A) \[x=0\]
B) \[x=4\]
C) \[x=1\]
D) \[x=3\]
Correct Answer: A
Solution :
Let\[f(x)={{x}^{2}}+\frac{1}{1+{{x}^{2}}}\] On differentiating w.r.t.\[x,\] we get \[f'(x)=2x-\frac{1}{{{(1+{{x}^{2}})}^{2}}}(2x)\] For maxima or minima, put\[f'(x)=0\] \[\therefore \] \[2x-\frac{2x}{{{(1+{{x}^{2}})}^{2}}}=0\] \[\Rightarrow \] \[2x\left[ 1-\frac{1}{{{(1+{{x}^{2}})}^{2}}} \right]=0\] \[\Rightarrow \] \[x=0\] \[\therefore \] The minimum value at\[x=0\].
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