JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f :R \to R\) be defined by \(f(x)\,\, = \,\,\frac{x}{{1 + {x^2}}},\,x\, \in \,R.\) Then the range of \(f\) is
- A \(\left[ { - \frac{1}{2},\frac{1}{2}} \right]\)
- B \(R\, - [ - 1,1]\)
- C \(R - \left[ { - \frac{1}{2},\frac{1}{2}} \right]\)
- D \(( - 1,1) - \{ 0\} \)
Answer & Solution
Correct Answer
(A) \(\left[ { - \frac{1}{2},\frac{1}{2}} \right]\)
Step-by-step Solution
Detailed explanation
\(y = \frac{x}{{{x^2} + 1}}\) \(y{x^2} - x + y = 0\) \(D \ge 0\) \( \Rightarrow 1 - 4{y^2} \ge 0\) \( \Rightarrow {y^2} \le \frac{1}{4}\) \(y \in \left[ { - \frac{1}{2}.\frac{1}{2}} \right]\)
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