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Let $\mathrm{S}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$ and $\mathrm{T}=\{1,2,3\}$. Find $\mathrm{F}^{-1}$ of the following functions $F$ from $S$ to $T$, if it exists.
(i) $\mathrm{F}=\{(\mathrm{a}, 3),(\mathrm{b}, 2),(\mathrm{c}, 1)\}$
(ii) $\mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\}$
(i) $\mathrm{F}=\{(\mathrm{a}, 3),(\mathrm{b}, 2),(\mathrm{c}, 1)\}$
(ii) $\mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\}$
Solution:
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Verified Answer
$\mathrm{S}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$ and $\mathrm{T}=\{1,2,3\}$
(i) $\mathrm{F}=\{(\mathrm{a}, 3),(\mathrm{b}, 2),(\mathrm{c}, 1)\}$ i.e. $\mathrm{F}(\mathrm{a})=3, \mathrm{~F}(\mathrm{~b})=2, \mathrm{~F}(\mathrm{c})=1$ As each element is having different images, so $\mathrm{F}$ is one-one and its inverse is possible.
$$
\Rightarrow \mathrm{F}^{-1}(3)=\mathrm{a}, \mathrm{F}^{-1}(2)=\mathrm{b}, \mathrm{F}^{-1}(1)=\mathrm{C}
$$
$\left.\therefore \mathrm{F}^{-1}=\{3, a),(2, b),(1, c)\right\}$
(ii) $\mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\} \mathrm{F}$ is not one-one since element $\mathrm{b}$ and $\mathrm{C}$ have the same image 1 .
$\therefore$ It is not a one-one function. Its inverse does not exists.
(i) $\mathrm{F}=\{(\mathrm{a}, 3),(\mathrm{b}, 2),(\mathrm{c}, 1)\}$ i.e. $\mathrm{F}(\mathrm{a})=3, \mathrm{~F}(\mathrm{~b})=2, \mathrm{~F}(\mathrm{c})=1$ As each element is having different images, so $\mathrm{F}$ is one-one and its inverse is possible.
$$
\Rightarrow \mathrm{F}^{-1}(3)=\mathrm{a}, \mathrm{F}^{-1}(2)=\mathrm{b}, \mathrm{F}^{-1}(1)=\mathrm{C}
$$
$\left.\therefore \mathrm{F}^{-1}=\{3, a),(2, b),(1, c)\right\}$
(ii) $\mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\} \mathrm{F}$ is not one-one since element $\mathrm{b}$ and $\mathrm{C}$ have the same image 1 .
$\therefore$ It is not a one-one function. Its inverse does not exists.
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