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in Mathematics by (71.7k points)

Let f be a one - one function with domain {x, y, z} and range {1, 2, 3}. It is given that exactly one of following statement is true and the remaining two are false

f(x) = 1, f(y) ≠ 1, f(z) ≠ 2 determine f-1(1).

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Best answer

It given three cases

Case I : When f(x) = 1 is true.

In this case remaining two are false

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.

Case II : when f(y) ≠ 1 is true.

If f(y) ≠ 1 is true then the remaining statement are false;

∴ f(x) ≠ 1 and f(z) = 2

ie, both x and y are not mapped to 1. so, either both associate to 2 or 3, Thus, it is not injective.

Case III : When f(z) ≠ 2 is true 

If f(z) ≠ 2 is true then remaining statement 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

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