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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Functions

NDA 2019 · Mathematics · Functions

A function f defined by f(x)= ( x^2+1 - x) is

A function f defined by f(x)=ln⁡\mathrm{f}(\mathrm{x})=\ln\left(\sqrt{x^2+1}-x\right) is

Options

  1. Aan even function
  2. Ban odd function
  3. CBoth even and odd function
  4. DNeither even nor odd function

Correct answer

D. Neither even nor odd function

Step-by-step solution

f(x)=ln⁡(x2+1−x)f(x)=\ln \left(\sqrt{x^2+1}-x\right)f(−x)=ln⁡((−x)2+1−(−x))=ln⁡(x2+1+x)=ln⁡((x2+1+x)(x2+1−x)x2+1−x)=ln⁡(x2+1−x2x2+1−x)=ln⁡(1x2+1−x)=−ln⁡(x2+1−x)=−f(x)\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}$So,So,\mathrm{f}(\mathrm{x})$ is odd function.
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