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

Options

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

Answer

A

Explanation

Injectivity: Let `x,y in R` be such that
f(x)=f(y)
`Rightarrow (x)/(1+x^(2))=(y)/(1+y^(2))`
`Rightarrow x-y+xy(y-x)=0`
`Rightarrow (x-y)(1-xy)=0`
`Rightarrow x=y or xy=1`
Putting x=2 in xy=1, we obtain y=1//2
`therefore f(2)=f((1)/(2))=(2)/(5)`
So, f is not injective.
Surjectivity: `"Let" y in [-1//2, 1//2]` such that
`f(x)=y`
`Rightarrow (x)/(1+x^(2))=y Rightarrowy+x^(2)y Rightarrow x^(2)y-x+y=0 Rightarrow x=(1+sqrt(1-4y^(2)))/(2y)`
Clearly, `x in R` for all `y in [-1//2,1//2]` except y=0
We observe that f(0)-0. So, all `y in [-1//2,1//2]` has its pre-image in R. Hence, f is surjective.