Question

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.} 
[I.] \( f(x) = 1 \Rightarrow f(y) = 1, f(z) = 2 \) 
[II.] \( f(x) = 2 \Rightarrow f(y) = 1, f(z) = 1 \) 
[III.] \( f(x) = 1, f(y) = 1, f(z) = 2 \) Then the value of \( f(1) \) is:

• A. \( x \)
• B. \( y \)
• C. \( z \)
• D. None of the above

Hint:

In injective mappings, no two elements in the domain can map to the same element in the range.

Answer 1

The Correct Option is D

We test each statement. Only one can be true. 

Statement I: If \( f(x) = 1 \Rightarrow f(y) = 1, f(z) = 2 \) ⇒ not injective since \( f(x) = f(y) \) 

Statement II: If \( f(x) = 2 \Rightarrow f(y) = 1, f(z) = 1 \) ⇒ Again not injective 

Statement III: Claims all values: \( f(x) = 1, f(y) = 1, f(z) = 2 \) ⇒ \( f(x) = f(y) \) ⇒ not injective None of the statements support injectiveness. So none are valid under the rule that only one is true. \[ {\text{None of the above}} \]