To solve the problem, we need to find the area bounded by the function \( f(x) = \min\{\sin^{-1} x, \cos^{-1} x, \frac{\pi}{6}\} \) over the interval \([0, 1]\). We will break this down step by step.
### Step 1: Identify the functions
We have three functions to consider:
1. \( y = \sin^{-1} x \)
2. \( y = \cos^{-1} x \)
3. \( y = \frac{\pi}{6} \)
### Step 2: Determine the intersection points
We need to find the points where these functions intersect within the interval \([0, 1]\).
1. **Intersection of \( \sin^{-1} x \) and \( \frac{\pi}{6} \)**:
\[
\sin^{-1} x = \frac{\pi}{6} \implies x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
\]
2. **Intersection of \( \cos^{-1} x \) and \( \frac{\pi}{6} \)**:
\[
\cos^{-1} x = \frac{\pi}{6} \implies x = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
\]
### Step 3: Define the function \( f(x) \)
From the intersections, we can define \( f(x) \) in segments:
- For \( x \in [0, \frac{1}{2}] \), \( f(x) = \sin^{-1} x \)
- For \( x \in [\frac{1}{2}, \frac{\sqrt{3}}{2}] \), \( f(x) = \frac{\pi}{6} \)
- For \( x \in [\frac{\sqrt{3}}{2}, 1] \), \( f(x) = \cos^{-1} x \)
### Step 4: Calculate the area under \( f(x) \)
The area \( A \) can be calculated using definite integrals:
\[
A = \int_0^{\frac{1}{2}} \sin^{-1} x \, dx + \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{\pi}{6} \, dx + \int_{\frac{\sqrt{3}}{2}}^1 \cos^{-1} x \, dx
\]
### Step 5: Evaluate each integral
1. **First integral**:
\[
\int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C
\]
Evaluating from \( 0 \) to \( \frac{1}{2} \):
\[
\left[ x \sin^{-1} x + \sqrt{1 - x^2} \right]_0^{\frac{1}{2}} = \left[ \frac{1}{2} \cdot \frac{\pi}{6} + \sqrt{1 - \left(\frac{1}{2}\right)^2} \right] - \left[ 0 + 1 \right]
\]
\[
= \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1
\]
2. **Second integral**:
\[
\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{\pi}{6} \, dx = \frac{\pi}{6} \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = \frac{\pi}{6} \cdot \frac{\sqrt{3} - 1}{2} = \frac{\pi(\sqrt{3} - 1)}{12}
\]
3. **Third integral**:
\[
\int \cos^{-1} x \, dx = x \cos^{-1} x + \sqrt{1 - x^2} + C
\]
Evaluating from \( \frac{\sqrt{3}}{2} \) to \( 1 \):
\[
\left[ x \cos^{-1} x + \sqrt{1 - x^2} \right]_{\frac{\sqrt{3}}{2}}^1 = \left[ 1 \cdot 0 + 0 \right] - \left[ \frac{\sqrt{3}}{2} \cdot \frac{\pi}{6} + \frac{1}{2} \right]
\]
\[
= 0 - \left[ \frac{\sqrt{3}\pi}{12} + \frac{1}{2} \right] = -\frac{\sqrt{3}\pi}{12} - \frac{1}{2}
\]
### Step 6: Combine the areas
Now, we can combine all the areas:
\[
A = \left( \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1 \right) + \frac{\pi(\sqrt{3} - 1)}{12} - \left( \frac{\sqrt{3}\pi}{12} + \frac{1}{2} \right)
\]
### Step 7: Simplify the expression
Combine like terms:
\[
A = \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1 + \frac{\pi\sqrt{3}}{12} - \frac{\pi}{12} - \frac{\sqrt{3}\pi}{12} - \frac{1}{2}
\]
\[
= \frac{\sqrt{3}}{2} - 1 - \frac{1}{2} = \frac{\sqrt{3}}{2} - \frac{3}{2} = \frac{\sqrt{3} - 3}{2}
\]
### Step 8: Express in the required form
We need to express \( A \) in the form \( \frac{a}{b(\sqrt{3}+1)} \):
\[
\frac{\sqrt{3} - 3}{2} = \frac{(\sqrt{3} - 3)(\sqrt{3} + 1)}{2(\sqrt{3} + 1)} = \frac{3 - 3\sqrt{3} - 3}{2(\sqrt{3} + 1)} = \frac{-3\sqrt{3}}{2(\sqrt{3} + 1)}
\]
From this, we can identify \( a = -3 \) and \( b = 2 \).
### Step 9: Find \( a - b \)
\[
a - b = -3 - 2 = -5
\]
### Final Answer
Thus, the final answer is:
\[
\boxed{-5}
\]
