Q84 / 225 · JEE Main 2003 · Mathematics

The function $f(x)=\log \left(x+\sqrt{x^2+1}\right)$, is

  • A
    neither an even nor an odd function
  • B
    an even function
  • Correct answer
    C
    an odd function
  • D
    a periodic function

Solution

$f(x)=\log \left(x+\sqrt{x^2+1}\right)$ $f(-x)=-\log \left(x+\sqrt{x^2+1}\right)$ $f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.

Hence, the correct answer is option: 'C'.

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