A) \[(-\infty ,\infty )\]
B) \[[-1,1]\]
C) \[\left[ -\frac{1}{2},\frac{1}{2} \right]\]
D) \[[-\sqrt{2},\sqrt{2}]\]
Correct Answer: C
Solution :
Let \[y=\frac{x}{1+{{x}^{2}}}\] \[\Rightarrow \] \[{{x}^{2}}y-x+y=0\] For \[x\]to be real, \[1-4{{y}^{2}}\ge 0\] \[\Rightarrow \]\[(1-2y)(1+2y)\ge 0\] \[\Rightarrow \]\[\left( \frac{1}{2}-y \right)\left( \frac{1}{2}+y \right)\ge 0\] \[\Rightarrow \]\[-\frac{1}{2}\le y\le \frac{1}{2}\] \[\therefore \] \[y=f(x)\in \left[ -\frac{1}{2},\frac{1}{2} \right]\]
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