Let ‘f’ be an injective function with domain {x, y, z} and range {1,2,3} such that exactly one of the following statements is correct and the remaining are false f(x) = 1. f(x)=1, f(y)≠1, f(z)≠2 the value of f−1(1) is

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x or z

answer is B.

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Detailed Solution

Suppose f(x) = 1, then f(y)≠1, f(z)=2⇒f is not an injection.Suppose f(y)≠1, thenf(z)=2, f(y)=3, f(x)≠1 A contradictionSuppose f(z)≠2, thenf(y)=1, f(z) = 3, f(x) = 2 this is true ∴ f−1(1)=y

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