The function `f(x)={{:(sinx/x+cosx", if "xne0),(k", if "x=0):}` is continuous at x = 0, then the value of 'k' is :

Options

A. 3
B. 2
C. 1
D. 1.5

Answer

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

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

is continuous at \( x = 0 \), we need to ensure that

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

### Step 1: Find \( f(0) \)

Since \( f(0) = k \), we have:

\[ f(0) = k. \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \), we need to evaluate:

\[ \lim_{x \to 0} \left( \frac{\sin x}{x} + \cos x \right). \]

### Step 3: Evaluate \( \lim_{x \to 0} \frac{\sin x}{x} \)

We know from limit properties that:

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

### Step 4: Evaluate \( \lim_{x \to 0} \cos x \)

Also, we know that:

\[ \lim_{x \to 0} \cos x = \cos(0) = 1. \]

### Step 5: Combine the limits

Now, we can combine these results:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\sin x}{x} + \cos x \right) = 1 + 1 = 2. \]

### Step 6: Set the limits equal for continuity

For the function to be continuous at \( x = 0 \), we set:

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

Thus, we have:

\[ 2 = k. \]

### Conclusion

The value of \( k \) is:

\[ \boxed{2}. \]