`f(x)={(x^2+e^(1/(2-x)))^(-1)k ,x=2, x!=2` is continuous from right at the point `x=2,` then `k` equals
a. `0`
b. `1//4`
c. `-1//4`
d. none of these

Options

A. 0
B. 1\4
C. `-1//4`
D. none of these

Explanation

To determine the value of \( k \) such that the function

\[ f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases} \]

is continuous from the right at the point \( x = 2 \), we need to ensure that the right-hand limit of \( f(x) \) as \( x \) approaches 2 is equal to \( f(2) = k \).

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

We need to calculate:

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

### Step 2: Substitute \( x = 2 + h \) where \( h \to 0^+ \).

Let \( x = 2 + h \), where \( h \) is a small positive number approaching 0. Then we have:

\[ \lim_{h \to 0^+} ((2 + h)^2 + e^{\frac{1}{2 - (2 + h)}})^{-1} \]

### Step 3: Simplify the expression.

Calculating \( (2 + h)^2 \):

\[ (2 + h)^2 = 4 + 4h + h^2 \]

Now, for \( e^{\frac{1}{2 - (2 + h)}} = e^{\frac{1}{-h}} \):

As \( h \to 0^+ \), \( \frac{1}{-h} \to -\infty \), thus:

\[ e^{\frac{1}{-h}} \to 0 \]

### Step 4: Combine the results.

Now substituting back into our limit:

\[ \lim_{h \to 0^+} ((4 + 4h + h^2) + 0)^{-1} = \lim_{h \to 0^+} (4 + 4h + h^2)^{-1} \]

As \( h \to 0 \):

\[ 4 + 4h + h^2 \to 4 \]

Thus, we have:

\[ \lim_{h \to 0^+} (4 + 4h + h^2)^{-1} = \frac{1}{4} \]

### Step 5: Set the limit equal to \( k \).

For continuity at \( x = 2 \):

\[ k = \lim_{x \to 2^+} f(x) = \frac{1}{4} \]

### Conclusion:

Thus, the value of \( k \) is

\[ \boxed{\frac{1}{4}} \]