A) does not exist
B) is equal to 1
C) is equal- to 0
D) is equal to \[-1\]
Correct Answer: D
Solution :
\[f(x)=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}\](for every real number \[x\]) On differentiating w.r.t. \[x,\]we get \[f(x)=\frac{({{x}^{2}}+1)(2x)-({{x}^{2}}-1)2x}{{{({{x}^{2}}+1)}^{2}}}\] \[=\frac{2x[{{x}^{2}}+1-{{x}^{2}}+1]}{{{({{x}^{2}}+1)}^{2}}}\] \[\Rightarrow \]\[f(x)=\frac{4x}{{{({{x}^{2}}+1)}^{2}}}\] ?.(i) Put \[f(x)=0\]for minima or maxima \[\therefore \] \[\frac{4x}{{{({{x}^{2}}+1)}^{2}}}=0\] \[\Rightarrow \] \[x=0\] On differentiating Eq. (i) w.r.t. \[x,\]we get \[f\,=\frac{{{({{x}^{2}}+1)}^{2}}4-(4x)2({{x}^{2}}+1)2x}{{{({{x}^{2}}+1)}^{4}}}\] \[=\frac{4({{x}^{2}}+1)[{{x}^{2}}+1-4{{x}^{2}}]}{{{({{x}^{2}}+1)}^{4}}}\] \[=\frac{4[{{x}^{2}}+1-4{{x}^{2}}]}{{{({{x}^{2}}+1)}^{3}}}\] \[\Rightarrow \] \[f\,(x)=\frac{-12{{x}^{2}}+4}{{{({{x}^{2}}+1)}^{3}}}\] At \[x=0\] \[f\,(0)=\frac{-12\times {{0}^{2}}+4}{{{({{0}^{2}}+1)}^{3}}}\] \[f\,(0)=4(+\,ve)\] Thus, the function is minimum at \[x=0.\] Then, minimum value of \[f(x)\]at \[x=0\]is \[f(0)=\frac{{{0}^{2}}-1}{{{0}^{2}}+1}\] \[=-1\]
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