The function of \( \Large f \left(x\right)= log \left(x+\sqrt{x^{2}+1}\right) \) is:
Correct Answer: Description for Correct answer:
We have \( \Large f \left(x\right)=log \left(x+\sqrt{x^{2}+1}\right) \)
=> \( \Large f \left(-x\right)=log \left(-x+\sqrt{x^{2}+1}\right) \)
\( \Large \therefore f \left(x\right)+f \left(x\right)=log \left(x+\sqrt{x^{2}+1}\right)+log \left(-x+\sqrt{x^{2}+1}\right) \)
= \( \Large log \left(1\right) = 0 \)
Hence, \( \Large f \left(x\right) \) is an odd function.
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