If `f(x)=sin^(-1)((2x)/(1+x^(2))),` then Statement I The value of `f(2)=sin^(-1)((4)/(5))`. Statement II `f(x)=sin^(-1)((2x)/(1+x^(2)))=-2,` for `xlt1`

Options

A. Both statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
C. Statement I is correct but Statement II is incorrect
D. Statement II is correct but Statement I is incorrect.

Answer

To solve the problem step by step, we need to evaluate the function \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) for the given statements.

### Step 1: Evaluate \( f(2) \)
We start by substituting \( x = 2 \) into the function:

\[ f(2) = \sin^{-1}\left(\frac{2 \cdot 2}{1 + 2^2}\right) \]

### Step 2: Simplify the expression
Now we simplify the expression inside the sine inverse:

\[ f(2) = \sin^{-1}\left(\frac{4}{1 + 4}\right) = \sin^{-1}\left(\frac{4}{5}\right) \]

### Step 3: Conclusion for Statement I
Since we have calculated \( f(2) = \sin^{-1}\left(\frac{4}{5}\right) \), we conclude that Statement I is correct.

### Step 4: Analyze Statement II
Now, we need to analyze Statement II, which claims that \( f(x) = -2 \) for \( x < 1 \).

### Step 5: Understanding the range of \( f(x) \)
The function \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) outputs values in the range of \( \sin^{-1} \), which is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). The sine inverse function cannot equal \(-2\) since \(-2\) is outside this range.

### Step 6: Conclusion for Statement II
Thus, Statement II is incorrect because the output of \( f(x) \) cannot be a constant value of \(-2\) for all \( x < 1 \).

### Final Conclusion
- Statement I is correct: \( f(2) = \sin^{-1}\left(\frac{4}{5}\right) \).
- Statement II is incorrect: \( f(x) \) cannot equal \(-2\) for \( x < 1 \).

### Answer
The correct option is C (Statement I is correct, Statement II is incorrect).

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