JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 9)
If the function ƒ : R – {1, –1} $$ \to $$ A defined by
ƒ(x) = $${{{x^2}} \over {1 - {x^2}}}$$ , is surjective, then A is equal to
ƒ(x) = $${{{x^2}} \over {1 - {x^2}}}$$ , is surjective, then A is equal to
R – (–1, 0)
R – {–1}
R – [–1, 0)
[0, $$\infty $$)
Explanation
Let ƒ(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).
$$ \Rightarrow $$ $$y\left( {1 - {x^2}} \right) = {x^2}$$
$$ \Rightarrow $$ $${x^2} = {y \over {1 + y}}$$
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).
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