The function defined by `f(x)={(((1)/(x^2+e^2-x))^(-1),xne2),(" "k" ,",x=2):}` ,is continuous from right at the point x = 2 ,then k is equal to

Options

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

Answer

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

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

is continuous from the right at the point \( x = 2 \), we need to ensure that

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

### Step 1: Calculate \( f(2) \)

Since \( f(2) = k \), we have:

\[ f(2) = k. \]

### Step 2: Calculate \( \lim_{x \to 2^+} f(x) \)

For \( x \to 2^+ \), we use the expression for \( f(x) \) when \( x \neq 2 \):

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

### Step 3: Substitute \( x = 2 \) into the limit expression

Now we substitute \( x = 2 \) into the limit:

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

### Step 4: Set the limit equal to \( f(2) \)

For continuity from the right, we need:

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

Thus, we have:

\[ \frac{1}{2 + e^2} = k. \]

### Final Answer

Therefore, the value of \( k \) is:

\[ k = \frac{1}{2 + e^2}. \]

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