Functions 3 Question 1
1. If $X$ and $Y$ are two non-empty sets where $f: X \rightarrow Y$, is function is defined such that
$$ \begin{aligned} f(C) & ={f(x): x \in C} \text { for } C \subseteq X \text { and } \\ f^{-1}(D) & ={x: f(x) \in D} \text { for } D \subseteq Y \end{aligned} $$
for any $A \subseteq Y$ and $B \subseteq Y$, then
(2005, 1M)
(a) $f^{-1}{f(A)}=A$
(b) $f^{-1}{f(A)}=A$, only if $f(X)=Y$
(c) $f{f^{-1}(B) }=B$, only if $B \subseteq f(x)$
(d) $f{f^{-1}(B) }=B$
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Answer:
Correct Answer: 1. (c)
Solution:
- Given, function $f: \mathbf{R}-{1,-1} \rightarrow A$ defined as
$$ \begin{aligned} & f(x)=\frac{x^{2}}{1-x^{2}}=y \\ & \Rightarrow \quad x^{2}=y\left(1-x^{2}\right) \\ & \Rightarrow \quad x^{2}(1+y)=y \\ & \left.\Rightarrow \quad x^{2}=\frac{y}{1+y} \quad \text { [provided } y \neq-1\right] \\ & \because \quad x^{2} \geq 0 \\ & \Rightarrow \quad \frac{y}{1+y} \geq 0 \Rightarrow y \in(-\infty,-1) \cup[0, \infty) \end{aligned} $$
Since, for surjective function, range of $f=$ codomain
$\therefore$ Set $A$ should be $\mathbf{R}-[-1,0)$.
