The function `f:R->[-1/2,1/2]` defined as `f(x)=x/(1+x^2)` is

Options

A. neither injective nor surjective.
B. invertible.
C. injective but not surjective.
D. Surjective but not injective.

Answer

D

Explanation

`f:R to [-(1)/(2),(1)/(2)], f(x)=(x)/(1+x^(2)) AA x in R`
For range of the function, `y=(x)/(1+x^(2))`
`implies yx^(2)-x+y=0`
Since x is real, `D ge 0`
`implies 1-4y^(2) ge 0`
`implies y in [-(1)/(2),(1)/(2)]`
Also, for `y=(1)/(4),(1)/(4)=(x)/(1+x^(2))`
`or x^(2)-4x+1=0,` which has two distinct roots.
Thus, `f(x)` is onto but not injective.