The minimum value of x ^ 2 + 1 1+ x ^ 2 is at:
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The minimum value of \(\mathbf{x}^{2}+\frac{1}{1+\mathbf{x}^{2}} \text { is at: }\)
- \(x=0\)
- \(x=4\)
- \(\mathbf{x}=1\)
- \(x=3\)
Answer
(A) x=0
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