RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

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

xy² + x = x²y + y

xy² - x²y + x - y = 0

-xy(-y + x) + 1(x - y) = 0

(x - y)(1 - xy) = 0

x = y or $\text{x}=\frac{1}{\text{y}}$

So, f is not an injection.

Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

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

yx² - x + y = 0

$\text{x}=\frac{-(-1)\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ if $\text{y}\neq0$

$=\frac{1\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ which may not be in R

For example, if y = 1, then

$\text{x}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\text{i}\sqrt{3}}{2},$ which is not in R

So, f is not surjection and f is not bijection.