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A function $\mathrm{f}$ defined by $\mathrm{f}(\mathrm{x})=\ln \left(\sqrt{x^{2}+1}-x\right)$ is
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The correct answer is:
an odd function
$\begin{aligned} & f(x)=\ln \left(\sqrt{x^{2}+1}-x\right) \\ & f(-x)=\ln \left(\sqrt{(-x)^{2}+1}-(-x)\right)=\ln \left(\sqrt{x^{2}+1}+x\right) \\ &=\ln \left(\frac{\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)}{\sqrt{x^{2}+1}-x}\right) \\ &=\ln \left(\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}-x}\right)=\ln \left(\frac{1}{\sqrt{x^{2}+1}-x}\right) \\ &=-\ln \left(\sqrt{x^{2}+1}-x\right)=-f(x) \end{aligned}$
So, $\mathrm{f}(\mathrm{x})$ is odd function.
So, $\mathrm{f}(\mathrm{x})$ is odd function.
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