If `a le 2 sin^(-1)x+cos^(-1)x le b`, then find the value of a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that:

\[ a \leq 2 \sin^{-1}(x) + \cos^{-1}(x) \leq b \]

### Step-by-Step Solution:

1. **Understand the Range of Inverse Functions**:
The function \( \sin^{-1}(x) \) (inverse sine) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, we can state that:
\[ -\frac{\pi}{2} \leq \sin^{-1}(x) \leq \frac{\pi}{2} \]

2. **Add \( \frac{\pi}{2} \) to All Parts**:
To analyze the expression \( 2 \sin^{-1}(x) + \cos^{-1}(x) \), we can add \( \frac{\pi}{2} \) to all parts of the inequality:
\[ 0 \leq \sin^{-1}(x) + \frac{\pi}{2} \leq \pi \]

3. **Use the Identity**:
We know the identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
Using this identity, we can express \( \cos^{-1}(x) \) as:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \]

4. **Substitute the Identity**:
Substitute \( \cos^{-1}(x) \) into the expression:
\[ 2 \sin^{-1}(x) + \cos^{-1}(x) = 2 \sin^{-1}(x) + \left(\frac{\pi}{2} - \sin^{-1}(x)\right) \]
Simplifying this gives:
\[ 2 \sin^{-1}(x) + \cos^{-1}(x) = \sin^{-1}(x) + \frac{\pi}{2} \]

5. **Establish the New Inequality**:
Now we can rewrite our original inequality:
\[ 0 \leq \sin^{-1}(x) + \frac{\pi}{2} \leq \pi \]
This implies:
\[ 0 \leq 2 \sin^{-1}(x) + \cos^{-1}(x) \leq \pi \]

6. **Identify Values of \( a \) and \( b \)**:
From the inequalities derived, we can compare:
\[ a \leq 2 \sin^{-1}(x) + \cos^{-1}(x) \leq b \]
Thus, we find:
\[ a = 0 \quad \text{and} \quad b = \pi \]

### Final Answer:
The values of \( a \) and \( b \) are:
\[ a = 0, \quad b = \pi \]