The set of values of k for which the equation `sin^(-1)x+cos^(-1)x +pi(|x|-2)=k pi` possesses real solution is [a,b] then the value of a + b is

Options

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

Explanation

To solve the equation \( \sin^{-1} x + \cos^{-1} x + \pi (|x| - 2) = k \pi \) for the set of values of \( k \) that allow for real solutions, we can follow these steps:

### Step 1: Simplify the Equation
We know that for any \( x \) in the domain of \( \sin^{-1} x \) and \( \cos^{-1} x \), the identity holds:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Thus, we can rewrite the equation as:
\[ \frac{\pi}{2} + \pi (|x| - 2) = k \pi \]

### Step 2: Rearranging the Equation
Now, we can simplify this equation:
\[ \frac{\pi}{2} + \pi |x| - 2\pi = k \pi \]
This simplifies to:
\[ \pi |x| - \frac{3\pi}{2} = k \pi \]

### Step 3: Dividing by \( \pi \)
Dividing the entire equation by \( \pi \) (assuming \( \pi \neq 0 \)):
\[ |x| - \frac{3}{2} = k \]
This gives us:
\[ |x| = k + \frac{3}{2} \]

### Step 4: Establishing the Domain for \( |x| \)
Since \( x \) must lie within the domain of \( \sin^{-1} x \) and \( \cos^{-1} x \), we have:
\[ -1 \leq x \leq 1 \]
Thus, the absolute value \( |x| \) must satisfy:
\[ 0 \leq |x| \leq 1 \]

### Step 5: Setting Up Inequalities
From the equation \( |x| = k + \frac{3}{2} \), we can set up the following inequalities:
\[ 0 \leq k + \frac{3}{2} \leq 1 \]

### Step 6: Solving the Inequalities
1. From \( 0 \leq k + \frac{3}{2} \):
\[ k \geq -\frac{3}{2} \]
2. From \( k + \frac{3}{2} \leq 1 \):
\[ k \leq 1 - \frac{3}{2} \implies k \leq -\frac{1}{2} \]

### Step 7: Combining the Results
Combining the results from the inequalities, we find:
\[ -\frac{3}{2} \leq k \leq -\frac{1}{2} \]
Thus, the set of values of \( k \) is the closed interval:
\[ [a, b] = \left[-\frac{3}{2}, -\frac{1}{2}\right] \]

### Step 8: Calculating \( a + b \)
Now, we can find \( a + b \):
\[ a + b = -\frac{3}{2} + -\frac{1}{2} = -2 \]

### Final Answer
The value of \( a + b \) is:
\[ \boxed{-2} \]