A) Does not exist because f is bounded
B) Is not attained even through f is bounded
C) Is equal to +1
D) Is equal to ?1
Correct Answer: D
Solution :
Let \[f(x)=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}=\frac{{{x}^{2}}+1-2}{{{x}^{2}}+1}=1-\frac{2}{{{x}^{2}}+1}\] \[\because {{x}^{2}}+1>1\]; \ \[\frac{2}{{{x}^{2}}+1}\le 2\] So \[1-\frac{2}{{{x}^{2}}+1}\ge 1-2\]; \ \[-1\le f(x)<1\] Thus \[f(x)\] has the minimum value equal to ?1.
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