Question

let the domain of x be the set {1} . which of the following function give value equal to 1 .​

Answer 1

Answer:

It gives three cases

Case I:

When f(x)=1 is true.

In this case remaining two are false i.e. f(y)=1 and f(z)=2

This means x and y have the same image, so f(x) is not an injective, which is a contradiction as the function is injective.

Case II:

When f(y) =1 is true then the remaining statements are false.

..f(x)

=1 and f(z)=2

→ Either of the x,y mapped to 2

Or both x and y are not mapped to 1.

So, either both associate to 2 or 3,

Thus, it is not injective which is a contradiction.

Thus, both the cases I and II are not true.

Case III:

When f(z)=2 is true then remaining statements are false

:. If f(x) =1 and f(y)=1

But f is injective

Thus, we have f(x)=2,f(y)=1 and f(z)=3

Hence, f-1

(1)=y