A) even function
B) odd function
C) periodic function
D) none of these
Correct Answer: B
Solution :
Given that, \[f(x)=\log (x+\sqrt{{{x}^{2}}+1})\] Now, \[f(-x)=\log (-x+\sqrt{{{(-x)}^{2}}+1})\] \[=\log (\sqrt{{{x}^{2}}+1}-x)\] \[=-\log \left( \frac{1}{\sqrt{{{x}^{2}}+1}-x} \right)\] \[=-\log \left[ \frac{\sqrt{{{x}^{2}}+1}+x}{(\sqrt{{{x}^{2}}+1}-x)(\sqrt{{{x}^{2}}+1}+x)} \right]\] \[=-\log \left[ \frac{\sqrt{{{x}^{2}}+1}+x}{{{x}^{2}}+1-{{x}^{2}}} \right]\] \[=-\log (x+\sqrt{{{x}^{2}}+1})\] \[\therefore \] \[f(-x)=-f(x)\forall x\] Thus, the given function is an odd function.
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