Statement-1: The function f(x) given by `f(x)=sin^(-1){log(x+sqrt(x^(2)+1))}` is an odd function.
Statement:2 The composition of two odd functions is an odd function.

Options

A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .
B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False , Statement-2 is True.

Explanation

To determine whether the function \( f(x) = \sin^{-1}(\log(x + \sqrt{x^2 + 1})) \) is an odd function, we will follow these steps:

### Step 1: Find \( f(-x) \)
We start by substituting \(-x\) into the function:
\[ f(-x) = \sin^{-1}(\log(-x + \sqrt{(-x)^2 + 1})) \]
This simplifies to:
\[ f(-x) = \sin^{-1}(\log(-x + \sqrt{x^2 + 1})) \]

### Step 2: Simplify the expression inside the logarithm
Next, we simplify the expression inside the logarithm:
\[ \sqrt{(-x)^2 + 1} = \sqrt{x^2 + 1} \]
Thus, we have:
\[ f(-x) = \sin^{-1}(\log(-x + \sqrt{x^2 + 1})) \]

### Step 3: Rationalize the expression
We can rationalize the expression:
\[ -x + \sqrt{x^2 + 1} = \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} \]
This gives:
\[ \sqrt{x^2 + 1} - x = \frac{1}{\sqrt{x^2 + 1} + x} \]
Thus:
\[ f(-x) = \sin^{-1}\left(\log\left(\frac{1}{\sqrt{x^2 + 1} + x}\right)\right) \]

### Step 4: Use logarithmic properties
Using the property of logarithms, we can rewrite:
\[ \log\left(\frac{1}{\sqrt{x^2 + 1} + x}\right) = -\log(\sqrt{x^2 + 1} + x) \]
So:
\[ f(-x) = \sin^{-1}(-\log(\sqrt{x^2 + 1} + x)) \]

### Step 5: Use the property of the inverse sine function
Using the property that \( \sin^{-1}(-y) = -\sin^{-1}(y) \), we have:
\[ f(-x) = -\sin^{-1}(\log(\sqrt{x^2 + 1} + x)) \]
Recognizing that:
\[ f(x) = \sin^{-1}(\log(x + \sqrt{x^2 + 1})) \]
We can conclude:
\[ f(-x) = -f(x) \]

### Conclusion
Since \( f(-x) = -f(x) \), we have shown that \( f(x) \) is an odd function.

### Step 6: Verify Statement 2
The second statement claims that the composition of two odd functions is an odd function. Since both the sine inverse function and the logarithm function are odd, their composition \( f(x) \) is also an odd function.

### Final Answer
Both statements are true.

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