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If the given functions satisfy both one-one and onto functions then it will have

inverse.

Given,

$

f:\left\{ {1,2,3,4} \right\} \to \{ 10\} with \\

f = \{ (1,10),(2,10),(3,10),(4,10)\} \\

$

Here, the domain of ‘f’ is $\{ 1,2,3,4\} $and co-domain is$\{ 10\} $.

As we know a function is said to be a one-one function if distinct elements of domain

mapped with distinct elements of co-domain.

$f = \{ (1,10),(2,10),(3,10),(4,10)\} $

But, in this case if we see the function ‘f’ each element from the domain is mapped with the

same element from co-domain i.e.., 10.Since, all the elements have the same image 10 which is not satisfying the one-one function condition. Hence, f is not a one-one function.

Therefore, f doesn’t have inverse.

ii.Given,

$

g:\left\{ {5,6,7,8} \right\} \to \{ 1,2,3,4\} with \\

g = \{ (5,4),(6,3),(7,4),(8,2)\} \\

$

Here, the domain of ‘g’ is $\{ 5,6,7,8\} $and co-domain is$\{ 1,2,3,4\} $.

As we know a function is said to be a one-one function if distinct elements of domain mapped with distinct elements of co-domain.

$g = \{ (5,4),(6,3),(7,4),(8,2)\} $

But, in this case if we see the function ‘g’, the elements 5 and 7 from the domain are mapped

with the same element from co-domain i.e... As ‘g’ is not satisfying the one-one function

condition. Hence,’g’ is not a one-one function.

Therefore, g doesn’t have inverse.

iii.Given,

$

h:\left\{ {2,3,4,5} \right\} \to \{ 7,9,11,13\} with \\

h = \{ (2,7),(3,9),(4,11),(5,13)\} \\

$

Here, the domain of ‘h’ is $\{ 2,3,4,5\} $and co-domain is$\{ 7,9,11,13\} $.

As we know a function is said to be a one-one function if distinct elements of domain mapped with distinct elements of co-domain.

$h = \{ (2,7),(3,9),(4,11),(5,13)\} $

Here, each element from the domain is mapped with the different element from the co-domain. Therefore, ‘h’ is a one-one function.

Now, let us check with the onto condition i.e.., each element in the co-domain has a pre-image from the domain.

Here, each element from the co-domain has a pre-image from the domain. Therefore ‘h’ is

an onto function. As, function ‘h’ is both one-one and onto functions.

Hence, the inverse of ‘h’ exists i.e..,

$

h = \{ (2,7),(3,9),(4,11),(5,13)\} \\

{h^{ - 1}} = \{ (7,2),(9,3),(11,4),(13,5)\} \\

$

Hence, among the functions ‘f’, ‘g’, ‘h’ only the function ‘h’ has the inverse.

Note: The alternate method to find whether a function is one-one is by horizontal line test

i.e. ., if a horizontal line intersects the original function in a single region, the function is a

one-to-one function otherwise it is not a one-one function.

inverse.

Given,

$

f:\left\{ {1,2,3,4} \right\} \to \{ 10\} with \\

f = \{ (1,10),(2,10),(3,10),(4,10)\} \\

$

Here, the domain of ‘f’ is $\{ 1,2,3,4\} $and co-domain is$\{ 10\} $.

As we know a function is said to be a one-one function if distinct elements of domain

mapped with distinct elements of co-domain.

$f = \{ (1,10),(2,10),(3,10),(4,10)\} $

But, in this case if we see the function ‘f’ each element from the domain is mapped with the

same element from co-domain i.e.., 10.Since, all the elements have the same image 10 which is not satisfying the one-one function condition. Hence, f is not a one-one function.

Therefore, f doesn’t have inverse.

ii.Given,

$

g:\left\{ {5,6,7,8} \right\} \to \{ 1,2,3,4\} with \\

g = \{ (5,4),(6,3),(7,4),(8,2)\} \\

$

Here, the domain of ‘g’ is $\{ 5,6,7,8\} $and co-domain is$\{ 1,2,3,4\} $.

As we know a function is said to be a one-one function if distinct elements of domain mapped with distinct elements of co-domain.

$g = \{ (5,4),(6,3),(7,4),(8,2)\} $

But, in this case if we see the function ‘g’, the elements 5 and 7 from the domain are mapped

with the same element from co-domain i.e... As ‘g’ is not satisfying the one-one function

condition. Hence,’g’ is not a one-one function.

Therefore, g doesn’t have inverse.

iii.Given,

$

h:\left\{ {2,3,4,5} \right\} \to \{ 7,9,11,13\} with \\

h = \{ (2,7),(3,9),(4,11),(5,13)\} \\

$

Here, the domain of ‘h’ is $\{ 2,3,4,5\} $and co-domain is$\{ 7,9,11,13\} $.

As we know a function is said to be a one-one function if distinct elements of domain mapped with distinct elements of co-domain.

$h = \{ (2,7),(3,9),(4,11),(5,13)\} $

Here, each element from the domain is mapped with the different element from the co-domain. Therefore, ‘h’ is a one-one function.

Now, let us check with the onto condition i.e.., each element in the co-domain has a pre-image from the domain.

Here, each element from the co-domain has a pre-image from the domain. Therefore ‘h’ is

an onto function. As, function ‘h’ is both one-one and onto functions.

Hence, the inverse of ‘h’ exists i.e..,

$

h = \{ (2,7),(3,9),(4,11),(5,13)\} \\

{h^{ - 1}} = \{ (7,2),(9,3),(11,4),(13,5)\} \\

$

Hence, among the functions ‘f’, ‘g’, ‘h’ only the function ‘h’ has the inverse.

Note: The alternate method to find whether a function is one-one is by horizontal line test

i.e. ., if a horizontal line intersects the original function in a single region, the function is a

one-to-one function otherwise it is not a one-one function.

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