A) Even function
B) Odd function
C) Neither even nor odd
D) Periodic function
Correct Answer: B
Solution :
\[f(x)=\sin \,\left( \log \,(x+\sqrt{1+{{x}^{2}}}) \right)\] Þ \[f(-x)=\sin \,[\log \,(-x+\sqrt{1+{{x}^{2}}})]\] Þ \[f(-x)=\sin \,\log \left( (\sqrt{1+{{x}^{2}}}-x)\frac{(\sqrt{1+{{x}^{2}}}+x)}{(\sqrt{1+{{x}^{2}}}+x)} \right)\] Þ \[f(-x)=\sin \,\log \,\left[ \frac{1}{(x+\sqrt{1+{{x}^{2}}})} \right]\] Þ \[f(-x)=\sin \left[ \log {{(x+\sqrt{1+{{x}^{2}}})}^{-1}} \right]\] Þ \[f(-x)=\sin \left[ -\log (x+\sqrt{1+{{x}^{2}}}) \right]\] Þ \[f(-x)=-\sin \left[ \log (x+\sqrt{1+{{x}^{2}}}) \right]\]Þ \[f(-x)=-f(x)\] \\[f(x)\] is odd function.
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