The function <span class="math-tex">\(\rm f(x)=\log \left(x+\sqrt{x^2 + 1}\right)\)</span> is:
\(f(-x) = \log\left(-x + \sqrt{x^2 + 1}\right)\). Multiply and divide by \(x + \sqrt{x^2+1}\):
\(-x + \sqrt{x^2+1} = \frac{(x^2 + 1) - x^2}{x + \sqrt{x^2+1}} = \frac{1}{x + \sqrt{x^2+1}}\)
So \(f(-x) = \log\frac{1}{x + \sqrt{x^2+1}} = -f(x)\). Therefore f is an odd function.
Answer: (B)
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