The domain off (x) = `sqrt""(cos(sin x)] + sin^(-1) ((x^2 +1)/( 2x)) =`

Options

A. `R- {1}`
B. `{-1,1}`
C. `(1,oo)`
D. none of these

Answer

To find the domain of the function \( f(x) = \sqrt{\cos(\sin x)} + \sin^{-1}\left(\frac{x^2 + 1}{2x}\right) \), we need to ensure that both parts of the function are defined.

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

1. The expression inside the square root, \( \cos(\sin x) \), must be non-negative:
\[ \cos(\sin x) \geq 0 \]
2. The cosine function is non-negative when its argument is in the range:
\[ \sin x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \]
However, since \( \sin x \) oscillates between -1 and 1, we can find the values of \( x \) that yield \( \sin x \) in this range.

### Step 2: Determine the values of \( x \) for \( \cos(\sin x) \geq 0 \)

1. The cosine function is non-negative for angles in the first and fourth quadrants. Thus, we need:
\[ \sin x \in [0, 1] \]
This occurs when:
- \( x = n\pi + (-1)^n \frac{\pi}{2} \) for integers \( n \).

### Step 3: Analyze the inverse sine part \( \sin^{-1}\left(\frac{x^2 + 1}{2x}\right) \)

1. The argument of the inverse sine function must be in the range \([-1, 1]\):
\[ -1 \leq \frac{x^2 + 1}{2x} \leq 1 \]

### Step 4: Solve the inequalities

1. **Upper Bound:**
\[ \frac{x^2 + 1}{2x} \leq 1 \]
This simplifies to:
\[ x^2 + 1 \leq 2x \implies x^2 - 2x + 1 \leq 0 \implies (x - 1)^2 \leq 0 \]
This implies:
\[ x = 1 \]

2. **Lower Bound:**
\[ -1 \leq \frac{x^2 + 1}{2x} \]
This simplifies to:
\[ -2x \leq x^2 + 1 \implies x^2 + 2x + 1 \geq 0 \implies (x + 1)^2 \geq 0 \]
This is always true for all \( x \).

### Step 5: Combine the results

From the analysis, we have:
- The square root part is defined for \( x \) such that \( \sin x \) is in the range where \( \cos(\sin x) \geq 0 \).
- The inverse sine part is defined for \( x = 1 \).

Thus, the only value of \( x \) that satisfies both conditions is:
\[ \text{Domain of } f(x) = \{1\} \]

### Final Answer
The domain of the function \( f(x) \) is \( \{1\} \).

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