`f:Nto[-sqrt2,sqrt2]:f(x)=sinx+cosx`, then f(x) is

Options

A. one-one.
B. onto
C. many-one
D. into

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin x + \cos x \) and determine its range when the domain is restricted to natural numbers.

### Step-by-Step Solution:

1. **Rewrite the Function**:
We start with the function:
\[ f(x) = \sin x + \cos x \]
We can rewrite this using the identity:
\[ \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) \]
This can be expressed as:
\[ f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \]

2. **Determine the Maximum and Minimum Values**:
The sine function oscillates between -1 and 1. Therefore, the maximum and minimum values of \( f(x) \) are:
\[ \text{Maximum: } f(x) = \sqrt{2} \cdot 1 = \sqrt{2} \]
\[ \text{Minimum: } f(x) = \sqrt{2} \cdot (-1) = -\sqrt{2} \]

3. **Identify the Range**:
Thus, the range of \( f(x) \) is:
\[ [-\sqrt{2}, \sqrt{2}] \]

4. **Consider the Domain**:
The domain of \( f(x) \) is restricted to natural numbers \( \mathbb{N} \). The natural numbers are \( 1, 2, 3, \ldots \).

5. **Evaluate \( f(x) \) at Natural Numbers**:
We will evaluate \( f(x) \) at a few natural numbers:
- For \( x = 1 \):
\[ f(1) = \sin(1) + \cos(1) \approx 0.8415 + 0.5403 \approx 1.3818 \]
- For \( x = 2 \):
\[ f(2) = \sin(2) + \cos(2) \approx 0.9093 - 0.4161 \approx 0.4932 \]
- For \( x = 3 \):
\[ f(3) = \sin(3) + \cos(3) \approx 0.1411 - 0.9899 \approx -0.8488 \]

6. **Conclusion**:
Since \( f(x) \) takes values in the range \( [-\sqrt{2}, \sqrt{2}] \) and the outputs for natural numbers \( x \) do not cover the entire range, we conclude that \( f(x) \) is not onto.

### Final Answer:
The function \( f(x) = \sin x + \cos x \) is a one-one function but not onto when the domain is restricted to natural numbers.