Question

The function `f: Rvec[-1/2,1/2]` defined as `f(x)=x/(1+x^2),` is : Surjective but not injective (2) Neither injective not surjective Invertible (4) Injective but not surjective
A. invertible
B. injective but not surjective
C. surjective but not injective
D. neither injective nor surjective

Answer 1

Correct Answer - C
We have, `f(x)=(x)/(1+x^(2))`
`therefore f((1)/(x))=(x)/(1+(1)/(x^(2)))=(x)/(1+x^(2))=f(x)`
` therefore f((1)/(2))=f(2) or f((1)/(3))=f(3) ` and so on.
So, f(x) is many-one function.
Again, let `y=f(x)rArr y=(x)/(1+x^(2))`
`rArr y+x^(2)y=x rArr yx^(2)-x+y=0`
As `x in R`
`therefore (-1)^(2)-4(y)(y) ge 0`
`rArr 1-4y^(2) ge 0`
`rArr y in [(-1)/(2),(1)/(2)]`
` therefore "Range" ="Codomain"=[(-1)/(2),(1)/(2)]`
So, f(x) is surjective.
Hence, f(x) is surjective but not injective.