Mark the correct alternative in each of the following: Let f be an injective map 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(y) ≠ 1, f(z) ≠ 2. The value of f –1 (1) is

Question

Philoid · Accepted 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

Mark the correct alternative in each of the following:Let f be an injective map 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(y) ≠ 1, f(z) ≠ 2.The value of f–1(1) is

Mark the correct alternative in each of the following:
Let f be an injective map 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(y) ≠ 1, f(z) ≠ 2.

The value of f^–1(1) is