Let `f(x) = sin^(-1)((2x)/(1+x^(2)))`Statement I `f'(2) = - 2/5 ` and
Statement II `sin^(-1)((2x)/(1 +x^(2))) = pi - 2 tan^(-1) x, AA x gt 1`

Options

A. Statement I is True, Statement II is True, Statement II is a correct explanation for statement I
B. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I
C. Statement I is True, Statement II is False
D. Statement I is False, Statement II is True.

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) and verify the two statements provided.

### Step 1: Differentiate \( f(x) \)

We start by differentiating \( f(x) \):
\[ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]

Using the chain rule, the derivative of \( \sin^{-1}(u) \) is given by:
\[ \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{2x}{1+x^2} \).

### Step 2: Find \( \frac{du}{dx} \)

Now, we find \( \frac{du}{dx} \):
\[ u = \frac{2x}{1+x^2} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} = \frac{2(1+x^2 - 2x^2)}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

### Step 3: Substitute \( u \) into the derivative formula

Next, we need to find \( 1 - u^2 \):
\[ u^2 = \left(\frac{2x}{1+x^2}\right)^2 = \frac{4x^2}{(1+x^2)^2} \]
Thus,
\[ 1 - u^2 = 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} \]

### Step 4: Find \( f'(x) \)

Now we can write \( f'(x) \):
\[ f'(x) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} = \frac{1}{\sqrt{\frac{1 - 2x^2 + x^4}{(1+x^2)^2}}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \]
This simplifies to:
\[ f'(x) = \frac{2(1-x^2)}{(1+x^2)^2 \sqrt{\frac{1 - 2x^2 + x^4}{(1+x^2)^2}}} = \frac{2(1-x^2)(1+x^2)}{(1+x^2)^2 \sqrt{1 - 2x^2 + x^4}} \]

### Step 5: Evaluate \( f'(2) \)

Now we evaluate \( f'(2) \):
\[ f'(2) = \frac{2(1-2^2)(1+2^2)}{(1+2^2)^2 \sqrt{1 - 2(2^2) + (2^2)^2}} = \frac{2(1-4)(1+4)}{(1+4)^2 \sqrt{1 - 8 + 16}} \]
Calculating each term:
\[ = \frac{2(-3)(5)}{25 \sqrt{9}} = \frac{-30}{25 \cdot 3} = \frac{-30}{75} = -\frac{2}{5} \]

### Conclusion

Thus, Statement I \( f'(2) = -\frac{2}{5} \) is true.

### Step 6: Verify Statement II

For Statement II, we need to show:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}(x) \quad \text{for } x > 1 \]
Using the identity \( \tan^{-1}(x) + \tan^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} \), we can derive that:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} - \tan^{-1}(x) \]
Thus,
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}(x) \quad \text{for } x > 1 \]
This confirms Statement II is also true.

### Final Answer

Both statements are true:
- Statement I: \( f'(2) = -\frac{2}{5} \)
- Statement II: \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}(x) \) for \( x > 1 \)