A) even function
B) odd function
C) neither even nor odd
D) periodic function
Correct Answer: B
Solution :
We have,\[f(x)=\log (x+\sqrt{{{x}^{2}}+1})\] \[\therefore \]\[f(-x)=\log (-x+\sqrt{{{x}^{2}}+1})\times \frac{(x+\sqrt{{{x}^{2}}+1})}{(x+\sqrt{{{x}^{2}}+1})}\] \[=\log \left( \frac{-{{x}^{2}}+{{x}^{2}}-1}{x+\sqrt{{{x}^{2}}+1}} \right)\] \[=-\log (x+\sqrt{{{x}^{2}}+1})\] \[\Rightarrow \] \[f(-x)=-f(x)\] \[\therefore \]\[f(x)\]is an odd function. Note: (i) If\[f(-x)=f(x)\], is an even function. (ii) If \[f(x+T)=f(x)\], is a periodic function with period\[T\].
You need to login to perform this action.
You will be redirected in
3 sec
