The graph of the function `y = log_(a) (x + sqrt(x^(2) + 1))` is not symmetric about the origin.

Options

A. True
B. False
C. Can't say
D. Insufficient data

Answer

To determine whether the function \( y = \log_a (x + \sqrt{x^2 + 1}) \) is symmetric about the origin, we need to check if \( f(-x) = -f(x) \).

### Step-by-Step Solution:

1. **Define the function**:
Let \( f(x) = \log_a (x + \sqrt{x^2 + 1}) \).

2. **Find \( f(-x) \)**:
Substitute \(-x\) into the function:
\[ f(-x) = \log_a (-x + \sqrt{(-x)^2 + 1}) = \log_a (-x + \sqrt{x^2 + 1}). \]

3. **Simplify \( f(-x) \)**:
We can rewrite \( f(-x) \):
\[ f(-x) = \log_a (-x + \sqrt{x^2 + 1}). \]

4. **Rationalize the expression**:
To simplify further, we can rationalize the term inside the logarithm:
\[ -x + \sqrt{x^2 + 1} = \frac{(-x + \sqrt{x^2 + 1})(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} = \frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x} = \frac{1}{\sqrt{x^2 + 1} + x}. \]
Thus,
\[ f(-x) = \log_a \left( \frac{1}{\sqrt{x^2 + 1} + x} \right). \]

5. **Use logarithmic properties**:
Using the property of logarithms, we can express this as:
\[ f(-x) = -\log_a (\sqrt{x^2 + 1} + x). \]

6. **Compare \( f(-x) \) with \(-f(x)\)**:
Recall that:
\[ f(x) = \log_a (x + \sqrt{x^2 + 1}). \]
Therefore, we have:
\[ -f(x) = -\log_a (x + \sqrt{x^2 + 1}). \]

7. **Conclusion**:
Since \( f(-x) = -\log_a (\sqrt{x^2 + 1} + x) \) and \(-f(x) = -\log_a (x + \sqrt{x^2 + 1})\), we see that:
\[ f(-x) \neq -f(x). \]
Thus, the function is not symmetric about the origin.

### Final Statement:
The statement that the graph of the function \( y = \log_a (x + \sqrt{x^2 + 1}) \) is not symmetric about the origin is **true**.