If the function f : R -- 1, --1 [ ] A defined by f(x) = [x^2 1 - x^2] , is surjective, the…

Question

If the function f : R -- {1, --1} $$ \to $$ A defined by

f(x) = $${{{x^2}} \over {1 - {x^2}}}$$ , is surjective, then A is equal to

Accepted Answer

Let f(x) = $${{{x^2}} \over {1 - {x^2}}}$$ = y

$$ \Rightarrow $$ $$y\left( {1 - {x^2}} \right) = {x^2}$$

$$ \Rightarrow $$ $${x^2} = {y \over {1 + y}}$$

As ${x^2}$ is always $$ \ge $$ 0.

$$ \therefore $$ $${y \over {1 + y}}$$ $$ \ge $$ 0

y $$ \in $$ $$\left( { - \infty , - 1} \right) \cup \left[ {0,\left. \infty  \right)} \right.$$

For surjective function co-domain = Range

$$ \therefore $$ A is R -- [--1, 0).

Solutions · 1