The minimum value of x which satisfies the inequality `sin^(-1)x ge cos^(-1)x` is `lambda`, then the value of `2lambda` is (use `sqrt2=1.41`)

Answer

To solve the inequality \( \sin^{-1} x \geq \cos^{-1} x \), we will follow these steps:

### Step 1: Use the relationship between inverse sine and cosine
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
From this, we can express \( \cos^{-1} x \) in terms of \( \sin^{-1} x \):
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]

### Step 2: Substitute into the inequality
Substituting this into the inequality gives:
\[ \sin^{-1} x \geq \frac{\pi}{2} - \sin^{-1} x \]

### Step 3: Rearrange the inequality
Rearranging the inequality, we have:
\[ \sin^{-1} x + \sin^{-1} x \geq \frac{\pi}{2} \]
This simplifies to:
\[ 2 \sin^{-1} x \geq \frac{\pi}{2} \]

### Step 4: Divide by 2
Dividing both sides by 2, we get:
\[ \sin^{-1} x \geq \frac{\pi}{4} \]

### Step 5: Find the value of \( x \)
To find the minimum value of \( x \), we take the sine of both sides:
\[ x \geq \sin\left(\frac{\pi}{4}\right) \]
Since \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), we have:
\[ x \geq \frac{1}{\sqrt{2}} \]

### Step 6: Define \( \lambda \)
Let \( \lambda = \frac{1}{\sqrt{2}} \).

### Step 7: Calculate \( 2\lambda \)
Now, we need to find \( 2\lambda \):
\[ 2\lambda = 2 \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Answer
Thus, the value of \( 2\lambda \) is:
\[ \sqrt{2} \]