Q. 135.0( 1 Vote )

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Answer :

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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