Question

question is in attachment

don't spam​

Answer 1

Answer is attached to the file

thank you

Answer 2

For x1, x2 ∈ R, consider

$f(\ x {}^{1} )= f(x { }^{2} )$

$= > \frac{x1}{{x}^{2} _{1} + 1} = \frac{x2}{ {x}^{2}_{2} + 1}$

$= > {x}^{1} \: {x}^{2} _{2} + {x}^{1} = {x}^{2} {x}^{1} _{2} + {x}^{2}$

$= > {x}^{1} {x}^{2}( {x}^{2} - {x}^{1} ) = {x}^{2} - {x}^{1}$

$= > {x}^{1} = {x}^{2} \: or \: {x}^{2} {x}^{1} = 1$

We note that there are point, x1 and x2 with x1 ≠ x2 and f (x1) = f (x2), for instance, if we take x1 = 2 and x2 = 1/2, then we have f (x1) =2/5 and f(x2) =2/5 but 2 ≠ 1/2 . Hence f is not one-one. Also, f is not onto for if so then for 1∈R ∃ x ∈ R such that f (x) = 1 which gives x/(x2+1) =1. But there is no such x in the domain R, since the equation x2 – x + 1 = 0 does not give any real value of x.