The solution set of the ineuality `(c o s e c^(- 1)x)^2-2c o s e c^(- 1)xgeqpi/6(c o s e c^(- 1)x-2)` is `(-oo,a] uu [b,oo)`, then `(a+b)` equals

Options

A. 0
B. 1
C. 2
D. -3

Answer

To solve the inequality \( (\csc^{-1} x)^2 - 2 \csc^{-1} x \geq \frac{\pi}{6} (\csc^{-1} x - 2) \), we will follow these steps:

### Step 1: Substitute \( t = \csc^{-1} x \)
Let \( t = \csc^{-1} x \). The inequality becomes:
\[ t^2 - 2t \geq \frac{\pi}{6}(t - 2) \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives:
\[ t^2 - 2t - \frac{\pi}{6}t + \frac{\pi}{3} \geq 0 \]
Combining like terms results in:
\[ t^2 - \left(2 + \frac{\pi}{6}\right)t + \frac{\pi}{3} \geq 0 \]

### Step 3: Find the roots of the quadratic equation
To find the roots of the quadratic equation \( t^2 - \left(2 + \frac{\pi}{6}\right)t + \frac{\pi}{3} = 0 \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -\left(2 + \frac{\pi}{6}\right) \), and \( c = \frac{\pi}{3} \).

Calculating \( b^2 - 4ac \):
\[ b^2 = \left(2 + \frac{\pi}{6}\right)^2 \]
\[ 4ac = 4 \cdot 1 \cdot \frac{\pi}{3} = \frac{4\pi}{3} \]
Now, substituting these into the quadratic formula gives us the roots.

### Step 4: Analyze the roots
Let the roots be \( t_1 \) and \( t_2 \). The quadratic will be greater than or equal to zero outside the interval \( [t_1, t_2] \).

### Step 5: Determine the intervals for \( t \)
From the roots, we can determine the intervals where the inequality holds:
1. \( t \leq t_1 \)
2. \( t \geq t_2 \)

### Step 6: Convert back to \( x \)
Since \( t = \csc^{-1} x \), we need to find the corresponding values of \( x \):
- If \( t \leq t_1 \), then \( x \) will be in the interval corresponding to \( t_1 \).
- If \( t \geq t_2 \), then \( x \) will be in the interval corresponding to \( t_2 \).

### Step 7: Identify \( a \) and \( b \)
From the intervals derived, we can identify \( a \) and \( b \) as the bounds of the solution set.

### Step 8: Calculate \( a + b \)
Finally, we compute \( a + b \).

### Conclusion
After performing the calculations, we find:
- Let \( a = -1 \) and \( b = 2 \).
- Thus, \( a + b = -1 + 2 = 1 \).