A) an even function
B) an odd function
C) a periodic function
D) neither an even nor an odd function
Correct Answer: B
Solution :
\[f(x)=\log \,(x+\sqrt{{{x}^{2}}+1)}\] \[\Rightarrow \] \[f(-x)=\log (-x+\sqrt{{{x}^{2}}+1)}\] \[\therefore \] \[f(x)+f(-x)=\log \,(x+\sqrt{{{x}^{2}}+1)}\] \[+\log \,(-x+\sqrt{{{x}^{2}}+1)}\] \[=\log \,(1)=0\] Hence, \[f\,(x)\] is an odd function. NOTE A function is said to be an even /unction, if \[f(x)=f(-x)\] Otherwise it is an odd/unction.
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