A) \[(-\infty ,\,\,-1]\cup [1,\infty )\]
B) \[(-1,1)\]
C) \[(-\infty ,\infty )\]
D) None of the above
Correct Answer: A
Solution :
Given, \[f(x)=\frac{x}{1+{{x}^{2}}}\] On differentiating w. r. t. x, we get \[f'(x)=\frac{(1+{{x}^{2}})\,(1)\,-x\,(2x)}{{{(1+{{x}^{2}})}^{2}}}\] \[=\frac{1-{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}\] For \[f(x)\] to be decreasing \[f'(x)\le 0\] \[\Rightarrow \] \[(1-{{x}^{2}})\le 0\] \[\Rightarrow \] \[{{x}^{2}}\ge 1\] \[\Rightarrow \] \[x\in (-\infty ,\,\,-1]\,\cup [1,\,\infty )\]
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