`f(x)={{:(sinx/x", " xne0),(k-1", " x=0):}` is continuous at x = 0, then 'k' is :

Options

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

Answer

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x \neq 0 \\ k - 1 & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0

We need to calculate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} \]

### Step 2: Use the known limit

From calculus, we know that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Step 3: Set the limit equal to \( f(0) \)

Since \( f(0) = k - 1 \), we set up the equation:

\[ 1 = k - 1 \]

### Step 4: Solve for \( k \)

Adding 1 to both sides gives:

\[ k = 1 + 1 \]
\[ k = 2 \]

### Conclusion

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{2} \]

---