Domain of the function `f(x)=sqrt(cos(sin x))+sin(x^2-1)i s` `[-1,1]` `[-2,2]` c.`[-pi,sqrt(2)]uu[sqrt(2,)pi]` `[-sqrt(2),-sqrt(2)]`

Options

A. [-1, 1]
B. [-2, 2]
C. `[-pi, -sqrt(2)] cup [ sqrt(2), pi] `
D. `[-sqrt(2), sqrt(2)]`

Answer

To find the domain of the function \( f(x) = \sqrt{\cos(\sin x)} + \sin(x^2 - 1) \), we need to analyze the two components of the function separately.

### Step 1: Analyze \( \sqrt{\cos(\sin x)} \)

The expression inside the square root must be non-negative:

\[ \cos(\sin x) \geq 0 \]

The cosine function is non-negative when its argument lies within the intervals where cosine is positive. Specifically, we know that:

\[ \cos(t) \geq 0 \quad \text{for} \quad t \in [2k\pi - \frac{\pi}{2}, 2k\pi + \frac{\pi}{2}] \quad \text{for any integer } k \]

Since \( t = \sin x \), we need to find the values of \( x \) such that \( \sin x \) falls within these intervals. The sine function ranges from -1 to 1, so we will focus on:

\[ \sin x \in [-1, 1] \]

In this range, \( \cos(\sin x) \) is non-negative when:

\[ \sin x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \]

### Step 2: Analyze \( \sin(x^2 - 1) \)

The sine function is defined for all real numbers, but we need to ensure that the argument \( x^2 - 1 \) is within the range where the sine function is defined:

\[ x^2 - 1 \in [-1, 1] \]

This leads to two inequalities:

1. \( x^2 - 1 \geq -1 \) which simplifies to \( x^2 \geq 0 \) (always true for real numbers).
2. \( x^2 - 1 \leq 1 \) which simplifies to \( x^2 \leq 2 \).

From \( x^2 \leq 2 \), we can derive:

\[ -\sqrt{2} \leq x \leq \sqrt{2} \]

### Step 3: Find the Common Domain

Now we need to find the intersection of the domains derived from both components:

1. From \( \sqrt{\cos(\sin x)} \): \( \sin x \in [-1, 1] \) (which is always true for \( x \in \mathbb{R} \)).
2. From \( \sin(x^2 - 1) \): \( x \in [-\sqrt{2}, \sqrt{2}] \).

Thus, the common domain is:

\[ x \in [-\sqrt{2}, \sqrt{2}] \]

### Conclusion

The domain of the function \( f(x) = \sqrt{\cos(\sin x)} + \sin(x^2 - 1) \) is:

\[ [-\sqrt{2}, \sqrt{2}] \]