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 can follow these steps:

### Step 1: Understand the relationship between \(\sin^{-1}x\) and \(\cos^{-1}x\)

We know that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
for \( x \) in the range \([-1, 1]\).

### Step 2: Rewrite the inequality

Using the relationship from Step 1, we can rewrite the inequality:
\[ \sin^{-1}x \geq \cos^{-1}x \implies \sin^{-1}x \geq \frac{\pi}{2} - \sin^{-1}x \]

### Step 3: Simplify the inequality

Rearranging the inequality gives:
\[ 2\sin^{-1}x \geq \frac{\pi}{2} \]
Dividing both sides by 2:
\[ \sin^{-1}x \geq \frac{\pi}{4} \]

### Step 4: Solve for \(x\)

Taking 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 5: Identify the minimum value of \(x\)

The minimum value of \(x\) that satisfies the inequality is:
\[ \lambda = \frac{1}{\sqrt{2}} \]

### Step 6: Calculate \(2\lambda\)

Now, we need to find \(2\lambda\):
\[ 2\lambda = 2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]
Using the approximation \(\sqrt{2} \approx 1.41\), we find:
\[ 2\lambda \approx 1.41 \]

### Final Answer

Thus, the value of \(2\lambda\) is approximately \(1.41\).

---