The function `f(x)=sinx+cosx` will be

Options

A. an even function
B. an odd function
C. a constant function
D. None of these

Answer

To determine the nature of the function \( f(x) = \sin x + \cos x \), we need to check if it is an even function, an odd function, or neither.

### Step 1: Define the function
The given function is:
\[ f(x) = \sin x + \cos x \]

### Step 2: Check for even function
A function is even if:
\[ f(-x) = f(x) \]
Let's calculate \( f(-x) \):
\[ f(-x) = \sin(-x) + \cos(-x) \]
Using the properties of sine and cosine:
\[ \sin(-x) = -\sin x \quad \text{and} \quad \cos(-x) = \cos x \]
So,
\[ f(-x) = -\sin x + \cos x \]

Now, we compare \( f(-x) \) with \( f(x) \):
\[ f(-x) = -\sin x + \cos x \quad \text{and} \quad f(x) = \sin x + \cos x \]
Since \( f(-x) \neq f(x) \), the function is not even.

### Step 3: Check for odd function
A function is odd if:
\[ f(-x) = -f(x) \]
We already have \( f(-x) = -\sin x + \cos x \) and \( f(x) = \sin x + \cos x \). Now, let's calculate \( -f(x) \):
\[ -f(x) = -(\sin x + \cos x) = -\sin x - \cos x \]

Now, we compare \( f(-x) \) with \( -f(x) \):
\[ f(-x) = -\sin x + \cos x \quad \text{and} \quad -f(x) = -\sin x - \cos x \]
Since \( f(-x) \neq -f(x) \), the function is not odd.

### Conclusion
Since \( f(x) \) is neither even nor odd, we conclude that:
\[ \text{The function } f(x) = \sin x + \cos x \text{ is neither even nor odd.} \]

Thus, the correct option is:
\[ \text{None of these.} \]

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