The function `f(x)=sqrt(cos(sinx))+sin^-1((1+x^2)/(2x))` is defined for :

Options

A. `x in {-1,1}`
B. `x in [-1,1]`
C. ` x in R`
D. ` x in (-1,1)`

Answer

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

### Step 1: Analyze the square root term \( \sqrt{\cos(\sin x)} \)

The square root function is defined when the expression inside the square root is non-negative:

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

The cosine function is non-negative in the intervals where its argument is within the range of \( [0, \frac{\pi}{2}] \) and \( [\frac{3\pi}{2}, 2\pi] \). However, since \( \sin x \) oscillates between -1 and 1, we need to check the values of \( \cos(y) \) where \( y = \sin x \).

- The range of \( \sin x \) is \( [-1, 1] \).
- Therefore, we need to evaluate \( \cos(y) \) for \( y \) in \( [-1, 1] \).

Since \( \cos(y) \) is positive for \( y \) in \( [-1, 1] \), we conclude that:

\[ \cos(\sin x) \geq 0 \quad \text{is always true for all } x. \]

### Step 2: Analyze the inverse sine term \( \sin^{-1}\left(\frac{1 + x^2}{2x}\right) \)

The inverse sine function is defined for values in the range \( [-1, 1] \):

\[ -1 \leq \frac{1 + x^2}{2x} \leq 1 \]

We will break this into two inequalities.

#### Inequality 1: \( \frac{1 + x^2}{2x} \leq 1 \)

Cross-multiplying (assuming \( x \neq 0 \)) gives:

\[ 1 + x^2 \leq 2x \]

Rearranging this, we get:

\[ x^2 - 2x + 1 \leq 0 \]

This simplifies to:

\[ (x - 1)^2 \leq 0 \]

The only solution is:

\[ x = 1 \]

#### Inequality 2: \( \frac{1 + x^2}{2x} \geq -1 \)

Cross-multiplying (again assuming \( x \neq 0 \)) gives:

\[ 1 + x^2 \geq -2x \]

Rearranging this, we have:

\[ x^2 + 2x + 1 \geq 0 \]

This simplifies to:

\[ (x + 1)^2 \geq 0 \]

This inequality is always true for all \( x \).

### Conclusion

From the analysis, we find that the only value of \( x \) that satisfies both conditions is:

\[ x = 1 \]

Thus, the function \( f(x) \) is defined only at:

\[ \boxed{1} \]