Given f (x) = sin [log(x+√x2+1)] ... (i) Now. replacing x by —x in (I), we get f (- x) = sin [log (- x + √(−x)2+1] =
sin[log(−x+√x2+1.
−x−√x2+1
−x−√x2+1
)]
= sin [log(
x2−x2−1
−(x+√x2+1)
)] = sin [log(
1
x+√x2+1
)] = sin [log 1 - log (x + √x+1)] [Properties of logarithmic functlon-2] = sin [−log(x+√x2+1)] [Properties of logarithmic function-4] = - sin [log (x + √x2+1)] = - f (x) Since, f (- x) = - f (x) ∴ f (x) is an odd function Hence, option 'B' is correct.