Let $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$. Find $F^{-1}$ of the following function $F$ from $S$ to $T$,if it exists: $F = \{(a, 2), (b, 1), (c, 1)\}$.
(D) Given $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$.
The function $F : S \rightarrow T$ is defined as $F = \{(a, 2), (b, 1), (c, 1)\}$.
$A$ function is invertible if and only if it is both one-one (injective) and onto (surjective).
Here,$F(b) = 1$ and $F(c) = 1$. Since $F(b) = F(c)$ but $b \neq c$,the function $F$ is not one-one.
Because $F$ is not one-one,it is not a bijection.
Therefore,$F$ is not invertible,which means $F^{-1}$ does not exist.
The function $F : S \rightarrow T$ is defined as $F = \{(a, 2), (b, 1), (c, 1)\}$.
$A$ function is invertible if and only if it is both one-one (injective) and onto (surjective).
Here,$F(b) = 1$ and $F(c) = 1$. Since $F(b) = F(c)$ but $b \neq c$,the function $F$ is not one-one.
Because $F$ is not one-one,it is not a bijection.
Therefore,$F$ is not invertible,which means $F^{-1}$ does not exist.
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Let $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.
Statement-$1$: $S = \{x : f(x) = f^{-1}(x)\} = \{1, 2\}$.
Statement-$2$: $f$ is a bijection and $f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$.
Statement-$1$: $S = \{x : f(x) = f^{-1}(x)\} = \{1, 2\}$.
Statement-$2$: $f$ is a bijection and $f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$.
DifficultAIEEE 2011
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Medium
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