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Question

The range of the function $$ f (x) = \dfrac {x}{1+x^2} $$ is


A
,
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B
[1,1]
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C
[12,12]
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D
[2,2]
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Solution

The correct option is C $$\left[ - \dfrac {1}{2} , \dfrac {1}{2}\right] $$
Let $$ y = \dfrac {x} {1+ x^2} $$
$$ \Rightarrow x^2 y- x + y = 0 $$
For $$x$$ to be real, $$ 1 - 4y^2 \ge 0 $$
$$ \Rightarrow  (1 - 2y )(1 + 2y ) \ge 0 $$
$$ \Rightarrow  \left(  \dfrac {1}{2} - y \right)  \left(  \dfrac {1}{2} + y \right) \ge 0 $$
$$ \Rightarrow - \dfrac {1}{2} \le y \le \dfrac {1}{2} $$
$$ \therefore y = f(x) \epsilon \left[ - \dfrac {1}{2} , \dfrac {1}{2} \right] $$

Mathematics

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