The function `f(x)=sin(log(x+sqrt(1+x^2)))` is (a) even function (b) odd function (c) neither even nor odd (d) periodic function

Options

A. even function
B. odd function
C. neither even nor odd
D. periodic function

Explanation

To determine the nature of the function \( f(x) = \sin(\log(x + \sqrt{1+x^2})) \), we will check if it is an even function, an odd function, or neither.

### Step 1: Define the function
We start with the given function:
\[ f(x) = \sin(\log(x + \sqrt{1+x^2})) \]

### Step 2: Find \( f(-x) \)
Next, we need to compute \( f(-x) \):
\[ f(-x) = \sin(\log(-x + \sqrt{1+(-x)^2})) \]
Since \( (-x)^2 = x^2 \), we have:
\[ f(-x) = \sin(\log(-x + \sqrt{1+x^2})) \]

### Step 3: Simplify \( f(-x) \)
Now, we simplify \( -x + \sqrt{1+x^2} \):
\[ f(-x) = \sin(\log(\sqrt{1+x^2} - x)) \]

### Step 4: Use logarithmic properties
We can use the property of logarithms:
\[ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \]
Thus, we can express \( f(-x) \) as:
\[ f(-x) = \sin\left(\log\left(\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2} + x}\right) + \log(\sqrt{1+x^2} + x)\right) \]
This simplifies to:
\[ f(-x) = \sin\left(\log\left(\frac{1}{\sqrt{1+x^2} + x}\right)\right) \]

### Step 5: Use properties of sine
Using the property of logarithms:
\[ \log\left(\frac{1}{a}\right) = -\log(a) \]
we can rewrite \( f(-x) \):
\[ f(-x) = -\sin(\log(\sqrt{1+x^2} + x)) \]
Thus:
\[ f(-x) = -f(x) \]

### Step 6: Conclusion
Since we have shown that \( f(-x) = -f(x) \), we conclude that the function \( f(x) \) is an odd function.

### Final Answer
The function \( f(x) = \sin(\log(x + \sqrt{1+x^2})) \) is an **odd function** (option b).

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