Question

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

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

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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