If the function $f: R-\{1,-1\} \rightarrow A$ defined by $f(x)=\frac{x^{2}}{1-x^{2}}$, is surjective, then $A$ is equal to:

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Sets Relations and Functions

If the function

f : R - {1, - 1} \to A

defined by

f (x) = \frac{x ^{2}}{1 - x ^{2}}

, is surjective, then

A

is equal to:

Answer: (A)

For

f

to be onto

A

, the set

A

is exactly the range. Put

y = \frac{x ^{2}}{1 - x ^{2}}

and solve:

y (1 - x^{2}) = x^{2}

gives

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

. A real

x

needs

\frac{y}{1 + y} \geq 0

, which holds for

y \geq 0

y < - 1

, and

y = - 1

is excluded. So the range, and hence

A

, is

R - [- 1, 0)

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option