Question

State with reason whether the following functions have inverse
i. f:\left\\{ {1,2,3,4} \right\\} \to \\{ 10\\} with \\\ f = \\{ (1,10),(2,10),(3,10),(4,10)\\} \\\
ii. g:\left\\{ {5,6,7,8} \right\\} \to \\{ 1,2,3,4\\} with \\\ g = \\{ (5,4),(6,3),(7,4),(8,2)\\} \\\
iii. h:\left\\{ {2,3,4,5} \right\\} \to \\{ 7,9,11,13\\} with \\\ h = \\{ (2,7),(3,9),(4,11),(5,13)\\} \\\

Answer 1

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.