The solution of the inequality `"log"_(2) sin^(-1) x gt "log"_(1//2) cos^(-1) x` is

Options

A. `x in [0, (pi)/(sqrt2)]`
B. `x in ((1)/(sqrt2), 1]`
C. `x in (0, (1)/(sqrt2))`
D. none of these

Explanation

To solve the inequality \( \log_{1/2}(\sin^{-1} x) > \log_{1/2}(\cos^{-1} x) \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \log_{1/2}(\sin^{-1} x) > \log_{1/2}(\cos^{-1} x) \]
Since the base of the logarithm \( \frac{1}{2} \) is less than 1, we can rewrite the inequality by switching the direction of the inequality:
\[ \sin^{-1} x < \cos^{-1} x \]

### Step 2: Analyze the Functions
The functions \( \sin^{-1} x \) and \( \cos^{-1} x \) are defined for \( x \) in the interval \( [0, 1] \). We know:
- \( \sin^{-1} x \) is increasing on \( [0, 1] \)
- \( \cos^{-1} x \) is decreasing on \( [0, 1] \)

### Step 3: Find the Intersection Point
To find when \( \sin^{-1} x = \cos^{-1} x \):
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This equality holds when:
\[ \sin^{-1} x = \cos^{-1} x \implies x = \frac{1}{\sqrt{2}} \text{ or } x = \frac{\sqrt{2}}{2} \]

### Step 4: Determine the Valid Range
We need to find the values of \( x \) for which \( \sin^{-1} x < \cos^{-1} x \). Since \( \sin^{-1} x \) is less than \( \cos^{-1} x \) for \( x < \frac{1}{\sqrt{2}} \) and equal at \( x = \frac{1}{\sqrt{2}} \), we conclude:
\[ x < \frac{1}{\sqrt{2}} \]

### Step 5: Combine with the Domain
The domain of \( x \) is \( [0, 1] \). Therefore, combining the conditions, we have:
\[ 0 \leq x < \frac{1}{\sqrt{2}} \]

### Final Solution
Thus, the solution to the inequality \( \log_{1/2}(\sin^{-1} x) > \log_{1/2}(\cos^{-1} x) \) is:
\[ x \in \left[0, \frac{1}{\sqrt{2}}\right) \]