Q. 135.0( 1 Vote )

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Given that f is an injective map with domain {x, y, z} and range {1, 2, 3}.

Case-1

Let us assume that f(x) =1 is true and f(y) ≠ 1, f(z) ≠ 2 is false.

Then f(x) = 1, f(y) = 1 and f(z) = 2.

This violates the injectivity of f because it is one-one.

Case-2

Let us assume that f(y) ≠1 is true and f(x)= 1, f(z) ≠ 2 is false.

Then f(x) ≠ 1, f(y) ≠ 1 and f(z) = 2.

This means there is no pre image of 1 which contradicts the fact that the range of f is {1, 2, 3}.

Case-3

Let us assume that f(z) ≠2 is true and f(x)= 1, f(y) ≠ 1 is false.

Then f(z) ≠ 2, f(y) = 1 and f(x) ≠1.

f–1(1) = y

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