A) Does not exist because f is unbounded
B) Is not attained even though f is bounded
C) Is equal to 1
D) Is equal to ?1
Correct Answer: D
Solution :
\[f(x)=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}=\frac{{{x}^{2}}+1-2}{{{x}^{2}}+1}=1-\frac{2}{{{x}^{2}}+1}\] \[\therefore f(x)<1\forall x\]and \[\ge -1\]as \[\frac{2}{{{x}^{2}}+1}\le 2\] \[\therefore -1\le f(x)<1\] Hence \[f(x)\] has minimum value ?1 and also there is no maximum value. Aliter : \[{f}'(x)=\frac{({{x}^{2}}+1)2x-({{x}^{2}}-1)2x}{{{({{x}^{2}}+1)}^{2}}}=\frac{4x}{{{({{x}^{2}}+1)}^{2}}}\] \[{f}'(x)=0\Rightarrow x=0\] \[{f}''\,(x)=\frac{{{({{x}^{2}}+1)}^{2}}4-4x.2({{x}^{2}}+1)2x}{{{({{x}^{2}}+1)}^{4}}}\] \[=\frac{({{x}^{2}}+1)4-16x(x)}{{{({{x}^{2}}+1)}^{3}}}=\frac{-12{{x}^{2}}+4}{{{({{x}^{2}}+1)}^{3}}}\] \[\therefore \,\,{f}''(0)>0\] \[\therefore \]There is only one critical point having minima. Hence \[f(x)\] has least value at \[x=0\]. \[{{f}_{\min }}=f(0)=\frac{-1}{1}=-1\].
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