Let `(a,b) subset (0,2 pi)` be the largest interval for which `sin^(-1)(sin theta)-cos^(-1)(sin theta)>0,theta in (0,2 pi),` holds.If `ax^(2)+beta x+sin^(-1)(x^(2)-6x+10)+cos^(-1)(x^(2)-6x+10)=0` and `alpha-beta=b-a,` then `alpha` is equal to,

Options

A. `(pi)/(8)`
B. `(pi)/(48)`
C. `(pi)/(16)`
D. `(pi)/(12)`

Answer

To solve the given problem, we will break it down into manageable steps.

### Step 1: Analyze the inequality
We need to find the largest interval \((a,b)\) for which:
\[ \sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0 \]
for \(\theta \in (0, 2\pi)\).

### Step 2: Simplify the expression
Recall that:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \]
Thus, we can rewrite the inequality as:
\[ \sin^{-1}(\sin \theta) - \left(\frac{\pi}{2} - \sin^{-1}(\sin \theta)\right) > 0 \]
This simplifies to:
\[ 2\sin^{-1}(\sin \theta) - \frac{\pi}{2} > 0 \]
or
\[ \sin^{-1}(\sin \theta) > \frac{\pi}{4} \]

### Step 3: Determine the values of \(\theta\)
The function \(\sin^{-1}(\sin \theta)\) behaves differently depending on the interval of \(\theta\):
- For \(\theta \in [0, \pi]\), \(\sin^{-1}(\sin \theta) = \theta\).
- For \(\theta \in (\pi, 2\pi)\), \(\sin^{-1}(\sin \theta) = \pi - \theta\).

Setting \(\theta > \frac{\pi}{4}\):
- In the interval \((0, \pi)\), this gives \(\theta > \frac{\pi}{4}\).
- In the interval \((\pi, 2\pi)\), we need \(\pi - \theta > \frac{\pi}{4}\) which simplifies to \(\theta < \frac{3\pi}{4}\).

### Step 4: Identify the interval \((a,b)\)
From the analysis, we find:
- In \((0, \pi)\): \(\theta \in \left(\frac{\pi}{4}, \pi\right)\)
- In \((\pi, 2\pi)\): \(\theta \in \left(\pi, \frac{3\pi}{4}\right)\) is not valid since \(\frac{3\pi}{4} < \pi\).

Thus, the largest interval is:
\[ \left(\frac{\pi}{4}, \frac{3\pi}{4}\right) \]
So, \(a = \frac{\pi}{4}\) and \(b = \frac{3\pi}{4}\).

### Step 5: Calculate \(b - a\)
\[ b - a = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Step 6: Set up the equation
We are given:
\[ \alpha - \beta = b - a \]
Thus:
\[ \alpha - \beta = \frac{\pi}{2} \]

### Step 7: Analyze the equation
We need to solve:
\[ \alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0 \]

### Step 8: Simplify the quadratic expression
The expression \(x^2 - 6x + 10\) can be rewritten as:
\[ (x - 3)^2 + 1 \]
This means \(x^2 - 6x + 10 \geq 1\) for all \(x\).

### Step 9: Find maximum values
The maximum value of \(\sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10)\) is:
\[ \sin^{-1}(1) + \cos^{-1}(1) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]

### Step 10: Set up the final equation
Thus, we can rewrite the equation:
\[ \alpha x^2 + \beta x + \frac{\pi}{2} = 0 \]

### Step 11: Solve for \(\alpha\)
Substituting \(x = 3\) (the vertex of the quadratic):
\[ \alpha(3^2) + \beta(3) + \frac{\pi}{2} = 0 \]
This gives:
\[ 9\alpha + 3\beta + \frac{\pi}{2} = 0 \]

### Step 12: Solve the system of equations
From the two equations:
1. \(\alpha - \beta = \frac{\pi}{2}\)
2. \(9\alpha + 3\beta + \frac{\pi}{2} = 0\)

Substituting \(\beta = \alpha - \frac{\pi}{2}\) into the second equation gives:
\[ 9\alpha + 3(\alpha - \frac{\pi}{2}) + \frac{\pi}{2} = 0 \]
This simplifies to:
\[ 12\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0 \]
\[ 12\alpha - \pi = 0 \implies \alpha = \frac{\pi}{12} \]

### Final Answer
Thus, the value of \(\alpha\) is:
\[ \boxed{\frac{\pi}{12}} \]