JEE MAIN - Mathematics (2003 - No. 2)
The function $$f\left( x \right)$$ $$ = \log \left( {x + \sqrt {{x^2} + 1} } \right)$$, is
neither an even nor an odd function
an even function
an odd function
a periodic function
Explication
$$f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$$
$$f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\}$$
$$ = \log \left\{ {{{ - {x^2} + {x^2} + 1} \over {x + \sqrt {{x^2} + 1} }}} \right\}$$
$$ = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\left( x \right)$$
$$ \Rightarrow f\left( x \right)\,\,\,\,$$ is an odd function.
$$f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\}$$
$$ = \log \left\{ {{{ - {x^2} + {x^2} + 1} \over {x + \sqrt {{x^2} + 1} }}} \right\}$$
$$ = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\left( x \right)$$
$$ \Rightarrow f\left( x \right)\,\,\,\,$$ is an odd function.