The function `f(x)={sinx/x +cosx , x!=0` and f(x) =k at x=0 is continuous at `x=0` then find the value of `k`

Options

A. 1
B. 2
C. 3
D. None of these

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 the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

### Step-by-Step Solution:

1. **Identify the condition for continuity**:
A function is continuous at a point if the following condition holds:
\[ \lim_{x \to 0} f(x) = f(0) \]
In our case, this means:
\[ \lim_{x \to 0} f(x) = k \]

2. **Calculate the limit as \( x \) approaches 0**:
We need to compute:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} + \cos x \right) \]

3. **Evaluate the limit of \( \frac{\sin x}{x} \)**:
It is a well-known limit that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

4. **Evaluate the limit of \( \cos x \)**:
The limit of \( \cos x \) as \( x \) approaches 0 is:
\[ \lim_{x \to 0} \cos x = \cos(0) = 1 \]

5. **Combine the limits**:
Now we can combine the results:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\sin x}{x} + \cos x \right) = 1 + 1 = 2 \]

6. **Set the limit equal to \( k \)**:
Since we want \( \lim_{x \to 0} f(x) = k \), we have:
\[ k = 2 \]

### Conclusion:
The value of \( k \) for which the function \( f(x) \) is continuous at \( x = 0 \) is:
\[ \boxed{2} \]