A) x = 0
B) x = 1
C) x = 4
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 a minimum put \[f'(x)=0\Rightarrow x=0\] So, the function has minimum value at x = 0.You need to login to perform this action.
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