A) \[|x|\,<1\]
B) \[x=1,-1\]
C) \[|x|\,>1\]
D) None of these
Correct Answer: B
Solution :
\[y'=\frac{1}{\sqrt{1-{{\left( \frac{2x}{1+{{x}^{2}}} \right)}^{2}}}}.\frac{2(1+{{x}^{2}})-4{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}=\frac{2(1-{{x}^{2}})}{\sqrt{{{(1-{{x}^{2}})}^{2}}.(1+{{x}^{2}})}}\] Þ \[y'=\left\{ \begin{align} & \frac{2}{1+{{x}^{2}}}\,\,\,\,\,\,\text{for}\,\,\,\,|x|<1 \\ & \frac{-2}{1+{{x}^{2}}}\,\,\,\,\,\,\text{for}\,\,\,\,|x|>1 \\ \end{align} \right.\] Hence for \[|x|=1\], the derivative does not exist.
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