Functions Ques 1
- If the function $f: \mathbf{R}-\{1,-1\} \rightarrow A$ defined by $f(x)=\frac{x^2}{1-x^2}$, is surjective, then $A$ is equal to
(2019 Main, 9 April I)
(a) $\mathrm{R}-\{-1\}$
(b) $[0, \infty)$
(c) $\mathbf{R}-[-1,0)$
(d) $\mathbf{R}-(-1,0)$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Given, function $f: \mathbf{R}-\{1,-1\} \rightarrow A$ defined as
$ \begin{aligned} & f(x)=\frac{x^2}{1-x^2}=y \\ & \Rightarrow \quad x^2=y\left(1-x^2\right) \\ & {\left[\because \quad x^2 \neq 1\right]} \\ & \Rightarrow \quad x^2(1+y)=y \\ & \text { [provided } y \neq-1 \text { ] } \\ & \Rightarrow \quad x^2=\frac{y}{1+y} \\ & \because \quad x^2 \geq 0 \\ & \Rightarrow \quad \frac{y}{1+y} \geq 0 \Rightarrow y \in(-\infty,-1) \cup[0, \infty) \\ \end{aligned} $
Since, for surjective function, range of $f=$ codomain
$\therefore$ Set $A$ should be $\mathbf{R}-[-1,0)$.
BETA
