If `2tan^(-1)x+sin^(-1)((2x)/(1+x^2) )` is independent of x then :

Options

A. `x gt 1`
B. `x lt -1`
C. `0 lt x lt 1`
D. `-1 lt x lt 0`

Explanation

To solve the problem, we need to analyze the expression \(2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right)\) and determine the conditions under which it is independent of \(x\).

### Step 1: Understanding the Inverse Trigonometric Functions
We know that:
\[ \tan^{-1}(x) = \theta \implies \tan(\theta) = x \]
From the double angle formula for tangent, we have:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)} = \frac{2x}{1-x^2} \]
Thus,
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

### Step 2: Relating \(2\tan^{-1}(x)\) and \(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\)
We also know that:
\[ \sin^{-1}(y) = \theta \implies \sin(\theta) = y \]
In our case, we have:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]
The expression \(\frac{2x}{1+x^2}\) can be derived from the tangent double angle formula.

### Step 3: Setting Up the Equation
We want to find conditions under which:
\[ 2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right) = C \]
where \(C\) is a constant (independent of \(x\)).

### Step 4: Analyzing the Conditions
We know the following:
- If \(|x| \leq 1\), then:
\[ 2\tan^{-1}(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]
- If \(x > 1\):
\[ 2\tan^{-1}(x) = \pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]
- If \(x < -1\):
\[ 2\tan^{-1}(x) = -\pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]

### Step 5: Conclusion
For the expression \(2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right)\) to be independent of \(x\), we need to consider the cases:
- For \(|x| \leq 1\), the expression simplifies to a constant.
- For \(x > 1\) and \(x < -1\), the expression also simplifies to constants based on the values of \(x\).

Thus, the conditions for independence of \(x\) are satisfied for the ranges of \(x\) mentioned.

### Final Answer
The values of \(x\) for which \(2\tan^{-1}(x) + \sin^{-1}\left(\frac{2x}{1+x^2}\right)\) is independent of \(x\) are:
- \(x \leq 1\)
- \(x \geq -1\)