differentiable
B. Non-differentiable
C. R.H.L at x=2 is 1
D. None of these

Answer

To find the value of the function \( f(x) \) at \( x = 2 \), we need to analyze the given piecewise function:

\[ f(x) = \begin{cases} (x-2) \cdot 2^{-\left(\frac{1}{|x-2|} + \frac{1}{x-2}\right)}, & x \neq 2 \\ 0, & x = 2 \end{cases} \]

We want to determine \( f(2) \) and check the behavior of \( f(x) \) as \( x \) approaches 2 from both sides to see if the function is differentiable at that point.

### Step 1: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches 2 from the left (denoted as \( x \to 2^- \)) is given by:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x-2) \cdot 2^{-\left(\frac{1}{|x-2|} + \frac{1}{x-2}\right)} \]

Since \( x < 2 \), we have \( |x-2| = 2-x \). Thus, we can rewrite the limit:

\[ = \lim_{x \to 2^-} (x-2) \cdot 2^{-\left(\frac{1}{2-x} + \frac{1}{x-2}\right)} \]

Notice that \( \frac{1}{x-2} \) approaches \( -\infty \) as \( x \to 2^- \) and \( \frac{1}{2-x} \) approaches \( +\infty \). Therefore, we can simplify the exponent:

\[ = \lim_{x \to 2^-} (x-2) \cdot 2^{-\left(\frac{1}{2-x} - \frac{1}{2-x}\right)} = \lim_{x \to 2^-} (x-2) \cdot 2^0 = \lim_{x \to 2^-} (x-2) = 0 \]

### Step 2: Calculate the Right-Hand Limit (RHL)

Now, we calculate the right-hand limit as \( x \) approaches 2 from the right (denoted as \( x \to 2^+ \)):

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x-2) \cdot 2^{-\left(\frac{1}{|x-2|} + \frac{1}{x-2}\right)} \]

Since \( x > 2 \), we have \( |x-2| = x-2 \). Thus, we can rewrite the limit:

\[ = \lim_{x \to 2^+} (x-2) \cdot 2^{-\left(\frac{1}{x-2} + \frac{1}{x-2}\right)} = \lim_{x \to 2^+} (x-2) \cdot 2^{-\left(\frac{2}{x-2}\right)} \]

As \( x \to 2^+ \), \( (x-2) \) approaches 0 and \( 2^{-\left(\frac{2}{x-2}\right)} \) approaches 0 as well. Therefore:

\[ = \lim_{x \to 2^+} (x-2) \cdot 0 = 0 \]

### Step 3: Conclusion

Now we have both limits:

\[ \lim_{x \to 2^-} f(x) = 0 \quad \text{and} \quad \lim_{x \to 2^+} f(x) = 0 \]

Since both the left-hand limit and right-hand limit are equal to 0, we conclude that:

\[ f(2) = 0 \]

Thus, the function is continuous at \( x = 2 \).

### Step 4: Check Differentiability

To check if \( f(x) \) is differentiable at \( x = 2 \), we need to find the left-hand derivative (LHD) and right-hand derivative (RHD):

1. **Left-Hand Derivative (LHD)**:
\[ LHD = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} \]
After substituting and simplifying, we find:
\[ LHD = 1 \]

2. **Right-Hand Derivative (RHD)**:
\[ RHD = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} \]
After substituting and simplifying, we find:
\[ RHD = \frac{1}{2} \]

Since \( LHD \neq RHD \), the function is not differentiable at \( x = 2 \).

### Final Answer

Thus, the value of \( f(2) \) is:

\[ \boxed{0} \]