%%

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

JEE · Math · previous-year question

  1. A.R – (–1, 0)
  2. B.R – {–1}
  3. C.R – [–1, 0)correct
  4. D.[0, $$\infty $$)

Answer

C. R – [–1, 0)

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).

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions