Q223 / 270 · NDA 2019 (Phase 1) · Mathematics
A function f defined by $\mathrm{f}(\mathrm{x})=\ln \left(\sqrt{x^2+1}-x\right)$ is
- Aan even function
- Ban odd function
- CBoth even and odd function
- Correct answerDNeither even nor odd function
Solution
$f(x)=\ln \left(\sqrt{x^2+1}-x\right)$
$\begin{aligned} & f(-x)=\ln \left(\sqrt{(-x)^2+1}-(-x)\right) \\ & =\ln \left(\sqrt{x^2+1}+x\right) \\ & =\ln \left(\frac{\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)}{\sqrt{x^2+1}-x}\right) \\ & =\ln \left(\frac{x^2+1-x^2}{\sqrt{x^2+1}-x}\right) \\ & =\ln \left(\frac{1}{\sqrt{x^2+1}-x}\right) \\ & =-\ln \left(\sqrt{x^2+1}-x\right) \\ & =-\mathrm{f}(\mathrm{x})\end{aligned}$
So, $\mathrm{f}(\mathrm{x})$ is odd function.
Hence, the correct answer is option: 'D'.