A) \[x=0\]
B) \[x=1\]
C) \[x=4\]
D) \[x=3\]
Correct Answer: A
Solution :
\[f(x)={{x}^{2}}+\frac{1}{1+{{x}^{2}}}\], \[{f}'(x)=2x-\frac{1}{{{(1+{{x}^{2}})}^{2}}}\,.\,2x\] Now \[{f}'(x)=0\] Þ \[x=0\] So the function has minimum value at \[x=0\].You need to login to perform this action.
You will be redirected in
3 sec