Key Concepts and Formulas

• Even Function: A function $f(x)$ is even if $f(-x) = f(x)$ for all $x$ in its domain. The domain must be symmetric about the origin.
• Odd Function: A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. The domain must be symmetric about the origin.
• Logarithm Properties: $\log(1/A) = -\log(A)$.

Step-by-Step Solution

1. Determine the Domain of the Function The given function is $f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$. For the logarithm to be defined, its argument must be strictly positive: $x + \sqrt {{x^2} + 1} > 0$. We know that $\sqrt{x^2 + 1} \ge \sqrt{x^2} = |x|$. Thus, $\sqrt{x^2 + 1} > -x$ for all real $x$. Adding $x$ to both sides, we get $x + \sqrt{x^2 + 1} > 0$. This inequality holds true for all real numbers $x$. Therefore, the domain of $f(x)$ is $\mathbb{R}$, which is symmetric about the origin. This means the function can potentially be even or odd.

2. Evaluate $f(-x)$ To check if the function is even or odd, we substitute $-x$ for $x$ in the function's expression: $f\left( { - x} \right) = \log \left( { - x + \sqrt {{{\left( { - x} \right)}^2} + 1} } \right)$ $f\left( { - x} \right) = \log \left( { - x + \sqrt {{x^2} + 1} } \right)$

3. Simplify the Expression for $f(-x)$ We can simplify the argument of the logarithm in $f(-x)$ by multiplying it by its conjugate, $\sqrt{x^2 + 1} + x$, and dividing by the same term to maintain equality: $- x + \sqrt {{x^2} + 1} = \left( {\sqrt {{x^2} + 1} - x} \right) \cdot \frac{{\sqrt {{x^2} + 1} + x}}{{\sqrt {{x^2} + 1} + x}}$ Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{x^2 + 1}$ and $b = x$: $= \frac{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2} - {x^2}}}{{\sqrt {{x^2} + 1} + x}} = \frac{{\left( {{x^2} + 1} \right) - {x^2}}}{{\sqrt {{x^2} + 1} + x}} = \frac{1}{{x + \sqrt {{x^2} + 1}}}$ Now, substitute this simplified expression back into $f(-x)$: $f\left( { - x} \right) = \log \left( {\frac{1}{{x + \sqrt {{x^2} + 1} }}} \right)$

4. Apply Logarithm Properties Using the logarithm property $\log\left(\frac{1}{A}\right) = -\log(A)$, we get: $f\left( { - x} \right) = - \log \left( {x + \sqrt {{x^2} + 1} } \right)$

5. Compare $f(-x)$ with $f(x)$ We observe that the expression for $f(-x)$ is the negative of the original function $f(x)$: $f\left( { - x} \right) = - f\left( x \right)$

6. Conclusion Since $f(-x) = -f(x)$ for all $x$ in the domain of $f(x)$, the function $f(x)$ is an odd function.

Common Mistakes & Tips

• Domain Check: Always verify that the domain of the function is symmetric about the origin before attempting to classify it as even or odd.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when using the conjugate to simplify expressions involving square roots. Sign errors are a common pitfall.
• Recognize Inverse Hyperbolic Functions: The expression $\log(x + \sqrt{x^2+1})$ is equivalent to $\text{arsinh}(x)$ (the inverse hyperbolic sine function), which is a known odd function. Recognizing this can provide a quick confirmation.

Summary We determined the domain of the function $f(x) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$ to be all real numbers, which is symmetric about the origin. By evaluating $f(-x)$ and simplifying the expression using algebraic manipulation (multiplying by the conjugate) and logarithm properties, we found that $f(-x) = -f(x)$. This confirms that the function is an odd function.

The final answer is \boxed{A} which corresponds to option (A).