2000
Problem - 3651
Is function $f(x)=\lg(x+\sqrt{x^2+1})$ an odd or even function?
It is an odd function because $$f(-x)=\lg(-x+\sqrt{(-x)^2+1})=\lg\Big(\frac{1}{x+\sqrt{x^2+1}}\Big)=-f(x)$$
Alternatively, we can also check
$$f(x)+f(-x)=\lg(x+\sqrt{x^2+1})+\lg(-x+\sqrt{(-x)^2+1})=\lg 1 = 0$$