Range of the function `f(x)=sqrt(cos^(- 1)(sqrt(log_4x))-pi/2)+sin^(- 1)((1+x^2)/(4x))` is equal to
(A) `(0,pi/2+sqrt(pi/2)]`
(B) `[pi/2,pi/2+sqrt(pi/2)]`
(C) `[pi/6,pi/4)`
(D) `{pi/6}`

Options

A. `(0,(pi)/(2)+sqrt((pi)/(2))]`
B. `[(pi)/(2),(pi)/(2)+sqrt((pi)/(2))]`
C. `[(pi)/(6),(pi)/(2)]`
D. `{(pi)/(6)}`

Explanation

To find the range of the function \( f(x) = \sqrt{\cos^{-1}(\sqrt{\log_4 x}) - \frac{\pi}{2}} + \sin^{-1}\left(\frac{1+x^2}{4x}\right) \), we will follow these steps:

### Step 1: Determine the Domain of the Function
The function \( f(x) \) consists of two parts: \( \sqrt{\cos^{-1}(\sqrt{\log_4 x}) - \frac{\pi}{2}} \) and \( \sin^{-1}\left(\frac{1+x^2}{4x}\right) \).

1. **For \( \sqrt{\log_4 x} \)**:
- The expression \( \log_4 x \) is defined for \( x > 0 \).
- We need \( \log_4 x \geq 0 \) for \( \sqrt{\log_4 x} \) to be real, which implies \( x \geq 1 \) (since \( 4^0 = 1 \)).

2. **For \( \cos^{-1}(\sqrt{\log_4 x}) \)**:
- The range of \( \cos^{-1} \) is defined for inputs in the interval \([-1, 1]\).
- Therefore, we require \( \sqrt{\log_4 x} \) to be in \([0, 1]\), which leads to:
\[ 0 \leq \log_4 x \leq 1 \implies 1 \leq x \leq 4 \]

3. **For \( \sin^{-1}\left(\frac{1+x^2}{4x}\right) \)**:
- The expression \( \frac{1+x^2}{4x} \) must be in the interval \([-1, 1]\).
- Since \( x > 0 \), we can analyze \( \frac{1+x^2}{4x} \):
\[ \frac{1+x^2}{4x} \geq 0 \text{ for } x > 0 \]
- We also need \( \frac{1+x^2}{4x} \leq 1 \):
\[ 1+x^2 \leq 4x \implies x^2 - 4x + 1 \leq 0 \]
- Solving the quadratic inequality \( x^2 - 4x + 1 = 0 \) gives:
\[ x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} \]
- The roots are \( 2 - \sqrt{3} \) and \( 2 + \sqrt{3} \). The inequality holds between the roots, thus:
\[ 2 - \sqrt{3} \leq x \leq 2 + \sqrt{3} \]

### Step 2: Determine the Intersection of Domains
From the analysis, we have:
- \( 1 \leq x \leq 4 \) from the first part.
- \( 2 - \sqrt{3} \leq x \leq 2 + \sqrt{3} \) from the second part.

Calculating \( 2 - \sqrt{3} \) and \( 2 + \sqrt{3} \):
- \( 2 - \sqrt{3} \approx 0.268 \)
- \( 2 + \sqrt{3} \approx 3.732 \)

Thus, the intersection of the two intervals is:
\[ [1, 2 + \sqrt{3}] \]

### Step 3: Evaluate the Function at the Endpoints
1. **At \( x = 1 \)**:
\[ f(1) = \sqrt{\cos^{-1}(0) - \frac{\pi}{2}} + \sin^{-1}\left(\frac{1+1^2}{4 \cdot 1}\right) = \sqrt{0} + \sin^{-1}\left(\frac{1}{2}\right) = 0 + \frac{\pi}{6} = \frac{\pi}{6} \]

2. **At \( x = 2 + \sqrt{3} \)**:
- Calculate \( f(2 + \sqrt{3}) \):
\[ f(2 + \sqrt{3}) = \sqrt{\cos^{-1}(1) - \frac{\pi}{2}} + \sin^{-1}(1) = \sqrt{0} + \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 4: Determine the Range of the Function
Since \( f(x) \) is continuous on the interval \( [1, 2+\sqrt{3}] \) and takes values from \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \), the range of \( f(x) \) is:
\[ \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \]

### Final Answer
The range of the function \( f(x) \) is \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \).