Given that ,
`tan^(-1) ((2x)/(1-x^(2))) = {{:(2 tan^(-1) x"," |x| le 1),(-pi +2 tan^(-1)x","x gt 1),(pi+2 tan^(-1)x"," x lt -1):}`
`sin^(-1)((2x)/(1+x^(2))) ={{:(2 tan^(-1)x","|x|le1),(pi -2 tan^(-1)x","x gt 1 and ),(-(pi+2tan^(-1))","x lt -1):}`
`sin^(-1) x + cos^(-1) x = pi//2 " for " - 1 le x le 1`
`sin^(-1) ((4x)/(x^(2)+4)) + 2 tan^(-1)( - x/2)` is independent of `x`, then

Options

A. `x in [-3,4]`
B. `x in [-2,2]`
C. `x in [-1,1]`
D. `x in [1,infty)`

Answer

To solve the problem step by step, we will analyze the given expression and derive the conditions under which it is independent of \( x \).

### Step 1: Rewrite the given expression
We start with the expression:
\[ \sin^{-1}\left(\frac{4x}{x^2 + 4}\right) + 2 \tan^{-1}\left(-\frac{x}{2}\right) \]
We know that \( \tan^{-1}(-x) = -\tan^{-1}(x) \), so we can rewrite the second term:
\[ = \sin^{-1}\left(\frac{4x}{x^2 + 4}\right) - 2 \tan^{-1}\left(\frac{x}{2}\right) \]

### Step 2: Use the double angle formula for tangent
We can use the identity for the sine of a double angle:
\[ \sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta} \]
Let \( \theta = \tan^{-1}\left(\frac{x}{2}\right) \). Then,
\[ \tan\theta = \frac{x}{2} \implies \sin(2\theta) = \frac{2\cdot\frac{x}{2}}{1 + \left(\frac{x}{2}\right)^2} = \frac{x}{1 + \frac{x^2}{4}} = \frac{4x}{x^2 + 4} \]
Thus, we can rewrite the expression as:
\[ \sin^{-1}\left(\frac{4x}{x^2 + 4}\right) - 2 \tan^{-1}\left(\frac{x}{2}\right) = 0 \]

### Step 3: Set the expression to zero
For the expression to be independent of \( x \), we need:
\[ \sin^{-1}\left(\frac{4x}{x^2 + 4}\right) = 2 \tan^{-1}\left(\frac{x}{2}\right) \]
This implies that the left-hand side must equal the right-hand side for all \( x \).

### Step 4: Analyze the conditions
We know that:
- The range of \( \sin^{-1}(y) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
- The range of \( \tan^{-1}(y) \) is also \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).

### Step 5: Find the intervals for \( x \)
To ensure that the expression is defined, we need:
1. The argument of \( \sin^{-1} \) must be in the interval \([-1, 1]\):
\[ -1 \leq \frac{4x}{x^2 + 4} \leq 1 \]
This gives us two inequalities to solve.

2. The argument of \( \tan^{-1} \) is defined for all real numbers, so we do not have additional restrictions from that term.

### Step 6: Solve the inequalities
1. **First inequality**:
\[ \frac{4x}{x^2 + 4} \leq 1 \implies 4x \leq x^2 + 4 \implies x^2 - 4x + 4 \geq 0 \implies (x-2)^2 \geq 0 \]
This is true for all \( x \).

2. **Second inequality**:
\[ \frac{4x}{x^2 + 4} \geq -1 \implies 4x \geq -x^2 - 4 \implies x^2 + 4x + 4 \geq 0 \implies (x + 2)^2 \geq 0 \]
This is also true for all \( x \).

### Step 7: Combine the results
Since both inequalities are satisfied for all \( x \), we can conclude that the expression is independent of \( x \) for:
\[ |x| \leq 2 \]
Thus, the final answer is:
\[ x \in [-2, 2] \]