# Show that the function defined by is neither one-one nor onto.

Given:

$f : R \rightarrow R$

To prove: f(x) is neither one-one nor onto.

For one-one

$\\ Let x_{1}, x_{2} \in R$ and

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

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

$\\ {x}_{1}-{x}_{2}={x}^2_{1}{x}_{2}-{x}_{1} {x}^2_{2}$

$\\ {x}_{1}-{x}_{2}={x}_{1}{x}_{2}({x}_{1} -{x}_{2})$

$\\ {x}_{1}-{x}_{2}\neq 0$

$\\ {x}_{1}{x}_{2}=1$

f(X) is not one-one

For onto

$y \in R$

$y=\frac{x}{1+x^2}$

$y({1+x^2})={x}$

$y+yx^2=x$

$x^2y-x+y=0$

$x = \frac{1\pm\sqrt{1^2-4\times y \times y}}{2 \times y}$

for any value of y x has two values hence f(x) is not onto.

## Related Chapters

### Preparation Products

##### Knockout NEET May 2023 (Easy Installments)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 5499/-
##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout NEET Aug 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-