Question-45

function

Consider:

\[ \begin{equation*} f( x) =\ln\left( x+\sqrt{1+x^{2}} \ \right) \end{equation*} \]

\(\displaystyle f\) is an even function
\(\displaystyle f\) is an odd function
\(\displaystyle f\) is neither even nor odd

Hint

Multiply the quantity inside the logarithm for \(f(x)\) and \(f(-x)\)

Answer

\(\displaystyle f\) is an even function
\(\displaystyle f\) is an odd function
\(\displaystyle f\) is neither even nor odd

Solution

Let us multiply the terms inside the logarithm for \(\displaystyle f( x)\) and \(\displaystyle f( -x)\). This seems like a reasonable thing to do since we know that \(\displaystyle \ln( ab) =\ln( a) +\ln( b)\):

\[ \begin{equation*} \left(\sqrt{1+x^{2}} +x\right)\left(\sqrt{1+x^{2}} -x\right) =1 \end{equation*} \]

This is convenient. Since both products are positive for all \(\displaystyle x\), we can take \(\displaystyle \ln\) on both sides:

\[ \begin{equation*} \ln\left(\sqrt{1+x^{2}} +x\right) =-\ln\left(\sqrt{1+x^{2}} -x\right) \end{equation*} \]

This is nothing but:

\[ \begin{equation*} f( x) =-f( -x) \end{equation*} \]

This shows that \(\displaystyle f\) is an odd function.